Math in Focus Grade 7 Chapter 5 Lesson 5.4 Answer Key Understanding Inverse Proportion

Practice the problems of Math in Focus Grade 7 Workbook Answer Key Chapter 5 Lesson 5.4 Understanding Inverse Proportion to score better marks in the exam.

Math in Focus Grade 7 Course 2 A Chapter 5 Lesson 5.4 Answer Key Understanding Inverse Proportion

Hands-On Activity

Materials

  • algebra tiles

RECOGNIZE INVERSE PROPORTION

Work in pairs.
There are 6 ways of forming a rectangle with 12 algebra tiles. The diagram shows two possible ways.
Math in Focus Grade 7 Chapter 5 Lesson 5.4 Answer Key Understanding Inverse Proportion 1
Step 1: Form a different rectangle by rearranging the 12 algebra tiles. Record your results in a table like the one shown.
Math in Focus Grade 7 Chapter 5 Lesson 5.4 Answer Key Understanding Inverse Proportion 2
Step 2: Complete the table by repeating Step 1:

Math Journal Write down your observations about the values of v • h. Describe the relationship between v and h.
Answer:

Math in Focus Grade 7 Chapter 5 Lesson 5.4 Guided Practice Answer Key

Copy and complete.

Question 1.
Some friends want to share the cost of buying a present. The table shows the amount of money that each person has to contribute, y dollars, and the number of people sharing the cost, x. Tell whether y is inversely proportional to x. If so, find the constant of proportionality.
Math in Focus Grade 7 Chapter 5 Lesson 5.4 Answer Key Understanding Inverse Proportion 3
For each pair of values, x and y:
1 • 180 = Math in Focus Grade 7 Chapter 5 Lesson 5.4 Answer Key Understanding Inverse Proportion 4 Math in Focus Grade 7 Chapter 5 Lesson 5.4 Answer Key Understanding Inverse Proportion 4 • 90 = Math in Focus Grade 7 Chapter 5 Lesson 5.4 Answer Key Understanding Inverse Proportion 4. Math in Focus Grade 7 Chapter 5 Lesson 5.4 Answer Key Understanding Inverse Proportion 4Math in Focus Grade 7 Chapter 5 Lesson 5.4 Answer Key Understanding Inverse Proportion 4 = Math in Focus Grade 7 Chapter 5 Lesson 5.4 Answer Key Understanding Inverse Proportion 4
The value of x increases as the value of y decreases, and the product of x and y is a Math in Focus Grade 7 Chapter 5 Lesson 5.4 Answer Key Understanding Inverse Proportion 4 value. So, y is inversely proportional to x.
The constant of proportionality is Math in Focus Grade 7 Chapter 5 Lesson 5.4 Answer Key Understanding Inverse Proportion 4.
Answer:
Math in Focus Grade 7 Chapter 5 Lesson 5.4 Answer Key Understanding Inverse Proportion 010
For each pair of values, x and y ;
1. 180 = 180
2. 90 = 180
3. 60 = 180
The va1te of increases as the value of y decreases, and the product of x and y is a constant value.
So y is inversely proportional to x.
constant of proportionality = 180
Yes, y is proportional to x
Constant of proportionality = 180

Question 2.
Henry drove from Town A to Town B. The table shows the time he took, y hours, if he traveled at various speeds, x miles per hour. Tell whether x and y are in inverse proportion. If so, find the constant of proportionality.
Math in Focus Grade 7 Chapter 5 Lesson 5.4 Answer Key Understanding Inverse Proportion 5
For each pair of values, x and y:
40 • 9 = Math in Focus Grade 7 Chapter 5 Lesson 5.4 Answer Key Understanding Inverse Proportion 4 50 • Math in Focus Grade 7 Chapter 5 Lesson 5.4 Answer Key Understanding Inverse Proportion 4 = Math in Focus Grade 7 Chapter 5 Lesson 5.4 Answer Key Understanding Inverse Proportion 4 Math in Focus Grade 7 Chapter 5 Lesson 5.4 Answer Key Understanding Inverse Proportion 4Math in Focus Grade 7 Chapter 5 Lesson 5.4 Answer Key Understanding Inverse Proportion 4 = Math in Focus Grade 7 Chapter 5 Lesson 5.4 Answer Key Understanding Inverse Proportion 4
The value of x increases as the value of y decreases, but the product of x and y is Math in Focus Grade 7 Chapter 5 Lesson 5.4 Answer Key Understanding Inverse Proportion 4 value. So, y is Math in Focus Grade 7 Chapter 5 Lesson 5.4 Answer Key Understanding Inverse Proportion 4 inversely proportional to x.
Answer:
Math in Focus Grade 7 Chapter 5 Lesson 5.4 Answer Key Understanding Inverse Proportion 011
For each pair of values, x and y :
40 ∙ 9 = 360
50 ∙ 7\(\frac{1}{3}\) = 366.66
60 ∙ 6 = 360
The value of x increases as the value of y decreases, but the product of x and y is not a constant value.
So y is not inversely proportional to z.
No, y is not proportional to x

Tell whether the equation represents an inverse proportion. If so, find the constant of proportionality.

Question 3.
\(\frac{3}{5}\)y = \(\frac{6}{x}\)
\(\frac{3}{5}\)y = \(\frac{6}{x}\)
\(\frac{3}{5}\) ∙ Math in Focus Grade 7 Chapter 5 Lesson 5.4 Answer Key Understanding Inverse Proportion 4 = \(\frac{6}{x}\) ∙ Math in Focus Grade 7 Chapter 5 Lesson 5.4 Answer Key Understanding Inverse Proportion 4 Multiply both sides by Math in Focus Grade 7 Chapter 5 Lesson 5.4 Answer Key Understanding Inverse Proportion 4.
y = \(\frac{?}{x}\) Simplify
y ∙ x = \(\frac{?}{x}\) ∙ x Multiply both sides by x.
xy = Math in Focus Grade 7 Chapter 5 Lesson 5.4 Answer Key Understanding Inverse Proportion 4 Simplify
The original equation Math in Focus Grade 7 Chapter 5 Lesson 5.4 Answer Key Understanding Inverse Proportion 4 be rewritten as two equivalent equations in the form y = \(\frac{k}{x}\) and xy = k. So, the equation Math in Focus Grade 7 Chapter 5 Lesson 5.4 Answer Key Understanding Inverse Proportion 4 an inverse proportion. The constant of proportionality is Math in Focus Grade 7 Chapter 5 Lesson 5.4 Answer Key Understanding Inverse Proportion 4.
Answer:
\(\frac{3}{5}\)y = \(\frac{6}{x}\)
\(\frac{3}{5}\)y ∙ \(\frac{5}{3}\) = \(\frac{6}{x}\) ∙ \(\frac{5}{3}\). Multiply both sides by \(\frac{5}{3}\)
y = \(\frac{10}{x}\) Simplify
y ∙ x = \(\frac{10}{x}\) ∙ x Multiply both sides by x
xy = 10 Simplify
The original equation can be rewritten as two equivalent equations in the form
y = \(\frac{k}{x}\) and xy = k. So, the equation represents an inverse proportion.
The constant of proportionality is 10
Yes, the equation represents an inverse proportion.
Constant of proportionality is 10

Question 4.
y – 3x = 5
y – 3x = 5
y – 3x + Math in Focus Grade 7 Chapter 5 Lesson 5.4 Answer Key Understanding Inverse Proportion 4 = 5 + Math in Focus Grade 7 Chapter 5 Lesson 5.4 Answer Key Understanding Inverse Proportion 4 Add Math in Focus Grade 7 Chapter 5 Lesson 5.4 Answer Key Understanding Inverse Proportion 4 to both sides.
Math in Focus Grade 7 Chapter 5 Lesson 5.4 Answer Key Understanding Inverse Proportion 4 = Math in Focus Grade 7 Chapter 5 Lesson 5.4 Answer Key Understanding Inverse Proportion 4 Simplify.
The original equation Math in Focus Grade 7 Chapter 5 Lesson 5.4 Answer Key Understanding Inverse Proportion 4 be rewritten as two equivalent equations in the form y = \(\frac{k}{x}\) and xy = k. So, the equation Math in Focus Grade 7 Chapter 5 Lesson 5.4 Answer Key Understanding Inverse Proportion 4 an inverse proportion.
Answer:
y – 3x = 5
y – 3x + 3x = 5 + 3x Add 3x to both sides
y = 5 + 3x Simplify
The original equation cannot be rewritten as two equivalent equations in the form
y = \(\frac{k}{x}\) and xy = k. So, the equation cannot represents an inverse proportion.
No, the equation cannot represents an inverse proportion.

Solve an inverse proportion problem graphically.

Question 5.
The amount of time needed for volunteers to pick up trash on a beach is inversely proportional to the number of volunteers. The graph shows the amount of time, y hours, needed by x volunteers.
Math in Focus Grade 7 Chapter 5 Lesson 5.4 Answer Key Understanding Inverse Proportion 6

a) Find the constant of proportionality graphically. Then write an inverse proportion equation.
Use the point (Math in Focus Grade 7 Chapter 5 Lesson 5.4 Answer Key Understanding Inverse Proportion 4, Math in Focus Grade 7 Chapter 5 Lesson 5.4 Answer Key Understanding Inverse Proportion 4) from the graph to find the constant of proportionality:
x • y = Math in Focus Grade 7 Chapter 5 Lesson 5.4 Answer Key Understanding Inverse Proportion 4Math in Focus Grade 7 Chapter 5 Lesson 5.4 Answer Key Understanding Inverse Proportion 4 Choose the point (Math in Focus Grade 7 Chapter 5 Lesson 5.4 Answer Key Understanding Inverse Proportion 4, Math in Focus Grade 7 Chapter 5 Lesson 5.4 Answer Key Understanding Inverse Proportion 4).
xy = Math in Focus Grade 7 Chapter 5 Lesson 5.4 Answer Key Understanding Inverse Proportion 4 Multiply.
The constant of proportionality is Math in Focus Grade 7 Chapter 5 Lesson 5.4 Answer Key Understanding Inverse Proportion 4.
The inverse proportion equation is Math in Focus Grade 7 Chapter 5 Lesson 5.4 Answer Key Understanding Inverse Proportion 4.
Answer:
Use the point (6, 10) from the graph to find the constant of proportionality.
x ∙ y = 6 10 Choose the point (6, 10)
xy = 60 Multiply
The constant of proportionality is 60
The inverse proportion equation is xy = 60

b) Explain what the point (6, 10) represents in this situation.
It means that Math in Focus Grade 7 Chapter 5 Lesson 5.4 Answer Key Understanding Inverse Proportion 4 volunteers can clean the beach in Math in Focus Grade 7 Chapter 5 Lesson 5.4 Answer Key Understanding Inverse Proportion 4 hours.
Answer:
It means that 6 volunteers can beach in 10 hours.

Copy and complete.

Question 6.
y is inversely proportional to x, and y = 3 when x = 5.
Answer:
y = 3 when x = 5
a) Find the value of the constant
Constant of proportionality:
x ∙ y = Math in Focus Grade 7 Chapter 5 Lesson 5.4 Answer Key Understanding Inverse Proportion 4Math in Focus Grade 7 Chapter 5 Lesson 5.4 Answer Key Understanding Inverse Proportion 4
The constant of proportionality is Math in Focus Grade 7 Chapter 5 Lesson 5.4 Answer Key Understanding Inverse Proportion 4.
Answer:
Constant of proportionality :
x ∙ y = 5 ∙ 3 = 15

b) Write an inverse proportion equation that relates x and y.
Inverse proportion equation:
Math in Focus Grade 7 Chapter 5 Lesson 5.4 Answer Key Understanding Inverse Proportion 4 = Math in Focus Grade 7 Chapter 5 Lesson 5.4 Answer Key Understanding Inverse Proportion 4 or Math in Focus Grade 7 Chapter 5 Lesson 5.4 Answer Key Understanding Inverse Proportion 4 = Math in Focus Grade 7 Chapter 5 Lesson 5.4 Answer Key Understanding Inverse Proportion 4
The inverse proportion equation is Math in Focus Grade 7 Chapter 5 Lesson 5.4 Answer Key Understanding Inverse Proportion 4 = Math in Focus Grade 7 Chapter 5 Lesson 5.4 Answer Key Understanding Inverse Proportion 4 or Math in Focus Grade 7 Chapter 5 Lesson 5.4 Answer Key Understanding Inverse Proportion 4 = Math in Focus Grade 7 Chapter 5 Lesson 5.4 Answer Key Understanding Inverse Proportion 4.
Answer:
Inverse proportion equation :
xy = 15 or y = \(\frac{15}{x}\)

c) Find the value of x when y = 10.
When x = 10 and y = \(\frac{?}{x}\), y = \(\frac{?}{?}\) Evaluate y = \(\frac{?}{x}\) when x = 10.
y = Math in Focus Grade 7 Chapter 5 Lesson 5.4 Answer Key Understanding Inverse Proportion 4. Simplify.
The value of y is Math in Focus Grade 7 Chapter 5 Lesson 5.4 Answer Key Understanding Inverse Proportion 4.
Answer:
When x = 10 and y = \(\frac{15}{x}\),
y = \(\frac{15}{10}\)
y = 1.5

Solve.

Question 7.
Trucks are used to paint dividing lines on a long highway. The number of hours, y, the trucks take to paint the lines is inversely proportional to the number of trucks, x. 15 trucks can paint the highway in 28 hours. How many trucks are needed to paint the same highway in 20 hours?
Let x be the number of trucks.
Let y be the number of hours.
Constant of proportionality:
x ∙ y = Math in Focus Grade 7 Chapter 5 Lesson 5.4 Answer Key Understanding Inverse Proportion 4Math in Focus Grade 7 Chapter 5 Lesson 5.4 Answer Key Understanding Inverse Proportion 4
= Math in Focus Grade 7 Chapter 5 Lesson 5.4 Answer Key Understanding Inverse Proportion 4
Inverse proportion equation:
xy = Math in Focus Grade 7 Chapter 5 Lesson 5.4 Answer Key Understanding Inverse Proportion 4 Write an inverse equation.
When y = 20 and xy = Math in Focus Grade 7 Chapter 5 Lesson 5.4 Answer Key Understanding Inverse Proportion 4,
20 ∙ x = Math in Focus Grade 7 Chapter 5 Lesson 5.4 Answer Key Understanding Inverse Proportion 4
20x = Math in Focus Grade 7 Chapter 5 Lesson 5.4 Answer Key Understanding Inverse Proportion 4
\(\frac{20 x}{?}=\frac{?}{?}\)
x = Math in Focus Grade 7 Chapter 5 Lesson 5.4 Answer Key Understanding Inverse Proportion 4
Math in Focus Grade 7 Chapter 5 Lesson 5.4 Answer Key Understanding Inverse Proportion 4 trucks are needed to paint the highway in 20 hours.
Answer:
Let x be the number of trucks.
Let y be the number of hours.
Constant of proportionality :
x ∙ y = 15 ∙ 28
xy = 420
Inverse proportion equation :
xy = 420 Write an inverse equat ion
When y = 20 and xy = 420 :
20 ∙ x = 420 Evaluate y = 420 when y = 20
20x = 420 Simplify
\(\frac{20 x}{20}\) = \(\frac{420}{20}\) Divide both sides by 20
x = 21 simplify
21 trucks are needed to paint the
xy = 420
x = 21

Math in Focus Course 2A Practice 5.4 Answer Key

Tell whether two quantities are in inverse proportion. If so, find the constant of proportionality.

Question 1.
Math in Focus Grade 7 Chapter 5 Lesson 5.4 Answer Key Understanding Inverse Proportion 7
Answer:
Math in Focus Grade 7 Chapter 5 Lesson 5.4 Answer Key Understanding Inverse Proportion 015
For each pair of values, x and y :
25 . 2 = 50
10 . 5 = 50
5 . 10 = 50
The value of x decreases as the value of y increases and the product of x and y is a constant value.
So y is inversely proportional to x.
Constant of Proportionality = 50
Yes, y is proportional to x
Constant of proportionality = 50

Question 2.
Math in Focus Grade 7 Chapter 5 Lesson 5.4 Answer Key Understanding Inverse Proportion 8
Answer:
Math in Focus Grade 7 Chapter 5 Lesson 5.4 Answer Key Understanding Inverse Proportion 016
For each pair of values, x and y :
7 . 30 = 210
5 . 60 = 300
3 . 70 = 210
The value of x decreases as the value of y increases, but the product of x and y is not a constant value.
So y is not inversely proportional to x.
No, y is not proportional to x

Question 3.
Math in Focus Grade 7 Chapter 5 Lesson 5.4 Answer Key Understanding Inverse Proportion 9
Answer:
Math in Focus Grade 7 Chapter 5 Lesson 5.4 Answer Key Understanding Inverse Proportion 017
For each pair of values, x and y :
4 ∙ 16 = 64
6 ∙ 24 = 144
8 ∙ 32 = 236
The value of x increases as well as the value of y increases, and the product of x and y is also not a constant value.
So y is not inversely proportional to x.
No, y is not proportional to x

Question 4.
Math in Focus Grade 7 Chapter 5 Lesson 5.4 Answer Key Understanding Inverse Proportion 10
Answer:
Math in Focus Grade 7 Chapter 5 Lesson 5.4 Answer Key Understanding Inverse Proportion 018
For each pair of values, x and y :
6 ∙ 2 = 12
3 ∙ 4 = 12
1 ∙ 12 = 12
The value of decreases as the value of y increases, and the product of x and y is a constant value.
So y is inversely proportional to x.
Constant of Proportionality = 12
Yes, y is proportional to x
Constant of proportionality = 12

Tell whether each equation represents an inverse proportion. If so, give the constant of proportionality.

Question 5.
10x = \(\frac{5}{y}\)
Answer:
10x = \(\frac{5}{y}\)
\(\frac{y}{10}\) ∙ 10x = \(\frac{y}{10}\) ∙ \(\frac{5}{y}\) Multiply both sides by \(\frac{5}{y}\)
xy = \(\frac{1}{2}\) Simplify
The original equation can be rewritten as two qui valent equations in the form
y = \(\frac{k}{x}\) and xy = k. So, the equation represents an inverse proportion.
The constant of proportionality is \(\frac{1}{2}\)
Yes, the equation represents an inverse proportion.
Constant of proportionality is \(\frac{1}{2}\)

Question 6.
\(\frac{y}{20}\) = x
Answer:
\(\frac{y}{20}\) = x
\(\frac{1}{y}\) ∙ \(\frac{y}{20}\) = \(\frac{1}{y}\) ∙ x Multiply both sides by \(\frac{1}{y}\)
\(\frac{1}{20}\) = \(\frac{x}{y}\) Simplify
The original equation cannot be rewritten as two equivalent equations in the form
y = \(\frac{k}{x}\) and xy = k. So. the equation cannot represents an inverse proportion.
No, the equation represents an inverse proportion.

Question 7.
y + \(\frac{1}{7}\)x = \(\frac{1}{2}\)
Answer:
y + \(\frac{1}{7}\)x = \(\frac{1}{2}\)
y + \(\frac{1}{7}\)x – \(\frac{1}{7}\)x = \(\frac{1}{2}\) – \(\frac{1}{7}\)x Subtract \(\frac{1}{7}\)x to both sides
y = \(\frac{1}{2}\) – \(\frac{1}{7}\)x Simplify
The original equation cannot be rewritten as two equivalent equations in the form
y = \(\frac{k}{x}\) and xy = k. So, the equation cannot represents an inverse proportion.
No, the equation cannot represents an inverse proportion.

Question 8.
0.1x = \(\frac{5}{y}\)
Answer:
0.1x = \(\frac{5}{y}\)
y ∙ 0.1x = \(\frac{5}{y}\) ∙ y Multiply both sides by y
0.1xy = 5 Simplify
\(\frac{0.1 x y}{0.1}\) = \(\frac{5}{0.1}\) Divide both sides by 0.1
xy = 50 Simplify
The original equation can be rewritten as two equivalent equations in the form
y = \(\frac{k}{x}\) and xy = k. So, the equation represents an inverse proportion.
The constant of proportionality is 50
Yes, the equation represents an inverse proportion.
Constant of proportionality is 50

Each graph represents an inverse proportion. Find the constant of proportionality.

Question 9.
Math in Focus Grade 7 Chapter 5 Lesson 5.4 Answer Key Understanding Inverse Proportion 11
Answer:
Find the constant of proportionality graphically. Then write an inverse proportion equation.
Use the point (4, 0.5) f rom the graph to find the constant of proportionality.
x ∙ y = 4 ∙ 0.5 Choose the point (4, 0.5)
xy = 2 Multiply
The constant of proportionality is 2
The inverse proportion eqnation is xy = 2
Constant proportionality = 2; xy = 2

Question 10.
Math in Focus Grade 7 Chapter 5 Lesson 5.4 Answer Key Understanding Inverse Proportion 12
Answer:
Find the constant of proportionality graphically. Then write an inverse proportion equation.
Use the paint (2, 5) from the graph to find the constant of proportionality.
x ∙ y = 2 ∙ 5 Choose the point (2, 5)
xy = 10 Multiply
The constant of proportionality is 10
The inverse proportion equation is xy = 10
Constant of proportionality = 10; xy = 10

Question 11.
Math in Focus Grade 7 Chapter 5 Lesson 5.4 Answer Key Understanding Inverse Proportion 13
Answer:
Find the constant of proportionality graphically. Then write an inverse proportion equation.
Use the point (0.5, 8) from the graph to find the constant of proportionality.
x ∙ y = 0.5 ∙ 8 Choose the point (0.5, 8)
xy = 4 Multiply
The constant of proportionality is 4
The inverse proportion equation is xy = 4
Constant of proportionality = 4 ; xy = 4

Question 12.
Math in Focus Grade 7 Chapter 5 Lesson 5.4 Answer Key Understanding Inverse Proportion 14
Answer:
Find the constant of proportionality graphically. Then write an inverse proportion equal ion.
Use the point (\(\frac{1}{2}\), 30) from the graph to find the constant of proportionality.
x ∙ y = \(\frac{1}{2}\) ∙ 30 Choosethe point (\(\frac{1}{2}\), 3o)
xy = 15 Multiply
The constant of proportionality is 15
The inverse proportion equation is xy = 15
Constant of proportionality = 15; xy = 15

Solve. Show your work.

Question 13.
Math Journal Describe how can you tell whether two quantities are in inverse proportion.
Answer:
The two quantities are in inverse proportion when :
1 → When x increases and y decreases or when y increases and x decreases.
2 → The product of both the quantities should remain constant.
When one quantity increases and other quantity decreases.
The product of two quantity should be constant.

Question 14.
The workers at a bakery must mix 12 batches of bagel dough every hour to meet the needs of customers. The number of batches of bagel dough, b, that each worker needs to mix in one hour is inversely proportional to the number of workers, n. The graph shows the relationship between b and n.
Math in Focus Grade 7 Chapter 5 Lesson 5.4 Answer Key Understanding Inverse Proportion 15
a) Find the constant of proportionality from the graph. Then write an inverse proportion equation.
Answer:
Use the point (6, 2) from the graph to find the on stunt of proportionality.
n ∙ b = 6 ∙ 2 Choose the point (6, 2)
bn = 12 Multiply
The constant of proportionality is 12
The inverse proportion equation is bn = 12

b) Explain what the constant of proportionality represents in this situation.
Answer:
It means that the workers should mix 12 bagel dough in every hours.

c) Explain what the point (6, 2) represents in this situation.
Answer:
It means that 6 workers should mix 2 batches of bagel dough.

Question 15.
A rectangle has a fixed area that does not change. The length, l, of the rectangle is inversely proportional to its width, w. The graph shows the relationship between l and w.
Math in Focus Grade 7 Chapter 5 Lesson 5.4 Answer Key Understanding Inverse Proportion 16
a) Find the constant of proportionality from the graph. Then write an inverse proportion equation.
Answer:
Use the point (3, 8) from the graph to find the constant of proportionality.
w ∙ l = 3 ∙ 8 Choose the point (3, 8)
lw = 24 Multiply
The constant of proportionality is 24
The inverse proportion equation is lw = 24

b) Explain what the constant of proportionality represents in this situation.
Answer:
It represent the Area of rectangle.

c) Explain what the point (3, 8) represents in this situation.
Answer:
It ‘means that 3 represents the width and 8 represent the length.

Question 16.
y is inversely proportional to x, and y = 2 when x = 5.
Answer:
y = 2 when x = 5
a) Find the constant of proportionality.
Answer:
Constant of proportionality :
x ∙ y = 5 ∙ 2 = 10

b) Write an inverse equation relating x and y.
Answer:
inverse proportion equation :
xy = 10 or y = \(\frac{10}{x}\)

c) Find the value of x when y = 4.
Answer:
When y = 4 and y = \(\frac{15}{x}\),
4 = \(\frac{15}{x}\)
\(\frac{x}{4}\) ∙ 4 = \(\frac{x}{4}\) ∙ \(\frac{15}{x}\)
x = \(\frac{15}{4}\)
x = 3.75

Question 17.
y is inversely proportional to x, and y = \(\frac{1}{3}\) when x = \(\frac{1}{2}\).
Answer:
y = \(\frac{1}{3}\) when x = \(\frac{1}{2}\)
a) Find the constant of proportionality.
Answer:
Constant of proportionality :
x ∙ y = \(\frac{1}{3}\) ∙ \(\frac{1}{2}\) = \(\frac{1}{6}\)

b) Write an inverse proportion equation relating x and y.
Answer:
Inverse proportion equation :
xy = \(\frac{1}{6}\) or y = \(\frac{1}{6 x}\)

c) Find the value of y when x = \(\frac{1}{5}\).
Answer:
when x = \(\frac{1}{5}\) and y = \(\frac{1}{6 x}\),
y = \(\frac{1}{6 \cdot \frac{1}{5}}\)
y = \(\frac{1}{6}\) ∙ 5
y = \(\frac{5}{6}\)

Question 18.
Math Journal y is inversely proportional to x. Describe how the value of y changes if the value of x is halved.
Answer:
Since y is proportional to x so the product of x and y will always be constant.
So, when the value of x is halved then the value of y will be doubted.
The value of y will be doubled.

Question 19.
The number of hours it takes to mow nine lawns is inversely proportional to the number of gardeners. It takes three gardeners four hours to mow nine lawns. How many hours would it take one gardener to mow the same nine lawns?
Math in Focus Grade 7 Chapter 5 Lesson 5.4 Answer Key Understanding Inverse Proportion 17
Answer:
Let x be the number of hours.
Let y be the number of gardeners.
The numbers of hours to mow nine lawns is inversely proportional to the number of gardeners.
Constant of proportionality :
x ∙ y = 4 ∙ 3
xy = 12
Inverse proportion equation :
xy = 12 write an inverse equation
When y = 1 and xy = 12 :
1 ∙ x = 12 Evaluate xy = 12 ‘when y = 1
x = 12 Simplify
It takes one gardeners 12 hours to mow nine lawns..
xy = 12 Write an inverse equation
12 hours

Question 20.
The number of minutes it takes to download a file is inversely proportional to the download speed. It takes Jolene 12 minutes to download a file when the download speed is 256 kilobytes per second. How long will it take her to download the same file if the download speed is 512 kilobytes per second?
Answer:
Let x be the number of minutes.
Let y be the download speed.
Number of minutes to download a file is inversely proportional to the download speed.
Constant of proportionality :
x ∙ y = 12 ∙ 256
xy = 3072
Inverse proportion equation :
xy = 3072 Write an inverse equation
When y = 512 and xy = 3072 :
512 ∙ x = 3072 Evaltate xy = 3072 when y = 512
512x = 3072 Simplify
\(\frac{512 x}{512}\) = \(\frac{3072}{512}\) Divide both sides by 512
x = 6 Simplify
It will take 6 minutes to download the same file if the download speed is 512 kilobytes per second.
6 minutes

Question 21.
Math journal Each table shows the price, y dollars, that x people have to pay to rent a guest house for one day. Describe how the two tables are alike, and how they are different. Be sure to discuss inverse proportion.
Math in Focus Grade 7 Chapter 5 Lesson 5.4 Answer Key Understanding Inverse Proportion 18
Answer:
Math in Focus Grade 7 Chapter 5 Lesson 5.4 Answer Key Understanding Inverse Proportion 019
For each pair of values, x and y ;
1 ∙ 240 = 240
2 ∙ 120 = 240
3 ∙ 80 = 240
4 ∙ 60 = 240
The value of x increases as the value of y decreases, and the product of x and y is a constant value.
So y is inversely proportional to x.
Constant of Proportionality = 240

Math in Focus Grade 7 Chapter 5 Lesson 5.4 Answer Key Understanding Inverse Proportion 020

For each pair of values, x and y ;
1 ∙ 240 = 240
2 ∙ 120 = 240
3 ∙ 85 = 255
4 ∙ 65 = 260
The value of x increases as the value of y decreases, but the product of x and y is not a constant value.
So y is not inversely proportional to x.
Alike : The two guest house are alike when the number of people are 1 and 2
Different : The two guest house are different because guest house A is inversely proportional whereas the guest house B is ‘not inversely proportional.
Guest house A: Number of people is Inversely proportional to the price of guest house.
Guest house B : Number of people is not Inversely proportional to the price of guest house.

Math Journal In questions 22 to 25, tell whether each relationship represents a direct or inverse proportion. Explain your answer.

Question 22.
A rectangle with length x inches and width y inches has an area of 50 square inches. The area is given by the equation xy = 50.
Answer:
Direct Proportion : y is directly proportional to x then they have a constant of proportionality k, and it can be rewritten as y = kx or \(\frac{y}{x}\) = k
Inverse Proportion : x and y are in Inverse proportion then their product is a constant of proportionality k, and it can be rewritten as xy = k or y = \(\frac{k}{x}\)
The given equation is inverse Proportion because it is written as xy = k
Inverse proportion

Question 23.
The density of a substance is the mass of the substance per unit of volume. A particular substance with a mass of m grams and a volume of v cubic centimeters has a density of 3 grams per cubic centimeter. An equation for the density of the substance is 3 = \(\frac{m}{v}\).
Answer:
Direct Proportion : y is directly proportional to x then they have a constant of proportionality k, and it. can be rewritten as y = kx or \(\frac{y}{x}\) = k
Inverse Proportion : x and y are in inverse proportion then their product is a constant of proportionality k. and it cain be rewritten as xy = k or y = \(\frac{k}{x}\)
3 = \(\frac{m}{v}\) The given equation is Direct Proportion because it is written as \(\frac{y}{x}\) = k
Direct Proportion

Question 24.
The music director at a school wants students to sell 200 tickets to the spring musical. The 200 tickets are distributed to the students in equal amounts so that each student gets y tickets to sell. The number of tickets each student gets is given by the equation y = \(\frac{200}{x}\).
Answer:
Direct Proportion : y is directly proportional to x then they have a constant of proportionality k, and it can be rewritten as y = kx or \(\frac{y}{x}\) = k
Inverse Proportion : x and y are in inverse proportion then their product is a constant of proportionality k, and it can be rewritten as y = k or y = \(\frac{k}{x}\)
y = \(\frac{200}{x}\)
xy = 200
The given equation is Inverse Proportion because it is written us xy = k
Inverse proportion

Question 25.
The sales tax you pay when you buy a new shirt is based on the amount the store charges for the shirt.
Answer:
The amount of sale tax depends on the number of shirt purchased.
Sale tax = Shirt price ∙ sale tax rate
Since sale tax is directly proportional to number of shirts
So, the equation is Directly Proportional
Directly Proportional

Brain @ Work

Question 1.
Tom is French, but lives in United States. On a visit to Germany, he saw a book that cost 25.99 euros plus 7% VAT (value-added tax). At that time, one euro was approximately equal to 0.726 U.S. dollars. In the United States, Tom could have bought the same book for 23.99 U.S. dollars plus 6% tax. Should Tom have bought the book in Germany? Explain your answer.
Answer:
Cost of book In Germany:

Price of book = 25.99 euros
VAT = 7%
Cost of book after VAT
= 25.99 + 25.99 ∙ 7%
= 25.99 + 25.99 ∙ \(\frac{7}{100}\)
= 25.99 + 1.81
= 27.80 euros
= 27.80 ∙ 0.726 Convert euros to dollars
= $20.18

Cost of book in United States:

Price of book = 23.99 dollars
Tax = 6%
Cost of book after VAT
= 23.99 + 23.99 ∙ 6%
= 23.99 + 23.99 ∙ \(\frac{6}{100}\)
= 23.99 + 1.43
= $25.09
Tom should bought the book from Germany because it was cheaper in Germany.
Germany

Question 2.
Johnny leaves Town P to drive to Town Q, a distance of 350 miles. He hopes to use only 12 gallons of gasoline. After traveling 150 miles, he checks his gauge and estimates that he has used 5 gallons of gasoline. At this rate, will he be able to reach Town Q before stopping for gasoline? Justify your answer.
Math in Focus Grade 7 Chapter 5 Lesson 5.4 Answer Key Understanding Inverse Proportion 19
Answer:
Distance between Town P and Town Q = 350 miles
Estimated gasoline he wili requin =12 galions
After traveling 150 miles he used 5 gallons of gasoline
Gasoline requires per miles = \(\frac{5}{150}\) = \(\frac{1}{30}\) gallons
Distance left to travel = 350 – 150 = 200 miles
Gasoline required for remaining distance
= 200 • \(\frac{1}{30}\)
= 6\(\frac{2}{3}\) gallons

Math in Focus Grade 7 Chapter 5 Lesson 5.3 Answer Key Solving Direct Proportion Problems

Practice the problems of Math in Focus Grade 7 Workbook Answer Key Chapter 5 Lesson 5.3 Solving Direct Proportion Problems to score better marks in the exam.

Math in Focus Grade 7 Course 2 A Chapter 5 Lesson 5.3 Answer Key Solving Direct Proportion Problems

Math in Focus Grade 7 Chapter 5 Lesson 5.3 Guided Practice Answer Key

Solve.
Question 1.
At a factory, the number of cars produced is directly proportional to the number of hours factory workers are making the cars. It takes 45 hours to make 60 cars. Use a proportion to find how long it will take to make 250 cars.
Method 1
Use a proportion.
Math in Focus Grade 7 Chapter 5 Lesson 5.3 Answer Key Solving Direct Proportion Problems 1

Method 2
Use a direct proportion equation.
Let x be the number of hours.
Let y be the number of cars. Define the variables.
Constant of proportionality:
\(\frac{y}{x}=\frac{?}{?}\)
Substitute y = Math in Focus Grade 7 Chapter 5 Lesson 5.3 Answer Key Solving Direct Proportion Problems 2 and x = Math in Focus Grade 7 Chapter 5 Lesson 5.3 Answer Key Solving Direct Proportion Problems 2
Simplify.

Direct proportion equation:
y = Math in Focus Grade 7 Chapter 5 Lesson 5.3 Answer Key Solving Direct Proportion Problems 2 x Write an equation.
When y = 250 and y = Math in Focus Grade 7 Chapter 5 Lesson 5.3 Answer Key Solving Direct Proportion Problems 2 x, 250 = Math in Focus Grade 7 Chapter 5 Lesson 5.3 Answer Key Solving Direct Proportion Problems 2 • x Evaluate y = Math in Focus Grade 7 Chapter 5 Lesson 5.3 Answer Key Solving Direct Proportion Problems 2 x when y = 250.
Math in Focus Grade 7 Chapter 5 Lesson 5.3 Answer Key Solving Direct Proportion Problems 2 • x = 250 Write an equivalent equation.
\(\frac{? \cdot x}{?}=\frac{250}{?}\) Divide both sides by Math in Focus Grade 7 Chapter 5 Lesson 5.3 Answer Key Solving Direct Proportion Problems 2.
x = Math in Focus Grade 7 Chapter 5 Lesson 5.3 Answer Key Solving Direct Proportion Problems 2 Simplify.
It takes Math in Focus Grade 7 Chapter 5 Lesson 5.3 Answer Key Solving Direct Proportion Problems 2 hours to produce 250 cars.
Answer:
It takes 187.5 hours to produce 250 cars.

Explanation:
Method 1:
Use a proportion.
Let number of hours for making 250 cars be x.
60 cars ÷ 45 hours = 250 cars ÷ x hours
=>60:45 = 250:x
=> 60 × x = 250 × 45
=> x = 11250 ÷ 60
=> x = 187.5 hours.

Method 2:
Use a direct proportion equation.
Let x be the number of hours.
Let y be the number of cars. Define the variables.
Constant of proportionality:
\(\frac{y}{x}=\frac{?}{?}\)
Substitute y = 60 and x = 45
\(\frac{y}{x}=\frac{60}{45}\)
y = 1.33 × x
=> y = 1.33x.
Direct proportion equation:
y = kx.
When y = 250
y = 1.33 x
=> 250 = 1.33 • x
=> 250 ÷ 1.33 = x
=> 187.5 hours.
It takes 187.5 hours to produce 250 cars.

Solve.
Question 2.
The number of pears for sale at an orchard, P, is directly proportional to the number of crates used to pack the pears, C. The table shows the relationship between the total number of pears for sale and the number of crates.
Math in Focus Grade 7 Chapter 5 Lesson 5.3 Answer Key Solving Direct Proportion Problems 3

a) Write a direct proportion equation that relates P and C.
Answer:
Direct proportion equation:
P = kC

Explanation:
direct proportion equation:
P = 200.
C = 10.
P = kC
200 = k × 10
200 ÷ 10 = k
20 = k.

b) Find the missing values in the table.
Answer:
Math-in-Focus-Grade-7-Course-2-A-Chapter-5-Lesson-5.3-Answer-Key-Solving-Direct-Proportion-Problems-Solve-2

Explanation:
P = ?? C = 8.
P = kC
=> P = 20 ×8
=> P = 160.
P = 500 C = ??.
P = kC
=> 500 = 20 × C
=> 500 ÷ 20 = C
=> 25 = C.

Question 3.
A store owner bought some handbags for $32 each from the manufacturer. Later, the store owner marked up the price of each handbag by $8. Use a proportion to find the percent increase in the price of the handbags.
Answer:
Percent increase in the price of the handbags = 20%.

Explanation:
Cost of the handbags a store owner bought from the manufacturer = $32 each.
Marked up price of the handbags a store owner bought later = $8 each.
Total price of the each handbag = Cost of the handbags a store owner bought from the manufacturer + Marked up price of the handbags a store owner bought later
= $32 + $8
= $40.
Percent increase in the price of the handbags = Marked up price of the handbags a store owner bought later ÷ Total price of the each handbag × 100
= 8 ÷ 40 × 100
= 0.2 × 100
= 20%.

Math in Focus Course 2A Practice 5.3 Answer Key

Write a direct variation equation and find the indicated value.
Question 1.
m varies directly as n, and m = 14 when n = 7.
a) Write an equation that relates m and n.
Answer:
Direct variation equation:
m = kn.

Explanation:
m = 14 when n = 7
m = kn
=> 14 = k × 7
=> 14 ÷ 7 = k
=> 2 = k.

b) Find m when n = 16.
Answer:
m = 32.

Explanation:
Direct variation equation:
m = kn.
=> m = 2 × 16
=> m = 32.

c) Find n when m = 30.
Answer:
n = 15.

Explanation:
m = 30.
Direct variation equation:
m = kn.
=> 30 = 2 × n
=> 30 ÷ 2 = n.
=> 15 = n.

 

Question 2.
p varies directly as q, and p = 6 when q = 30.
a) Write an equation that relates p and q.
Answer:
Direct variation equation:
q = kp.

Explanation:
p = 6 when q = 30.
Direct variation equation:
q = kp.
=> 30 = k × 6
=> 30 ÷ 6 = k
=> 5 = k.

b) Find q when p = 10.
Answer:
q = 50.

Explanation:
p = 10.
Direct variation equation:
q = kp.
=> q = 5 × 10
=> q = 50.

c) Find p when q = 7.
Answer:
p = 1.4.

Explanation:
q = 7.
Direct variation equation:
q = kp.
=> 7 = 5 × p
=> 7 ÷ 5 = p
=> 1.4 = p.

 

In each table, b is directly proportional to a. Copy and complete the table.
Question 3.
Math in Focus Grade 7 Chapter 5 Lesson 5.3 Answer Key Solving Direct Proportion Problems 4
Answer:
Math-in-Focus-Grade-7-Course-2-A-Chapter-5-Lesson-5.3-Answer-Key-Solving-Direct-Proportion-Problems-In each table, b is directly proportional to a-Copy and complete the table-3

Explanation:
b = 12. a = 4.
Direct variation equation:
b = ka
=> 12 = k × 4
=> 12 ÷ 4 = k
=> 3 = k.

b = 15  a = ??
b = ka
=> 15 = 3 × a
=> 15 ÷ 3 = a
=> 5 = a.

a = 19 b = ??
b = ka
=> b = 3 × 19
=> b = 57.

Question 4.
Math in Focus Grade 7 Chapter 5 Lesson 5.3 Answer Key Solving Direct Proportion Problems 5
Answer:
Math-in-Focus-Grade-7-Course-2-A-Chapter-5-Lesson-5.3-Answer-Key-Solving-Direct-Proportion-Problems-In each table, b is directly proportional to a-Copy and complete the table-4

Explanation:
b = 10. a = 4.
Direct variation equation:
b = ka
=> 10 = k × 4
=> 10 ÷ 4 = k
=> 2.5 = k.

b = 25  a = ??
b = ka
=> 25 = 2.5 × a
=> 25 ÷ 2.5 = a
=> 10 = a.

a = 16 b = ??
b = ka
=> b = 2.5 × 16
=> b = 40.

 

Solve. Show your work.
Question 5.
The amount of blood in a person’s body, a quarts, is directly proportional to his or her body weight, w pounds. A person who weighs 128 pounds has about 4 quarts of blood.
a) Find the constant of proportionality.
Answer:
Constant of proportionality: 32.

Explanation:
The amount of blood in a person’s body, a quarts, is directly proportional to his or her body weight, w pounds.
Direct variation equation:
w = k a.
=> 128 = k × 4
=> 128 ÷ 4 = k
=> 32 = k.

b) Write an equation that relates the amount of blood in a person’s body to his or her body weight.
Answer:
Equation: w = ka.

Explanation:
Direct variation equation:
w = k a.
=> 128 = k × 4
=> 128 ÷ 4 = k
=> 32 = k.

c) Find the weight of a person whose body has about 5 quarts of blood.
Answer:
160 is the weight of a person whose body has about 5 quarts of blood.

Explanation:
a = 5 quarts.
Direct variation equation:
w = k a.
=> w = 32 × 5
=> w = 160.

Question 6.
The height of a stack of books, H inches, is directly proportional to the number of books, n. The height of a stack of 10 books is 12 inches.
a) Find the constant of proportionality.
Answer:
Constant of proportionality:1.2

Explanation:
The height of a stack of books = H inches.
The number of books = n.
Direct variation equation:
H = kn
=> 12 = k × 10
=> 12 ÷ 10 = k.
=> 1.2 = k.

b) Write an equation that relates H and n.
Answer:
Direct variation equation:
H = kn

Explanation:
The height of a stack of books = H inches.
The number of books = n.
Direct variation equation:
H = kn.

c) Find the height of a stack of 24 books.
Answer:
28.8 inches is the height of a stack of 24 books.

Explanation:
The height of a stack of books = 24 inches.
The number of books = 24.
Direct variation equation:
H = kn
=> H = 1.2 × 24
=> H = 28.8 inches.

 

Question 7.
The total weight of n soccer balls is m ounces, m is directly proportional to n, and n = 12 when m = 54.
a) Find the weight per soccer ball.
Answer:
4.5 ounces is the weight per soccer ball.

Explanation:
The total weight of n soccer balls is m ounces.
Direct variation equation:
m = kn
=> 54 = k × 12
=> 54 ÷ 12 = k
=> 4.5 ounces.

b) Write an equation that relates n and m.
Answer:
m = kn is an equation that relates n and m.

Explanation:
The total weight of n soccer balls is m ounces.
Direct variation equation:
m = kn.

c) Find the value of m when n = 30.
Answer:
m = 135 ounces when n = 30.

Explanation:
n = 30.
Direct variation equation:
m = kn.
=> m = 4.5 × 30
=> m = 135 oounces.

 

Question 8.
The cost of CD cases, C, is directly proportional to the number of CD cases, n. The cost of 6 CD cases is $2.34.
a) Find the cost per CD case.
Answer:
Cost per CD case = $0.39.

Explanation:
The cost of CD cases = C.
Number of CD cases = n.
Total Cost of 6 CD cases = $2.34.
Cost per CD case = Total Cost of 6 CD cases ÷ Number of CD cases
= $2.34 ÷ 6
= $0.39

b) Write an equation that relates C and n.
Answer:
C = kn is an equation that relates C and n.

Explanation:
Direct variation equation:
C = kn.

c) Find the value of C when n is 7.
Answer:
C = $2.73 ; n = 7.

Explanation:
n = 7.
Direct variation equation:
C = kn.
=> C = $0.39 × 7
=> C = $2.73.

 

Use a proportion to solve each question. Show your work.
Question 9.
Five oranges cost $2. Find the cost of two dozen oranges.
Answer:
Cost of two dozen oranges = $0.96.

Explanation:
Number of oranges = 5.
Cost of 5 oranges = $0.2.
Cost of each orange = Cost of 5 oranges ÷ Number of oranges
= $0.2 ÷ 5
= $0.04.
dozen oranges = 12.
2 dozen oranges = 2 × 12 = 24.
Cost of two dozen oranges = Cost of each orange × 24
= $0.04 × 24
= $0.96.

 

Question 10.
It costs $180 to rent a car for 3 days. Find the cost of renting a car for 1 week.
Answer:
Cost of renting a car for 1 week = $420.

Explanation:
Cost of car for 3 days = $180.
Number of days = 3.
Cost of renting a car for a day = Cost of car for 3 days ÷ Number of days
=> $180 ÷ 3
=> $60.
Number of days in a week = 7.
Cost of renting a car for 1 week = Cost of renting a car for a day × Number of days in a week
= $60 × 7
= $420.

 

Question 11.
John drove 48 miles and used 2 gallons of gasoline. How many gallons of gasoline will he use if he drives 78 miles?
Answer:
Number of gallons of gasoline will he use if he drives 78 miles = 3.25.

Explanation:
Number of miles John drove = 48.
Number of gallons of gasoline = 2.
Number of gallons of gasoline used per mile = Number of miles John drove ÷ Number of gallons of gasoline
= 48 ÷ 2
= 24 miles per gasoline.
Number of miles if John drives = 78.
Number of gallons of gasoline will he use if he drives 78 miles = Number of miles if John drives ÷ Number of gallons of gasoline used per mile
= 78 ÷ 24
= 3.25.

 

Question 12.
Based on past experience, a caterer knows that the ratio of the number of glasses of juice to the number of people at a party should be 3 : 1. If 15 people are coming to a party, how many glasses of juice should the caterer have ready?
Answer:
Number of people are coming to a party = 45.

Explanation:
A caterer knows that the ratio of the number of glasses of juice to the number of people at a party should be 3 : 1.
=> Ratio of Number of glasses of juice to the number of people at a party = 3:1.
Number of people are coming to a party = 15.
Number of glasses of juices = ??
=> 3 ÷ 1 = ?? ÷ 15
=> 3 × 15 = ?? × 1
=> 45 = ??.

 

Question 13.
A recipe for meatloaf requires 10 ounces of ground beef. The recipe serves five people, and you would like to make enough for 8. How much ground beef should you use?
Answer:
Quantity of ground beef should be used = 16 ounces.

Explanation:
Quantity of ground beef a recipe for meatloaf requires = 10 ounces.
Number of people the recipe serves = 5.
Quantity of ground beef  for one person = Quantity of ground beef a recipe for meatloaf requires ÷ Number of people the recipe serves
= 10 ÷ 5
= 2 ounces.
Number of people the recipe serves would to make enough = 8.
Quantity of ground beef should be used = Number of people the recipe serves would to make enough × Quantity of ground beef  for one person
= 8 × 2 ounces
= 16 ounces.

 

Question 14.
George has to pay $30 in taxes for every $100 that he earns. Last summer he earned $3,680. How much did he pay in taxes?
Answer:
Amount of money he pays for tax = $1,104.

Explanation:
Amount of money George has to pay in taxes = $30.
Amount of money he earns = $100.
Amount of money he earns last summer = $3680.
Ratio:
Amount of money he earns ÷ Amount of money George has to pay in taxes
= $30 ÷ $100 = 30%
Amount of money he pays for tax = Amount of money he earns last summer × 30%
= $3680 × 30 ÷ 100
= $110400 ÷ 100
= $1,104.

Question 15.
Ti Marina wants to buy a sound system that costs $540. The sales tax rate in ‘ her state is 8.25%. How much sales tax must she pay?
Answer:
Amount of sales tax she must pay = $44.55.

Explanation:
Cost of sound system Ti Marina wants to buy = $540.
Sales percentage = 8.25%
Amount of sales tax she must pay = Cost of sound system Ti Marina wants to buy × Sales percentage
= $540 × 8.25%
= $4455 ÷ 100
= $44.55.

 

Question 16.
Jason mixes cans of yellow and blue paint to make green paint. The ratio of the number of cans of yellow paint to the number of cans of blue paint is 4 : 3. Jason needs to make more paint. He has 2 cans of yellow paint. How many cans of blue paint does he need to make the same shade of green?
Answer:
Number of blue paint cans he needs = 1.5.

Explanation:
The ratio of the number of cans of yellow paint to the number of cans of blue paint is 4 : 3.
Number of yellow cans he has = 2.
Number of blue paint cans he needs = ??
=> 4 ÷ 3 = 2 ÷ ??
=> 4 × ?? = 2 × 3
=> ?? = 6 ÷ 4
=> ?? = 1.5.

 

Question 17.
The area, A square feet, of the wall Ivan is painting is directly proportional to the time he spends painting the wall, T hours. It takes Ivan 4 hours to paint 113.6 square feet of the wall. How long will he take to paint 227.2 square feet of the wall?
Answer:
Time taken to paint 227.2 square feet = 8 hours.

Explanation:
Length of the wall Ivan is painting = A square feet.
Time he spends to paint the wall = T hours.
It takes Ivan 4 hours to paint 113.6 square feet of the wall.
Time taken to paint 227.2 square feet = ??
=> Ratio: 113.6 ÷ 4 = 227.2 ÷ ??
=> 113.6 × ?? = 227.2 × 4
=> ?? = 908.8 ÷ 113.6
=> ?? = 8 hours.

 

Question 18.
It takes Christy 2 hours to paint 5 model boats.
a) How long will it take her to paint 10 model boats?
Answer:
Number of hours she paints 10 model boats = 4.

Explanation:
Number of model boats Christy paints = 5.
Number of hours she paints = 2.
Number of hours she paints one model boats = Number of model boats Christy paints ÷ Number of hours she paints
= 5 ÷ 2
= 2.5.
Number of hours she paints 10 model boats =  10 ÷ Number of hours she paints one model boats
= 10 ÷  2.5
= 4.

b) How many model boats can she paint in 10 hours?
Answer:
Number of model boats Christy paints in 10 hours = 25.

Explanation:
Number of hours she paints one model boats = Number of model boats Christy paints ÷ Number of hours she paints
= 5 ÷ 2
= 2.5.
Number of model boats Christy paints in 10 hours = Number of hours she paints one model boats  × 10hours
= 2.5 × 10
= 25.

Question 19.
A commission is an amount of money earned by a sales person, based on the amount of sales the person makes. James works at a shop and earns 5.5% commission on his sales. Last month, he earned $265.32 in commission. Calculate his sales for that month.
Answer:
Sales he made that month  = $4,824.

Explanation:
Commission James works at a shop and earns on his sales= 5.5%
Amount of money he earned last month = $265.32.
Let the sales be X.
Sales he made that month  × 5.5% = Amount of money he earned last month
=> x × 5.5% = $265.32
=> x = $265.32 ÷ 5.5%
=> x = $26532 ÷ 5.5
=> x = $4,824.

 

Question 20.
An initial amount of money deposited in a bank account that earns interest is called the principal. In the table below, P stands for principal, and P stands for the interest earned by that principal for a period of one year at a particular bank. At this bank, the interest earned for a period of one year is directly proportional to the principal amount deposited.
Math in Focus Grade 7 Chapter 5 Lesson 5.3 Answer Key Solving Direct Proportion Problems 6
a) Write a direct proportion equation that relates l and P.
Answer:
Direct proportion equation: P = kI

Explanation:
Principal = $600.
Interest earned = $15
Direct proportion equation: P = kI
=> $600 = k × $15
=> $600 ÷ $15 = k
=> $40 = k.

b) Copy and complete the table.
Answer:
Math-in-Focus-Grade-7-Course-2-A-Chapter-5-Lesson-5.3-Answer-Key-Solving-Direct-Proportion-Problems-In each table, b is directly proportional to a-Copy and complete the table-20

Explanation:
Principal = $600.
Interest earned = $15
Direct proportion equation: P = kI
=> $600 = k × $15
=> $600 ÷ $15 = k
=> $40 = k.

Principal = $1000.
Interest earned = $??
Direct proportion equation: P = kI
=> $1000 = $40 × I
=> $1000 ÷ $40 = I
=> $25 = I.

Principal = $??
Interest earned = $56.
Direct proportion equation: P = kI
=> P = $40 × $56
=> P = $2,240.

Question 21.
Math journal y varies directly as x. Describe how the value of y changes when the value of x is tripled.
Answer:
The value of y changes when the value of x is tripled because y is directly proportional to x.

Explanation:
Direct variation describes a simple relationship between two variables.
y = kx.

Question 22.
Math Journal Jenny wants to buy some blackberries. Three stores sell blackberries at different prices:
Math in Focus Grade 7 Chapter 5 Lesson 5.3 Answer Key Solving Direct Proportion Problems 7
Which store has the best deal? Give your reasons.
Answer:
Store A sells the best deal because cost of 16 strawberries is $0.15 ounces less than other two stores.

Explanation:
1 pound(lb) is equal to 16 ounces(oz).
Cost of blackberries Store A sells = $2.40 per lb.
=> $2.40 ÷ 16 = $0.15 per ounces.
Cost of blackberries Store B sells = $1.28 per 8 ounces.
=> $1.28 ÷ 8 = $0.16 per ounces.
Cost of blackberries Store C sells = $1.08 per 6 ounces.
=> $1.08 ÷ 6 = $0.18 per ounces.

Math in Focus Grade 7 Chapter 5 Lesson 5.2 Answer Key Representing Direct Proportion Graphically

Practice the problems of Math in Focus Grade 7 Workbook Answer Key Chapter 5 Lesson 5.2 Representing Direct Proportion Graphically to score better marks in the exam.

Math in Focus Grade 7 Course 2 A Chapter 5 Lesson 5.2 Answer Key Representing Direct Proportion Graphically

Math in Focus Grade 7 Chapter 5 Lesson 5.2 Guided Practice Answer Key

Tell whether each graph represents a direct proportion. If so, find the constant of proportionality. Then write a direct proportion equation.

Question 1.
Math in Focus Grade 7 Chapter 5 Lesson 5.2 Answer Key Representing Direct Proportion Graphically 1
Answer:
The given graph is a straight line through the origin, and
It does not lie along the x — axis or y — axis.
So, It represents a direct proportion.

Because the graph passes through (1, 20),
The constant of proportionality is 20

The direct proportion equation Is :
\(\frac{y}{x}\) = \(\frac{20}{1}\)
x ∙ \(\frac{y}{x}\) = x ∙ \(\frac{20}{1}\)
y = 20x
The graph represent a Direct Proportion

Question 2.
Math in Focus Grade 7 Chapter 5 Lesson 5.2 Answer Key Representing Direct Proportion Graphically 2
Answer:
The given graph is a straight line that does not lie along
the x — axis or y — axis but does not pass through the origin.
So, the given graph does not represents a direct proportion.
The graph does not represent a Direct Proportion

Question 3.
Math in Focus Grade 7 Chapter 5 Lesson 5.2 Answer Key Representing Direct Proportion Graphically 3
Answer:
3.
The given graph passes through the origin that does not lie along
the x — axis or y — axis but It is not a straight line.
So, the given graph does not represents a direct proportion.
The graph does not represent a Direct Proportion

Complete.

Question 4.
Ms. Gray is driving on a long distance trip. The distance she travels is directly proportional to time she travels. The graph shows the distance she travels, y miles, after t hours.
Math in Focus Grade 7 Chapter 5 Lesson 5.2 Answer Key Representing Direct Proportion Graphically 4

a) Find the constant of proportionality. What is Ms. Gray’s driving speed in miles per hour?
Constant of proportionality: \(\frac{?}{?}\) = Math in Focus Grade 7 Chapter 5 Lesson 5.2 Answer Key Representing Direct Proportion Graphically 5
The constant of proportionality is Math in Focus Grade 7 Chapter 5 Lesson 5.2 Answer Key Representing Direct Proportion Graphically 5. So, Ms. Gray’s driving speed is Math in Focus Grade 7 Chapter 5 Lesson 5.2 Answer Key Representing Direct Proportion Graphically 5 miles per hour.
Answer:
Math in Focus Grade 7 Chapter 5 Lesson 5.2 Answer Key Representing Direct Proportion Graphically 12

constant of Proportionality :\(\frac{y}{t}\) = \(\frac{50}{1}\) = 50
The Constant of Proportionality is 50.
So, Ms. Gray’s driving speed is 50 miles per hour.

b) Write a direct proportion equation.
The direct proportion equation is y = Math in Focus Grade 7 Chapter 5 Lesson 5.2 Answer Key Representing Direct Proportion Graphically 5 t.
Answer:
\(\frac{y}{t}\) = \(\frac{50}{1}\)
t ∙ \(\frac{y}{t}\) = t ∙ \(\frac{50}{1}\)

c) Explain what the point (7, 350) represents in this situation.
It means that Ms. Gray travels Math in Focus Grade 7 Chapter 5 Lesson 5.2 Answer Key Representing Direct Proportion Graphically 5 miles in Math in Focus Grade 7 Chapter 5 Lesson 5.2 Answer Key Representing Direct Proportion Graphically 5 hours.
Answer:
It means that Ms. Gray travels 350 miles in 7 hours.

d) Find the distance traveled in 3 hours.
From the graph, the distance traveled is Math in Focus Grade 7 Chapter 5 Lesson 5.2 Answer Key Representing Direct Proportion Graphically 5 miles.
Answer:
From the graph, the distance traveled is 150 miles.

e) How long does it take Ms. Gray to travel 400 miles?
From the graph, it takes her Math in Focus Grade 7 Chapter 5 Lesson 5.2 Answer Key Representing Direct Proportion Graphically 5 hours to travel 400 miles.
Answer:
From the graph, it takes her 8 hours to travel 400 miles.

Math in Focus Course 2A Practice 5.2 Answer Key

Tell whether each graph represents a direct proportion. If so, find the constant of proportionality. Then write a direct proportion equation.

Question 1.
Math in Focus Grade 7 Chapter 5 Lesson 5.2 Answer Key Representing Direct Proportion Graphically 6
Answer:
The given graph is a straight line that does not lie along
the x — axis or y — axis but does not pass through the origin.
So, the given graph does not represents a direct proportion.
The graph does not represent a Direct Proportion

Question 2.
Math in Focus Grade 7 Chapter 5 Lesson 5.2 Answer Key Representing Direct Proportion Graphically 7
Answer:
The given graph is a straight line through the origin, and
It does not lie along the x — axis or y — axis.
So, it represents a direct proportion.
Because the graph passes through (1, 500),
The constant of proportionality is 500
The direct proportion equation is :
\(\frac{y}{x}\) = \(\frac{500}{1}\)
x ∙ \(\frac{y}{x}\) = x ∙ \(\frac{500}{1}\)
y = 500x
The graph represents a Direct Proportion

Question 3.
Math in Focus Grade 7 Chapter 5 Lesson 5.2 Answer Key Representing Direct Proportion Graphically 8
Answer:
In the given graph there is no straight line which show that there is no relation between x — axis and y — axis.
So, the given graph does not represents a direct proportion.
The graph does not represent a Direct Proportion

Question 4.
Math in Focus Grade 7 Chapter 5 Lesson 5.2 Answer Key Representing Direct Proportion Graphically 9
Answer:
The given graph is a straight line through the origin, and
It does not lie along the x — axis or y – axis.
So, It represents a direct proportion.
Because the graph passes through (1, 100),
The constant of proportionality is 100
The direct proportion equation is :
x ∙ \(\frac{y}{x}\) = x ∙ \(\frac{10}{1}\)
y = 100x
The graph represent a Direct Proportion

Solve. Show your work.

Question 5.
The cost of staying at a motel is directly proportional to the number of nights you stay. The graph shows the cost of staying at a motel, y dollars, for x nights.
Math in Focus Grade 7 Chapter 5 Lesson 5.2 Answer Key Representing Direct Proportion Graphically 10
a) Find the constant of proportionality. What does this value represent in this situation?
Answer:
\(\frac{y}{x}\) = \(\frac{50}{1}\)
\(\frac{y}{x}\) = 50
The Constant of Proportionality is 50

b) How much does it cost to stay at the motel for one week?
Answer:
From the graph, staying at monotel for 7 days is $350

Question 6.
MathJournal Explain how you can tell whether a line represents a direct proportion.
Answer:
A line represent a direct proportion when it follows all the below conditions :
1. → It is a straight line.
2. → The line should pass through the origin.
3. → The line should riot lie along x — axis or y — axis.

The line should be Straight line passes through the origin.

Question 7.
When you travel to another country, you can exchange U.S. dollars for the currency of that country. The amount of the new currency you get for your dollars depends on the exchange rate. The graph shows the amount of Mexican pesos, y, you could get if you were to exchange x U.S. dollars for pesos.
Math in Focus Grade 7 Chapter 5 Lesson 5.2 Answer Key Representing Direct Proportion Graphically 11
Answer:
Math in Focus Grade 7 Chapter 5 Lesson 5.2 Answer Key Representing Direct Proportion Graphically 10

a) Is the amount of pesos directly proportional to the amount of U.S. dollars?
Answer:
\(\frac{y}{x}\) = \(\frac{12}{1}\) = 12
\(\frac{y}{x}\) = \(\frac{24}{2}\) = 12
\(\frac{y}{x}\) = \(\frac{36}{3}\) = 12
Yes the the amount of pesos is directly proportional to the anzount of U. S. Dollars

b) How many pesos do you get for 3 U.S. dollars?
Answer:
From the graph, for 3 U. S. Dollars we get 36 pesos

c) Convert 24 pesos to U.S. dollars.
Answer:
$From\ the\ graph,\ when\ y=24\ then\ x=2$
24 pesos is equal to 2 U. S. Dollar.

d) What is the exchange rate when you convert dollars to pesos?
Answer:
$From the\ graph when\ x=1\ then\ y=12$
So, exchange rate is 12 pesos for 1 U. S. Dollar.

e) Write the direct proportion equation.
Answer:
\(\frac{y}{x}\) = \(\frac{12}{1}\)
x ∙ \(\frac{y}{x}\) = x ∙ \(\frac{12}{1}\)
y = 12x

Use graph paper. Solve.

Question 8.
Beth works at a pottery studio. She is making ceramic pots to sell at a craft fair. Graph the relationship between the number of ceramic pots she makes, y, and the number of days she works at the studio, x. Use 1 unit on the horizontal axis to represent 1 day and 1 unit on the vertical axis to represent 5 ceramic pots.
Math in Focus Grade 7 Chapter 5 Lesson 5.2 Answer Key Representing Direct Proportion Graphically 12

a) Determine whether the graph represents a direct proportion. If so, find the constant of proportionality and write the direct proportion equation.
Answer:

Math in Focus Grade 7 Chapter 5 Lesson 5.2 Answer Key Representing Direct Proportion Graphically 11

Yes, the given graph is direct proporhon
\(\frac{y}{x}\) = \(\frac{5}{1}\) = 5
\(\frac{y}{x}\) = \(\frac{10}{2}\) = 5
\(\frac{y}{x}\) = \(\frac{15}{3}\) = 5
\(\frac{y}{x}\) = \(\frac{20}{4}\) = 5
\(\frac{y}{x}\) = \(\frac{25}{5}\) = 5
\(\frac{y}{x}\) = \(\frac{30}{6}\) = 5
Constant of Proportionality = 5
Direct Proportion Equation :
\(\frac{y}{x}\) = 5
x ∙ \(\frac{y}{x}\) = x ∙ 5

b) Explain what the point (4, 20) represents in this situation.
Answer:
(4, 20) represent that in 4 days she can make 20 pots

c) How many pots can Beth make in 3 days?
Answer:
$From\ the\ graph,\ when\ x=3\ then\ y=l5$
In 3 days Beth can make 15 pots

d) Beth will not start selling pots until she has made at least 30. How long will it take her to make that many pots?
Answer:
$From\ the\ graph,\ when\ y=30\ then\ x=6$
So, Beth can make 30 pots in 6 days.

Math in Focus Grade 7 Chapter 5 Lesson 5.1 Answer Key Understanding Direct Proportion

Practice the problems of Math in Focus Grade 7 Workbook Answer Key Chapter 5 Lesson 5.1 Understanding Direct Proportion to score better marks in the exam.

Math in Focus Grade 7 Course 2 A Chapter 5 Lesson 5.1 Answer Key Understanding Direct Proportion

Hands-On Activity

Identify Direct Proportion In An Experiment
Math in Focus Grade 7 Chapter 5 Lesson 5.1 Answer Key Understanding Direct Proportion 2
Work in pairs
Math in Focus Grade 7 Chapter 5 Lesson 5.1 Answer Key Understanding Direct Proportion 3
Step 1: Make a table like the one shown.
Math in Focus Grade 7 Chapter 5 Lesson 5.1 Answer Key Understanding Direct Proportion 4
Step 2: Tape a yardstick to the
Step 3: Stand 1 foot away from the yardstick. Look at the yardstick through the cardboard tube. How many inches of the yardstick can you see? Record the number of inches in the table.
Step 4: Repeat Step 3 for the other values of L shown in the table. Then complete the table.

Math Journal What happens to H as L increases? Based on your observations, do you think H is directly proportional to L? Explain your thinking.

Math in Focus Grade 7 Chapter 5 Lesson 5.1 Guided Practice Answer Key

Copy and complete to determine whether y is directly proportional to x.
Question 1.
The table shows the distance traveled by a school bus, y miles, after x hours.
Math in Focus Grade 7 Chapter 5 Lesson 5.1 Answer Key Understanding Direct Proportion 5
For each pair of values, x and y:
Math in Focus Grade 7 Chapter 5 Lesson 5.1 Answer Key Understanding Direct Proportion 6
So, the distance traveled by the school bus is Math in Focus Grade 7 Chapter 5 Lesson 5.1 Answer Key Understanding Direct Proportion 7 to the number of hours it has traveled.
The constant of proportionality is Math in Focus Grade 7 Chapter 5 Lesson 5.1 Answer Key Understanding Direct Proportion 7. and represents the speed of the bus.
The direct proportion equation is Math in Focus Grade 7 Chapter 5 Lesson 5.1 Answer Key Understanding Direct Proportion 7.
Answer:
The constant of proportionality is 50 miles per hour and represents the speed of the bus.
The direct proportion equation: Distance = speed × time.

Explanation:
Distance travelled in 2 hours = 100 miles.
Distance = speed × time
=> 100 = speed × 2
=> 100 ÷ 2 = Speed
=> Speed = 50 miles per hour.
Distance = speed × time
=> 150 = speed × 3
=> 150 ÷ 3 = Speed
=> Speed = 50 miles per hour.
Distance = speed × time
=> 200 = speed × 4
=> 200 ÷ 4 = Speed
=> Speed = 50 miles per hour.
Total distance travelled = 100 + 150 + 200 = 450 miles.

 

Question 2.
The table shows the number of pitches made, y, in x innings of a baseball game.
Math in Focus Grade 7 Chapter 5 Lesson 5.1 Answer Key Understanding Direct Proportion 8
For each pair of values, x and y:
Math in Focus Grade 7 Chapter 5 Lesson 5.1 Answer Key Understanding Direct Proportion 9
So, the number of pitches made is Math in Focus Grade 7 Chapter 5 Lesson 5.1 Answer Key Understanding Direct Proportion 7 to the number of innings of a baseball game.
Answer:
The number of pitches made is 95  to the number of innings of a baseball game.

Explanation:
Number of pitches in 1st innings = 15.
Number of pitches in 2nd innings = 30.
Number of pitches in 3rd innings = 50.
Total number of pitches made = Number of pitches in 1st innings + Number of pitches in 2nd innings + Number of pitches in 3rd innings
= 15 + 30 + 50
= 45 + 50
= 95.

 

Tell whether each equation represents a direct proportion. If so, identify the constant of proportionality.
Question 3.
0.4y = x
0.4y = x
\( \frac{0.4 y}{?}=\frac{x}{?}\)
Divide both sides by Math in Focus Grade 7 Chapter 5 Lesson 5.1 Answer Key Understanding Direct Proportion 7.
y = Math in Focus Grade 7 Chapter 5 Lesson 5.1 Answer Key Understanding Direct Proportion 7 Simplify.
Because the original equation 0.4y = x Math in Focus Grade 7 Chapter 5 Lesson 5.1 Answer Key Understanding Direct Proportion 7 be rewritten as an equivalent equation in the form y = kx, it Math in Focus Grade 7 Chapter 5 Lesson 5.1 Answer Key Understanding Direct Proportion 7 a direct proportion. The constant of proportionality is Math in Focus Grade 7 Chapter 5 Lesson 5.1 Answer Key Understanding Direct Proportion 7.
Answer:
Because the original equation 0.4y = x direct proportion be rewritten as an equivalent equation in the form y = kx, it y n x are a direct proportion. The constant of proportionality is k.

Explanation:
0.4y = x
Divide both sides by 0.4.
0.4y ÷ 0.4 = x ÷ 0.4
=> y = x ÷ 0.4.
Because the original equation 0.4y = x direct proportion be rewritten as an equivalent equation in the form y = kx, it yn x are a direct proportion. The constant of proportionality is k.

Tell whether each equation represents a direct proportion. If so, find the constant of proportionality.
Question 4.
x = 1 – 2y
x = 1 – 2y
x + 2y = 1 – 2y + 2y Add 2y to both sides.
x + 2y – Math in Focus Grade 7 Chapter 5 Lesson 5.1 Answer Key Understanding Direct Proportion 7 = 1 – Math in Focus Grade 7 Chapter 5 Lesson 5.1 Answer Key Understanding Direct Proportion 7 Subtract Math in Focus Grade 7 Chapter 5 Lesson 5.1 Answer Key Understanding Direct Proportion 7 from both sides
2y = 1 – Math in Focus Grade 7 Chapter 5 Lesson 5.1 Answer Key Understanding Direct Proportion 7 Simplify.
\(\frac{2 y}{?}=\frac{1}{?}-\frac{?}{?}\) . Divide both sides by Math in Focus Grade 7 Chapter 5 Lesson 5.1 Answer Key Understanding Direct Proportion 7 .
y = Math in Focus Grade 7 Chapter 5 Lesson 5.1 Answer Key Understanding Direct Proportion 7 Simplify.
Because the original equation x = 1 – 2y Math in Focus Grade 7 Chapter 5 Lesson 5.1 Answer Key Understanding Direct Proportion 7 be rewritten as an equivalent equation in the form y = kx, it Math in Focus Grade 7 Chapter 5 Lesson 5.1 Answer Key Understanding Direct Proportion 7 a direct proportion.
Answer:
Because the original equation x = 1 – 2y not a direct proportion be rewritten as an equivalent equation in the form y = kx, it not a direct proportion.

Explanation:
x = 1 – 2y
Add both sides 2y.
=> x + 2y = 1 – 2y + 2y
=> x + 2y = 1.
=> 2y = 1 – x.
Divide both sides by 2.
=>2y ÷ 2 = (1 – x) ÷ 2
=> y = (1 – x) ÷ 2.

 

Copy and complete.
Question 5.
The table shows the number of baseballs, y, made in x days. The number of baseballs made is directly proportional to the number of days of production. Find the constant of proportionality and tell what it represents in this situation. Then write a direct proportion equation.
Math in Focus Grade 7 Chapter 5 Lesson 5.1 Answer Key Understanding Direct Proportion 10
Constant of proportionality: Math in Focus Grade 7 Chapter 5 Lesson 5.1 Answer Key Understanding Direct Proportion 7
The constant of proportionality is Math in Focus Grade 7 Chapter 5 Lesson 5.1 Answer Key Understanding Direct Proportion 7 and represents Math in Focus Grade 7 Chapter 5 Lesson 5.1 Answer Key Understanding Direct Proportion 7.
The direct proportion equation is Math in Focus Grade 7 Chapter 5 Lesson 5.1 Answer Key Understanding Direct Proportion 7.
Answer:
Constant of proportionality: 56.
The constant of proportionality is 56 and represents baseballs produced.
The direct proportion equation: y = kx.

Explanation:
Number of baseballs produced  on 1st day = 56.
=> y = kx
=> 56 = k × 1
=> 56 ÷ 1 = k
=> 56 = k.
Number of baseballs produced  on 2nd day = 112.
y = kx
=> 112 = k × 2
=> 112 ÷ 2 = k
=> 56 = k.
Number of baseballs produced  on 3rd day = 168.
y = kx
=> 168 = k × 3
=> 168 ÷ 3 = k
=> 56 = k.
Total number of baseballs produced = (Number of baseballs produced  on 1st day + Number of baseballs produced  on 2nd day + Number of baseballs produced  on 3rd day)
= 56 + 112 + 168
= 336.

Question 6.
A cafeteria sells sandwiches for $4 each. The amount Jason pays for some sandwiches is directly proportional to the number he buys. Write an equation that represents the direct proportion.
Let Math in Focus Grade 7 Chapter 5 Lesson 5.1 Answer Key Understanding Direct Proportion 7 be the number of sandwiches.
Let Math in Focus Grade 7 Chapter 5 Lesson 5.1 Answer Key Understanding Direct Proportion 7 be the amount Jason pays.
Cost per sandwich: $ Math in Focus Grade 7 Chapter 5 Lesson 5.1 Answer Key Understanding Direct Proportion 7 per sandwich
The direct proportion equation is Math in Focus Grade 7 Chapter 5 Lesson 5.1 Answer Key Understanding Direct Proportion 7 = Math in Focus Grade 7 Chapter 5 Lesson 5.1 Answer Key Understanding Direct Proportion 7.
Answer:
Cost per sandwich: $4 per sandwich
The direct proportion equation: Cost of sandwiches a cafeteria sells × Number of sandwiches Jason buys = Total amount of sandwiches Jason buys.

Explanation:
Cost of sandwiches a cafeteria sells = $4 each.
Let,
Number of sandwiches Jason buys = 10.
Total amount of sandwiches Jason buys = Cost of sandwiches a cafeteria sells × Number of sandwiches Jason buys
= $4 × 10
= $40.

Question 7.
q is directly proportional to p, and p = 12 when q = 24. Find the constant of proportionality. Then write a direct proportion equation.
Constant of proportionality: \(\frac{q}{p}=\frac{?}{?}\)
= Math in Focus Grade 7 Chapter 5 Lesson 5.1 Answer Key Understanding Direct Proportion 7 Write in simplest form.
The constant of proportionality is Math in Focus Grade 7 Chapter 5 Lesson 5.1 Answer Key Understanding Direct Proportion 7 .
The direct proportion equation is Math in Focus Grade 7 Chapter 5 Lesson 5.1 Answer Key Understanding Direct Proportion 7 .
Answer:
The constant of proportionality: 2.
The direct proportion equation: q = kp.

Explanation:
p = 12 when q = 24.
Direct proportion equation:
q = kp
=> 24 = k × 12
=> 24 ÷ 12 = k
=> 2 = k.
\(\frac{q}{p}=\frac{24}{12}\) = 2.

Solve.
Question 8.
w is directly proportional to h, and w = 18 when h = 3. Find the constant of proportionality. Then write a direct proportion equation.
Answer:
Constant of proportionality: 6.
Direct proportion equation: w = kh.

Explanation:
w = 18 when h = 3.
Direct proportion equation:
w = kh
=> 18 = k × 3
=> 18 ÷ 3 = k
=> 6 = k.
\(\frac{w}{h}=\frac{18}{3}\) = 6.

Math in Focus Course 2A Practice 5.1 Answer Key

Tell whether y is directly proportional to x. If so, find the constant of proportionality. Then write a direct proportion equation.
Question 1.
Math in Focus Grade 7 Chapter 5 Lesson 5.1 Answer Key Understanding Direct Proportion 11
Answer:
Constant of proportionality: 5.

Explanation:
Direct proportion equation:
y = kx
=> 5 = k × 1
=> 5 ÷ 1 = k
=> 5 = k.
y = kx
=> 10 = k × 2
=> 10 ÷ 2 = k
=> 5 = k.
y = kx
=> 15 = k × 3
=> 15 ÷ 3 = k
=> 5 = k.

Question 2.
Math in Focus Grade 7 Chapter 5 Lesson 5.1 Answer Key Understanding Direct Proportion 12
Answer:
Constant of proportionality: 65, 25, 11.67.

Explanation:
Direct proportion equation:
y = kx
=> 130 = k × 2
=> 130 ÷ 2 = k
=> 65 = k.
y = kx
=> 100 = k × 4
=> 100 ÷ 4 = k
=> 25 = k.
y = kx
=> 70 = k × 6
=> 70 ÷ 6 = k
=> 11.67 = k.

 

Question 3.
Math in Focus Grade 7 Chapter 5 Lesson 5.1 Answer Key Understanding Direct Proportion 13
Answer:
Constant of proportionality: 6.67, 6.67, 5.56.

Explanation:
Direct proportion equation:
y = kx
=> 20 = k × 3
=> 20 ÷ 3 = k
=> 6.67 = k.
y = kx
=> 40 = k × 6
=> 40 ÷ 6 = k
=> 6.67 = k.
y = kx
=> 50 = k × 9
=> 50 ÷ 9 = k
=> 5.56 = k.

 

Question 4.
Math in Focus Grade 7 Chapter 5 Lesson 5.1 Answer Key Understanding Direct Proportion 14
Answer:
Constant of proportionality: 25.

Explanation:
Direct proportion equation:
y = kx
=> 50 = k × 2
=> 50 ÷ 2 = k
=> 25 = k.
y = kx
=> 100 = k × 4
=> 100 ÷ 4 = k
=> 25 = k.
y = kx
=> 150 = k × 6
=> 150 ÷ 6 = k
=> 25 = k.

 

Tell whether each equation represents a direct proportion. If so, identify the constant of proportionality.
Question 5.
3y = \(\frac{1}{2}\) x
Answer:
Constant of proportionality: \(\frac{1}{6}\)

Explanation:
3y = \(\frac{1}{2}\) x
=>y = [\(\frac{1}{2}\) x] ÷ 3
=> y = \(\frac{1}{6}\) x

Question 6.
2y – 5 = x
Answer:
No, 2y – 5 = x is not direct proportion equation.

Explanation:
2y – 5 = x
=> 2y = x + 5
=> y = (x + 5) ÷ 2

Question 7.
p = 0.25q
Answer:
Constant of proportionality: 0.25

Explanation:
p = 0.25q
=> y = kx

Question 8.
4.5a = b + 12
Answer:
No, 4.5a = b + 12 is not direct proportion equation.

Explanation:
4.5a = b + 12
=> a = (b + 12) ÷ 4.5

Solve. Show your work.
Question 9.
The table shows the distance traveled, d miles, and the amount of gasoline used, n gallons. Tell whether d is directly proportional to n. If so, give the constant of proportionality and tell what it represents in this situation. Then write a direct proportion equation.
Math in Focus Grade 7 Chapter 5 Lesson 5.1 Answer Key Understanding Direct Proportion 15
Answer:
Constant of proportionality = 20.
Direct proportion equation: d = kn.

Explanation:
y = kx
=> d = kn
=> 20 = k × 1
=> 20 ÷ 1 = k
=> 20 = k.
d = kn
=> 40 = k × 2
=> 40 ÷ 2 = k
=> 20 = k.
d = kn
=> 60 = k × 3
=> 60 ÷ 3 = k
=> 20 = k.

Question 10.
The table shows the number of points scored, y, in x basketball games. Tell whether y is directly proportional to x. If so, give the constant of proportionality and tell what it represents in this situation. Then write a direct proportion equation.
Math in Focus Grade 7 Chapter 5 Lesson 5.1 Answer Key Understanding Direct Proportion 16
Answer:
Constant of proportionality: 24, 24, 26.67.
Direct proportion equation: y = kx.

Explanation:
y = kx
=> 24 = k × 1
=> 24 ÷ 1 = k
=> 24 = k.
y = kx
=> 48 = k × 2
=> 48 ÷ 2 = k
=> 24 = k.
y = kx
=> 80 = k × 3
=> 80 ÷ 3 = k
=> 26.67 = k.

Question 11.
The table shows the number of tennis balls produced, y, by x machines. Tell whether y is directly proportional to x. If so, give the constant of proportionality and tell what it represents in this situation. Then write a direct proportion equation.
Math in Focus Grade 7 Chapter 5 Lesson 5.1 Answer Key Understanding Direct Proportion 17
Answer:
Constant of proportionality: 20.
Direct proportion equation: y = kx.

Explanation:
y = kx
=> 20 = k × 1
=> 20 ÷ 1 = k
=> 20 = k.
y = kx
=> 60 = k × 3
=> 60 ÷ 3 = k
=> 20 = k.
y = kx
=> 100 = k × 5
=> 100 ÷ 5 = k
=> 20 = k.

Question 12.
Math Journal Describe how can you tell whether two quantities are in direct proportion.
Answer:
Two quantities a and b are said to be in direct proportion if they increase or decrease together. In other words, the ratio of their corresponding values remains constant.

Explanation:
Two quantities a and b are said to be in direct proportion if they increase or decrease together. In other words, the ratio of their corresponding values remains constant. This means that,
a/ b = k where k is a positive number, then the quantities a and b are said to vary directly.

 

Question 13.
Math Journal An equilateral triangle with a side length of c inches has a perimeter of P inches. The perimeter of the equilateral triangle is described by the equation P = 3c. Tell whether P is directly proportional to c. Explain your reasoning.
Answer:
P is directly proportional to c having k = 3.

Explanation:
Length of equilateral triangle side = c inches.
Perimeter of equilateral triangle side = p inches.
Equation of perimeter of the equilateral triangle: P = 3c.
Direct proportion equation: y = kx.
=> P = 3c.

 

Question 14.
Jim rode his bike at a steady rate of 20 miles per hour. Given that his distance, d miles, is directly proportional to the time he rides, t hours, identify the constant of proportionality and write a direct proportion equation.
Answer:
Constant of proportionality: 20.
Direct proportion equation: d = 20t.

Explanation:
Speed of  Jim rode his bike = 20 miles per hour.
Distance = d miles.
Time = t hours.
Distance = speed × time
= d = 20 × t
=> d = 20t miles.

 

Question 15.
Emily worked in a florist shop and earned $12 per hour. Given that the amount she earned, w dollars, is directly proportional to the time she worked, t hours, identify the constant of proportionality and write a direct proportion equation.
Answer:
Constant of proportionality: $12.
Direct proportion equation: w = S12t.

Explanation:
Amount of money Emily worked in a florist shop and earned = $12 per hour.
Number of hours = t.
Number of dollars she earned = w.
Amount of money she earned = Number of hours × Number of dollars
Number of dollars she earned = Number of hours × Amount of money Emily worked in a florist shop and earned
=>w = t × $12
=> w = S12t.

Question 16.
y is directly proportional to x, and y = 10 when x = 15. Write a direct proportion equation that relates x and y.
Answer:
Direct proportion equation: y = kx.

Explanation:
y = 10 when x = 15.
y = kx
=> 10 = k × 15
=> 10 × 15 = k
=> 2 ÷ 3  or \(\frac{2}{3}\) = k.

 

Question 17.
y is directly proportional to x, and y = 33 when x = 11. Write a direct proportion equation that relates x and y.
Answer:
Direct proportion equation: y = kx.

Explanation:
y = 33 when x = 11.
Direct proportion equation:
y = kx
=> 33 = k × 11
=> 33 ÷ 11 = k
=> 3 = k.

 

Question 18.
Karl hikes 3 miles in 45 minutes. Given that his distance is directly proportional to the time he walks, find the constant of proportionality and write an equation to represent the direct proportion.
Math in Focus Grade 7 Chapter 5 Lesson 5.1 Answer Key Understanding Direct Proportion 18
Answer:
Constant of proportionality: speed = 0.67 miles per hour.
Direct proportion equation: Distance = speed × time.

Explanation:
Distance Karl hikes = 3 miles.
Time Karl hikes = 45 minutes.
Let speed be x.
Distance = speed × time
=> 3 = x × 45
=> 3 ÷ 45 = x.
=> 0.67 miles per hour = x.

 

Question 19.
Paul pays $20 to download 16 songs. Given that the amount he pays is directly proportional to the number of songs he downloads, find the constant of proportionality and write a direct proportion equation.
Answer:
Constant of proportionality: Cost of each song = $1.25.
Direct proportion equation: Total amount of songs = Number of songs download × Cost of each song.

Explanation:
Amount of money Paul pays = $20.
Number of songs download = 16.
Cost of each song =
Total amount of songs = Number of songs download × Cost of each song
=> $20 = 16 × Cost of each song
=> $20 ÷ 16 = Cost of each song
=> $1.25 = Cost of each song.

 

Question 20.
Math journal Each table shows the cost of placing an advertisement in a newspaper, C dollars, for t days. Describe how the two tables are alike, and how they are different. Be sure to discuss direct proportion in your answer.
Math in Focus Grade 7 Chapter 5 Lesson 5.1 Answer Key Understanding Direct Proportion 19
Answer:
Constant of proportionality: Cost of placing an advertisement in a newspaper per day = 20 dollars.
Direct proportion equation: Total cost of advertisement ( c dollars) = Cost of placing an advertisement in a newspaper per day × Number of days.

Explanation:
Number of days of daily post = t = 5.
Total cost of advertisement ( c dollars) = 100.
Let Cost of placing an advertisement in a newspaper per day be x.
Total cost of advertisement ( c dollars) = Cost of placing an advertisement in a newspaper per day × Number of days of daily post
=> 100 = x × 5
=> 100 ÷ 5 = x
=> 20 dollars = x.
Number of days of evening star(t) = 5.
Total cost of advertisement (c dollars) = 80.
Let Cost of placing an advertisement in a newspaper per day be x.
Total cost of advertisement ( c dollars) = Cost of placing an advertisement in a newspaper per day × Number of days of evening star
=> 80 = x × 5
=> 80 ÷ 5 = x
=> 16 dollars = x.

Math in Focus Grade 7 Chapter 5 Answer Key Direct and Inverse Proportion

Practice the problems of Math in Focus Grade 7 Workbook Answer Key Chapter 5 Direct and Inverse Proportion to score better marks in the exam.

Math in Focus Grade 7 Course 2 A Chapter 5 Answer Key Direct and Inverse Proportion

Math in Focus Grade 7 Chapter 5 Quick Check Answer Key

Write a ratio to compare quantities.
A store sells 60 headphones, 45 sets of earbuds, and 80 speakers. Write a ratio in simplest form to compare each of the following.

Question 1.
The number of speakers to the number of sets of earbuds.
Answer:
Ratio of earbuds and speakers = 9:16.

Explanation:
Number of headphones a store sells = 60.
Number of earbuds a store sells = 45.
Number of speakers a store sells = 80.
Ratio of earbuds and speakers = Number of earbuds a store sells: Number of speakers a store sells
= 45:80
= 9:16.

Question 2.
The number of headphones to the number of speakers.
Answer:
Ratio of headphones and speakers = 3:4.

Explanation:
Number of headphones a store sells = 60.
Number of speakers a store sells = 80.
Ratio of headphones and speakers = Number of headphones a store sells: Number of speakers a store sells
= 60:80
= 6:8
= 3:4.

 

Tell whether each pair of ratios are equivalent.
Question 3.
9 : 11 and 18 : 22
Answer:
9 : 11 and 18 : 22 are equivalent pairs of ratios.

Explanation:
9:11
18:22 = 9:11.

Question 4.
\(\frac{1}{33}\) and \(\frac{33}{1}\)
Answer:
\(\frac{1}{33}\) and \(\frac{33}{1}\) are not equivalent pairs of ratios.

Explanation:
\(\frac{1}{33}\) = 1:33.
\(\frac{33}{1}\) = 33:1.

Question 5.
3 to 6 and 9 to 18
Answer:
3 to 6 and 9 to 18 are equivalent pairs of ratios.

Explanation:
3 to 6 = 3:6 = 1:2.
9 to 18 = 9:18 = 1:2.

 

Tell whether each ratio is in simplest form. Then write two ratios that are equivalent to the given ratio.
Question 6.
4 : 5
Answer:
4:5 ratio is in simplest form.

Explanation:
4:5
16:20 = 4:5.
12:15 = 4:5.

Question 7.
\(\frac{15}{100}\)
Answer:
\(\frac{15}{100}\) ratio is not in simplest form.

Explanation:
\(\frac{15}{100}\) = 15:100 = 3:20.

 

Question 8.
7 to 14
Answer:
7 to 14 ratio is not in simplest form.

Explanation:
7 to 14 = 7:14 = 1:2.

Find the unit rate.
Question 9.
The winner of the first Tour de France bicycle race in 1903 was Maurice Garin. It took him over 94 hours to complete 2,428 kilometers. Find his approximate average speed. Round your answer to the nearest whole number.
Answer:
Average speed of Maurice Garin = 30 kilometer/hour.

Explanation:
Number of kilometers Maurice Garin completes = 2,428.
Number of hours Maurice Garin takes = 94.
Average speed of Maurice Garin = Number of kilometers Maurice Garin completes ÷ Number of hours Maurice Garin takes
= 2,428 ÷ 94
= 25.83 kilometer/hour.
Nearest whole number of Average speed of Maurice Garin = 30 kilometer/hour.

Find and compare unit rates.
The cost of a food item at two different stores is shown. Find the unit price at each store and tell where the item costs less.

Question 10.
Store A: $3.20 for 16 oz of walnuts.
Store B: $2.30 for 10 oz of walnuts.
Answer:
Unit price at store A = $0.2.
Unit price at store B = $2.3.
Store A items costs less.

Explanation:
Store A: $3.20 for 16 oz of walnuts.
Store B: $2.30 for 10 oz of walnuts.
Unit price at store A = Cost of walnuts ÷ Number of walnuts
= $3.20 ÷ 16
= $0.2.
Unit price at store B = Cost of walnuts ÷ Number of walnuts
= $2.30 ÷ 10
= $2.3.

Question 11.
Store C: $2.13 for 3 Ib of potatoes.
Store D: $3.35 for 5 Ib of potatoes.
Answer:
Unit price at store C = $0.71.
Unit price at store D =$0.67.
Store D items costs less.

Explanation:
Store C: $2.13 for 3 Ib of potatoes.
Store D: $3.35 for 5 Ib of potatoes.
Unit price at store C = Cost of potatoes ÷ Number of potatoes
= $2.13 ÷ 3
= $0.71.
Unit price at store D = Cost of potatoes ÷ Number of potatoes
= $3.35 ÷ 5
= $0.67.

Use the coordinate plane below.

Math in Focus Grade 7 Chapter 5 Answer Key Direct and Inverse Proportion 1
Question 12.
Give the coordinates of points P, Q, R. S, and T.
Answer:
coordinates of point of  P = (1,2)
coordinates of point of Q = (2,4)
coordinates of point of R = (5,6)
coordinates of point of  S = (5,2)
coordinates of point of T = (6,3)

Explanation:
coordinates of point of  P = (1,2)
coordinates of point of Q = (2,4)
coordinates of point of R = (5,6)
coordinates of point of  S = (5,2)
coordinates of point of T = (6,3)

Solve word problems involving percent.
Question 13.
45% of the beads in a box are blue. If there are 36 blue beads in the box, how many beads are there altogether?
Answer:
Total number of beads altogether = 80.

Explanation:
45% of the beads in a box are blue.
Number of blue beads in the box = 36.
Let total beads be x.
=> 45% = 36
100% = x
=> x × 45% = 36 × 100%
=> x = (36 × 100%) ÷ 45%
=> x = (36 × 100) ÷ 45
=> x = 3600/45
=> x = 80.

Question 14.
Tabitha bought an antique model car priced at $72. She also had to pay 5% sales tax. What was the total amount she paid?
Answer:
Total amount she paid = $75.6.

Explanation:
Cost of an antique model car Tabitha bought = $72.
Sales tax an antique model car Tabitha bought = 5%
Total amount she paid = Cost of an antique model car Tabitha bought + (Sales tax an antique model car Tabitha bought × Cost of an antique model car Tabitha bought)
= $72 + (5% × $72)
= $72 + (1­ ÷ 20 × $72)
= $72 + ($3.6)
= $72 + $3.6
= $75.6.

Math in Focus Grade 1 Chapter 15 Practice 3 Answer Key Telling Time to the Half Hour

Go through the Math in Focus Grade 1 Workbook Answer Key Chapter 15 Practice 3 Telling Time to the Half Hour to finish your assignments.

Math in Focus Grade 1 Chapter 15 Practice 3 Answer Key Telling Time to the Half Hour

Match the clock to the correct time.

Question 1.
Math in Focus Grade 1 Chapter 15 Practice 3 Answer Key Telling Time to the Half Hour 1
Answer:
Math-in-Focus-Grade-1-Chapter-15-Practice-3-Answer-Key-Telling-Time-to-the-Half-Hour-1

Color the clock faces that show the correct time.

Question 2.
Math in Focus Grade 1 Chapter 15 Practice 3 Answer Key Telling Time to the Half Hour 2
Answer:
Math-in-Focus-Grade-1-Chapter-15-Practice-3-Answer-Key-Telling-Time-to-the-Half-Hour-2

Question 3.
Math in Focus Grade 1 Chapter 15 Practice 3 Answer Key Telling Time to the Half Hour 3
Answer:
Math-in-Focus-Grade-1-Chapter-15-Practice-3-Answer-Key-Telling-Time-to-the-Half-Hour-3

Question 4.
Math in Focus Grade 1 Chapter 15 Practice 3 Answer Key Telling Time to the Half Hour 4
Answer:
Math-in-Focus-Grade-1-Chapter-15-Practice-3-Answer-Key-Telling-Time-to-the-Half-Hour-4

Question 5.
Math in Focus Grade 1 Chapter 15 Practice 3 Answer Key Telling Time to the Half Hour 5
Answer:
Math-in-Focus-Grade-1-Chapter-15-Practice-3-Answer-Key-Telling-Time-to-the-Half-Hour-5

Question 6.
Math in Focus Grade 1 Chapter 15 Practice 3 Answer Key Telling Time to the Half Hour 6
Answer:
Math-in-Focus-Grade-1-Chapter-15-Practice-3-Answer-Key-Telling-Time-to-the-Half-Hour-6

The children go to the zoo with their parents. Look at the pictures. Write the correct times.

Math in Focus Grade 1 Chapter 15 Practice 3 Answer Key Telling Time to the Half Hour 7
Answer: The children go to the zoo with their parents at ten o’clock

Question 7.
They visit the bird and butterfly area at ___________
Math in Focus Grade 1 Chapter 15 Practice 3 Answer Key Telling Time to the Half Hour 8
Answer:

Question 8.
They look at the bears at ___________
Answer: They look at the bears at half past eleven

The pictures show what each child does on a Sunday. Look at the pictures. Then fill in the blanks.

Example

Math in Focus Grade 1 Chapter 15 Practice 3 Answer Key Telling Time to the Half Hour 9
Question

Math in Focus Grade 1 Chapter 15 Practice 3 Answer Key Telling Time to the Half Hour 10
Answer: At half past twelve (12:30) joe’s had her dinner

Question 9.
At ____ Ashley enjoys lunch with her mother.

Math in Focus Grade 1 Chapter 15 Practice 3 Answer Key Telling Time to the Half Hour 11

Answer: At 3 o’clock (03:00) Ashley enjoys lunch with her mother.

Question 10.
Pedro goes to the bowling alley with his Grandpa at ____
Answer:

Math in Focus Grade 1 Chapter 15 Practice 3 Answer Key Telling Time to the Half Hour 12

Question 11.
John walks his dog at
_____.
Answer: John walks his dog at half past five
05:30

Math in Focus Grade 7 Chapter 4 Review Test Answer Key

Practice the problems of Math in Focus Grade 7 Workbook Answer Key Chapter 4 Review Test to score better marks in the exam.

Math in Focus Grade 7 Course 2 A Chapter 4 Review Test Answer Key

Concepts and Skills

Solve each equation.

Question 1.
8x – 7 = 17
Answer:
x = 3

Explanation:
Given, 8x – 7 = 17

Add 7 on both sides.

8x – 7 + 7 = 17 + 7

8x = 24

Divide 8 on both sides.

8x ÷ 8 = 24 ÷ 8

x = 3

Question 2.
4 – 6x = 8
Answer:
x = -(2/3)

Explanation:
Given, 4 – 6x = 8

Subtract 4 on both sides.

4 – 6x – 4 = 8 – 4

-6x = 4

Divide -6 on both sides.

-6x ÷ -6 = 4 ÷ -6

x = -(2/3)

Question 3.
6 – \(\frac{y}{3}\) = 0
Answer:
y = 18

Explanation:
Give, 6 – (y/3) = 0

Subtract 6 on both sides.

6 – (y/3) – 6= 0 – 6

-(y/3) = -6

(y/3) = 6

Multiply 3 on both sides.

(y/3) × 3 = 6 × 3

y = 18

Question 4.
3 – 3.6x = 4.2
Answer:
x = -0.333

Explanation:
Given, 3 – 3.6x = 4.2

Subtract 3 on both sides.

3 – 3.6x – 3  = 4.2 – 3

-3.6x = 1.2

Divide  -3.6 on both sides.

-3.6x ÷ -3.6 = 1.2 ÷ -3.6

x = -0.333

Question 5.
7x – 5 = 3x + 4
Answer:
x = (3/2)

Explanation:
Given, 7x – 5 = 3x + 4

Add 5 on both sides.

7x – 5 + 5 = 3x + 4 + 5

7x = 3x + 9

Subtract 3x on both sides.

7x – 3x = 3x + 9 – 3x

4x  = 9

Divide 4  on both sides.

4x ÷ 4 = 9 ÷ 4

x = (9/4)

x = (3/2)

Question 6.
\(\frac{7}{10}\)y – \(\frac{1}{5}\) = \(\frac{3}{5}\)y + \(\frac{6}{5}\)
Answer:
y = 14

Explanation:
(7/10)y – (1/5) = (3/5)y + (6/5)

Add (1/5) on the both sides.

(7/10)y – (1/5) + (1/5) = (3/5)y + (6/5) + (1/5)

(7/10)y = (3/5)y + (7/5)

Subtract (3/5)y on both sides.

(7/10)y – (3/5)y = (3/5)y +(7/5) – (3/5)y

(7/10)y – ((3×2)/10)y =(7/5)

(7/10)y – (6/10)y =(7/5)

(y/10) = (7/5)

Multiply 1y = 140 on both sides.

(y/10) × 10 = (7/5) × 10

y = 14

Question 7.
3.4y – 5.2 – 3y = 2
Answer:
y = 18

Explanation:
Given, 3.4y – 5.2 – 3y = 2

0.4y – 5.2 = 2

Add 5.2 on both sides.

0.4y – 5.2 + 5.2 = 2 + 5.2

0.4y = 7.2

Divide 0.4 on both sides.

0.4y ÷ 0.4 = 7.2 ÷ 0.4

y = 18

Question 8.
15y – 4(2y – 3) = -2
Answer:
y = -2

Explanation:
Given, 15y – 4(2y – 3) = -2

15y – 8y + 12 = -2

7y + 12 = -2

Subtract 12 on both sides.

7y + 12 – 12= -2 – 12

7y = -14

Divide 7 on both sides.

7y ÷ 7= -14 ÷ 7

y = -2

Question 9.
\(\frac{1}{4}\)(x + 3) + \(\frac{3}{8}\)x = \(\frac{13}{4}\)
Answer:
x = 4

Explanation:
Given, (1/4)(x + 3) + (3/8)x = (13/4)

(1/4)x + (3/4) + (3/8)x = (13/4)

Subtract (3/4) on both sides.

(1/4)x + (3/4) + (3/8)x – (3/4) = (13/4) – (3/4)

(1/4)x + (3/8)x =(13-3)/4

(2 + 3)x/8 = 10/4

(5/8)x = (10/4)

Multiply 8 on both sides.

(5/8)x × 8 = (10/4) × 8

5x = 20

Divide 5 on both sides.

5x ÷ 5 = 20 ÷ 5

x = 4

Question 10.
0.4(x + 0.7) = 0.6x – 4.2
Answer:
x = 22.4

Explanation:
Given, 0.4(x + 0.7) = 0.6x – 4.2

0.4x + 0.28 =0.6x – 4.2

Add 4.2 on the both sides.

0.4x +0.28 + 4.2 =0.6x – 4.2 + 4.2

0.4x + 4.48 = 0.6x

Subtract 0.4x on the both sides.

0.4x + 4.48 – 0.4x = 0.6x – 0.4x

4.48 = 0.2x

0.2x = 4.48

Divide o.2 on both sides.

0.2x ÷ 0.2 = 4.48 ÷ 0.2

x = 22.4

Solve each inequality. Graph each solution set.

Question 11.
4x – 3 > 1
Answer:
x > 1

Explanation:
Given, 4x – 3 > 1

Add 3 on both sides.

4x – 3 + 3 > 1 + 3

4x > 4

x > 1

Question 12.
6 ≤ 1 – 5x
Answer:
-1 ≤ x

Explanation:
Given, 6 ≤ 1 – 5x

Subtract 6 on both sides.

6 – 6 ≤ 1 – 5x – 6

0 ≤ -5x – 5

0 ≤ -5 (x + 1)

Divide -5 on both sides.

0 ÷ -5  ≤ -5 (x + 1) ÷ -5

0 ≤ (x + 1)

-1 ≤ x

Question 13.
\(\frac{2}{3}\) – \(\frac{x}{6}\) ≥ –\(\frac{1}{2}\)
Answer:
x ≤ 1

Explanation:
(2/3) – (x/6) ≥ (1/2)

Subtract (2/3) on both sides.

(2/3) – (x/6) – (2/3) ≥ (1/2) – (2/3)

-(x/6) ≥ (3 – 4/6)

-(x/6) ≥ -(1/6)

Multiply -6 on both sides.

-(x/6) × (-6) ≥ -(1/6) × (-6)

x ≤ 1

Question 14.
-6.9 < 8.1 – 1.5x
Answer:
x > 10

Explanation:
Given, -6.9 < 8.1 – 1.5x

Subtract 8.1 on both sides.

-6.9 – 8.1 < 8.1 – 1.5x – 8.1

-15 < -1.5x

1.5x > 15

Divide 1.5 on both sides.

1.5x ÷ 1.5 > 15 ÷ 1.5

x > 10

Question 15.
9y – 5 ≤ 4y + 15
Answer:
y ≤  4

Explanation:
Given, 9y – 5 ≤ 4y + 15

Subtract 4y on both sides.

9y – 5 – 4y ≤ 4y + 15 – 4y

5y – 5 ≤ 15

Add 5 on both sides.

5y – 5 + 5 ≤ 15 + 5

5y ≤ 20

Divide 5 on both sides.

5y ÷ 5 ≤ 20 ÷ 5

y ≤  4

Question 16.
\(\frac{7}{9}\)x – \(\frac{2}{3}\) > \(\frac{1}{6}\)x + 3
Answer:
x > 6

Explanation:
Given, (7/9)x – (2/3) > (1/6)x + 3

Subtract (1/6)x on both sides.

(7/9)x – (2/3) – (1/6)x > (1/6)x + 3 – (1/6)x

(14 – 3)/18x – (2/3) > 3

11/18x – (2/3) > 3

Add (2/3) on both sides.

11/18x- (2/3) + (2/3) > 3 + (2/3)

11/18x > (11/3)

Divide 11 on both sides.

11/18x ÷ 11 > (11/3) ÷ 11

(1/18)x > (1/3)

Multiply 3 on both sides.

(1/18)x × 3  > (1/3) × 3

(1/6)x > 1

Multiply 6 on both sides.

(1/6)x × 6 > 1 × 6

x > 6

Question 17.
12.9 < 0.3(5.3 – x)
Answer:

Question 18.
3(x + 1) > 5x + 7
Answer:

Question 19.
\(\frac{1}{5}\)(4x – 1) ≥ \(\frac{2}{3}\)x + \(\frac{3}{5}\)
Answer:
x ≥ 6

Explanation:
Given, (1/5)(4x – 1) ≥ (2/3)x + (3/5)

Multiply 5 on both sides

(1/5)(4x – 1) × 5  ≥ ((2/3)x + (3/5)) × 5

(4x – 1) ≥  (10/3)x + 3

Add 1 on both sides.

(4x – 1) + 1 ≥  (10/3)x + 3 + 1

4x ≥  (10/3)x + 4

Subtract (10/3)x on both sides.

4x – (10/3)x ≥  (10/3)x + 4 – (10/3)x

(12 – 10)x/3 ≥ 4

(2/3)x ≥ 4

Multiply 3 on both sides.

(2/3)x × 3 ≥ 4 × 3

2x ≥ 12

Divide 2 on both sides.

2x ÷ 2  ≥ 12 ÷ 2

x ≥ 6

Question 20.
4(3 – 0.1x) ≤ 15 – 0.6x
Answer:
x ≤ 15

Explanation:
Given, 4(3 – 0.1x) ≤ 15 – 0.6x

12 -0.4x ≤ 15 – 0.6x

Subtract 12 on both sides.

12 -0.4x – 12 ≤ 15 – 0.6x – 12

-0.4x ≤ 3 -0.6x

Add 0.6x on both sides.

-0.4x + 0.6x ≤ 3 -0.6x + 0.6x

0.2x ≤ 3

Divide 0.2 on both sides.

0.2x ÷ 0.2 ≤ 3 ÷ 0.2

x ≤ 15

Problem Solving

Write an equation for questions 21 to 25. Solve and show your work.

Question 21.
Aiden wrote a riddle: Five less than \(\frac{1}{5}\) times a number is the same as the sum of the number and \(\frac{1}{3}\). Find the number.
Answer:
x = -(20/3)

Explanation:
Let us consider the number be ‘x’.

(1/5)x – 5 = x + (1/3)

Add 5 on both sides.

(1/5)x – 5 + 5 = x + (1/3) + 5

(1/5)x = x + (16/3)

Subtract (1/5)x on both sides.

(1/5)x – (1/5)x  = x + (16/3) – (1/5)x

0 = (4/5)x + (16/3)

(4/5)x = (- 16/3)

Multiply 5 on both sides.

(4/5)x × 5 = (- 16/3) × 5

4x = – 80/3

Divide 4 on both sides.

4x ÷ 4 = (- 80/3) ÷ 4

x = -(20/3)

Question 22.
Mary is 6 years older than her sister Kelly. The sum of their ages is 48. How old is Kelly?
Answer:
Kelly’s age, x = 21

Explanation:
Let us consider the age of the Kelly be ‘x’.

Age of Mary is  x +6

Sum of their ages = 48

x +x + 6 = 48

2x + 6 = 48

Subtract 6 on both sides.

2x + 6 – 6 = 48 – 6

2x = 42

Divide 2 on  both sides.

2x ÷ 2 = 42 ÷ 2

x = 21

Question 23.
The sum of the page numbers of two facing pages in a book is 145. What are the page numbers?
Answer:
x = 72; x + 1 = 73.

72, 73 are the page numbers.

Explanation:
Let us consider the numbers be x and x+1.

x + x +1 = 145

2x + 1 = 145

Subtract 1 on both sides.

2x + 1 – 1 = 145 – 1

2x = 144

Divide 2 on both sides.

2x ÷ 2 = 144 ÷ 2

x = 72

x + 1 = 73

The sum of numbers of two facing pages in a book = 145

Question 24.
The perimeter of an equilateral triangle is 6\(\frac{3}{4}\) inches. Find the length of each side of the equilateral triangle.
Answer:
Length of each side, a = (3/2)

Explanation:
Given, The perimeter of an equilateral triangle is 6(3/4).

3a = 6(3/4)

Divide 3 on both sides.

3a ÷ 3 = 6(3/4) ÷ 3

a = (6/4)

a = (3/2)

Question 25.
The sum of the interior angle measures of a quadrilateral is 360°. The measure of angle A is three times the measure of angle D. The measure of angle 6 is four times that of angle D. The measure of angle C is 24° more than angle B. Find the measure of each angle of the quadrilateral.
Math in Focus Grade 7 Chapter 4 Review Test Answer Key 1
Answer:

Write an inequality for each question. Solve and show your work.

Question 26.
Laura wants the average amount of money she spends each day on her four-day vacation to be no more than $64. On the first three days, she spends $71, $62, and $59. What is the greatest amount of money she can spend on the fourth day?
Answer:
64

Explanation:
Given, Mean = ax=n(Mean)(a1+...+an)

= 4(64) – (71 + 62 + 59)

= 256 – 192

= 64

Question 27.
Kevin plans to sign up for p hours of training at a culinary school. The school offers two payment options as shown below.
Math in Focus Grade 7 Chapter 4 Review Test Answer Key 2
For how many hours of training is Option B less expensive than Option A?
Answer:
x < 20, 20 hours.

Explanation:
Let us consider number of hours be ‘x’.

18x + 0 < 8x +200

18x < 8x + 200

Subtract 8x on both sides.

18x – 8x < 8x + 200 – 8x

10x < 200

Divide 10 on both sides.

10x ÷ 10 < 200 ÷ 10

x < 20

Question 28.
Peter has found a job in a computer store. As shown below, he has two options for how he will be paid. The commission he makes for Option B is based on his weekly sales. For example, if his sales total $1,000 a week, he receives his base salary of $250 plus 8% of $1,000.
Math in Focus Grade 7 Chapter 4 Review Test Answer Key 3
Peter is thinking about Option B. What would his weekly sales need to be for him to make at least as much as he would for Option A?
Answer:
The sales should be nearly above 4000$ per week to get more than Option A. Option A is better than Option B.

Explanation:

Let us consider ‘x’ be the number of sales.

600 ≥ 250 + 0.08x

Subtract 250 on both sides.

600 – 250 ≥ 250 + (0.08×1000)x – 250

350 ≥ 80x

Divide 80 on both sides.

350 ÷ 80 ≥ 80x ÷ 80

4 ≥ x

The sales should be nearly above 4000$ per week to get more than Option A.

so Option A is better than Option B.

Question 29.
The school events committee is planning to buy a banner and some helium balloons for graduation night. A store charges them $35 for the banner and $3.50 for each helium balloon. If the committee has at most $125 to spend, how many helium balloons can they buy?
Answer:
x ≤ 25

Explanation:
Let us consider number of helium balloons be ‘x’.

35 + 3.5x ≤ 125

Subtract 35 on both sides.

35 + 3.5x – 35 ≤ 125 – 35

3.5x ÷ 3.5  ≤ 90 ÷ 3.5

Divide 3.5 on both sides.

3.5x ≤ 90

x ≤ 25

Question 30.
The coach of the field hockey team can spend at most $475 on new team uniforms. The coach will order the uniforms online and pay a mailing cost of $6.50. If each uniform costs $29, how many uniforms can the coach order?
Answer:
x ≤ 16, 16 uniforms.

Explanation:
Let us consider the number of uniforms be ‘x’.

6.5 + 29x ≤ 475

Subtract 6.5 on both sides.

6.5 + 29x – 6.5 ≤ 475 – 6.5

29x ≤ 468.5

Divide 29 on both sides.

29x ÷ 29 ≤ 468.5 ÷ 29

x ≤ 16

Math in Focus Grade 7 Chapter 4 Lesson 4.5 Answer Key Real-World Problems: Algebraic Inequalities

Practice the problems of Math in Focus Grade 7 Workbook Answer Key Chapter 4 Lesson 4.5 Real-World Problems: Algebraic Inequalities to score better marks in the exam.

Math in Focus Grade 7 Course 2 A Chapter 4 Lesson 4.5 Answer Key Real-World Problems: Algebraic Inequalities

Math in Focus Grade 7 Chapter 4 Lesson 4.5 Guided Practice Answer Key

Solve.

Question 1.
The average of 87, 90, 89, and a fourth number is at least 90. Describe the value of the fourth number.
Answer:
x = 94

Explanation:
The average of the 4 number is 90.

The sum of the numbers = 4 × 90

= 360

Let us  consider the fourth number be ‘x’

The sum of the numbers = 360

87 +90 + 89 + x = 360

266 + x = 360

x = 360 – 266

x = 94

Copy and complete.

Question 2.
Grace is at the bookstore with $75 to spend. She plans to buy a reference book that costs $18 and some novels that cost $12 each. Find how many novels Grace can buy along with the reference book.
Let x be the number of novels Grace can buy. Define the variable.
Math in Focus Grade 7 Chapter 4 Lesson 4.5 Answer Key Real-World Problems Algebraic Inequalities 1 + 12x ≤ Math in Focus Grade 7 Chapter 4 Lesson 4.5 Answer Key Real-World Problems Algebraic Inequalities 1 Write an inequality.
Math in Focus Grade 7 Chapter 4 Lesson 4.5 Answer Key Real-World Problems Algebraic Inequalities 1 + 12x – Math in Focus Grade 7 Chapter 4 Lesson 4.5 Answer Key Real-World Problems Algebraic Inequalities 1Math in Focus Grade 7 Chapter 4 Lesson 4.5 Answer Key Real-World Problems Algebraic Inequalities 1Math in Focus Grade 7 Chapter 4 Lesson 4.5 Answer Key Real-World Problems Algebraic Inequalities 1 Subtract Math in Focus Grade 7 Chapter 4 Lesson 4.5 Answer Key Real-World Problems Algebraic Inequalities 1 from both sides.
12x ≤ Math in Focus Grade 7 Chapter 4 Lesson 4.5 Answer Key Real-World Problems Algebraic Inequalities 1 Simplify.
12 ÷ Math in Focus Grade 7 Chapter 4 Lesson 4.5 Answer Key Real-World Problems Algebraic Inequalities 1Math in Focus Grade 7 Chapter 4 Lesson 4.5 Answer Key Real-World Problems Algebraic Inequalities 1 ÷ Math in Focus Grade 7 Chapter 4 Lesson 4.5 Answer Key Real-World Problems Algebraic Inequalities 1 Divide both sides by Math in Focus Grade 7 Chapter 4 Lesson 4.5 Answer Key Real-World Problems Algebraic Inequalities 1.
x ≤ Math in Focus Grade 7 Chapter 4 Lesson 4.5 Answer Key Real-World Problems Algebraic Inequalities 1 Simplify.
Grace can buy at most Math in Focus Grade 7 Chapter 4 Lesson 4.5 Answer Key Real-World Problems Algebraic Inequalities 1 novels.
Answer:
x ≤ 4.75

Explanation:
The equation is 12x+18 ≤ 75

Subtract 18 on both sides.

12x+18 – 18 ≤ 75 – 18

12x ≤ 57

Divide 12 on both sides.

12x ÷ 12 ≤ 57 ÷ 12

x ≤ 4.75

Question 3.
While Cheryl is on vacation, she wants to put her dog Cocoa in a kennel that offers an obedience class. She calls two boarding kennels to find their fees for their services.
Math in Focus Grade 7 Chapter 4 Lesson 4.5 Answer Key Real-World Problems Algebraic Inequalities 2
After how many days will Best Dog Kennel be the cheaper option?
Let x be the number of days.
Math in Focus Grade 7 Chapter 4 Lesson 4.5 Answer Key Real-World Problems Algebraic Inequalities 3
For Best Dog Kennel to be the cheaper option,
80 + Math in Focus Grade 7 Chapter 4 Lesson 4.5 Answer Key Real-World Problems Algebraic Inequalities 1 > 100 + Math in Focus Grade 7 Chapter 4 Lesson 4.5 Answer Key Real-World Problems Algebraic Inequalities 1 Write an inequality.
80 + Math in Focus Grade 7 Chapter 4 Lesson 4.5 Answer Key Real-World Problems Algebraic Inequalities 1Math in Focus Grade 7 Chapter 4 Lesson 4.5 Answer Key Real-World Problems Algebraic Inequalities 1 > 100 + Math in Focus Grade 7 Chapter 4 Lesson 4.5 Answer Key Real-World Problems Algebraic Inequalities 1Math in Focus Grade 7 Chapter 4 Lesson 4.5 Answer Key Real-World Problems Algebraic Inequalities 1 Subtract Math in Focus Grade 7 Chapter 4 Lesson 4.5 Answer Key Real-World Problems Algebraic Inequalities 1 from both sides.
80 + Math in Focus Grade 7 Chapter 4 Lesson 4.5 Answer Key Real-World Problems Algebraic Inequalities 1 > 100 Simplify.
80 + Math in Focus Grade 7 Chapter 4 Lesson 4.5 Answer Key Real-World Problems Algebraic Inequalities 1 – 80 > 100 – 80 Subtract 80 from both sides.
Math in Focus Grade 7 Chapter 4 Lesson 4.5 Answer Key Real-World Problems Algebraic Inequalities 1 > 20 Simplify.
Math in Focus Grade 7 Chapter 4 Lesson 4.5 Answer Key Real-World Problems Algebraic Inequalities 1 ÷ Math in Focus Grade 7 Chapter 4 Lesson 4.5 Answer Key Real-World Problems Algebraic Inequalities 1 > 20 ÷ Math in Focus Grade 7 Chapter 4 Lesson 4.5 Answer Key Real-World Problems Algebraic Inequalities 1 Divide both sides by Math in Focus Grade 7 Chapter 4 Lesson 4.5 Answer Key Real-World Problems Algebraic Inequalities 1.
x > Math in Focus Grade 7 Chapter 4 Lesson 4.5 Answer Key Real-World Problems Algebraic Inequalities 1 Simplify.
Anytime after Math in Focus Grade 7 Chapter 4 Lesson 4.5 Answer Key Real-World Problems Algebraic Inequalities 1 days, Best Dog Kennel will be the cheaper option.
Answer:
x > 4

Anytime after 4 days, Best Dog Kennel will be the cheaper option.

Explanation:
Happy dogs kernel costs 80 + 20x

Best dogs kernel costs  100 + 15x

80 + 20x > 100 + 15x

Subtract 15x on both sides.

80 + 20x – 15x > 100 + 15x – 15x

80 + 5x > 100

Subtract 80 on both sides.

80 + 5x – 80 > 100 – 80

5x > 20

Divide 5 on both sides.

5x ÷ 5 > 20 ÷ 5

x > 4

Anytime after 4 days, Best Dog Kennel will be the cheaper option.

Math in Focus Course 2A Practice 4.5 Answer Key

Solve. Show your work.

Question 1.
The perimeter of an equilateral triangle is at most 45 centimeters. Find the possible length of each side.
Answer:
The length of each side, a = 15

Explanation:
Perimeter of an equilateral triangle = 45 cms

Perimeter = 3a = 45cms

3a = 45

Divide 3 on both sides.

3a ÷ 3 = 45 ÷ 3

a = 15

Question 2.
Roger scored 1,800 points in four rounds of a debate competition. His opponent, Sawyer, scored 324 points in the first round, 530 points in the second round, and 619 points in the third round. How many points must Sawyer score in the final round to surpass Roger’s score?
Answer:
Sawyer should score more than 327 in the fourth round to surpass Roger’s score.

Explanation:
Roger scored 1800 points in four rounds.

Sawyer scored 324 + 530 + 619 + x in four rounds

‘x’ be the points sawyer scored in the fourth round.

Sawyer score in the final round to surpass Roger’s score:
324 + 530 + 619 + x > 1800

1473 + x > 1800

Subtract 1473 on both sides.

1473 + x – 1473 > 1800 – 1473

x > 327

Sawyer should score more than 327 in the fourth round.

Question 3.
Ben plans to sign up for a language class that will cost at least $195. His father gives him $75 and he earns $28 from mowing the lawn for his neighbors. Write and solve an inequality to find out how much more money he needs to save before he can sign up for the class.
Answer:
x + 75 + 28 ≥ 195; Ben needs to save, x ≥ 92 

Explanation:
The equation for the above problem is, x + 75 + 28 ≥ 195

x + 103 ≥ 195

Subtract 103 on both sides.

x + 103 -103 ≥ 195 -103

x ≥ 92

Question 4.
In her last basketball game, Casey scored 46 points. In the current game, she has scored 24 points so far. How many more two-point baskets must she make if she wants her total score in her current game to be at least as great as her score in the last game?
Answer:
11 two-point baskets Casey should make to want her total score in the current game to be at least as great as the score in the last game.

Explanation:
Casey scored in the last basket ball game = 46

Casey scored in the current basket ball game = 24

Number of baskets she must make be ‘x’.

Let the equation be 24 + 2x ≥ 46

Subtract 24 on both sides.

24 + 2x – 24 ≥ 46 – 24

2x ≥ 22

Divide 2 on both sides.

2x ÷ 2 ≥ 22 ÷ 2

x ≥ 11

Question 5.
At Middleton Middle School, Marianne must score an average of at least 80 points on 4 tests before she can apply for the scholarship. If she scored 79, 81, and 77 for the first three tests, what must she score on her last test?
Answer:
The score of Marianne in her last test, x = 83

Explanation:
The average Marianne scored in the 4 tests = 80

The three tests scores are 79, 81 and 77

Let us consider the fourth test be ‘x’.

The total score =  the no of tests × average of Marianne scores

= 4 × 80

= 320

79 + 81 + 77 + x = 320

237 + x = 320

Subtract 237 on both sides.

237 + x – 237 = 320 – 237

x = 83

Question 6.
At the movies, a bag of popcorn costs $3.50 and a bottle of mineral water costs $2.75. If Madeline has $18 and bought only 2 bottles of water, how many bags of popcorn can she buy at most?
Math in Focus Grade 7 Chapter 4 Lesson 4.5 Answer Key Real-World Problems Algebraic Inequalities 4
Answer:
3.5 > x, Madeline can buy more than 3 bags.

Explanation:
Let the number of popcorn bags be  x’

Given,   18 > 2 × 2.75 + 3.50x

18 > 5.5 +3.50x

Subtract 5.5 on both sides.

18 – 5.5 > 5.5 +3.50x – 5.5

12.5 > 3.50x

Divide 3.50 on both sides.

12.5 ÷ 3.50 > 3.50x ÷ 3.50

3.5 > x

Question 7.
Party favors are on sale for $2.40 each. You have $380 to spend on the decorations and gifts, and you have already spent $270 on decorations. Write and solve an inequality to find the number of party favors you can buy.
Answer:
approximately, we can buy 45 party favors.

Explanation:
Money you have to spend on decorations and gifts = 380

Money already spent = 270

Remaining money = Money you have to spend on decorations and gifts – Money already spent

= 380 – 270

=110

Party favors are on sale for $2.40 each.

Number of party favor = 2.40x

‘x’ be the number of party favors.

2.40x < 110

Divide 2.40 on both sides.

2.40x ÷ 2.40 < 110 ÷ 2.40

x < 45.8

So approximately, we can buy 45 party favors.

Question 8.
Charlie wants to join a golf club. He finds two clubs that have fees as shown in the table.
Math in Focus Grade 7 Chapter 4 Lesson 4.5 Answer Key Real-World Problems Algebraic Inequalities 5
After how many months will Golf Club B be less expensive than Golf Club A?
Answer:
2 months.

Explanation:
Let us consider the number of months be ‘x’

80 + 45x > 110 + 30x

Subtract 30x on both sides.

80 + 45x – 30x > 110 + 30x – 30x

80 + 15x > 110

Subtract 80 on both sides.

80 + 15x – 80 > 110 – 80

15x > 30

Divide 15 on both sides.

15x ÷ 15 > 30 ÷ 15

x > 2

Question 9.
Molly can either take her lunch or buy it at school, It costs $1 .95 to buy lunch. If she wants to spend no more than $30 each month, how many lunches can she buy at most?
Answer:
15 days

Explanation:
Let us consider the number of days ‘x’

1.95x ≤ 30

Divide 1.95 on both sides.

1.95x ÷ 1.95 ≤ 30 ÷ 1.95

x ≤ 15.38

Approximately, 15 days

Question 10.
Tyson always likes to have at least $150 in his savings account. Currently he has $800 in the account. If he withdraws $35 each week, after how many weeks will the amount n his savings account be less than $150?
Answer:
18.5 weeks

Explanation:
Let us consider the number if weeks be ‘x’.

800 + 35x < 150

Subtract 150 on both sides.

800 + 35x – 150 < 150 – 150

35x < – 650

Divide 35 on both sides.

35x ÷ 35 < – 650 ÷ 35

x > 18.5

Question 11.
A cab company charges $0.80 per mile plus $2 for tolls. Melissa has at most $16 to spend on her cab fare. Write and solve an inequality for the maximum distance she can travel if she has at most $16 for cab fare. Can she afford to take a cab from her home to an airport that is 25 miles away?
Answer:
The maximum distance Melissa can travel is 17.5 miles. she cannot afford to travel to airport as 25 miles id greater than 17.5

Explanation:
Given, A cab company charges $0.80 per mile plus $2 for tolls.

Melissa has at most $16 to spend on her cab fare.

Let us consider the miles travelled e ‘x’.

0.8x + 2 ≤ 16

Subtract 2 on both sides.

0.8x + 2 – 2 ≤ 16 – 2

0.8x ≤ 14

Divide 0.8 on both sides.

0.8x ÷ 0.8 ≤ 14 ÷ 0.8

x ≤ 17.5

Question 12.
Nine subtracted from four times a number is less than or equal to fifteen. Write an inequality and solve it.
Answer:
4x-9≤15; x ≤ 6 

Explanation:
Give, Nine subtracted from four times a number is less than or equal to fifteen

i.e; 4x-9≤15

Add 9 on both sides.

4x-9 + 9 ≤ 15 + 9

4x ≤ 24

Divide 4 on both sides.

4x ÷ 4 ≤ 24 ÷ 4

x ≤ 6

Question 13.
Sixteen plus five times a number is more than the number minus eight. Write an inequality and solve it.
Answer:
16 + 5x > x – 8 ; x > -6

Explanation:
Given, Sixteen plus five times a number is more than the number minus eight.

i.e; 16 + 5x > x – 8

Add 8 on both sides.

16 + 5x + 8 > x – 8 + 8

24 + 5x > x

Subtract x on both sides.

24 + 5x – x > x -x

24 +4x > 0

4x > -24

Divide 4 on both sides.

4x ÷ 4 > -24 ÷ 4

x > -6

Question 14.
Math Journal Write a word problem that can be solved using an inequality. Write the inequality that represents your problem. Then solve it.
Answer:
x = 2

Explanation:
Problem: four less than three times a number is equal to the same number.

3x – 4 = x

Subtract x on both sides.

3x – 4 – x = x – x

2x -4 = 0

Add 4 on both sides.

2x -4 + 4 = 0 + 4

2x = 4

Divide 2 on both sides.

2x ÷ 2 = 4 ÷ 2

x = 2

Brain @ Work

A father said, “My son is five times as old as my daughter. My wife is five times as old as my son and I am twice as old as my wife. Grandmother here, who is as old as all of us put together, is celebrating her 81st birthday today.” What ¡s the age of the man’s son?
Math in Focus Grade 7 Chapter 4 Lesson 4.5 Answer Key Real-World Problems Algebraic Inequalities 6
Answer:

The age of man’s son is 5  years old.

Explanation:
Let us consider daughter’s age be ‘x’.

So, the son’s age is 5x

The wife’s age is “5 times old as son”, so 5 × 5x = 25x

His age is twice of this wife’s age, 2 × 25x = 50x

The sum of all  the ages  = 81

x + 5x + 25x + 50x = 81

81x = 81

Divide 81 on both sides.

81x ÷ 81 = 81 ÷ 81

x = 1

The age of man’s son is 5  years old.

Math in Focus Grade 7 Chapter 4 Lesson 4.4 Answer Key Solving Algebraic Inequalities

Practice the problems of Math in Focus Grade 7 Workbook Answer Key Chapter 4 Lesson 4.4 Solving Algebraic Inequalities to score better marks in the exam.

Math in Focus Grade 7 Course 2 A Chapter 4 Lesson 4.4 Answer Key Solving Algebraic Inequalities

Math in Focus Grade 7 Chapter 4 Lesson 4.4 Guided Practice Answer Key

Copy and complete. Solve each inequality and graph the solution set on a number line.

Question 1.
0.2x + 3 + 0.8x ≤ 4
0.2x + 3 + 0.8x ≤ 4
Math in Focus Grade 7 Chapter 4 Lesson 4.4 Answer Key Solving Algebraic Inequalities 1 + 3 ≤ 4 Add the like terms.
Math in Focus Grade 7 Chapter 4 Lesson 4.4 Answer Key Solving Algebraic Inequalities 1 + 3 Math in Focus Grade 7 Chapter 4 Lesson 4.4 Answer Key Solving Algebraic Inequalities 1 ≤ 4 – Math in Focus Grade 7 Chapter 4 Lesson 4.4 Answer Key Solving Algebraic Inequalities 1 Subtract Math in Focus Grade 7 Chapter 4 Lesson 4.4 Answer Key Solving Algebraic Inequalities 1 from both sides.
Math in Focus Grade 7 Chapter 4 Lesson 4.4 Answer Key Solving Algebraic Inequalities 1Math in Focus Grade 7 Chapter 4 Lesson 4.4 Answer Key Solving Algebraic Inequalities 1 Simplify.
Answer:
x ≤ 1

Explanation:
Given, 0.2x + 3 + 0.8x ≤ 4
x + 3 ≤ 4

Subtract 3 on both sides.

x + 3 – 3 ≤ 4 – 3

x ≤ 1

Question 2.
\(\frac{1}{4}\)x – 1 + \(\frac{3}{4}\)x > 0
\(\frac{1}{4}\)x – 1 + \(\frac{3}{4}\)x > 0
Math in Focus Grade 7 Chapter 4 Lesson 4.4 Answer Key Solving Algebraic Inequalities 1 – 1 > 0 Add the like terms.
Math in Focus Grade 7 Chapter 4 Lesson 4.4 Answer Key Solving Algebraic Inequalities 1 – 1 + Math in Focus Grade 7 Chapter 4 Lesson 4.4 Answer Key Solving Algebraic Inequalities 1 > 0 + Math in Focus Grade 7 Chapter 4 Lesson 4.4 Answer Key Solving Algebraic Inequalities 1 Add Math in Focus Grade 7 Chapter 4 Lesson 4.4 Answer Key Solving Algebraic Inequalities 1 to both sides.
Math in Focus Grade 7 Chapter 4 Lesson 4.4 Answer Key Solving Algebraic Inequalities 1 > Math in Focus Grade 7 Chapter 4 Lesson 4.4 Answer Key Solving Algebraic Inequalities 1 Simplify.
Answer:
x > 1

Explanation:
Given, (1/4)x – 1 + (3/4)x >0

x – 1 > 0

Add 1 on both sides.

x – 1 + 1 > 0 + 1

x > 1

Solve each inequality and graph the solution set on a number line.

Question 3.
2x + 3 < 13 + x
Answer:
x < 10

Explanation:
Given, 2x + 3 < 13 + x

Subtract x and 3 on both sides.

2x + 3 – x – 3 < 13 + x – x – 3

x < 10

Question 4.
1.5x – 3 ≥ 4 + 0.5x
Answer:
x ≥ 7

Explanation:
Given, 1.5x – 3 ≥ 4 + 0.5x

Subtract 0.5x o both sides.

1.5x – 3 – 0.5x ≥ 4 + 0.5x – 0.5x

1x – 3 ≥ 4

Add 3 on both sides.

x – 3 + 3 ≥ 4 + 3

x ≥ 7

Question 5.
4 + \(\frac{1}{3}\)x > 8 + \(\frac{4}{3}\)x
Answer:
x < 4

Explanation:
Given, 4 + (1/3)x > 8 + (4/3)x

Subtract (1/3)x on both sides.

4 + (1/3)x – (1/3)x > 8 + (4/3)x – (1/3)x

4 > 8 -x

Subtract 8 on both sides.

4 – 8 > 8 -x – 8

-x < -4

x < 4

Hands-On Activity

EXPLORE DIVISION AND MULTIPLICATIVE PROPERTIES OF AN INEQUALITY

Work individually.

Step 1: Use a copy of this table. Complete each Math in Focus Grade 7 Chapter 4 Lesson 4.4 Answer Key Solving Algebraic Inequalities 2 using the symbols > or <.
Math in Focus Grade 7 Chapter 4 Lesson 4.4 Answer Key Solving Algebraic Inequalities 3

Symbols

Math Journal What happens to the direction of the inequality symbol when you divide by a positive number? Based on your observation, write a rule for dividing both sides of an inequality by a positive number.

Math Journal What happens to the direction of the inequality symbol when you divide by a negative number? Based on your observation, write a rule for dividing both sides of an inequality by a negative number.

Step 2: Use a copy of this table. Complete each Math in Focus Grade 7 Chapter 4 Lesson 4.4 Answer Key Solving Algebraic Inequalities 2 using the symbols > or <.
Math in Focus Grade 7 Chapter 4 Lesson 4.4 Answer Key Solving Algebraic Inequalities 4

symbols2

Math Journal What happens to the direction of the inequality symbol when you multiply by a positive number? Based on your observation, write a rule for multiplying both sides of an inequality by a positive number.

Math Journal What happens to the direction of the inequality symbol when you multiply by a negative number? Based on your observation, write a rule for multiplying both sides of an inequality by a negative number.

Solve each inequality and graph the solution set on a number line.

Question 6.
–\(\frac{1}{5}\)w ≤ 2
Answer:
x ≥ -10

Explanation:
Given, -(1/5)x ≤ 2

Multiply 5 on both sides

-(1/5)x × 5  ≤ 2 × 5

-x ≤ 10

x ≥ -10

Question 7.
-7m > 21
Answer:
m < -3

Explanation:
Given, -7m > 21

Divide 7 on both sides.

-7m ÷ 7 > 21 ÷ 7

-m > 3

m < -3

Question 8.
6 > -0.3y
Answer:
y > -20

Explanation:
Given, 6 > -0.3y

Divide 0.3 on both sides.

6 ÷ 0.3 > -0.3y ÷ 0.3

20 > -y

y > -20

Solve each inequality and graph the solution set on a number line.

Question 9.
4y + 7 < 27
4y + 7 < 27
4y + 7 – Math in Focus Grade 7 Chapter 4 Lesson 4.4 Answer Key Solving Algebraic Inequalities 1 < 27 – Math in Focus Grade 7 Chapter 4 Lesson 4.4 Answer Key Solving Algebraic Inequalities 1 Subtract Math in Focus Grade 7 Chapter 4 Lesson 4.4 Answer Key Solving Algebraic Inequalities 1 from both sides.
4y < Math in Focus Grade 7 Chapter 4 Lesson 4.4 Answer Key Solving Algebraic Inequalities 1 Simplify.
4y ÷ Math in Focus Grade 7 Chapter 4 Lesson 4.4 Answer Key Solving Algebraic Inequalities 1 < Math in Focus Grade 7 Chapter 4 Lesson 4.4 Answer Key Solving Algebraic Inequalities 1 ÷ Math in Focus Grade 7 Chapter 4 Lesson 4.4 Answer Key Solving Algebraic Inequalities 1 Divide both sides by Math in Focus Grade 7 Chapter 4 Lesson 4.4 Answer Key Solving Algebraic Inequalities 1 .
y < Math in Focus Grade 7 Chapter 4 Lesson 4.4 Answer Key Solving Algebraic Inequalities 1 Simplify.
Answer:
y < 5

Explanation:
Given,  4y + 7 < 27

Subtract 7 on  both sides.

4y + 7 – 7 < 27 – 7

4y < 20

Divide 4 on both sides

4y ÷ 4 < 20 ÷ 4

y < 5

Question 10.
-5y – 9 > 21
-5y – 9 > 21
-5y – 9 + Math in Focus Grade 7 Chapter 4 Lesson 4.4 Answer Key Solving Algebraic Inequalities 1 ≥ 21 + Math in Focus Grade 7 Chapter 4 Lesson 4.4 Answer Key Solving Algebraic Inequalities 1 Add Math in Focus Grade 7 Chapter 4 Lesson 4.4 Answer Key Solving Algebraic Inequalities 1 to both sides.
-5y ≥ Math in Focus Grade 7 Chapter 4 Lesson 4.4 Answer Key Solving Algebraic Inequalities 1 Simplify.
– 5y ÷ Math in Focus Grade 7 Chapter 4 Lesson 4.4 Answer Key Solving Algebraic Inequalities 1Math in Focus Grade 7 Chapter 4 Lesson 4.4 Answer Key Solving Algebraic Inequalities 1 ÷ Math in Focus Grade 7 Chapter 4 Lesson 4.4 Answer Key Solving Algebraic Inequalities 1 Divide both sides by Math in Focus Grade 7 Chapter 4 Lesson 4.4 Answer Key Solving Algebraic Inequalities 1 and reverse the inequality symbol.
y ≤ Math in Focus Grade 7 Chapter 4 Lesson 4.4 Answer Key Solving Algebraic Inequalities 1 Simplify.
Answer:
y ≤ -6

Explanation:
Given, -5y – 9 > 21

Add 9 on both sides.

-5y – 9 + 9 > 21 + 9

– 5y > 30

Divide 5 on both sides.

– 5y ÷ 5 > 30 ÷ 5

-y < 6

y ≤-6

Question 11.
\(\frac{1}{2}\)x + \(\frac{3}{4}\) ≥ 5
Answer:
x ≥  (17/2)

Explanation:
Given, (1/2)x  + (3/4) ≥ 5

Subtract (3/4) on both sides.

(1/2)x  + (3/4) – (3/4) ≥ 5 – (3/4)

(1/2)x ≥  (20 – 3)/4

(1/2)x ≥  (17/4)

Multiply 2 on both sides

(1/2)x × 2 ≥  (17/4) × 2

x ≥  (17/2)

Question 12.
1.5 – 0.3y > 3.6
Answer:
y < -7

Explanation:
Given, 1.5 – 0.3y > 3.6

Subtract 1.5 on both sides.

1.5 – 0.3y – 1.5 > 3.6 – 1.5

-0.3y > 2.1

Divide -0.3 on both sides

-0.3y ÷ (-0.3) > 2.1 ÷ (-0.3)

y < -7

Question 13.
-8y + 32 ≤ -17 – y
Answer:
y ≥ 7

Explanation:
Given, -8y + 32 ≤ -17 – y

Add y on both sides

-8y + 32 + y ≤ -17 – y + y

-7y +32 ≤ -17

Subtract 32 on both sides.

-7y +32 – 32 ≤ -17 -32

-7y ≤ -49

Divide -7 on both sides.

-7y ÷ (-7) ≤ -49 ÷ (-7)

y ≥ 7

Question 14.
4(2 – y) ≥ 20
Answer:
y ≤ -3

Explanation:
Given, 4(2 – y) ≥ 20

8 – 4y ≥ 20

Subtract  8 on both sides

8 – 4y – 8 ≥ 20 – 8

-4y ≥ 12

Divide -4 on both sides

-4y ÷ (-4) ≥ 12 ÷ (-4)

y ≤ -3

Math in Focus Course 2A Practice 4.4 Answer Key

Solve each inequality using addition and subtraction. Then graph each solution set on a number line.

Question 1.
x + 8 > 14
Answer:
x > 6

Explanation:
Given, x + 8 > 14

Subtract 8 on both sides.

x + 8 – 8 > 14 – 8

x > 6

Question 2.
2 ≥ x – 12
Answer:
x ≤14

Explanation:
Given, 2 ≥ x – 12

Add 12 on both sides.

2 + 12 ≥ x – 12 + 12

14 ≥ x

x ≤14

Question 3.
-7x + 5 + 8x > 3
Answer:
x > -2

Explanation:
Given, -7x + 5 + 8x > 3

x + 5 > 3

Subtract 5 on both sides.

x + 5 – 5 > 3 – 5

x > -2

Question 4.
– 2x – 3 + 3x ≥ 12
Answer:
x ≥ 15

Explanation:
Given, -2x – 3 + 3x ≥ 12

x – 3 ≥ 12

Add 3 on both sides.

x – 3 + 3 ≥ 12 + 3

x ≥ 15

Question 5.
29 < \(\frac{2}{3}\)x + 14 + \(\frac{1}{3}\)x
Answer:
x>15

Explanation:
29 < (2/3)x + 14 + (1/3)x

Subtract 14 on both sides.

29 – 14 < (2/3)x + 14 + (1/3)x – 14

15 < x

x>15

Question 6.
\(\frac{1}{5}\)x + 9 + \(\frac{4}{5}\)x > -11
Answer:
x > -20

Explanation:
Given, (1/5)x + 9 + (4/5)x > -11

Subtract 9 on both sides.

(1/5)x + 9 + (4/5)x – 9 > -11 – 9

x > -20

Question 7.
0.7x + 4 + 0.3x ≤ 10
Answer:
x ≤ 6

Explanation:
Given, 0.7x + 4 + 0.3x ≤ 10

Subtract 4 on both sides.

0.7x + 4 + 0.3x – 4 ≤ 10 – 4

x ≤ 6

Question 8.
0.4x – 6 + 0.6x ≥ 19
Answer:
x ≥ 25

Explanation:
Given , 0.4x – 6 + 0.6x ≥ 19

x – 6 ≥ 19

Add 6 on both sides.

x – 6 + 6 ≥ 19 + 6

x ≥ 25

Question 9.
3x + 4 < 2x + 9
Answer:
x < 5

Explanation:
Given, 3x + 4 < 2x + 9

Subtract 4 on both sides.

3x + 4 – 4 < 2x + 9 – 4

3x < 2x + 5

Subtract 2x on both sides

3x – 2x < 2x + 5 – 2x

x < 5

Question 10.
8 – 4x > 12 – 3x
Answer:
x < -4

Explanation:
Given, 8 – 4x > 12 – 3x

Add 3x on both sides

8 – 4x + 3x > 12 – 3x +3x

8 -x > 12

Subtract 8 on both sides.

8 -x – 8 > 12 – 8

-x > 4

x < -4

Question 11.
\(\frac{2}{3}\)x + 2 ≥ 9 – \(\frac{1}{3}\)x
Answer:
x ≥ 7

Explanation:
Given, (2/3)x + 2 ≥ 9 – (1/3)x

Add (1/3)x on both sides

(2/3)x + 2 + (1/3)x ≥ 9 – (1/3)x + (1/3)x

x +2 ≥ 9

Subtract 2 on both sides.

x +2 – 2 ≥ 9 – 2

x ≥ 7

Question 12.
13 + 1\(\frac{3}{5}\)x ≥ 18 + \(\frac{3}{5}\)x
Answer:
x ≥ 5

Explanation:
Given, 13 + 1(3/5)x ≥ 18 + (3/5)x

13 + (8/5)x ≥ 18 + (3/5)x

Subtract (3/5)x on both sides.

13 + (8/5)x – (3/5)x ≥ 18 + (3/5)x – (3/5)x

13 + x ≥ 18

Subtract 13 on both sides.

13 + x – 13 ≥ 18 – 13

x ≥ 5

Question 13.
1.7x + 5 < 16 + 0.7x
Answer:
x < 11

Explanation:
Given, 1.7x + 5 < 16 + 0.7x

Subtract 0.7x on both sides.

1.7x + 5 – 0.7x < 16 + 0.7x – 0.7x

5  + x < 16

Subtract 5 on both sides.

5  + x – 5 < 16 – 5

x < 11

Question 14.
8.5 – 0.9x > 9.8 – 1.9x
Answer:
x < 1.3

Explanation:
Given, 8.5 – 0.9x > 9.8 – 1.9x

Add 0.9x on both sides.

8.5 – 0.9x + 0.9x > 9.8 – 1.9x + 0.9x

8.5 > 9.8 – x

Subtract 9.8 on both sides.

8.5  – 9.8 > 9.8 – x – 9.8

-1 .3 > – x

x < 1.3

Question 15.
Math Journal Solve the inequality 8 + 2x ≥ 12 and show your work. What value is a solution of 8 + 2x ≥ 12 but is not a solution of 8 + 2x > 12?
Answer:
x ≥ 2, x > 2 Both  the solutions are not same.

Explanation:
Given, two equations: 8 + 2x ≥ 12

8 + 2x > 12

Let us solve 1st equation: 8 + 2x ≥ 12

Subtract 8 on both sides.

8 + 2x – 8 ≥ 12 – 8

2x ≥ 4

Divide 2 on both sides

2x ÷ 2 ≥ 4 ÷ 2

x ≥ 2

Let us solve 2nd equation:
8 + 2x > 12

Subtract 8 on both sides.

8 + 2x – 8 > 12 – 8

2x > 4

Divide 2 on both sides

2x ÷ 2 > 4 ÷ 2

x > 2

Question 16.
Math Journal Eric solved the inequality 6y ≤ -18 as shown below:
6y ≤ -18
6y ÷ 6 ≥ -18 ÷ 6
y ≥ -3
Describe and correct the error that Eric made.
Answer:
y ≥ 3

Explanation:
Given, 6y ≤ -18

Divide 6 on both sides.

6y ÷ 6 ≥ -18 ÷ 6

y ≥ 3

Solve each inequality using division and multiplication. Then graph the solution set on a number line.

Question 17.
3 ≥ -3x
Answer:
x ≥ -1

Explanation:
Given, 3 ≥ -3x

Divide 3 on both sides.

3 ÷ 3 ≥ -3x ÷ 3

1 ≥ -x

x ≥ -1

Question 18.
-4x > 12
Answer:
x < -3

Explanation:
Given, -4x > 12

Divide -4 on both sides.

-4x ÷ – 4 > 12 ÷ – 4

x < -3

Question 19.
–\(\frac{x}{5}\) ≤ 2
Answer:
x ≥ -10

Explanation:
Given, -(x/5) ≤ 2

Multiply -5 on both sides.

-(x/5)  ≤ 2

-(x/5) × -5 ≤ 2 × -5

x ≥ -10

Question 20.
–\(\frac{2}{3}\)x > 8
Answer:
x < -12

Explanation:
Given, -(2/3)x > 8

Multiply 3 on both sides.

-(2/3)x × 3 > 8 × 3

-2x > 24

Divide -2 on both sides.

-2x ÷ -2 > 24 ÷ -2

x < -12

Question 21.
-0.2x ≥ 6
Answer:
x ≤ -30

Explanation:
Given, -0.2x ≥ 6

Divide -0.2 on both sides.

-0.2x ÷ -0.2 ≥ 6 ÷ -0.2

x ≤ -30

Question 22.
9 > -0.5x
Answer:
x > – 18

Explanation:
Given, 9 > -0.5x

Divide – 0.5 on both sides.

9 ÷ -0.5 > -0.5x ÷ -0.5

-18 >x

x > – 18

Solve each inequality using the four operations. Then graph each solution set on a number line.

Question 23.
7y – 3 > 11
Answer:
y > 2

Explanation:
Given, 7y – 3 > 11

Add 3 on both sides.

7y – 3 + 3 > 11 + 3

7y > 14

Divide 7 on both sides.

7y ÷ 7 > 14 ÷ 7

y > 2

Question 24.
-3a + 5 < -7
Answer:
a > 4

Explanation:
Given, -3a + 5 < -7

Subtract 5 on both sides.

-3a + 5 – 5 < -7 – 5

-3a < -12

Divide -3 on both sides.

-3a ÷ -3 < -12 ÷ -3

a > 4

Question 25.
\(\frac{x}{4}\) + \(\frac{3}{16}\) ≥ 1
Answer:
x ≥ (13/4)

Explanation:
(x/4) + (3/16) ≥ 1

multiply 4 on both sides.

((x/4) + (3/16) ) ×  4 ≥ 1 ×  4

x + (3/4) ≥ 4

Subtract (3/4) on both sides.

x + (3/4) – (3/4) ≥ 4 –  (3/4)

x ≥ (13/4)

Question 26.
\(\frac{3}{5}\)a – \(\frac{4}{5}\) < \(\frac{7}{10}\)
Answer:
a < (5/2)

Explanation:
Given, (3/5)a – (4/5) < (7/10)

Add (4/5) on both sides.

(3/5)a – (4/5) + (4/5) < (7/10) + (4/5)

(3/5)a < ((7+(4 × 2))/10)

(3/5)a < (7+8)/10

(3/5)a < 15/10

Divide 3 on both sides.

(3/5)a ÷ 3 < 15/10 ÷ 3

(a/5) < (5/10)

(a/5) < (1/2)

Multiply 5 on both sides.

(a/5) × 5 < (1/2) × 5

a < (5/2)

Question 27.

7 – 0.3x > 4
Answer:
x < 10

Explanation:
Given, 7 – 0.3x > 4

Subtract 7 on both sides.

7 – 0.3x – 7 > 4 – 7

-0.3x > -3

Divide -0.3 on both sides.

-0.3x ÷ -0.3 > -3 ÷ -0.3

x < 10

Question 28.
2.4y + 5 < 29
Answer:
y < 10

Explanation:
Given, 2.4y + 5 < 29

Subtract 5 on both sides.

2.4y + 5 – 5 < 29 – 5

2.4y < 24

Divide 2.4 on both sidea.

2.4y ÷ 2.4 < 24 ÷ 2.4

y < 10

Question 29.
5x + 3 < 7 + 7x
Answer:
x > -2

Explanation:
Given, 5x + 3 < 7 + 7x

Subtract 3 on both sides.

5x + 3 – 3 < 7 + 7x – 3

5x < 4 + 7x

subtract 5x on both sides.

5x -5x < 4 + 7x – 5x

0 < 4 + 2x

-4 < 2x

2x > -4

Divide 2 on both sides.

2x ÷ 2 > -4 ÷ 2

x > -2

Question 30.
11 – 7x ≤ 20 – 8x
Answer:
x ≤ 9

Explanation:
Given, 11 – 7x ≤ 20 – 8x

Subtract 11 on both sides.

11 – 7x – 11 ≤ 20 – 8x – 11

-7x ≤ 9 – 8x

8x – 7x ≤ 9

x ≤ 9

Question 31.
\(\frac{4}{3}\) – \(\frac{5}{6}\)x ≥ –\(\frac{1}{6}\) – \(\frac{2}{3}\)x
Answer:
x ≤ 7

Explanation:
Given, (4/3) – (5/6)x ≥ -(1/6) – (2/3)x

Add (2/3)x on both sides.

(4/3) – (5/6)x + (2/3)x ≥ -(1/6) – (2/3)x + (2/3)x

(4/3) – ((5x – 4x)/6) ≥ -(1/6)

(4/3) -(x/6) ≥ -(1/6)

Add (1/6) on both sides.

(4/3) -(x/6) + (1/6) ≥ -(1/6) + (1/6)

((6 + 1)/6) – x/6 ≥  0

(7/6) – x/6 ≥  0

7 – x ≥  0

7 ≥  x

x ≤ 7

Question 32.
\(\frac{2}{5}\)x + 4 ≤ \(\frac{7}{10}\)x – 8
Answer:
x ≥ 40

Explanation:
Given, (2/5)x + 4 ≤ (7/10)x – 8

Subtract 4 on both sides.

(2/5)x + 4 – 4 ≤ (7/10)x – 8 – 4

(2/5)x ≤ (7/10)x – 12

12 ≤ (7/10)x  – (2/5)x

12 ≤  (7 – 4)x/10

12 ≤ 3x/10

Multiply 10 on both sides.

12 × 10 ≤ 3x/10 × 10

120 ≤ 3x

Divide 3 on both sides.

120 ÷ 3 ≤ 3x ÷ 3

40 ≤ x

x ≥ 40

Question 33.
5.4x + 4.2 – 3.8x > 9
Answer:
x > 3

Explanation:
Given, 5.4x + 4.2 – 3.8x > 9

1.6x + 4.2 > 9

Subtract 4.2 on both sides.

1.6x + 4.2 – 4.2 > 9 – 4.2

1.6x > 4.8

Divide 1.6 on both sides.

1.6x > 4.8

1.6x ÷ 1.6 > 4.8 ÷ 1.6

x > 3

Question 34.
6.6 + 1.3x – 5.2x ≤ 14
Answer:

Explanation:
Given, 6.6 + 1.3x – 5.2x ≤ 14

Subtract 6.6 on both sides.

6.6 + 1.3x – 5.2x – 6.6 ≤ 14 – 6.6

-3.9x ≤ 7.4

Divide -3.9 on both sides.

-3.9x ÷ -3.9 ≤ 7.4 ÷ -3.9

x ≥ -(74/39)

Solve each inequality with parentheses using the four operations.

Question 35.
3(y + 2) ≤ 18
Answer:
y ≤ 4

Explanation:
Given, 3(y + 2) ≤ 18
Divide 3 on the both sides.

3(y + 2) ÷ 3 ≤ 18 ÷ 3

(y + 2) ≤ 6

Subtract 2 on both sides.

y + 2 – 2 ≤ 6 – 2

y ≤ 4

Question 36.
8(y – 1) > 24
Answer:
y > 4

Explanation:
Given, 8(y – 1) > 24

Divide 8 on both sides.

8(y – 1) ÷ 8 > 24 ÷ 8

y – 1 > 3

Add 1 on both sides.

y – 1 + 1 > 3 + 1

y > 4

Question 37.
\(\frac{1}{2}\)(a + 1) ≤ 4
Answer:
a ≤ 7

Explanation:
Given, (1/2) (a + 1) ≤ 4

Multiply 2 on both sides.

(1/2) (a + 1) × 2 ≤ 4 × 2

a +1 ≤ 8

Subtract 1 on both sides.

a +1 -1 ≤ 8 – 1

a ≤ 7

Question 38.
\(\frac{2}{3}\)(3 – a) < 3
Answer:
a > -(3/2)

Explanation:
Given, (2/3)(3 – a) < 3

Divide (2/3) on both sides.

(2/3)(3 – a) ÷ (2/3) < 3 ÷ (2/3)

(3 – a) < (9/2)

Subtract  3 on both sides.

3 – a – 3 < (9/2) – 3

-a  < (9-6)/2

a > -(3/2)

Question 39.
1.3(2 – x) > 3.9
Answer:
x < -1

Explanation:
Given, 1.3(2 – x) > 3.9

Divide 1.3 on both sides.

1.3(2 – x) ÷ 1.3 > 3.9 ÷ 1.3

2 – x > 3

Subtract 2 on both sides.

2 – x – 2 > 3 – 2

-x > 1

x < -1

Question 40.
3.6(5x – 1) < 5.4
Answer:
x < (1/2)

Explanation:
Given, 3.6(5x – 1) < 5.4

Divide 3.6 on both sides.

3.6(5x – 1) ÷ 3.6 < 5.4 ÷ 3.6

5x -1 < 1.5

Add 1 on both sides.

5x -1 + 1 < 1.5 + 1

5x < 2.5

Divide 5 on both sides.

5x ÷ 5 < 2.5 ÷ 5

x < (1/2)

Question 41.
4 + 2(1 – 3y) < 36
Answer:
y > -5

Explanation:
Given, 4 + 2(1 – 3y) < 36

Subtract 4 on both sides.

4 + 2(1 – 3y) – 4 < 36 – 4

2(1 – 3y) < 32

Divide 2 on both sides.

2(1 – 3y) ÷ 2 < 32 ÷ 2

1 – 3y < 16

Subtract 1 on both sides.

1 – 3y – 1 < 16 – 1

-3y < 15

Divide -3 on both sides.

-3y ÷ -3 < 15 ÷ -3

y > -5

Question 42.
2(3 – x) > 5x – 1
Answer:
x < 1

Explanation:
Given, 2(3 – x) > 5x – 1

6 – 2x > 5x – 1

Add 1 on both sides.

6 – 2x + 1 > 5x – 1 + 1

7 -2x > 5x

Add 2x on both sides.

7 -2x + 2x > 5x + 2x

7 > 7x

Divide 7 on both sides.

7 ÷ 7 > 7x ÷ 7

1 > x

x < 1

Question 43.
\(\frac{5}{9}\)(x + 1) ≥ \(\frac{2}{3}\)
Answer:
x ≥ (1/5)

Explanation:
Given, (5/9)(x + 1) ≥ (2/3)

Multiply 9 on both sides.

(5/9)(x + 1) × 9 ≥ (2/3) × 9

5 (x + 1) ≥  2 × 3

5 (x + 1) ≥ 6

Divide 5 on both sides.

5 (x + 1) ÷ 5  ≥ 6 ÷ 5

x + 1 ≥ (6/5)

Subtract 1 on both sides.

x + 1 – 1 ≥ (6/5) – 1

x ≥ (6-5)/5

x ≥ (1/5)

Question 44.
\(\frac{2}{3}\)(1 – 3x) > \(\frac{1}{6}\)
Answer:
x < (1/4)

Explanation:
Given,  (2/3) (1 – 3x) > (1/6)

Multiply 3 on both sides.

(2/3) (1 – 3x) × 3 > (1/6) × 3

2 (1 – 3x) > (1/2)

Divide 2 on both sides.

2 (1 – 3x) ÷ 2 > (1/2) ÷ 2

1 – 3x > (1/4)

Subtract 1 on both sides.

1 – 3x – 1 > (1/4) – 1

-3x > -(3/4)

Divide  -3 on both sides.

-3x ÷ -3 > -(3/4) ÷ -3

x < (1/4)

Question 45.
1.7 + 0.2(1 – x) ≥ 2.7
Answer:
x ≤ -4

Explanation:
Given, 1.7 + 0.2(1 – x) ≥ 2.7

1.7 + 0.2 -0.2x ≥ 2.7

1.9 – 0.2x ≥ 2.7

Subtract 1.9 on both sides.

1.9 – 0.2x – 1.9 ≥ 2.7 – 1.9

-0.2x ≥ 0.8

Divide -0.2 on both sides.

-0.2x ÷ -0.2 ≥ 0.8 ÷ -0.2

x ≤ -4

Question 46.
2.5(3 – 2x) + 1 ≥ 29
Answer:
x ≤ 4.1

Explanation:
Given, 2.5(3 – 2x) + 1 ≥ 29

7.5 – 5x + 1 ≥ 29

8.5 -5x ≥ 29

Subtract 8.5 on both sides.

8.5 -5x – 8.5 ≥ 29 – 8.5

-5x ≥ 20.5

Divide -5 on both sides.

-5x ÷ -5  ≥ 20.5 ÷ -5

x ≤ 4.1

Question 47.
Math Journal Compare solving the inequality -5(x + 6) < 10 with solving the equation -5(x + 6) = 10. Describe the similarities and differences between solving the inequality and solving the equation. Explain how the solution set of the inequality -5(x + 6) < 10 is different from the solution of the equation -5(x + 6) = 10.
Answer:
The solution of the equation -5(x + 6) < 10 is  x > -8, which represents the number x can be any number greater than -8.

The solution of the equation -5(x + 6) = 10 is x = -8, here the number x will be only -8 not any other number.

Explanation:
Given two equations are, -5(x + 6) < 10 and  -5(x + 6) = 10

Let us consider -5(x + 6) < 10

Divide -5 on both sides.

-5(x + 6) ÷ -5 < 10 ÷ -5

(x + 6) > -2

Subtract 6 on both sides.

x + 6 – 6 > -2 – 6

x > -8

Let us consider the second equation -5(x + 6) = 10

Divide -5 on both sides.

-5(x + 6) ÷ -5 = 10 ÷ -5

(x + 6) = -2

Subtract 6 on both sides.

x + 6 – 6 = -2 – 6

x = -8

Solve each inequality using the four operations.

Question 48.
10 – 3(4a – 3) < 2(3a – 4) – 9.2
Answer:
a > 2.01

Explanation:
Given, 10 – 3(4a – 3) < 2(3a – 4) – 9.2

10 – 12a + 9 < 6a – 8 – 9.2

19 – 12a < 6a – 17.2

Subtract  6a on both sides.

19 – 12a – 6a < 6a – 17.2 – 6a

19 – 18a < -17.2

Subtract 19 on both sides.

19 – 18a – 19 < -17.2 – 19

-18a <-36.2

Divide -18 on both sides.

-18a ÷ -18 <-36.2 ÷ -18

a > 2.01

Question 49.

7(2a – 3) ≤ 5 – 2(3a – 1)
Answer:
a ≤  (7/5)

Explanation:
Given, 7(2a – 3) ≤ 5 – 2(3a – 1)

14a – 21 ≤ 5 – 6a + 2

14a -21 ≤ 7 – 6a

Subtract 7 on both sides.

14a -21 – 7 ≤ 7 – 6a – 7

14a -28 ≤ -6a

Add 6a on both sides.

14a -28 + 6a ≤ -6a + 6a

20a ≤ 28

Divide 20 on both sides.

20a ÷ 20 ≤ 28 ÷ 20

a ≤  (7/5)

Math in Focus Grade 7 Chapter 4 Lesson 4.3 Answer Key Real-World Problems: Algebraic Equations

Practice the problems of Math in Focus Grade 7 Workbook Answer Key Chapter 4 Lesson 4.3 Real-World Problems: Algebraic Equations to score better marks in the exam.

Math in Focus Grade 7 Course 2 A Chapter 4 Lesson 4.3 Answer Key Real-World Problems: Algebraic Equations

Math in Focus Grade 7 Chapter 4 Lesson 4.3 Guided Practice Answer Key

Solve. Copy and complete.

Question 1.
Mark wrote a riddle: A negative number is \(\frac{2}{5}\) of another negative number. If the sum of the two negative numbers is -35, find the two negative numbers.
Method 1
Use bar models.
Math in Focus Grade 7 Chapter 4 Lesson 4.3 Answer Key Real-World Problems Algebraic Equations 1
The two negative numbers are Math in Focus Grade 7 Chapter 4 Lesson 4.3 Answer Key Real-World Problems Algebraic Equations 2 and Math in Focus Grade 7 Chapter 4 Lesson 4.3 Answer Key Real-World Problems Algebraic Equations 2.

Method 2
Use algebraic reasoning.
Let one of the numbers be x.
Then the other number is \(\frac{2}{5}\)x.
Math in Focus Grade 7 Chapter 4 Lesson 4.3 Answer Key Real-World Problems Algebraic Equations 3
Math in Focus Grade 7 Chapter 4 Lesson 4.3 Answer Key Real-World Problems Algebraic Equations 2 + Math in Focus Grade 7 Chapter 4 Lesson 4.3 Answer Key Real-World Problems Algebraic Equations 2 = -35 Write an equation.
Math in Focus Grade 7 Chapter 4 Lesson 4.3 Answer Key Real-World Problems Algebraic Equations 2 x = -35 Add the like terms.
Math in Focus Grade 7 Chapter 4 Lesson 4.3 Answer Key Real-World Problems Algebraic Equations 2 x ÷ Math in Focus Grade 7 Chapter 4 Lesson 4.3 Answer Key Real-World Problems Algebraic Equations 2 = Math in Focus Grade 7 Chapter 4 Lesson 4.3 Answer Key Real-World Problems Algebraic Equations 2 ÷ Math in Focus Grade 7 Chapter 4 Lesson 4.3 Answer Key Real-World Problems Algebraic Equations 2 Divide both sides by Math in Focus Grade 7 Chapter 4 Lesson 4.3 Answer Key Real-World Problems Algebraic Equations 2
x = Math in Focus Grade 7 Chapter 4 Lesson 4.3 Answer Key Real-World Problems Algebraic Equations 2 Simplify.
The other number: \(\frac{2}{5}\)x = \(\frac{2}{5}\) ∙ Math in Focus Grade 7 Chapter 4 Lesson 4.3 Answer Key Real-World Problems Algebraic Equations 2 Evaluate \(\frac{2}{5}\)x when x = Math in Focus Grade 7 Chapter 4 Lesson 4.3 Answer Key Real-World Problems Algebraic Equations 2.
= Math in Focus Grade 7 Chapter 4 Lesson 4.3 Answer Key Real-World Problems Algebraic Equations 2
The two negative numbers are Math in Focus Grade 7 Chapter 4 Lesson 4.3 Answer Key Real-World Problems Algebraic Equations 2 and Math in Focus Grade 7 Chapter 4 Lesson 4.3 Answer Key Real-World Problems Algebraic Equations 2.
Answer:
x = -25, (2/5)x = -10

Explanation:
Let one of the numbers be x.
Then the other number is (2/5)x.

x + (2/5)x = -35

(5x + 2x)/5 = -35

7x/5 = -35

Divide both sides by 7

7x/5 ÷ 7 = -(35) ÷ 7

(x/5) = -5

Multiply both sides by 5

(x/5) × 5 = -5 × 5

x = -25

(2/5)x = (2/5) × (-25)

= 2 × -5

= -10

Solve.

Question 2.
At an auditorium, tickets are sold for “circle seats” and “row seats.” There are 220 circle seats, and the rest of the seats are row seats. Each circle seat ticket costs $100 and each row seat ticket costs $60.
a) Write an expression for the total amount collected from the sale of all the seats at the auditorium.
Answer:
100(220) + 60(y – 220) = x

Explanation:
Let us consider the amount collected from the sale be ‘x’

Number of seats be ‘y’

Number of circle seats =220

Number of row seats = y – 220

Cost of each circle seat = $100

Cost of each row seat = $60

100(220) + 60(y – 220) = x

b) The total amount collected when all the tickets are sold is $68,800. How many row seat tickets are sold?
Answer:
x = 780

Explanation:
Number of row tickets be ‘x’

The total amount collected when all the tickets are sold = $68,800

With the above given details

100(220) + 60(x) = 68800

22000 + 60x =68800

subtract 22000 on both sides.

22000 + 60x – 22000 =68800 – 22000

60x = 46800

Divide 6 on both sides.

60x ÷ 60 = 46800 ÷ 60

x = 780

Copy and complete each Math in Focus Grade 7 Chapter 4 Lesson 4.3 Answer Key Real-World Problems Algebraic Equations 2 with a value and each Math in Focus Grade 7 Chapter 4 Lesson 4.3 Answer Key Real-World Problems Algebraic Equations 4 with +, -, ×, or ÷.

Question 3.
James has 16 more game cards than Fay. If they have 48 game cards altogether, find the number of game cards James has.
Math in Focus Grade 7 Chapter 4 Lesson 4.3 Answer Key Real-World Problems Algebraic Equations 5
Let the number of cards that Fay has be x.
Then the number of cards that James has is x Math in Focus Grade 7 Chapter 4 Lesson 4.3 Answer Key Real-World Problems Algebraic Equations 4 16.
Because they have 48 cards altogether,
x Math in Focus Grade 7 Chapter 4 Lesson 4.3 Answer Key Real-World Problems Algebraic Equations 4 (x Math in Focus Grade 7 Chapter 4 Lesson 4.3 Answer Key Real-World Problems Algebraic Equations 4 16) = Math in Focus Grade 7 Chapter 4 Lesson 4.3 Answer Key Real-World Problems Algebraic Equations 2
Math in Focus Grade 7 Chapter 4 Lesson 4.3 Answer Key Real-World Problems Algebraic Equations 2x Math in Focus Grade 7 Chapter 4 Lesson 4.3 Answer Key Real-World Problems Algebraic Equations 4 16 – Math in Focus Grade 7 Chapter 4 Lesson 4.3 Answer Key Real-World Problems Algebraic Equations 2
Math in Focus Grade 7 Chapter 4 Lesson 4.3 Answer Key Real-World Problems Algebraic Equations 2x Math in Focus Grade 7 Chapter 4 Lesson 4.3 Answer Key Real-World Problems Algebraic Equations 4 16 Math in Focus Grade 7 Chapter 4 Lesson 4.3 Answer Key Real-World Problems Algebraic Equations 4 Math in Focus Grade 7 Chapter 4 Lesson 4.3 Answer Key Real-World Problems Algebraic Equations 2 = Math in Focus Grade 7 Chapter 4 Lesson 4.3 Answer Key Real-World Problems Algebraic Equations 2 Math in Focus Grade 7 Chapter 4 Lesson 4.3 Answer Key Real-World Problems Algebraic Equations 4 Math in Focus Grade 7 Chapter 4 Lesson 4.3 Answer Key Real-World Problems Algebraic Equations 2
Math in Focus Grade 7 Chapter 4 Lesson 4.3 Answer Key Real-World Problems Algebraic Equations 2 x = Math in Focus Grade 7 Chapter 4 Lesson 4.3 Answer Key Real-World Problems Algebraic Equations 2
Math in Focus Grade 7 Chapter 4 Lesson 4.3 Answer Key Real-World Problems Algebraic Equations 2 x Math in Focus Grade 7 Chapter 4 Lesson 4.3 Answer Key Real-World Problems Algebraic Equations 4 Math in Focus Grade 7 Chapter 4 Lesson 4.3 Answer Key Real-World Problems Algebraic Equations 2 = Math in Focus Grade 7 Chapter 4 Lesson 4.3 Answer Key Real-World Problems Algebraic Equations 2 Math in Focus Grade 7 Chapter 4 Lesson 4.3 Answer Key Real-World Problems Algebraic Equations 4 Math in Focus Grade 7 Chapter 4 Lesson 4.3 Answer Key Real-World Problems Algebraic Equations 2
x = Math in Focus Grade 7 Chapter 4 Lesson 4.3 Answer Key Real-World Problems Algebraic Equations 2
Number of cards that James has:
x Math in Focus Grade 7 Chapter 4 Lesson 4.3 Answer Key Real-World Problems Algebraic Equations 4 16 Math in Focus Grade 7 Chapter 4 Lesson 4.3 Answer Key Real-World Problems Algebraic Equations 2 Math in Focus Grade 7 Chapter 4 Lesson 4.3 Answer Key Real-World Problems Algebraic Equations 4 16
= Math in Focus Grade 7 Chapter 4 Lesson 4.3 Answer Key Real-World Problems Algebraic Equations 2
James has Math in Focus Grade 7 Chapter 4 Lesson 4.3 Answer Key Real-World Problems Algebraic Equations 2 game cards.
Answer:
James has 32 game cards

Explanation:
Let the number of cards that Fay has be x.

Then the number of cards that James has is x + 16

Because they have 48 cards altogether,

x + x+ 16 = 48

2x + 16 = 48

Subtract 16 on both sides.

2x + 16 – 16 = 48 -16

2x = 32

Divide 2 on both sides

2x ÷ 2 = 32 ÷ 2

x = 16

number of cards that Fay, x = 16

the number of game cards James has x + 16 = 16 + 16

 

Math in Focus Course 2A Practice 4.3 Answer Key

Solve. Show your work.

Question 1.
Two sections of a garden are shaped like identical isosceles triangles. The base of each triangle is 50 feet, and the other two sides are each x feet long. If the combined perimeter of both gardens is 242 feet, find the value of x.
Math in Focus Grade 7 Chapter 4 Lesson 4.3 Answer Key Real-World Problems Algebraic Equations 6
Answer:
a = 35.5

Explanation:
Let us consider the sides be x

The perimeter of both the gardens = 242

2 × garden areas = 242

one garden are perimeter = 242 ÷ 2

one garden are perimeter = 121

Perimeter of a isosceles triangle = 2a +b

b =50

2a + 50 =121

Subtract 50 on both sides.

2a +50 – 50= 121 -50

2a = 71

Divide 2 on both sides

2a ÷ 2 = 71 ÷ 2

a = 35.5

Question 2.
Richard has a rectangular plot of land that is 525 feet long and y feet wide. He decides to build a fence around the plot. If the perimeter of the plot is 1,504 feet, find the value of y.
Answer:
y = 227

Explanation:
Given, perimeter of the plot = 1504

Length of the plot, l = 525

Breadth of the plot, b = y

Perimeter of the plot = 2(l + b)

1504 = 2 (525 + y)

Divide 2 on the both sides

1504 ÷ 2 = 2 (525 + y)÷ 2

752 = 525 + y

Subtract 525 on both sides.

752 – 525 = 525 + y -525

y = 227

Question 3.
The diagram shows an artificial lake. When Amanda jogged twice around the lake, she jogged a distance of 2,700 meters. Find the value of x.
Math in Focus Grade 7 Chapter 4 Lesson 4.3 Answer Key Real-World Problems Algebraic Equations 7
Answer:
x = 120

Explanation:

As Amanda jogged twice around the lake.

2(x + 140 + x + x+ 250+x +x + 3x) = 2700

2 (8x + 390) = 2700

Divide 2 on both sides

2 (8x + 390) ÷ 2 = 2700 ÷ 2

8x + 390 = 1350

Subtract 390 on both sides.

8x + 390 – 390 = 1350 -390

8x = 960

Divide 8 on both sides

8x ÷ 8= 960 ÷ 8

x = 120

Question 4.
Olivia wants to trim a lampshade with braid. The lampshade is shaped like a rectangular prism. The length of the base of the lampshade is 4 inches greater than its width. If the perimeter of the base is 54 inches, find the length of the base.
Math in Focus Grade 7 Chapter 4 Lesson 4.3 Answer Key Real-World Problems Algebraic Equations 8
Answer:

Explanation:
Given, perimeter of the base = 54 inches

Consider length of the lamp, l = (x +4)

x = (l – 4)

Consider breadth of the lamp, b = l

perimeter of the base = 54

2 (l + b) = 54

2 ((l – 4) + l) = 54

Divide 2 on both sides.

2 ((l – 4) + l) ÷ 2 = 54 ÷ 2

2l – 4 =  27

add 4 on both sides.

2l – 4 + 4 =  27 + 4

2l = 31

Divide 2 on both sides

2l ÷ 2 = 31 ÷ 2

l = 15.5

breadth, l = 15.5

Length, x = l -4 = 15.5 – 4 = 11.5

Verification:
2(l + b ) = 54

2(11.5 + 15.5)

2 (27)

54

Hence proved

Question 5.
Daphne was given a riddle to solve: The sum of two consecutive positive integers is 71. Find the two positive integers.
Answer:

Explanation:
Consider the two consecutive positive numbers be x, x +1

Given, The sum of two consecutive positive integers = 71

x + x +1 = 71

2x +1 = 71

Subtract 1 on both sides.

2x +1 -1 = 71 – 1

2x = 70

Divide 2 on both sides.

2x ÷ 2 = 70 ÷ 2

x = 35

since, x =35  ==>  x + 1 = 35 + 1 = 36

Verification:
x + x +1 = 71

35 + 36 = 71

Hence proved.

Question 6.
The sum of a negative number, \(\frac{1}{4}\) of the negative number, and \(\frac{7}{16}\) of the negative number is -13 \(\frac{1}{2}\). What is the negative number?
Answer:
-x = -9 (5/11)

Explanation:
-x(1/4) – x(7/16) =  -13 (1/2)

-11x/16 = -13(1/2)

Multiply 2 on both sides.

(-11x/16) × 2 = -13(1/2) × 2

(-11x/8) = -13

Multiply 8 on both sides.

(-11x/8) × 8 =-13 × 8

x = 9 (5/11)

Question 7.
3.5 times a positive number is equal to the sum of the positive number and 0.5. What is the positive number?
Answer:
x = 0.2

Explanation:
Let us consider the positive number be ‘x’

3.5x = x + 0.5

Subtract x on both sides.

3.5x – x = x + 0.5 – x

2.5x = 0.5

Divide 2.5 on both sides.

2.5x ÷ 2.5 = 0.5 ÷ 2.5

x = 0.2

Verification:

Positive Number x = 0.2

3.5x = x + 0.5

Substitute 0.2 in the equation.

3.5(0.2) = 0.2 + 0.5

0.7 = 0.7

Hence Proved.

Question 8.
Eugene wrote a riddle: A positive number is 5 less than another positive number. Six times the lesser number minus 3 times the greater number is 3. Find the two positive numbers.
Answer:
x = 11

Explanation:
Let us consider the two positive numbers be  ‘x’ and  ‘x -5’

Given, Six times the lesser number minus 3 times the greater number is 3

6 (x – 5) – 3(x) = 3

6x – 30 -3x = 3

3x – 30 = 3

Add 30 on both sides.

3x – 30 + 30 = 3 + 30

3x = 33

Divide 3 on both sides.

3x ÷ 3 = 33 ÷ 3

x = 11

Verification:
6 (x – 5) – 3(x) = 3

Substitute x = 11

6 (11 – 5) – 3(11)

= 6 × 6 – 33

= 36 – 33

= 3

Hence proved.

Question 9.
At a charity basketball game, 450 tickets were sold to students at a school. The remaining x tickets were sold to the public. The prices of the two types of tickets are shown. When all the tickets were sold, $10,500 was collected. How many tickets were sold to the public?
Math in Focus Grade 7 Chapter 4 Lesson 4.3 Answer Key Real-World Problems Algebraic Equations 9
Answer:
x = 150

Explanation:
Consider the number of tickets sold to public be ‘x’

The cost of each ticket sold to the student = 15$

Number of students at school the tickets were sold = 450

Cost of ticket to the public = 25$

Total amount collected when all the tickets were sold = $10,500

25(x) + 15(450) = 10500

25x + 6750 = 10500

Subtract 6750 on both sides.

25x + 6750 – 6750 = 10500 – 6750

25x = 3750

Divide 25 on both sides.

25x ÷ 25 = 3750 ÷ 25

x = 150

Verification:
25(x) + 15(450) = 10500

Substitute x = 150

25(x) + 15(450)

= 25(150) + 15(450)

= 3750 + 6750

= 10,500

Hence proved.

Question 10.
Henry ordered pizzas for a party and organized the information into a table. If Henry paid a total of $93.65, how many large cheese pizzas did he order?
Math in Focus Grade 7 Chapter 4 Lesson 4.3 Answer Key Real-World Problems Algebraic Equations 10
Answer:

Explanation:
Total amount Henry paid = 93.65$

Number of medium pepperoni pizza = 2

Price of one medium pepperoni pizza = $11.95

Number of large cheese be ‘x’

Price of one large cheese = $13.95

2(11.95) + x(13.95) = 93.65

23.9 + 13.95x = 93.65

Subtract 23.9 on both sides.

23.9 + 13.95x – 23.9 = 93.65 – 23.9

13.95x = 69.75

Divide 13.95 on both sides.

13.95x ÷ 13.95 = 69.75 ÷ 13.95

x = 5

Verification:
2(11.95) + x(13.95) = 93.65

23.9 + 13.95x

Substitute x = 5

23.9 + 13.95(5)

= 23.9 + 69.75

= 93.65

Hence proved.

Question 11.
Marvin saved dimes and quarters in his piggy bank to buy a gift for his mother. He counted his savings and organized the information in a table.
Math in Focus Grade 7 Chapter 4 Lesson 4.3 Answer Key Real-World Problems Algebraic Equations 11
If he saved $11, how many dimes and quarters did Marvin have?
Answer:
Dime, x = 40 ; Quarter, x- 12 = 28

Explanation:
Given, Number of dime coins = x

Number of quarter coins = x – 12

Value of one dime coin = 0.10$

Value of one quarter coin = 0.25$

0.10x + 0.25(x – 12) =11

0.10x + 0.25x – 3 =11

Add 3 on both sides.

0.10x + 0.25x – 3 + 3 =11 + 3

0.35x =14

Divide 0.35 on both sides.

0.35x ÷ 0.35 =14 ÷ 0.35

x = 40

x – 12 = 40 -12 = 28

Verification:
0.10x + 0.25(x – 12) =11

Substitute x = 40

0.10(40)+ 0.25(40- 12)

= 4 + 7

= 11

Question 12.
A bike shop charges x dollars to rent a bike for half a day. It charges (x + 40) dollars to rent a bike for a full day. The table shows the shop’s bike rentals for one day. On that day, the shop made a total of $600 from bike rentals.
Math in Focus Grade 7 Chapter 4 Lesson 4.3 Answer Key Real-World Problems Algebraic Equations 12
How much does it cost to rent a bike for a full day?
Answer:
Amount for full day = x + 40 =  60 + 40 = 100$

Explanation:
Given, Amount for half day = x

Number of bikes taken for rental for half day = 5

Amount for full day = x + 40

Number of bikes taken for rental for full day = 3

Total number of amount the bike shop made on rental on that day = $600

5(x) + 3(x +40) = 600

5x + 3x +120 = 600

Subtract 120 on both sides.

5x + 3x +120 – 120 = 600 – 120

8x = 480

Divide 8 on both sides.

8x ÷ 8 = 480 ÷ 8

x = 60

Amount for full day = x + 40 = 60 + 40 = 100

Verification:
5(x) + 3(x +40) = 600

= 5x + 3x +120

= 8x + 120

Substitute x = 60

8 (60) +120

= 480 + 120

= 600

Hence proved

Question 13.
An artist is weaving a rectangular wall hanging. The wall hanging is already 18 inches long, and the artist plans to weave an additional 2 inches each day. The finished wall hanging will be 60 inches long. How many days will it take the artist to finish the wall hanging?
Math in Focus Grade 7 Chapter 4 Lesson 4.3 Answer Key Real-World Problems Algebraic Equations 13
Answer:
x = 21 days

Explanation:
Given, total length of the wall hanging = 60 inches

Artist finished 18 inches, so the remaining would be 60 – 18 = 42

the artist plans to weave an additional 2 inches each day

i.e; 2 inches – 1 day

42 inches – ?

Let us consider the no. of days for weaving 42 inches be ‘x’

2 inches – 1 day

42 inches – x days

Let us cross multiply

2 × (x) = 42 × 1

2x = 42

Divide 2 on both sides.

2x ÷ 2 = 42 ÷ 2

x = 21 days

Question 14.
Ms. Kendrick plans to buy a laptop for $1,345 in 12 weeks. $he has already saved $145. How much should she save each week so she can buy the laptop?
Answer:
Amount she should save each week, x = 100

Explanation:
Given, Ms. Kendrick plans to buy a laptop for $1,345 in 12 weeks

she already saved $145.

Let us consider the amount per week in the 12 weeks = x

12x + 145 = 1345

Subtract 145 in both sides.

12x + 145 – 145 = 1345 – 145

12x = 1200

Divide 12 on both sides.

12x ÷ 12 = 1200 ÷ 12

x = 100

Question 15.
A plant grows at a rate of 4.5 centimeters per week. It is now 12 centimeters tall. 5uppose that the plant continues to grow at the same rate. In how many weeks will it reach a height of 48 centimeters?
Answer:
x = 8 weeks

Explanation:
We can apply the formula, y = mx + b

y = 48

m = 4.5

b= 12

48 = 4.5x + 12

Subtract 12 on both sides.

48 – 12 = 4.5x + 12 – 12

4.5x = 36

Divide 4.5 on both sides.

4.5x  ÷ 4.5  = 36 ÷ 4.5

x = 8

Question 16.
Mr. Johnson is currently four times as old as his son, David. If Mr. Johnson was 46 years old two years ago, how old is David now?
Answer:
David age, x = 12

Explanation:
Let us consider, David’s age be ‘x’

Mr. Johnson is currently four times as old as his son, s0 4x

As Mr. Johnson was 46 years old two years ago, at present his age is 46 +2 = 48

4x = 48

Divide 4 on both sides

4x ÷ 4 = 48 ÷ 4

x = 12

Question 17.
Mr. Warren drove from Townsville to Villaville and back again at the speeds shown. His total driving time was 12 hours. How far apart are the two towns?
Math in Focus Grade 7 Chapter 4 Lesson 4.3 Answer Key Real-World Problems Algebraic Equations 14
Answer:
Distance, x = 350 miles

Explanation:
Distance from townsville to villaville = x/70 miles per hour

Distance from villavile to townsville = x/50 miles per hour

it takes 12 hours

x/70 + x/50 = 12

(7x + 5x)/350 = 12

12x/350 = 12

Divide 12 on both sides.

12x/350 ÷ 12 = 12 ÷ 12

x/350 = 1

x = 350 miles

Question 18.
A factory made 845 pairs of shoes in January. These shoes were sent to three shoe stores and one outlet mall. The number of pairs of shoes sent to each store was four times the number sent to the outlet mall. How many pairs of shoes were sent to the outlet mall in January?
Answer:
Pairs of shoes were sent to the outlet mall in January, x = 65

Explanation:
Consider the number of pairs shoes sent to outlet mall be ‘x’

Number of pairs shoes sent to each store = 4x + 4x +4x

4x + 4x +4x +x = 845

13x =845

Divide 13 on both sides

13x ÷ 13 =845 ÷ 13

x = 65

Question 19.
The cost of seeing a weekday show is \(\frac{2}{3}\) the cost of a weekend show. In one month, Andy spent $42.50 for 4 weekday shows and 3 weekend shows. Find the price of a weekday show and the price of a weekend show.
Answer:
The cost of seeing a weekday show, x = 5

The cost of seeing weekend show, y = 7.5

Explanation:
Let us consider, x be the cost of week day show

y be the cost of weekend show

x = (2/3)y

In one month, Andy spent $42.50 for 4 weekday shows and 3 weekend shows

4x + 3y = 42.50

4(2/3)y + 3y = 42.50

Multiply 3 on both sides

4(2/3)y × 3+ 3y × 3 = 42.50 × 3

8y + 9y = 127.5

17y = 127.5

Divide 17 on both sides

17y ÷ 17 = 127.5 ÷ 17

y = 7.5

x = (2/3)y = (2/3) × 7.5

= 5