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Practice the problems of Math in Focus Grade 6 Workbook Answer Key Chapter 8 Equations and Inequalities to score better marks in the exam.

Math in Focus Grade 6 Course 1 B Chapter 8 Answer Key Multiplying and Dividing Fractions and Decimals

Math in Focus Grade 6 Chapter 8 Quick Check Answer Key

Complete with =, >, or <.

Question 1.
25 -26
Answer:
25 > -26,

Explanation:
Given to compare 25 and -26 as 25 is greater than -26 therefore 25 > -26.

Question 2.
12 + 12 + 12 3 ∙ 12
Answer:
12 + 12 + 12 = 36,

Explanation:
As 12 + 12 + 12 = 36 and 3 X 12 is 36 so both are equal therefore 12  + 12 + 12 = 36.

Question 3.
40 ÷ 8 8 ÷ 40
Answer:
40 ÷ 8 = 8 ÷ 40,

Explanation:
As 40 ÷ 8 is 5 and 8 ÷ 40 is 5 therefore 40 ÷ 8 = 8 ÷ 40.

Question 4.
-16 -7
Answer:
-16 < -7,

Explanation:
Given to compare between – 16 and – 7 as – 16 is lessthan -7 so -16 < -7.

Write an algebraic expression for each of the following.

Question 5.
The sum of 15 and p
Answer:
x = 15 + p,

Explanation:
An algebraic expression for the sum of 15 and p, let it be x so x = 15 + p.

Question 6.
The difference “q less than 10”
Answer:
x – q < 10,

Explanation:
Given the difference “q less than 10” is let it be x so  x – q < 10.

Question 7.
The product of r and 23
Answer:
x = r X 23,

Explanation:
Given the product of r and 23 is let it be x so x = r X 23.

Question 8.
Divide s by 11.
Answer:
x = s ÷ 11,

Explanation:
Given divide s by 11 so let it be x so x = s ÷ 11.

Evaluate each expression for the given values of the variable.

Question 9.
3x + 5 when x = 9 and x = 12
Answer:
3x + 5 = 32, when x = 9,
3x + 5 = 41, when x = 12,

Explanation:
Given to find 3x + 5 when x = 9, so 3 X 9 + 5 = 27 + 5 = 32, when x = 12 we get 3x + 5 = 3 x 12 + 5 = 36 + 5 = 41.

Question 10.
28 – 4x when x = 4 and x = 7
Answer:
28 – 4x = 12, when x = 4,
28 – 4x = 0 when x = 7,

Explanation:
Given when x = 4 the equation 28 – 4x is 28 – 16 = 12, when x = 7 then the equation becomes 28 – 4 X 7 = 28 – 28 = 0.

Plot the points on a coordinate plane.

Question 11.
K (2, 1), L (3, 3), M (0, 6), and N (7, 5)
Answer:

Explanation:
Given to plot the points on a coordinate plane as K (2, 1), L (3, 3), M (0, 6), and N (7, 5) above.

Practice the problems of Math in Focus Grade 6 Workbook Answer Key Chapter 8 Lesson 8.1 Solving Algebraic Equations to score better marks in the exam.

Math in Focus Grade 6 Course 1 B Chapter 8 Lesson 8.1 Answer Key Solving Algebraic Equations

Math in Focus Grade 6 Chapter 8 Lesson 8.1 Guided Practice Answer Key

Complete each with = or ≠, and each with the correct value.

Question 1.
For what value of x will x + 3 = 7 be true?
If x = 1, x + 3 = + 3
= ( 7)
If x = 2, x + 3 = + 3
= ( 7)
If x = 4, x + 3 = + 3
=
x + 3 = 7 is true when x = .
Answer:
x + 3 = 7 is true when x = 4,

Explanation:
Given to find for what value of x will x + 3 = 7 be true, so
if x =1, x + 3 = 1 + 3 = 4 ≠ 7,
If x = 2, x + 3 = 2 + 3 = 5 ≠ 7,
if x = 3, x + 3 = 3 + 3 = 6 ≠ 7,
if x = 4, x + 3 = 4 + 3 = 7 = 7, therefore x + 3 = 7 is true when x = 4.

Solve each equation using the substitution method.

Question 2.
p + 6 = 13
Answer:
p = 7, p + 6 = 13 is true,

Explanation:
Given equation p + 6 = 13, so solving p = 13 – 6 = 7, So when p = 7 means 7 + 6 = 13 is true therefore p = 7, p + 6 = 13 is true.

Question 3.
r + 4 = 12
Answer:
r = 8, r + 4 = 12 is true,

Equation::
Given equation r + 4 = 12 so solving r = 12 – 4 = 8 therefore when r = 8 means 8 + 4 = 12 is true, therefore r = 8, r + 4 = 12 is true.

Question 4.
k – 10 = 7
Answer:
k = 17 means k – 10 = 7 is true,

Explanation:
Given equation k – 10 = 7 so solving k = 7 + 10 = 17 therefore when k = 17 means 17 – 10 = 7 is true so k = 17., k – 10 = 7 is true.

Question 5.
2m = 6
Answer:
m = 3 means 2m = 6 is true,

Explanation:
Given equation 2m = 6 so solving m =6/2 = 3 therefore when m = 3 means 2 x 3 = 6 is true so m = 3.

Question 6.
4n = 20
Answer:
n = 5 means 4n = 20 is true,

Explanation:
Given equation 4n = 20 so solving n = 20/4 = 5 therefore when n = 5 means 4 X 5 = 20 is true so n = 5.

Question 7.
\(\frac{1}{5}\)z = 3
Answer:
z = 15 means \(\frac{1}{5}\)z = 3,

Explanation:
Given equation \(\frac{1}{5}\)z = 3 so solving x = 3 X 5 = 15 therefore when z = 15 means \(\frac{1}{5}\) X 15 = 3 is true so z = 3.
Complete each with + or —, and each with the correct value.

Question 8.
Solve x + 8 = 19.
x + 8 = 19
x + 8 =
x =
Answer:
x = 11,

Explanation:
Given to solve x + 8 = 19,
x + 8 – 19 = 19 – 19,
x – 11 = 0,
x = 11.

Solve each equation.

Question 9.
f + 5 = 14
Answer:
f = 9,

Explanation:
Given to solve f + 5 = 14,
subtracting both sides by 12 as f + 5 – 14 = 14 – 14,
f – 9 = 0,
therefore f = 9.

Question 10.
26 = g + 11
Answer:
g = 15,

Explanation:
Given to solve 26 = g + 11,
subtracting by 11 on the both sides as 26 – 11 = g + 11 – 11,
15 = g or g = 15.

Question 11.
w – 6 = 10
Answer:
w = 16,

Explanation:
Given to solve w – 6 = 10,
Adding 6 both the sides we get w – 6 + 6 = 10 + 6,
therefore w = 16.

Question 12.
z – 9 = 21
Answer:
z = 30,

Explanation:
Given to solve z – 9 = 21,
Adding 9 both the sides we get z – 9 + 9 = 21 + 9,
therefore z = 30.

Complete each with × or ÷, and with the correct value.

Question 13.
Solve 3x = 27.
3x = 27
3x = 27
x =
Answer:
x = 9,

Explanation:
Given to solve 3x = 27,
Dividing both sides by 3 as 3x ÷ 3 = 27 ÷ 3,
therefore x = 9.

Solve each equation.

Question 14.
6a = 42
Answer:
a = 7,

Explanation:
Given to solve 6a = 42,
Dividing both sides by 6 as 6a ÷ 6 = 42 ÷ 6,
we get a = 7.

Question 15.
65 = 13b
Answer:
b = 5,

Explanation:
Given to solve 65 = 13b,
Dividing both sides by 13 as 65 ÷ 13 = 13b ÷ 13,
we get 5 = b or b = 5.

Question 16.
\(\frac{m}{8}\) = 9
Answer:
m = 72,

Explanation:
Given to solve \(\frac{m}{8}\) = 9,
Multiplying both sides by 8,
\(\frac{m}{8}\) X 8 = 9 X 8,
We get m = 72.

Question 17.
12 = \(\frac{n}{7}\)
Answer:
n = 84,

Explanation:
Given to solve 12 = \(\frac{n}{7}\),
Multiplying both sides by 7,
12 X 7 = \(\frac{n}{7}\) X 7,
we get 84 = n or n = 84.

Complete each with +, —, × or ÷, and with the correct value.

Question 18.
Solve x + \(\frac{3}{7}\) = \(\frac{5}{7}\)
x + \(\frac{3}{7}\) = \(\frac{5}{7}\)
x + \(\frac{3}{7}\) = \(\frac{5}{7}\)
x =
Answer:
x = \(\frac{2}{7}\),

Explanation:
Given to solve x + \(\frac{3}{7}\) = \(\frac{5}{7}\) subtracting \(\frac{3}{7}\) both sides as x + \(\frac{3}{7}\) – \(\frac{3}{7}\) = \(\frac{5}{7}\) – \(\frac{3}{7}\) we have common denominator so x = \(\frac{5 – 3}{7}\) = \(\frac{2}{7}\).

Solve each equation. First tell which operation you will perform on each side of the equation. Write your answer in simplest form.

Question 19.
k + \(\frac{1}{8}\) = \(\frac{7}{8}\)
Answer:
k = \(\frac{6}{8}\) or \(\frac{3}{4}\),

Explanation:
Given to solve k + \(\frac{1}{8}\) = \(\frac{3}{4}\) subtracting \(\frac{1}{8}\) both sides as k + \(\frac{1}{8}\) – \(\frac{1}{8}\) = \(\frac{7}{8}\) – \(\frac{1}{8}\) we have common denominators so x = \(\frac{7 – 1}{8}\) = \(\frac{6}{8}\) further can be divided as both numerator and denominator goes by 2 we get \(\frac{3}{4}\).

Question 20.
4p = \(\frac{3}{4}\)
Answer:
p = \(\frac{3}{16}\),

Explanation:
Given to solve 4p = \(\frac{3}{4}\), Dividing both sides by 4 as
4p ÷ 4 = \(\frac{3}{4}\) ÷ 4,
p = \(\frac{3}{4 X 4}\) = \(\frac{3}{16}\) .

Math in Focus Course 1B Practice 8.1 Answer Key

Solve each equation using the substitution method.

Question 1.
b + 7 = 10
Answer:
b = 3 means b + 7 = 10 is true,

Explanation:
Given equation b + 7 = 10 so solving b = 10 – 7 = 3 when b = 3 means 3 + 7 = 10 is true therefore b = 3.

Question 2.
17 = e + 9
Answer:
e = 8,

Explanation:
Given 17 = e + 9 so solving e = 17 – 9 = 8 when e = 8 means 17 = 8 + 9  is true therefore e = 8.

Question 3.
k – 4 = 11
Answer:
k = 15,

Explanation:
Given k – 4 = 11 so solving k = 11 + 4 = 15 when k = 15 means 15 – 4 = 11 is true therefore k = 15.

Question 4.
42 = 3p
Answer:
p = 14,

Explanation:
Given 42 = 3p so solving 42/3 = p when p = 14 means 42 = 3 X 14 is true therefore p = 14.

Question 5.
8t = 56
Answer:
t = 7,

Explanation:
Given 8t = 56 so solving t = 56/8 = 7 when t = 7 means 8 X 7 = 56 is true therefore t = 7.

Question 6.
\(\frac{1}{4}\)v = 5
Answer:
v = 20,

Explanation:
Given \(\frac{1}{4}\)v = 5 solving v = 5 X 4 = 20 when v = 4 means \(\frac{1}{4}\) X 4 = 5 x 4 = 20 is true therefore v = 4.

Solve each equation using the concept of balancing.

Question 7.
k + 12 = 23
Answer:
k = 11,

Explanation:
Given to solve k + 12 = 23,
subtracting by 12 on the both sides as k + 12 – 12 = 23 – 12,
k = 11.

Question 8.
x- 8 = 17
Answer:
x = 25,

Explanation:
Given to solve x – 8 = 17,
adding 8 by both sides as x – 8 + 8 = 17 + 8,
x = 25.

Question 9.
24 = f – 16
Answer:
f = 40,

Explanation:
Given to solve 24 = f – 16,
adding both sides by 16 as 24 + 16 = f – 16 + 16,
40 = f or f =40.

Question 10.
5j = 75
Answer:
j = 15,

Explanation:
Given to solve 5j = 75,
dividing both sides by 5 as 5j/5 = 75/5,
j = 15.

Question 11.
81 = 9m
Answer:
m = 9,

Explanation:
Given to solve 81 = 9m,
dividing both sides by 9 as 81/9 = 9m/9,
9 = m or m = 9.

Question 12.
\(\frac{r}{6}\) = 11
Answer:
r = 66,

Explanation:
Given to solve \(\frac{r}{6}\) = 11,
multiplying both sides by 6 as \(\frac{r}{6}\) X 6 = 11 X 6,
r = 66.

Solve each equation using the concept of balancing. Write all fraction answers in simplest form.

Question 13.
\(\frac{5}{6}\) = c + \(\frac{1}{6}\)
Answer:
c = \(\frac{4}{6}\) or \(\frac{2}{3}\),

Explanation:
Given to solve \(\frac{5}{6}\) = c + \(\frac{1}{6}\) so subtracting both sides by \(\frac{1}{6}\) we get \(\frac{5}{6}\) – \(\frac{1}{6}\) = c + \(\frac{1}{6}\) – \(\frac{1}{6}\) as we have common denominators we get \(\frac{5 – 1}{6}\) = c,
c = \(\frac{4}{6}\) we further divide as both numerator and denominators goes by 2 we get c = \(\frac{2 X 2}{3 X 2}\), c = \(\frac{2}{3}\).

Question 14.
h + \(\frac{5}{14}\) = \(\frac{11}{14}\)
Answer:
h = \(\frac{6}{14}\) or \(\frac{3}{7}\),

Explanation:
Given to solve h + \(\frac{5}{14}\) = \(\frac{11}{14}\) so subtracting both sides by \(\frac{5}{14}\) we get h + \(\frac{5}{14}\) – \(\frac{5}{14}\) = \(\frac{11}{14}\) – \(\frac{5}{14}\) as we have common denominators we get h = \(\frac{11 – 5}{6}\),
h = \(\frac{6}{14}\) we further divide as both numerator and denominators goes by 2 we get h = \(\frac{3 X 2}{7 X 2}\), h = \(\frac{3}{7}\).

Question 15.
q – \(\frac{3}{10}\) = \(\frac{7}{10}\)
Answer:
q = 1,

Explanation:
Given to solve q – \(\frac{3}{10}\) = \(\frac{7}{10}\) so adding both sides by \(\frac{3}{10}\) we get q – \(\frac{3}{10}\) + \(\frac{3}{10}\) = \(\frac{7}{10}\) + \(\frac{3}{10}\) as we have common denominators we get q = \(\frac{7 + 3}{10}\),
q = \(\frac{10}{10}\) we further divide as both numerator and denominators goes by 10 we get q = \(\frac{1 X 10}{1 X 10}\), q = 1.

Question 16.
7k = \(\frac{4}{7}\)
Answer:
k = \(\frac{4}{49}\),

Explanation:
Given to solve 7k = \(\frac{4}{7}\) dividing both sides by \(\frac{1}{7}\) we get 7k X \(\frac{1}{7}\) = \(\frac{4 X 1}{7 X 7}\), k = \(\frac{4}{49}\).

Question 17.
\(\frac{5}{12}\) = 5d
Answer:
d = \(\frac{1}{12}\),

Explanation:
Given to solve \(\frac{5}{12}\) = 5d so dividing both sides by \(\frac{1}{5}\) we get \(\frac{5}{12}\) X \(\frac{1}{5}\) = 5d X \(\frac{1}{5}\), d = \(\frac{1}{12}\).

Question 18.
\(\frac{1}{2}\)x = \(\frac{1}{4}\)
Answer:
x = \(\frac{1}{2}\),

Explanation:
Given to solve \(\frac{1}{2}\)x = \(\frac{1}{4}\) so multiplying both sides by 2 we get x X \(\frac{2}{2}\) = \(\frac{2}{4}\) we get x = \(\frac{2}{4}\) we further divide as both numerator and denominators goes by 2 we get x = \(\frac{1 X 2}{2 X 2}\), x = \(\frac{1}{2}\).

Question 19.
\(\frac{8}{9}\) = \(\frac{1}{3}\)f
Answer:
f = \(\frac{8}{3}\),

Explanation:
Given to solve \(\frac{8}{9}\) = \(\frac{1}{3}\)f so multiplying both sides by 3 we get \(\frac{8}{9}\) = \(\frac{1}{3}\)f X 3, \(\frac{8 X 3}{3 X 3}\)= f, So f = \(\frac{8}{3}\).

Question 20.
r + 2.1 = 4.7
Answer:
r = 2.6,

Explanation:
Given to solve r + 2.1 = 4.7 subtracting both sides by 2.1 we get r + 2.1 – 2.1 = 4.7 – 2.1, so r = 2.6.

Question 21.
9.9 = x + 5.4
Answer:
x = 4.5,

Explanation:
Given to solve 9.9 = x + 5.4 subtracting both sides by 5.4 we get 9.9 – 5.4 = x – 5.4, so 4.5 = x.

Question 22.
11.2 = f – 1.8
Answer:
f = 13,

Explanation:
Given to solve 11.2 = f – 1.8 adding both sides by 1.8 as 11.2 + 1.8 = f – 1.8 + 1.8 we get 13 = f or f = 13.

Question 23.
j – 3.7 = 20.4
Answer:
j = 24.1,

Explanation:
Given to solve j – 3.7 = 20.4 adding both sides by 3.7 as j – 3.7 + 3.7 = 20.4 + 3.7 we get j = 24.1.

Question 24.
4w = 6.8
Answer:
w = 1.7,

Explanation:
Given to solve 4w = 6.8, dividing both sides by 4 as 4w/4 = 6.8/4 we get w = 1.7.

Question 25.
13.9 = 2.5z
Answer:
z = 5.56,

Explanation:
Given to solve 13.9 = 2.5z, dividing both sides by 2.5 as 13.9/2.5 = 2.5z/2.5 we get 5.56 = z or z = 5.56.

Question 26.
3.2d = 40.8
Answer:
d = 12.75,

Explanation:
Given to solve 3.2d = 40.8 dividing both sides by 3.2 as 3.2d/3.2 = 40.8/3.2 we get d = 12.75.

Question 27.
x + \(\frac{1}{2}\) = 1\(\frac{3}{4}\)
Answer:
x = \(\frac{5}{4}\) or 1\(\frac{1}{4}\),

Explanation:
Given to solve x + \(\frac{1}{2}\) = 1\(\frac{1}{4}\) so subtracting both sides by \(\frac{1}{2}\) we get x = \(\frac{1 X 4 + 3}{4}\) – \(\frac{1}{2}\) we get x = \(\frac{7}{4}\) – \(\frac{1}{2}\) , x = \(\frac{7 –  1 X 2}{4}\) , x = \(\frac{5}{4}\) as numerator is more we write in mixed fraction as x = \(\frac{1 X 4 + 1}{4}\), x = 1\(\frac{1}{4}\).

Question 28.
g + \(\frac{5}{3}\) = 3\(\frac{2}{3}\)
Answer:
g = 2,

Explanation:
Given to solve g + \(\frac{5}{3}\) = 3\(\frac{2}{3}\) so subtracting both sides by \(\frac{5}{3}\) we get g = \(\frac{3 X 3 + 2}{3}\) – \(\frac{5}{3}\) we get g = \(\frac{11}{3}\) – \(\frac{5}{3}\) , g = \(\frac{11 –  5}{3}\) , g = \(\frac{6}{3}\) as numerator and denominator both goes by 3 we get g = \(\frac{2 X 3}{3 X 1}\), g = 2.

Question 29.
2\(\frac{5}{7}\) = p – \(\frac{2}{7}\)
Answer:
p = 3,

Explanation:
Given to solve 2\(\frac{5}{7}\) = p – \(\frac{2}{7}\) so adding both sides by \(\frac{2}{7}\) we get \(\frac{2 X 7 + 5}{7}\) + \(\frac{2}{7}\)= p – \(\frac{2}{7}\) + \(\frac{2}{7}\), we have common denominators so \(\frac{19 + 2}{7}\) = p, \(\frac{21}{7}\) = p as numerator and denominator both goes by 7 we get p = \(\frac{7 X 3}{7 X 1}\), p = 3.

Question 30.
e – \(\frac{18}{11}\) = 1\(\frac{6}{11}\)
Answer:
e = \(\frac{35}{11}\) or 3\(\frac{2}{11}\),

Explanation:
Given to solve e – \(\frac{18}{11}\) = 1\(\frac{6}{11}\) so adding both sides by \(\frac{18}{11}\) we get e = \(\frac{11 X 1 + 6}{11}\) + \(\frac{18}{11}\) we get e = \(\frac{17}{11} \) + \(\frac{18}{11}\) both have common denominators we get e = \(\frac{17 + 18}{11}\) , e = \(\frac{35}{11}\) as numerator is greater than denominator we write in mixed fraction as e = \(\frac{3 X 11 + 2}{11}\), so e = 3\(\frac{2}{11}\).

Question 31.
\(\frac{4}{3}\)y = 36
Answer:
y = 27,

Explanation:
Given to solve \(\frac{4}{3}\)y = 36 multiplying both sides by \(\frac{3}{4}\) we get y = 36 X \(\frac{3}{4}\), y = \(\frac{36 X 3}{4}\) = \(\frac{4 X 9 X 3}{4}\) we get y = 9 X 3 = 27.

Question 32.
\(\frac{9}{10}\) = \(\frac{5}{6}\)v
Answer:
v = \(\frac{54}{50}\) or 1\(\frac{4}{50}\),

Explanation:
Given to solve \(\frac{9}{10}\) = \(\frac{5}{6}\)v multiplying both sides by \(\frac{6}{5}\) we get \(\frac{9}{10}\) X \(\frac{6}{5}\) =v, v = \(\frac{9 X 6}{10 X 5}\) = \(\frac{54}{50}\) as numerator is greater we can further write in mixed fraction as \(\frac{1 X 50 + 4}{50}\) = 1\(\frac{4}{50}\).

Question 33.
\(\frac{2}{3}\)k = 28 ∙ \(\frac{4}{9}\)
Answer:
k = \(\frac{56}{3}\) or 18\(\frac{2}{3}\),

Explanation:
Given to solve \(\frac{2}{3}\)k = 28 . \(\frac{4}{9}\) multiplying both sides by \(\frac{3}{2}\) we get k = \(\frac{28 X 4}{9}\) X \(\frac{3}{2}\), k = \(\frac{28 X 2}{3}\), k = \(\frac{56}{3}\) as numerator is greater we can further write in mixed fraction as \(\frac{18 X 3 + 2}{3}\) = 18\(\frac{2}{3}\).

Solve.

Question 34.
Find five pairs of whole numbers, such that when they are inserted into the equation below, the solution of the equation is 3,
x + =
Answer:
(0,3), (1,4), (2,5), (3,6) and (4,7),

Explanation:
Given to find five pairs of whole numbers such that when they are inserted into the equation x + __ =___  the solution of the equation is 3 so
1. When (0,3) substituting x + 0 = 3 yes true as x = 3,
2. When (1,4) substituting x + 1 = 4 solving we get x = 4 – 1 = 3 which is true,
3. When (2,5) substituting x + 2 = 5 solving we get x = 5 – 2 = 3 which is true,
4. When (3,6) substituting x + 3 = 6 solving we get x = 6 – 3 = 3 which is true,
5. When (4,7) substituting x + 4 = 7 solving we get x = 7 – 4 = 3 which is true therefore 5 pairs of whole numbers are (0,3), (1,4), (2,5), (3,6) and (4,7).

Question 35.
Find five pairs of numbers, such that when they are inserted into the equation below, the solution of the equation is \(\frac{2}{5}\).
x =
Answer:
(\(\frac{2}{5}\), 1), (\(\frac{1}{5}\), 2), (\(\frac{2}{15}\), 3), (\(\frac{2}{20}\),4), (\(\frac{2}{25}\), 5),

Explanation:
Given to find five pairs of numbers such that when they are inserted into the equation ___x = ____ , the solution of the equation is \(\frac{2}{5}\) so
1. When (\(\frac{2}{5}\),1) substituting \(\frac{2}{5}\) X 1 = \(\frac{2}{5}\),
2. When (\(\frac{1}{5}\), 2) substituting \(\frac{1}{5}\) X 2 = \(\frac{2}{5}\),
3. When (\(\frac{2}{15}\), 3) substituting \(\frac{2}{15}\) X 3 = \(\frac{2}{5}\),
4. When (\(\frac{2}{20}\),4) substituting \(\frac{2}{20}\) X 4 = \(\frac{2}{5}\),
5. When (\(\frac{2}{5}\),5) substituting \(\frac{2}{25}\) X 5 = \(\frac{2}{5}\) therefore 5 pairs of numbers are (\(\frac{2}{5}\), 1), (\(\frac{1}{5}\), 2), (\(\frac{2}{15}\), 3), (\(\frac{2}{20}\),4), (\(\frac{2}{25}\), 5).

Practice the problems of Math in Focus Grade 6 Workbook Answer Key Chapter 8 Lesson 8.2 Writing Linear Equations to score better marks in the exam.

Math in Focus Grade 6 Course 1 B Chapter 8 Lesson 8.2 Answer Key Writing Linear Equations

Math in Focus Grade 6 Chapter 8 Lesson 8.2 Guided Practice Answer Key

Complete.

Question 1.
Isaiah has h baseball cards. Miguel has 7 more baseball cards than Isaiah.
a) Write an expression for the number of baseball cards that Miguel has in terms of h.

Miguel has baseball cards.

b) If Miguel has k baseball cards, express k in terms of h.
k = +
c) State the independent and dependent variables.
Independent variable: , dependent variable:
Answer:
a) Miguel has h + 7 baseball cards,
b) k = h + 7,
c) Independent variable: h, dependent variable: k,

Explanation:
Given Isaiah has h baseball cards. Miguel has 7 more baseball cards than Isaiah,
a) Wrote an expression for the number of baseball cards that Miguel has in terms of h as Miguel has h + 7 baseball cards.
b) If Miguel has k baseball cards expressing k in terms of h as k = h + 7.
c) The independent and dependent variables are independent variable is h and dependent variable are k which depends on value of h.

Write an equation for each of the following. Then state the independent and dependent variables for each equation.

Question 2.
Hannah took p minutes to jog around a park. Sofia took 12 minutes longer to jog around the park. If Sofia took t minutes to jog around the park, express t in terms of p.
Answer:
t = p + 12,
Independent  variable : p, Dependent variable: t,

Explanation:
Given Hannah took p minutes to jog around a park. Sofia took 12 minutes longer to jog around the park. If Sofia took t minutes to jog around the park,To jog around the park it is p minutes, and Sofia took 12 more minutes if Sofia took t minutes expressing t in terms of p it is t = p + 12. Here p is independent variable and t is dependent variable on the value of p.

Question 3.
A bouquet of roses costs $30. A bouquet of tulips costs m dollars less. If the cost of one bouquet of tulips is n dollars, express n in terms of m.
Answer:
n = $30 – m,
Independent variable : m, Dependent variable: n,

Explanation:
Given a bouquet of roses costs $30. A bouquet of tulips costs m dollars less. If the cost of one bouquet of tulips is n dollars, expressing n in terms of m as n = $30 – m, Here m is independent variable and n is dependent variable on value of m.

Question 4.
Nathan has 7 boxes of marbles. Each box contains b marbles. If he has c marbles altogether, express c in terms of b.
Answer:
c = 7b,
Independent variable : b, Dependent variable: c,

Explanation:
Given Nathan has 7 boxes of marbles. Each box contains b marbles. If he has c marbles altogether, expressing c in terms of b as c = 7b as Nathan has 7 boxes of b marbles each here b is independent variable and c is dependent variable on the value of b.

Question 5.
A motel charges Mr. Kim x dollars for his stay. Mr. Kim stayed at the motel for 12 nights. If the rate per night for a room is y dollars, express y in terms of x.
Answer:
x = 12y,
Independent variable : b, Dependent variable: c,

Explanation:
Given a motel charges Mr. Kim x dollars for his stay. Mr. Kim stayed at the motel for 12 nights. If the rate per night for a room is y dollars expressing y in terms of x as it is x = 12y where y is independent variable and x is dependent variable on the value of y.

Copy and complete the table. Then use the table to answer the questions.

Question 6.
The width of a rectangular tank is 2 meters less than its length.
a) If the length is p meters and the width is q meters, write an equation relating p and q.

Answer:
Equation: q = p – 2,

Explanation:
Given The width of a rectangular tank is 2 meters less than its length.
a) If the length is p meters and the width is q meters so an equation relating p and q is q = p – 2,
If p =3 then q = 3 – 2 = 1,
If p = 4 then q = 4 – 2 = 2,
If p = 5 then q = 5 – 2 = 3,
If p = 7 then q = 7 – 2 = 5,
If p = 8 then q = 8 – 2 = 6,
Completed the table as shown above.

b) Use the data from a) to plot the points on a coordinate plane. Connect the points with a line.
Answer:

Explanation:
Used the data from a) to plot the points on a coordinate plane as length p meters on x – axis and width q meters on y axis. Connected the points with a line as shown above.

c) The point (5.5, 3.5) is on the line you drew in b). Does this point make sense in the situation?
Answer:
Yes,

Explanation:
The point(5.5,3.5) is on the line which i drew yes this point makes sense as p = 5.5 we have q = p – 2 = 5.5 – 2 = 3.5 which is true.

Copy and complete each table. Then express the relationship between the two variables as an equation.

Question 7.
Paul and Lee went to the library to borrow some books. Paul borrowed 6 more books than Lee.

Answer:
Equation: y = x + 6,

Explanation:
Given Paul and Lee went to the library to borrow some books. Paul borrowed 6 more books than Lee. If x is number of books Lee borrowed and y be books Paul borrowed so the equation is y = x + 6, so
if x = 1 then y = 1 + 6 = 7 books,
if x = 2 then y = 2 + 6 = 8 books,
if x = 3 then y = 3 + 6 = 9 books,
if x = 4 then y = 4 + 6 = 10 books and
if x = 5 then y = 5 + 6 = 11 books shown in the above table.

Question 8.
At a crafts store, Zoey bought some boxes of red beads and some boxes of blue beads. The number of boxes of red beads was 4 times the number of boxes of blue beads.

Answer:
Equation: r = 4b,

Explanation:
Given at a crafts store Zoey bought some boxes of red beads and some boxes of blue beads. The number of boxes of red beads was 4 times the number of boxes of blue beads. Let read beads be r and blue beads be b so the equation is r = 4b,
if b = 2 then r = 4 X 2 = 8 beads,
if b = 3 then r = 4 X 3 = 12 beads,
if b = 4 then r = 4 X 4 = 16 beads,
if b = 5 then r = 4 X 5 = 20 beads shown above in the table.

Use the data in the table to plot points on a coordinate plane. Connect the points to form a line. Then write an equation to show the relationship between the variables.

Question 9.

Answer:

Equation:
d = 50t,

Explanation:
Used the data in the table to plot points on a coordinate plane. Connected the points to form a line as shown above  here d is the distance traveled in miles on y – axis and t is the time taken in t hours on x – axis. As distance traveled is increasing by 50 each hour an equation to show the relationship between the variables is d = 50t.

Math in Focus Course 1B Practice 8.2 Answer Key

Solve.

Question 1.
Joshua is w years old. His brother is 3 years older than he is.
a) If his brother is x years old, express x in terms of w.
Answer:
x = w + 3,

Explanation:
Given Joshua is w years old. His brother is 3 years older than he is means brother is w + 3, So if his brother is x years old expressing x in terms of w as x = w + 3.

b) State the independent and dependent variables in the equation.
Answer:
Independent variable : w,
Dependent variable : x,

Explanation:
Here Joshua is w years old his brother is 3 years older than he is and  if his brother is x years old then the independent variable is w and dependent variable is x on the value of w.

Question 2.
Rita has b markers. Sandy has 11 fewer markers than she has.

a) If Sandy has h markers, express h in terms of b.
Answer:
h = b – 11,

Explanation:
As Rita has b markers. Sandy has 11 fewer markers than she has means Sandy has b – 11 markers, If Sandy has h markers expressing h in terms of b as h = b – 11.

b) State the independent and dependent variables in the equation.
Answer:
Independent variable : b,
Dependent variable : h,

Explanation:
As Rita has b markers. Sandy has 11 fewer markers than she has means Independent variable is b and dependent variable is h on the value of b.

Question 3.
A small box of cereal weighs k grams. A jumbo box of cereal weighs 5 times as much.

a) If the weight of the jumbo box of cereal is m grams, express m in terms of k.
Answer:

m = 5k,

Explanation:
Given a small box of cereal weighs k grams. A jumbo box of cereal weighs 5 times as much means jumbo box is 5k, If the weight of the jumbo box of cereal is m grams, expressing m in terms of k as m = 5k.

b) State the independent and dependent variables in the equation.
Answer:
Independent variable : k,
Dependent variable : m,

Explanation:
As a small box of cereal weighs k grams and a jumbo box of cereal weighs 5 times as much here the independent variable is k and dependent variable is m on the value of k.

Question 4.
The area of Hank’s farm is n acres. The area of Hank’s farm is 8 times as large as the area of Stan’s farm.

a) If s represents the area of Stan’s farm, express s in terms of n.
Answer:
n = 8s,

Explanation:
Given the area of Hank’s farm is n acres. The area of Hank’s farm is 8 times as large as the area of Stan’s farm means Hank’s farm is 8 X Stan’s farm if s represents the area of Stan’s farm expressing s in terms of n is n = 8s.

b) State the independent and dependent variables in the equation.
Answer:
Independent variable : s,
Dependent variable : n,

Explanation:
Given the area of Hank’s farm is n acres. The area of Hank’s farm is 8 times as large as the area of Stan’s farm means Hank’s farm is 8 X Stan’s farm if s represents the area of Stan’s farm expressing s in terms of n here Independent variable is s and dependent variable is n which depends on s.

Question 5.
Ethan scored x points in a game. His younger sister scored 8 points when she played the same game. Their combined score was y points.

a) Write an equation relating x and y.
Answer:
y = x + 8,

Explanation:
Given Ethan scored x points in a game. His younger sister scored 8 points when she played the same game. Their combined score was y points. So an equation relating x and y is y = x + 8.

b) Copy and complete the table to show the relationship between x and y.

Answer:

Explanation:
Completed the table to show the relationship between x and y as y = x + 8,
if x = 10 then y = 10 + 8 = 18,
if x = 11 then y = 11 + 8 = 19,
if x = 12 then y = 12 + 8 = 20,
if x = 13 then y = 13 + 8 = 21,
if x = 14 then y = 14 + 8 = 22 and
if x = 15 then y = 15 + 8 = 23 as shown above.

Question 6.
There are x sparrows in a tree. There are 50 sparrows on the ground beneath the tree. Let y represent the total number of sparrows in the tree and on the ground.
a) Express y in terms of x.
Answer:
y = x + 50,

Explanation:
Given there are x sparrows in a tree. There are 50 sparrows on the ground beneath the tree. Let y represent the total number of sparrows in the tree and on the ground.So expressing y in terms of x is y = x + 50.

b) Make a table to show the relationship between y and x. Use values of x = 10, 20, 30, 40, and 50 in your table.
Answer:

Explanation:
Asked to make a table to show the relationship between y and x. Used values of x = 10, 20, 30, 40, and 50 in my table and completed as shown above in the equation y = x + 50,
When x = 10 then y = 10 + 50 = 60,
when x = 20 then y = 20 + 50 = 70,
when x = 30 then y = 30 + 50 = 80,
when x = 40 then y = 40 + 50 = 90,
when x = 50 then y = 50 + 50 = 100.

c) Graph the relationship between y and x in a coordinate plane.
Answer:

Explanation:
Graphed the relationship between y and x in a coordinate plane as shown above as sparrows x on x -axis and total number of sparrows y on y – axis.

Question 7.
A rectangle has a perimeter of P centimeters. Its width is b centimeters. Its length is double its width.
a) Express P in terms of b.
Answer:
P = 6b,

Explanation:
Given a rectangle has a perimeter of P centimeters. Its width is b centimeters. Its length is double its width.
So expressing P in terms of b as P = 2(2b + b). = 2(3b) = 6b.

b) Copy and complete the table to show the relationship between P and b.

Answer:

Explanation:
Given a rectangle has a perimeter of P centimeters. Its width is b centimeters. Its length is double its width.
So expressed P in terms of b as P = 6b completed the table to show the relationship between P and b above as
if b= 1 then P = 6 X 1 = 6,
if b = 2 then P = 6 X 2 = 12,
if b = 3 then P = 6 X 3 = 18,
if b= 4 then P = 6 X 4 = 24,
if b = 5 then P = 6 X 5 = 30 and
if b = 6 then P = 6 X 6 = 36.

Question 8.
Every month, Amaan spends 60% of what he earns and saves the rest. Amaan earns n dollars and saves r dollars each month.
a) Express r in terms of n.
Answer:

r = n – 0.6n,

Explanation:
Given every month, Amaan spends 60% of what he earns and saves the rest. Amaan earns n dollars and saves r dollars each month. Expressing r in terms of n as r = n – n X 60/100 = n – 0.6n.

b) Make a table to show the relationship between r and n. Use values of n = 100, 200, 400, and 500 in your table.
Answer:

Explanation:
Made a table above to show the relationship between r and n. Using values of n = 100, 200, 400, and 500 in my table and substituting in r = n – 0.6n as
if n = 100 then r = 100 – 0.6 X 100 = 100 – 60 = 40,
if n = 200 then r = 200 – 0.6 X 200 = 200 – 120 = 80,
if n = 300 then r = 300 – 0.6 X 300 = 300 – 180 = 120,
if n = 400 then r = 400 – 0.6 X 400 = 400 – 240 = 160 and
if n = 500 then r = 500 – 0.6 X 500 = 500 – 300 = 200.

c) Graph the relationship between n and r in a coordinate plane.
Answer:

Explanation:
Graphed the relationship between n and r in a coordinate plane as  r = n – 0.6n where earnings n dollars on x – axis and savings r dollars on y – axis as shown above.

d) The point (287.5, 115) is on the line you drew in c). Does this point make sense in the situation? Explain.
Answer:
Yes,

Explanation:
Given the point (287.5, 115) is on the line I drew in c) checking in r = n – 0.6n, if n = 287.5 then r =  287.5 – 0.6 X 287.5 = 287.5 – 172.5 = 115 which matches so this point make sense in the situation.

Question 9.
The side length of a square is t inches. The perimeter of the square is z inches.
a) Express z in terms of t.
Answer:
z = 4t,

Explanation:
Given the side length of a square is t inches. The perimeter of the square is z inches. Expressing z in terms of t as perimeter of sqaure is 4 X side length so it is z = 4t.

b) Make a table to show the relationship between z and t. Use whole number values of t from 1 to 10.
Answer:

Explanation:
Made a table to show the relationship between z and t. Used whole number values of t from 1 to 10 in z – 4t as
if t = 1 then z = 4 X 1 = 4,
if t = 2 then z = 4 X 2 = 8,
if t = 3 then z = 4 X 3 = 12,
if t = 4 then z = 4 X 4 = 16,
if t = 5 then z = 4 X 5 = 20,
if t = 6 then z = 4 X 6 = 24,
if t = 7 then z = 4 X 7 = 28,
if t = 8 then z = 4 X 8 = 32,
if t = 9 then z = 4 X 9 = 36 and
if t = 10 then z = 4 X 10 = 40 shown above.

c) Graph the relationship between z and t in a coordinate plane.
Answer:

Explanation:
Graphed the relationship between z and t in a coordinate plane as the side length of a square is t inches on x – axis and the perimeter of the square is z inches on y – axis as shown above.

d) Use your graph to find the perimeter of the square when the length is 3.5 inches and 7.5 inches.
Answer:
1) 14 inches,
2) 30 inches,

Explanation:
Using my graph the perimeter of the square when the length is 3.5 inches is 4 X 3.5 = 14 inches and when the length is 7.5 inches the perimeter of the square is 4 X 7.5 inches = 30 inches respectively.

Practice the problems of Math in Focus Grade 6 Workbook Answer Key Chapter 8 Lesson 8.3 Solving Simple Inequalities to score better marks in the exam.

Math in Focus Grade 6 Course 1 B Chapter 8 Lesson 8.3 Answer Key Solving Simple Inequalities

Math in Focus Grade 6 Chapter 8 Lesson 8.3 Guided Practice Answer Key

Use substitution to determine the solutions of each inequality. Then represent the solution set of each inequality on a number line.

Question 1.
h > 8
Answer:
h = 8.1,

Explanation:
Let us take h = 8.1 and substituting h as 8.1 we get 8.1 > 8 which is true so we can substitute h as 8.1, shown h > 8 inequality on a number line as h is greater than 8 so it starts at 8.1 and red line moves towards the right side on the number line as shown above.

Question 2.
y < 1o
Answer:
y = 9.9,

Explanation:
Let us take y as 9.9 substituting y as 9.9 we get 9.9 < 10 which is true so we can substitute y as 9.9, shown y < 10 inequality on a number line as y is less than 10 so it starts at 9.9 and red line moves towards the left side on the number line as shown above.

Question 3.
p > 23
Answer:
p = 23.1,

Explanation:
Let us take p as 23.1 substituting p as 23.1 we get 23.1 > 23 which is true so we can substitute p as 23.1, shown p > 23 inequality on a number line as p is greater than 23 so it starts at 23.1 and red line moves towards the right side on the number line as shown above.

Question 4.
e < 14
Answer:
e = 13.9,

Explanation:
Let us take e as 13.9 substituting e as 13.9 we get 13.9 < 14 which is true so we can substitute e as 13.9 shown e < 14 inequality on a number line as e is less than 14 so it starts at 13.9 and red line moves towards the left side on the number line as shown above.

Question 5.
m > 30
Answer:
m = 30.1,

Explanation:
Let us take m as 30.1 substituting m as 30.1 we get 30.1 > 30 which is true so we can substitute m as 30.1 shown m > 30 inequality on a number line as m is greater than 30 so it starts at 30.1 and red line moves towards the right side on the number line as shown above.

Question 6.
n < 5
Answer:
n = 4.9,

Explanation:
Let us take n as 4.9 substituting n as 4.9 we get 4.9 < 5 which is true so we can substitute n as 4.9 shown n < 5 inequality on a number line as n is less than 5 so it starts at 4.9 and red line moves towards the left side on the number line as shown above.

Hands-On Activity

Writing Inequalities

Step 1.
The figure shows a balance scale.

Write the equation that this figure represents.

Step 2.
counters are added to the right side. Draw what the balance scale looks like now. Then write an inequality to represent the relationship between x and the counters on the right side of the balance scale.

Step 3.
3 counters are then removed from the right side. Draw what the balance scale looks like after removing the 3 counters. Then write an inequality to represent the relationship between x and the counters on the right side of the balance scale.

Step 4.
Now, if y > x, write an inequality to represent the solutions of y > x. Explain how x and y are related using a balance scale.
Answer:
x = 5,

Explanation:
The equation that this figure represents is as  both are equal so x = 5,

Use substitution to find three solutions of each inequality. Then represent the solutions of each inequality on a number line.

Question 7.
q ≥ 3
Answer:
q = 3, q = 4, q = 5,

Explanation:
Let us take q as 3,4,5 substituting q as 3,4,5  if q = 3 then q is equal to 3 only, q = 4 means 4 is greater than 3 and q = 5 means 5 is greater than 3 we get in all 3 substitues q ≥ 3 which is true so we can substitute q as 3,4,5, shown q ≥ 3 inequality on a number line as q is greater than and equal to 3
1. It starts at 3 and red line moves towards the right side on the number line,
2. It starts at 4 and red line moves towards the right side on the number line,
3. It starts at 5 and red line moves towards the right side on the number line as shown above.

Question 8.
d ≤ 12
Answer:
d = 12, d = 11, d = 10,

Explanation:
Let us take d as 12,11,10 substituting d as 12,13,14  if d = 12 then d is equal to 12 only, d = 13 means 13 is less than 12 and q = 11 means 11 is less than 12 we get in all 3 substitues d ≤ 12 which is true so we can substitute d as 12,11,10 shown d ≤ 12 inequality on a number line as d is less than and equal to 12
1. It starts at 12 and red line moves towards the left side on the number line,
2. It starts at 11 and red line moves towards the leftt side on the number line,
3. It starts at 10 and red line moves towards the left side on the number line as shown above.

Question 9.
k ≤ 25
Answer:
k = 25, k = 24, k = 23,

Explanation:
Let us take k as 25,24,23 substituting k as 25,24,23 if k = 25 then k is equal to 25 only, d = 24 means 24 is less than 25 and d = 23 means 23 is less than 25 we get in all 3 substitues k ≤ 25 which is true so we can substitute d as 25,24,23 shown k ≤ 25 inequality on a number line as d is less than and equal to 25
1. It starts at 25 and red line moves towards the left side on the number line,
2. It starts at 24 and red line moves towards the leftt side on the number line,
3. It starts at 23 and red line moves towards the left side on the number line as shown above.

Question 10.
m ≥ -28
Answer:
m = -28, m = -27, m = -26,

Explanation:
Let us take m as -28,-27,-26 substituting m as -28,-27,-26  if m = -28 then m is equal to -28 only, m = -27 means -27 is greater than -27 and m = -26 means -26 is greater than -28 we get in all 3 substitues m ≥ -28 which is true so we can substitute m as -28,-27,-26 shown m ≥ -28 inequality on a number line as m is greater than and equal to -28
1. It starts at -28 and red line moves towards the right side on the number line,
2. It starts at -27 and red line moves towards the right side on the number line,
3. It starts at -26 and red line moves towards the right side on the number line as shown above.

Match each inequality to its graph.

a) x < 10

b) x ≤ 10

c) x > 10

d) x ≥ 10

Question 11.

Answer:
b) x ≤ 10,

Explanation:
As the point on the graph starts at 10 and moves left side so it is less than and equal to 10 so it is x ≤ 10 matches with b.

Question 12.

Answer:
c) x > 10,

Explanation:
As the point on the graphs moves from 10 towards right side it is greater than 10 so it is x > 10 matches with c.

Question 13.

Answer:
d) x ≥ 10,

Explanation:
As the point on the graph starts at 10 and moves right side it is greater than and equal 10 so it is d) x ≥ 10 matches with d.

Question 14.

Answer:
a) x < 10,

Explanation:
As the point on the graphs moves from 10 towards left side it is lesser than 10 so it is x < 10 matches with a.

Math in Focus Course 1B Practice 8.3 Answer Key

Rewrite each statement using > <, ≥, or ≤.

Question 1.
k is less than 12.
Answer:
k < 12,

Explanation:
Given k is less than 12 so it is k < 12.

Question 2.
d is greater than 10.
Answer:
d > 10,

Explanation:
Given d is greater than 10 so it is d > 10.

Question 3.
w is greater than or equal to 17.
Answer:
w ≥ 17,

Explanation:
Given w is greater than or equal to 17 so it is w ≥ 17.

Question 4.
p is less than or equal to 36.
Answer:
p ≤ 36,

Explanation:
Given p is less than or equal to 36 so it is p ≤ 36.

Question 5.
A sack of potatoes weighs at least 20 pounds. Write an inequality to represent the weight of the sack of potatoes. Answer:
x ≥ 20,

Explanation:
Given a sack of potatoes weighs at least 20 pounds. Atleast means ≥ let x be the weight of the sack of potatoes so an inequality to represent the weight of the sack of potatoes is x ≥ 20.

Question 6.
The maximum number of shirts Amanda can buy is 9. Write an inequality to represent the number of shirts that she can buy.

Answer:
x ≤ 9,

Explanation:
Given the maximum number of shirts Amanda can buy is 9. An inequality to represent the number of shirts that  she can buy is as maximum only is 9 let it be x so x ≤ 9.

Represent the solutions of each inequality on a number line.

Question 7.
x > 5
Answer:

Explanation:
Given to show x > 5 inequality on a number line as x is greater than 5 so it starts at 5 and red line moves towards the right side on the number line as shown above.

Question 8.
r ≥ 8
Answer:

Explanation:
Given to show r ≥ 8 inequality on a number line as r is greater than or equal to 8 so it starts at 8 and red line moves towards the right side on the number line as shown above.

Question 9.
m < 22
Answer:

Explanation:
Given to show m < 22 inequality on a number line as m is less than 22 so it starts at 22 and red line moves towards the left side on the number line as shown above.

Question 10.
q ≤ 13
Answer:

Explanation:
Given to show q ≤ 13 inequality on a number line as q is less than or equal to 13 so it starts at 13 and red line moves towards the left side on the number line as shown above.

Write an inequality for each graph on a number line.

Question 11.

Answer:
x < 9,

Explanation:
Let the point be x on the graph which moves from 9 towards left side means it is lesser than 9 so the inequality is x < 9.

Question 12.

Answer:
x > 14,

Explanation:
Let the point be x on the graphs x moves from 14 towards right side means it is greater than 14 so the inequality is x > 14.

Question 13.

Answer:
x ≤ 11,

Explanation:
Let the point be x on the graph x starts at 11 and moves left side means it is less than and equal to 11 so the inequality is  x ≤ 11.

Question 14.

Answer:
x ≥ 7,

Explanation:
Let the point be x on the graph x starts at 7 and moves right side it is greater than and equal 7 means the inequality is x ≥ 7.

Represent the solutions of each inequality on a number line. Then give three possible integer solutions of each inequality.

Question 15.
p < 9\(\frac{1}{2}\)

Answer:

Explanation:
Given p < 9\(\frac{1}{2}\) means p < \(\frac{19}{2}\), p < 9.5, now substituting p as 9.4,9.3,9.2 substituting p as 9.4,9.3,9.2 if p = 9.4 then 9.4 is less than 9.5, d = 9.3 means 9.3 is less than 9.5 and d = 9.2 means 9.2 is less than 9.5 we get in all 3 substitues p < 9.5 which is true so we can substitute p as 9.4,9.3,9.2 shown p < 9.5 inequality on a number line as p is less than 9.5,
1. It starts at 9.4 and red line moves towards the left side on the number line,
2. It starts at 9.3 and red line moves towards the leftt side on the number line,
3. It starts at 9.2 and red line moves towards the left side on the number line as shown above.

Question 16.
y > \(\frac{37}{5}\)
Answer:

Explanation:
Given y > \(\frac{37}{5}\) means y >7.4, Let us take y as 7.5, 7.6, 7.7 substituting y as 7.5, 7.6, 7.7  if y = 7.5 then y is greater than 7.4, y = 7.6 means 7.6 is greater than 7.4 and y = 7.7 means 7.7 is greater than 7.4 we get in all 3 substitues y > 7.4 which is true so we can substitute y as 7.5,7.6,7.7, shown y > \(\frac{37}{5}\) inequality on a number line as y is greater than to 7.4,
1. It starts at 7.5 and red line moves towards the right side on the number line,
2. It starts at 7.6 and red line moves towards the right side on the number line,
3. It starts at 7.7 and red line moves towards the right side on the number line as shown above.

Question 17.
b ≤ \(\frac{23}{4}\)
Answer:

Explanation:
Given b ≤ \(\frac{23}{4}\), b ≤ 5.75, Let us take b as 5.75, 5.6, 5.5 substituting b as 5.75, 5.6, 5.5 if b = 5.75 then b is equal to 5.75 only, b = 5.6 means 5.6 is less than 5.75 and d = 5.5 means 5.5 is less than 5.75 we get in all 3 substitues b ≤ 5.75 which is true so we can substitute b as 5.75,5.6,5.5 shown b ≤ 5.75 inequality on a number line as b is less than and equal to 5.75,
1. It starts at 5.75 and red line moves towards the left side on the number line,
2. It starts at 5.6 and red line moves towards the leftt side on the number line,
3. It starts at 5.5 and red line moves towards the left side on the number line as shown above.

Question 18.
s ≥ 6\(\frac{3}{7}\)
Answer:

Explanation:
Given s ≥ 6\(\frac{3}{7}\) , s ≥ 6.42, Let us take s as 6.42, 6.5, 6.6 substituting s as 6.42,6.5,6.6  if s = 6.42 then s is equal to 6.42 only, s = 6.5 means 6.5 is greater than 6.42 and s = 6.6 means s is greater than 6.42 we get in all 3 substitues s ≥ 6.42 which is true so we can substitute s as 6.42,6.5,6.6 shown s ≥ 6.42 inequality on a number line as s is greater than and equal to 6.42,
1. It starts at 6.42 and red line moves towards the right side on the number line,
2. It starts at 6.5 and red line moves towards the right side on the number line,
3. It starts at 6.6 and red line moves towards the right side on the number line as shown above.

Question 19.
g > 1.5
Answer:

Explanation:
Given g >1.5, Let us take g as 2.5, 3.5,4.5 substituting g as 2.5, 3.5, 4.5  if g = 2.5 then g is greater than 1.5, g = 3.5 means 3.5 is greater than 1.5 and g = 4.5 means 4.5 is greater than 1.5 we get in all 3 substitues g > 1.5 which is true so we can substitute y as 2.5,3.5,4.5, shown g > 1.5 inequality on a number line as g is greater than 1.5
1. It starts at 2.5 and red line moves towards the right side on the number line,
2. It starts at 3.5 and red line moves towards the right side on the number line,
3. It starts at 4.5 and red line moves towards the right side on the number line as shown above.

Question 20.
m ≥ 4.8
Answer:

Explanation:
Given m ≥ 4.8, Let us take m as 4.8, 5.0, 6.0 substituting m as 4.8,5,6 if m = 4.8 then m is equal to 4.8 only, m = 5 means 5 is greater than 4.8 and m = 6 means m is greater than 4.8 we get in all 3 substitues m ≥ 4.8 which is true so we can substitute m as 4.8,5,6 shown m ≥ 4.8 inequality on a number line as m is greater than and equal to 4.8,
1. It starts at 4.8 and red line moves towards the right side on the number line,
2. It starts at 5 and red line moves towards the right side on the number line,
3. It starts at 6 and red line moves towards the right side on the number line as shown above.

Question 21.
z ≤ 9.1
Answer:

Explanation:
Given z ≤ 9.1, Let us take z as 9.1, 9, 8 substituting z as 9.1,9,8 if z = 9.1 then z is equal to 9.1 only, z =9 means 9 is less than 9.1 and z = 8 means 8 is less than 9.1 we get in all 3 substitues z ≤ 9.1 which is true so we can substitute b as 9.1,9,8 shown z ≤ 9.1 inequality on a number line as b is less than and equal to ,
1. It starts at 5.75 and red line moves towards the left side on the number line,
2. It starts at 5.6 and red line moves towards the leftt side on the number line,
3. It starts at 5.5 and red line moves towards the left side on the number line as shown above.

Question 22.
r < 16.6
Answer:

Explanation:
Given r < 16.6, now substituting r as 16, 15, 14 substituting r as 16, 15 , 14 if r = 16 then 16 is less than 16.6, r = 15 means 15 is less than 16.6 and r = 14 means 14 is less than 16.6 we get in all 3 substitues r < 16.6 which is true so we can substitute r as 16,15,14 shown r < 16.6 inequality on a number line as r is less than 16.6,
1. It starts at 16 and red line moves towards the left side on the number line,
2. It starts at 15 and red line moves towards the leftt side on the number line,
3. It starts at 14 and red line moves towards the left side on the number line as shown above.

Solve.

Question 23.
In the inequality x > 9, x represents the number of restaurants along a Street.

a) Is 9 a possible value of c? Explain.
Answer:
No,

Explanation:
Given the inequality x > 9, x represents the number of restaurants along a Street as x is above 9 so 9 is not possible value of c.

b) Is 9\(\frac{2}{5}\) a possible value of x? Explain.
Answer:
Yes,

Explanation:
Yes it is possible value of as given the inequality x > 9, x represents the number of restaurants along a Street as x is above 9 so 9.4 is not possible value of x.

c) Use a number line to represent the solution set of the inequality. Then state the least possible number of restaurants on the street.
Answer:

The least possible number of restaurants on the street is more than 9,

Explanation:
Used a number line to represent the solution set of the inequality as shown above. Then stated the least possible number of restaurants on the street is above 9.

Question 24.
In the inequality q ≤ 24.3, q represents the possible weights, in pounds, of a package.

a) Is 24.4 a possible value of q? Explain.
Answer:
No,

Explanation:
Given the inequality q ≤ 24.3, q represents the possible weights in pounds of a package. As 24.4 is not a possible value of q as q = 24.4 is not less than 24.3.

b) Is 20\(\frac{7}{10}\) a possible value of q. Explain.
Answer:
Yes,

Explanation:
Given the inequality q ≤ 24.3, q represents the possible weights in pounds of a package. As 20\(\frac{7}{10}\) = \(\frac{207}{10}\) =  20.7 is a possible value of q as q = 20.7 is less than 24.3.

c) Use a number line to represent the solution set of the inequality. Then state the greatest possible weight of the package.
Answer:

Explanation:
Used a number line to represent the solution set of the inequality. The greatest possible weight of the package is 24.3.

Each inequality has the variable on the right side of the Inequality symbol. Graph each solution set on a number line.

Question 25.
11 ≤ d
Answer:

Explanation:
Given inequality 11 ≤ d, Shown d is lessthan or equal to 11 on the number line above, d starts at point 11 and moves towards left side as shown above.

Question 26.
7\(\frac{3}{4}\) > q
Answer:

Explanation:
Given inequality 7\(\frac{3}{4}\) > q, Shown q is greater than 7\(\frac{3}{4}\) = 7.75 on the number line above q starts above 7.75 and moves towards right side of 7.75.

Question 27.
2.5 < h
Answer:

Explanation:
Given inequality 2.5 < h, Shown h is lessthan to 2.5 on the number line above h starts at 2.4 and moves towards left side as shown above.

Question 28.
-6 ≥ w
Answer:

Explanation:
Given inequality -6 ≥ w, Shown w is greaterthan or equal to -6 on the number line above w starts at point -6 and moves right side.

Question 29.
5.7 < m
Answer:

Explanation:
Given inequality 5.7 < m, Shown m is less than 5.7 on the number line above m starts below 5.7 and moves towards left side as shown above.

Question 30.
8.1 ≥ n
Answer:

Explanation:
Given inequality 8.1 ≥ n, Shown n is greater than or equal to 8.1 on the number line above n starts at point 8.1 and moves right side.

Practice the problems of Math in Focus Grade 6 Workbook Answer Key Chapter 8 Lesson 8.4 Real-World Problems: Equations and Inequalities to score better marks in the exam.

Math in Focus Grade 6 Course 1 B Chapter 8 Lesson 8.4 Answer Key Real-World Problems: Equations and Inequalities

Math in Focus Grade 6 Chapter 8 Lesson 8.4 Guided Practice Answer Key

Complete.

Question 1.
On Monday, Wendy had some leaves in a collection she was making for biology class. After she collected 23 more leaves on Tuesday, she had 41 leaves. Find the number of leaves Wendy had on Monday.
Let r represent the number of leaves Wendy had on Monday.
r + 23 =
r + 23 – = –
r =
Wendy had leaves on Monday.
Answer:
Wendy had 18 leaves on Monday,

Explanation:
Given on Monday, Wendy had some leaves in a collection she was making for biology class. After she collected 23 more leaves on Tuesday, she had 41 leaves. The number of leaves Wendy had on Monday. Let r represent the number of leaves Wendy had on Monday so r + 23 = 41, r = 41 – 23 = 18 leaves.

Write an algebraic equation for each problem. Then solve.

Question 2.
Carlos thinks of a number. When he adds 17 to it, the result is 45. What is the number that Carlos thought of?
Answer:
The number is 28,

Explanation:
Given Carlos thinks of a number. When he adds 17 to it, the result is 45. Let the number that Carlos thought of is x so x + 17 = 45, therefore x = 45 – 17 = 28.

Question 3.
Sylvia bought some blouses and T-shirts. She paid a total of $63. The T-shirts cost $29. How much did the blouses cost?
Answer:
Blouses cost $34,

Explanation:
Given Sylvia bought some blouses and T-shirts. She paid a total of $63. The T-shirts cost $29. Let the blouses be b so it is $29 + b = $63, So b = $63 – $29 = $34.

Question 4.
Felicia used 153 yellow beads and some green beads for her art project. She used 9 times as many yellow beads as green beads. How many green beads did she use for the project?
Answer:
Green beads did she use for the project are 1,377 beads,

Explanation:
Given Felicia used 153 yellow beads and some green beads for her art project. She used 9 times as many yellow beads as green beads. Let g be green beads did she use for the project so we have green beads as 9 X 153 = 1,377 beads.

Question 5.
Ivan had saved some quarters. He spent 50 quarters, which was \(\frac{2}{5}\) of the quarters he started out with. How many quarters did he start out with?
Answer:
Quarters did Ivan start out with is 125,

Explanation:
Given Ivan had saved some quarters. He spent 50 quarters, which was \(\frac{2}{5}\) of the quarters he started out with. Let x quarters did he starts so x X \(\frac{2}{5}\) = 50, x = 50 X \(\frac{5}{2}\) =
125.

Complete.

Question 6.
The figure shows a speed limit sign on a highway.
a) Let x represent the speed in miles per hour.

Write an inequality to represent the situation.
The inequality is .
Answer:
The inequality is x ≤ 55,

Expanation:
Given fiqure shows a speed limit sign on highway, Let x represent the speed in miles per hour so an inequality to represent the situation is x ≤ 55.
b) Give the maximum legal driving speed on the highway. The maximum legal driving speed is miles per hour.
Answer:
The maximum legal driving speed is 55 miles per hour,

Explanation:
Given fiqure shows a speed limit sign on highway, So the maximum legal driving speed is 55 miles per hour.

Solve.

Question 7.
More than 35 guests came to Katrina’s birthday party last Sunday.

a) Write an inequality to represent the number of guests who turned up for the birthday party.
Answer:
x>35,

Explanation:
Given more than 35 guests came to Katrina’s birthday party last Sunday, Let x represent the number of guests turned up for the birthday party, so the inequality to represent the number of guests who turned up for the birthday party are  x > 35.

b) What is the least possible number of guests who could have come to the party?
Answer:
35,

Explanation:
Given more than 35 guests came to Katrina’s birthday party last Sunday, So the least possible number of guests who could have come to the party are 35.

Question 8.
In Mr. Boyle’s class, the students are required to summarize a passage in less than 50 words.
a) Write an inequality to represent the number of words that the students can use to summarize the passage.
Answer:
x < 50,

Explanation:
Given in Boyle’s class, the students are required to summarize a passage in less than 50 words. So the inequality to represent the number of words that the students can use to summarize the passage , let it be x so x < 50.

b) What is the maximum number of words that a student can use?
Answer:
50,

Explanation:
Given in Boyle’s class, the students are required to summarize a passage in less than 50 words. So the maximum number of words that a student can use are 50.

Question 9.
A cargo elevator has a load limit of 240 tons.
a) Write an inequality to represent the load limit of the cargo elevator.
Answer:
x ≤ 240,

Explanation:
Given a cargo elevator has a load limit of 240 tons. Let load limit be x so an inequality to represent the load limit of the cargo elevator is x ≤ 240.

b) What is the greatest possible load the cargo elevator can carry?
Answer:
240 tons,

Explanation:
Given a cargo elevator has a load limit of 240 tons. So the greatest possible load the cargo elevator can carry is 240 tons.

Question 10.
To get a discount coupon at a bookstore, you need to spend at least $50 at the store.
a) Write an inequality to represent the amount of money that you must spend in order to get a discount coupon.
Answer:
x ≥ 50,

Explanation:
Given to get a discount coupon at a bookstore we need to spend at least $50 at the store. Let amount spend be x so an inequality to represent the amount of money that must be spent in order to get a discount coupon is x ≥ 50.

b) Andrea has spent $45 at the store, and her friend Alex has spent $55. Which person can get a discount coupon?
Answer:
Alex will get a discount coupon,

Explanation:
Given to get a discount coupon at a bookstore we need to spend at least $50 at the store as Andrea has spent $45 which is less than $50 will not get the discount and Alex has spent $55 which is more than $50 so Alex will get a discount coupon.

Math in Focus Course 1B Practice 8.4 Answer Key

Write and solve an algebraic equation for each problem. Show your work.

Question 1.
Damien thinks of a number. When he adds 32 to it, the sum is 97. What is the number that Damien thought of?
Answer:
65,

Explanation:
Given Damien thinks of a number. Let it be x when he adds 32 to it, the sum is 97. So the number that Damien thought of is x + 32 = 97, x = 97 -32 = 65.

Question 2.
A baker made some bagels in the morning. After selling 85 bagels, there were 64 left. How many bagels did the baker make in the morning?
Answer:

149 bagels,

Explanation:
Given a baker made some bagels in the morning. Let it be x after selling 85 bagels there were 64 left. So many bagels did the baker make in the morning are x – 85 bagels= 64 bagels, therefore x = 64 bagels + 85 bagels = 149 bagels.

Question 3.
Claudia can text 3 times as fast as Fiona. Claudia can text 78 words per minute. How many words per minute can Fiona text?
Answer:
26 words per minute,

Explanation:
Given Claudia can text 3 times as fast as Fiona. Claudia can text 78 words per minute means Fiona = Claudia/3 so many words per minute can Fiona text are 78/3 = 26 words per minute.

Question 4.
Eric spent \(\frac{2}{5}\) of his allowance on a jacket. The jacket cost him $12. How much was his allowance?
Answer:
Allowance is $30,

Explanation:
Given Eric spent \(\frac{2}{5}\) of his allowance on a jacket. The jacket cost him $12. Let A be the allowance so much was his allowance is A = $12 X \(\frac{5}{2}\) = \(\frac{$12  X  5}{2}\) = $30.

Write and solve an algebraic inequality for each problem.

Question 5.
In a science competition, students have to score more than 40 points in order to move on to the next round.
a) Write an inequality to represent this situation. Use a number line to represent the inequality.
Answer:

 

 

 

Explanation:
Given in a science competition students have to score more than 40 points in order to move on to the next round,
So an inequality to represent this situation is x > 40. Used a number line to represent the inequality as shown above.

b) What is the least number of points a student needs to score in order to move on to the next round? Only whole numbers of points are awarded to students.
Answer;
41,

Explanation:
As given in a science competition students have to score more than 40 points in order to move on to the next round,
So the least number of points a student needs to score in order to move on to the next round if only whole numbers of points are awarded to students is 41.

Question 6.
A stadium has a seating capacity of 65,000 spectators.
a) What is the maximum number of spectators the stadium can hold?
Answer:
65,000 spectators,

Explanation:
Given a stadium has a seating capacity of 65,000 spectators. So the maximum number of spectators the stadium can hold is 65,000 spectators.

b) Write an inequality to represent this situation. Then use a number line to represent the inequality.
Answer:
x ≤ 65,000,

 

 

 

Explanation:
Given a stadium has a seating capacity of 65,000 spectators. Inequality to represent this situation is x ≤ 65,000. Then used a number line to represent the inequality as shown above.

Write and solve an algebraic equation or inequality for each problem.
Show your work.

Question 7.
A bicycle store sells \(\frac{4}{7}\) of the mountain bikes in the store. Then only 24 mountain bikes are left. How many mountain bikes were there originally?
Answer:
42 mountain bikes,

Explanation:
Given a bicycle store sells \(\frac{4}{7}\) of the mountain bikes in the store. Then only 24 mountain bikes are left. Let x be mountain bikes were there originally so x X \(\frac{4}{7}\) = 24 mountain bikes, x = 24 mountain bikes X \(\frac{7}{4}\) = \(\frac{24 X 7}{4}\) mountain bikes = 6 X 7 = 42 mountain bikes.

Question 8.
Mabel has a total of 54 decorative beads. Some are black and some are white. The ratio of the number of black beads to the number of white beads is 7 : 2. How many more black beads than white beads are there?
Answer:
More black beads than white beads are there 30 beads,

Explanation:
Given Mabel has a total of 54 decorative beads. Some are black and some are white. The ratio of the number of black beads to the number of white beads is 7 : 2. Let x be number of beads so 7x + 2x = 54 beads, 9x =54 beads, x = 54/9 beads = 6 beads black beads are 7 X 6 = 42 beads and white beads are 2 X 6 = 12 beads so more black beads than white beads are there 42 beads – 12 beads = 30 beads.

Question 9.
Gary has a collection of comic books. After selling 70% of his comic books, he has 42 comic books left. How many comic books did he start with?
Answer:
Gary start with 60 comic books,

Explanation:
Given Gary has a collection of comic books. After selling 70% of his comic books, he has 42 comic books left. So many comic books did he start with be x is x X 70% = 42 comic books,  x = 42 X 100/70 = 60 comic books.

Question 10.
There are 30 students in the gym. If there are at least 16 girls, write an inequality to represent the number of boys in the gym.
Answer:
x ≤ 14,

Explanation:
Given there are 30 students in the gym. If there are at least 16 girls, So an inequality to represent the number of boys in the gym let it be x as number of boys + number of girls = 30 so number of boys are 30 – 16 = 14 therefore it is less than or equal to 14 so x ≤ 14.

Question 11.
The marbles in a box are repackaged in equal numbers into 6 smaller bags. If each bag has more than 8 marbles, what is the least possible number of marbles that could have been in the box?
Answer:
48 marbles,

Explanation:
Given the marbles in a box are repackaged in equal numbers into 6 smaller bags. If each bag has more than 8 marbles, let x be the least possible number of marbles that could have been in the box so it is x = 6 X 8 = 48 marbles.

Question 12.
Mr. Edwards is now 3 times as old as his daughter. In 15 years’ time, the sum of their ages will be 86.
a) Find their ages now.
Answer:
Father 42 years, Daughter 14 years,

EXplanation:
Given Mr. Edwards is now 3 times as old as his daughter. In 15 years’ time, the sum of their ages will be 86. Let the age of the daughter be x so 3x + 15 + x + 15 = 86, 4x + 30 = 86, 4x = 86 – 30 = 56, So x = 56/4 = 14 years. So fathers age is 3 X 14 years = 42 years.

b) How old was Mr. Edwards when his daughter was born?
Answer:
Mr.Edwards was 28 years when his daughter was born,

Explanation:
If the ages of father is 42 years and daughter age is 14 years then Mr.Edwards age when his daughter was born is 42 years – 14 years = 28 years.

Question 13.
In a competition, each school is allowed to send a team with at least 5 members, but not more than 8 members. 12 schools participated in the competition.
a) Find the least possible number of participants in the competition.
Answer:
60 participants,

Explanation:
Given in a competition, each school is allowed to send a team with at least 5 members. 12 schools participated in the competition. Let x be the least possible number of participants in the competition so x = 12 X 5 = 60 participants.

b) Find the greatest possible number of participants in the competition.
Answer:
96 participants,

Explanation:
In a competition each school is allowed to send a team but not more than 8 members. 12 schools participated in the competition. Let y be the greatest possible number of participants in the competition so y = 12 X 8 = 96 participants.

Brain @ Work

A rectangular photograph is mounted on a rectangular card. There is a border of equal width around the photograph. The perimeter of the card is 40 centimeters longer than that of the photograph. Find the width of the border in centimeters.

Answer:
Given a rectangular photograph is mounted on a rectangular card. There is a border of equal width around the photograph. The perimeter of the card is 40 centimeters longer than that of the photograph. As the perimeter is 2(length + width), let the perimeter of the photo is p and given the perimeter of the card as 40 + p, So the width of the border in centimeters.

Practice the problems of Math in Focus Grade 6 Workbook Answer Key Chapter 8 Review Test to score better marks in the exam.

Math in Focus Grade 6 Course 1 B Chapter 8 Review Test Answer Key

Concepts and Skills

Solve each equation using the concept of balancing. Write all fraction answers in simplest form.

Question 1.
x + 8 = 27
Answer:
x = 19,

Explanation:
Given x + 8 = 27, So x = 27 – 8 = 19.

Question 2.
\(\frac{10}{11}\) = a + \(\frac{4}{11}\)
Answer:
a=\(\frac{6}{11}\),

Explanation:
Given \(\frac{10}{11}\) = a + \(\frac{4}{11}\), So a = \(\frac{10}{11}\) – \(\frac{4}{11}\)  both have denominators common so we subtract only numerators so we get a = \(\frac{10 – 4}{11}\) = \(\frac{6}{11}\).

Question 3.
f + 3.8 = 9.2
Answer:
f = 5.4,

Explanation:
Given f + 3.8 = 9.2, So f = 9.2 – 3.8 = 5.4.

Question 4.
42 = y – 14
Answer:
y = 56,

Explanation:
Given 42 = y – 14, y = 42 + 14 = 56.

Question 5.
k – \(\frac{7}{8}\) = 2\(\frac{11}{24}\)
Answer:
k = 3\(\frac{1}{3}\)[/latex]

Explanation:
Given k – \(\frac{7}{8}\) = 2\(\frac{11}{24}\), k = 2\(\frac{11}{24}\) + \(\frac{7}{8}\) = \(\frac{59}{24}\) + \(\frac{7}{8}\) as denominators are different first we make them same
so we multiply \(\frac{7}{8}\) numerator and denominator by 3 as \(\frac{7 X 3}{8 X 3}\) = \(\frac{21}{24}\) now both denominators are same we add numerators \(\frac{59 + 21}{24}\) = \(\frac{80}{24}\) both have common factors so \(\frac{8 X 10}{8 X 3}\) we get \(\frac{10}{3}\) as numerator is greater than denominator we can further write as mixed fraction as \(\frac{3 X 3 + 1}{3} = 3[latex]\frac{1}{3}\)[/latex].

Question 6.
n – 2.7 = 13.4
Answer:
n = 16.1,

Explanation:
Given n- 2.7 = 13.4, So n = 13.4 + 2.7 = 16.1.

Question 7.
6h = 84
Answer:
h = 14,

Explanation:
Given 6h = 84, So h = 84/6,
6)84(14
   60
24
24
0
So h = 14.

Question 8.
75.6 = 7.2r
Answer:
r = 10.5,

Explanation:
Given 75.6 = 7.2r, So r = 75.6/7.2,
7.2)75.6(10.5
      72
        3.6
3.6
0
So r = 10.5.

Question 9.
\(\frac{4}{5}\)p = 10
Answer:
p = 12.5,

Explanation:
Given \(\frac{4}{5}\)p = 10, p = 10/\(\frac{4}{5}\), p= \(\frac{10 X 5}{4}\) = \(\frac{50}{4}\) =
4)50(12.5
   40
    10
      8
      20
      20
       0
So p = 12.5.

Question 10.
9 ∙ \(\frac{3}{5}\) = \(\frac{8}{11}\)w
Answer:
w = 13\(\frac{8}{40}\),

Explanation:
Given 9 ∙ \(\frac{3}{5}\) = \(\frac{8}{11}\)w, w = \(\frac{9 X 5 + 3}{5}\) X \(\frac{11}{8}\) = \(\frac{48 X 11}{5 X 8}\) = \(\frac{528}{40}\) as numerator is greater than denominator so we write in mixed fraction \(\frac{40 X 13 + 8}{40}\) = 13\(\frac{8}{40}\).

Represent the solution set of each inequality on a number line.

Question 11.
b < 7
Answer:

 

 

 

Explanation:
Given to show b < 7 inequality on a number line as b is less than 7 so it starts at 7 and red line moves towards the left side on the number line as shown above.

Question 12.
b > 13
Answer:

Explanation:
Given to show b > 13 inequality on a number line as b is greater than 13 so it starts at 13 and red line moves towards the right side on the number line as shown above.

Question 13.
m ≥ 24
Answer:

 

 

 

Explanation:
Given to show m ≥ 24 inequality on a number line as m is greater than or equal to 24 so it starts at 24 and red line moves towards the right side on the number line as shown above.

Question 14.
m ≤ 38
Answer:

Explanation:
Given to show m ≤ 38 inequality on a number line as m is less than or equal to 38 so it starts at 38 and red line moves towards the left side on the number line as shown above.

Question 15.
g > \(\frac{2}{3}\)
Answer:

 

 

 

 

Explanation:
Given to show g > \(\frac{2}{3}\) = 0.666 inequality on a number line as g is greater than 0.66 so it starts at 0.66 and red line moves towards the right side on the number line as shown above.

Question 16.
g ≤ 5\(\frac{3}{5}\)
Answer:

 

 

Explanation:
Given to show g ≤ 5\(\frac{3}{5}\) = \(\frac{5 X 5 + 3}{5}\) = \(\frac{28}{5}\) = 5.6 inequality on a number line as g is less than or equal to 5.6 so it starts at 5.6 and red line moves towards the left side on the number line as shown above.

Question 17.
z < 7.1
Answer:

Explanation:
Given to show z < 7.1 inequality on a number line as z is less than 7.1 so it starts at 7.1 and red line moves towards the left side on the number line as shown above.

Question 18.
z ≥ 10.4
Answer:

 

Explanation:
Given to show z ≥ 10.4 inequality on a number line as m is greater than or equal to 10.4 so it starts at 10.4 and red line moves towards the right side on the number line as shown above.

Write an inequality for each number line.

Question 19.

Answer:
x ≥ 9,

Explanation:
As seen on the number line let us take the point as x it starts exactly at 9 and moves towards right side means it is greater so the inequality is x ≥ 9.

Question 20.

Answer:
x < 16,

Explanation:
As seen on the number line let us take the point as x it starts from 16 and moves towards left side means lesser so the inequality is x <16.

Question 21.

Answer:
x < \(\frac{7}{10}\),

Explanation:
As seen on the number line let us take the point as x it starts from \(\frac{7}{10}\) and moves towards left side means lesser so the inequality is x < \(\frac{7}{10}\).

Question 22.

Answer:
x ≥ 12.5,

Explanation:
As seen on the number line let us take the point as x it starts exactly at 12.5 and moves towards right side means it is greater so the inequality is x ≥ 12.5.

This handy Math in Focus Grade 6 Workbook Answer Key Chapter 11 Review Test detailed solutions for the textbook questions.

Math in Focus Grade 6 Course 1 B Chapter 11 Review Test Answer Key

Chapter Review/Test

Concepts and Skills

Find the circumference and area of each circle. Use \(\frac{22}{7}\) as an approximation for π.

Question 1.

Answer:
Circumfrence: 307.72 cm,
Area: 7,539.14 square centimeters,

Explanation:
Given diameter of the circle as 98 cm so radius is 98 cm/2 = 49 cm, Circumfrence is 2πr = 2 X 3.14 X 49 cm = 307.72 cm, Area of the circle is πr²  = 3.14 X 49 cm X 49 cm = 7,539.14 square centimeters.

Question 2.

Answer:
Circumfrence: 351.68 cm,
Area: 9,847.04 square centimeters,

Explanation:
Given diameter of the circle as 98 cm so radius is 112 cm/2 = 56 cm, Circumfrence is 2πr = 2 X 3.14 X 56 cm = 351.68 cm, Area of the circle is πr²  = 3.14 X 56 cm X 56 cm = 9,847.04 square centimeters.

Find the distance around each semicircle. Use \(\frac{22}{7}\) as an approximation for π.

Question 3.

Answer:
The distance around the semicircle is 21.98 ft,

Explanation:
Given diameter of the circle as 14 ft so radius is 14 ft/2 = 7ft, Distance around the semicircle is \(\frac{1}{2}\)  X 2πr = 3.14 X 7 ft = 21.98 ft.

Question 4.

Answer:
The distance around the semicircle is 98.91 in,

Explanation:
Given diameter of the circle as 63 in so radius is 63 in/2 = 31.5 in, Distance around the semicircle is \(\frac{1}{2}\)  X 2πr = 3.14 X 31.5 in = 98.91 in.

Find the distance around each quadrant. Round your answer to the nearest tenth. Use 3.14 as an approximation for π.

Question 5.

Answer:
The distance around the quadrant is 7.85 m,

Explanation:
Given radius of the circle is 5 m, Distance around the quadrant is \(\frac{1}{4}\) X 2πr = \(\frac{1}{2}\) X πr  = \(\frac{1}{2}\) X 3.14 X 5 m = 7.85 m.

Question 6.

Answer:
The distance around the quadrant is 23.55 ft,

Explanation:
Given radius of the circle is 15 ft, Distance around the quadrant is \(\frac{1}{4}\) X 2πr = \(\frac{1}{2}\) X πr  = \(\frac{1}{2}\) X 3.14 X 15 ft = 23.55 ft.

Solve. Show your work.

Question 7.
The diameter of a flying disc is 10 inches. Find the circumference and area of the disc. Use 3.14 as an approximation for π.
Answer:
Circumfrence of the flying disc is 31 inches and area of the flying disc is 78.5 square inches,

Explanation:
Given diameter of the flying disc as 10 inches so radius is 10 in/2 = 5 in, Circumfrence is 2πr = 2 X 3.14 X 5 in = 31.4 in, Area of the circle is πr² = 3.14 X 5 in X 5 in = 78.5 square inches.

Question 8.
The area of a compact disc is 452\(\frac{4}{7}\) square centimeters. What is the diameter of the compact disc? Use \(\frac{22}{7}\) as an approximation for π.
Answer:
The diameter of the compact disc is 26.22 centimeters,

Explanation:
Given the area of a compact disc is 452\(\frac{4}{7}\) square centimeters. Let d be the diameter of the compact disc so 452\(\frac{4}{7}\) square centimeters = πr² = π X (d/2)² ,
3.14 X d² = 4 X 452\(\frac{4}{7}\) square centimeters,
d² = 4 X 3,168 sq cms/7 X 3.14,
d² = 12,672 sq cms /21.98 = 576.524 therefore d = square root of 576.524 sq cms = 26.22 centimeters.

Question 9.
The circumference of a circular table is 816.4 centimeters. Find the radius of the table. Use 3.14 as an approximation for π.
Answer:
The radius of the table is 130 cntimeters,

Explanation:
Given the circumference of a circular table is 816.4 centimeters means it is 2πr, So radius = 816.4 cms/2 X 3.14= 816.4 cms/6.28 = 130 centimeters

Problem Solving

Solve. Show your work.

Question 10.
A water fountain shoots up a jet of water. The water falls back down onto the ground in the shape of a circle. Michelle wants the circle of water on the ground to be 0.7 meter wider on each side. She gradually increases the strength of the water jet. The area of the circle of water increases at 0.2 square meter per second. Use \(\frac{22}{7}\) as an approximation for n.
a) Find the area of the original circle of water.
Answer:
The area of the original circle of water is 13.8474 square meters,

Explanation:
The radius of the original circle of water  as diameter 4.2 m is 4.2 m/2 = 2.1 m, So area is  3.14 X 2.1 m X 2.1 m = 13.8474 square meters.

b) Find the area of the larger circle of water.
Answer:
The area of the larger circle of water is 24.6176 square meters,

Explanation:
The radius of the original circle of water  as diameter 4.2 m + 0.7 m + 0.7 m = 5.6 m is 5.6 m/2 = 2.8 m, So area is 2 X 3.14 X 2.8 m X 2.8 m =  24.6176 square meters.

c) How long does it take for the original circle of water to become the larger circle of water? Round your answer to the nearest second.

Answer:
It take for the original circle of water to become the larger circle of water to the nearest second is 54 seconds,

Explanation:
The area increased from the orignal circle to larger circle is 24.6176 square meters – 13.8474 square meters = 10.7702 square meters, if the area of the circle of water increases at 0.2 square meter per second long does it take for the original circle of water to become the larger circle of water is 10.7702/ 0.2 = 53.851 seconds nearest second is 54 seconds.

Question 11.
A machine in an assembly line stamps pieces of metal. The stamping plate on the machine travels in a path shaped like the arc of a quadrant as the stamping plate opens and closes. It takes the machine 5 seconds to open and close the stamping plate one time. Use \(\frac{22}{7}\) as an approximation for π.
a) Find the total distance the outside edge of the stamping plate travels when the machine opens and closes one time.
Answer:
The total distance the outside edge of the stamping plate travels when the machine opens and closes one time is 60.442 cms,

Explanation:
Given a machine in an assembly line stamps pieces of metal. The stamping plate on the machine travels in a path shaped like the arc of a quadrant as the stamping plate opens and closes, So the total distance the outside edge of the stamping plate travels when the machine opens and closes one time is distance around the quadrant is \(\frac{1}{4}\) X 2πr we have radius as 38.5 cm so the distance is \(\frac{1}{4}\) X 2 X 3.14 X 38.5 cm = 60.445 cms.

b) Find the speed of the stamping plate’s outside edge in centimeters per second.
Answer:
The speed of the stamping plate’s outside edge in centimeters per second is 12.089 centimeters per second,

Explanation:
Given it takes the machine 5 seconds to open and close the stamping plate one time, So the speed of the stamping plate’s outside edge in centimeters per second is 60.445 cms/5 seconds = 12.089 centimeters per second.

c) Assume the machine starts and ends in an open position. How many seconds will it take the machine to stamp 500 pieces of metal?

Answer:

Answer:
Seconds will it take the machine to stamp 500 pieces of metal is 6,044.5 seconds,

Explanation:
Assuming the machine starts and ends in an open position. So seconds will it take the machine to stamp 500 pieces of metal is 12.089 centimeters per second X 500 = 6,044.5 seconds.

Question 12.
The figure shows four identical quadrants enclosed in a square. The side length of the square is 20 inches. Find the area of the blue part. Use 3.14 as an approximation for π.

Answer:
The area of the blue part is 321.5 square inches,

Explanation:
The figure shows four identical quadrants enclosed in a square. The side length of the square is 20 inches, The area of the squar is 20 in X 20 in = 400 sq in. As diameter is 20 in radius is 20 in/2 = 10 in so area of four quadrants is \(\frac{1}{4}\) X πr² = \(\frac{1}{4}\) X 3.14 X 10 in X 10 in = 78.5 square inches. Now area of the blue part is 400 square inches – 78.5 square inches = 321.5 square inches.

Question 13.
The figure shows 3 identical circles. X, Y, and Z are the centers of the circles, and the radius of each circle is 15 feet. \(\frac{1}{6}\) of each circle is shaded. What is the total area of the shaded portion? Round your answer to the nearest tenth of a foot. Use 3.14 as an approximation for π.

Answer:
The total area of the shaded portion is 350 square feet,

Explanation:
Given the figure shows 3 identical circles. X, Y, and Z are the centers of the circles, and the radius of each circle is 15 feet. \(\frac{1}{6}\) of each circle is shaded. So the total area of the shaded portion first we calculate area of the circle with radius 15 feet is πr² = 3.14 X 15 feet X 15 feet = 706.5 square feet so the area of  shaded circle is \(\frac{1}{6}\) X 706.5 square feet = 117.75 square feet so we have 3 identical circles shaded so its total area is 3 X 117.75 square feet is 353.25 nearest to 350 square feet.

Question 14.
The figure is made up of one semicircle and two quadrants. The distance around the figure is 97.29 inches. Find the value of k. Use 3.14 as an approximation for π.

Answer:
The value of k is 15.492 inches,

Explanation:
Given the figure is made up of one semicircle and two quadrants. The distance around the figure is 97.29 inches and the diameter is k in. first the distance of semi circle is \(\frac{1}{2}\) X 2πr = 3.14 X k in. = 3.14k in. Now the area of 2 quadrants is 2 X \(\frac{1}{4}\) X 2πr =3.14k in. Given the distance of fiqure as 97.29 inches = 3.14k inches + 3.14 k inches = 6.28k inches, Therefore k = 97.29/6.28 = 15.492 inches.

This handy Math in Focus Grade 6 Workbook Answer Key Cumulative Review Chapters 8-11 detailed solutions for the textbook questions.

Math in Focus Grade 6 Course 1 B Cumulative Review Chapters 8-11 Answer Key

Concepts and Skills

Represent the solution set of each Inequality on a number line. (Lesson 8.3)

Question 1.
p ≤ 35
Answer:

 

 

Explanation:
Shown p ≤ 35 inequality with red line on the number line above the point starts from 35 and moves towards left side values less than or equal to 35.

Question 2.
q ≥ 12.6
Answer:

 

 

 

Explanation:
Shown q ≥ 12.6 inequality with green line on the number line above, the points starts from 12.6 on the number lines and moves towards right side greater than or equal to 12.6.

Question 3.
r < \(\frac{4}{5}\)
Answer:

Explanation:
Shown r < \(\frac{4}{5}\) = 0.8 inequality with green line on the number line above the point starts from 0.8 and moves towards left side values less than r < \(\frac{4}{5}\) = 0.8.

Question 4.
s > 13\(\frac{1}{2}\)
Answer:

 

 

 

Explanation:
Shown s > 13\(\frac{1}{2}\) = 13.5 inequality with green line on the number line above, the points starts from 13.5 on the number lines and moves towards right side greater than 13.5.

Use graph paper. Plot the points on a coordinate plane and answer the question. (Lesson 9.1)

Question 5.
a) Plot points A (-2, -2) and B (-10, -2) on a coordinate plane. Connect the two points to form a line segment.
Answer:
Explanation:
Plotted points A (-2, -2) and B (-10, -2) on a coordinate plane and connected the two points to form a line segment as shown above.

b) Point C lies above \(\overline{A B}\), and is 5 units away from the x-axis. If triangle ABC is an isosceles triangle with base \(\overline{A B}\), find the coordinates of point C.
Answer:
If Point C lies above \(\overline{A B}\), and is 5 units away from the x-axis. If triangle ABC is an isosceles triangle with base \(\overline{A B}\) and points A (-2, -2) and B (-10, -2)  as AC = BC, Let Point be (x,y) so Square root of (x-(-2))²  +(y-(-2))²  = square root of (x -(-10))²  + (y-(-2))² ,
(x + 2)²  + (y + 2)²  = (x + 10)²  + (y + 2)² ,
4x + 4  = 20x + 100,
16x = 4 – 100 = -96,
x = -96/4 = -24 and y = 0

c) Points D and E lie below \(\overline{A B}\) such that ABDE is a rectangle. If BD is 7 units, find the coordinates of points D and E.

Answer:

Find the area of each figure. (Lesson 10.2)

Question 6.

Answer:
The area is 42 square meters,

Explanation:
Given figure is trapezoid and area of trapezoid is \(\frac{1}{2}\)(a +b) X h where a and b are bases of parallel sides so a = 9 m, b = 5 m and h = 6 m, therefore area = \(\frac{1}{2}\)(9 m + 5 m) X 6 m = 7 m X 6 m = 42 square meters.

Question 2.

Answer:
The area is 120 square feet,

Explanation:
Given figure is trapezoid and area of trapezoid is \(\frac{1}{2}\)(a +b) X h where a and b are bases of parallel sides so a = 16 ft, b = 8 ft and h = 10 ft, therefore area = \(\frac{1}{2}\)(16 ft + 8 ft) X 10 ft = 12 ft X 10 ft = 120 square feet.

Solve. Show your work. (Lessons 8.2, 11.1, 11.2)

Question 8.
The cost of a shirt is p dollars. The cost of a pair of pants is twice the cost of the shirt. If the cost of the pair of pants is t dollars, express t in terms of p.
Answer:
t = 2p dollars,

Explanation:
Given the cost of a shirt is p dollars. The cost of a pair of pants is twice the cost of the shirt. If the cost of the pair of pants is t dollars, So t in terms of p is t = 2p dollars.

Question 9.
A can has a circular base of diameter 8 centimeters. Find the area of this base. Use 3.14 as an approximation for π.
Answer:
The area of the base is 50.24 square centimeters,

Explanation:
Given A can has a circular base of diameter 8 centimeters. As the area is πr² where radius = 8 cms/2 = 4 cms, So area = 3.14 X 4 cms X 4 cms = 50.24 square centimeters.

Question 10.
The cross section of a bowl is in the shape of a semicircle. The area of the semicircle is 77 square centimeters. Find its radius. Use \(\frac{22}{7}\) as an approximation for π.
Answer:
Radius is approximately 7 cms,

Explanation:
Given the cross section of a bowl is in the shape of a semicircle. The area of the semicircle is 77 square centimeters.
As area of semicircle is \(\frac{1}{2}\) X πr² so 77 sq cms = \(\frac{1}{2}\) X 3.14 X r² ,
r² = 77 sq cms X 2/3.14 = 154 sq cms/3.14 = 49.04 sq cms, therefore r = square root of 49.04 approximately 7 cms.

Question 11.
The circumference of a platinum ring is 44 millimeters. Find its radius. Use \(\frac{22}{7}\) as an approximation for π.
Answer:
The radius of the platinum ring is 7.006 millimeters,

Explanation:
Given the circumference of a platinum ring is 44 millimeters as circumference is 2πr so r = 44 millimeter/2π, r = 44 mm/6.28 = 7.006 millimeters.

Problem Solving

Solve. Show your work.

Question 12.
Emily weighs x pounds. Jonathan weighs 3 times as much as Emily. If Jonathan weighs 81 pounds, write an equation in terms of x and solve it. (chapter 8)
Answer:
Equation is 3x = 81, Emily weighs 27 pounds,

Explanation:
Given Emily weighs x pounds. Jonathan weighs 3 times as much as Emily. If Jonathan weighs 81 pounds, So equation is 3x = 81 pounds so x = 81/3 = 27 pounds.

Question 13.
Kim has a bag of yo-yos. Some of the yo-yos are red. The rest are yellow. The ratio of the number of red yo-yos to the number of yellow yo-yos is 5 : 9. If Kim has a total of b yo-yos, how many more yellow yo-yos than red yo-yos are there? (Chapter 8)
Answer:
Many more yellow yo-yos than red yo-yos are r = (b -9y)/5,

Explanation:
Given Kim has a bag of yo-yos. Some of the yo-yos are red. The rest are yellow. The ratio of the number of red yo-yos to the number of yellow yo-yos is 5 : 9. If Kim has a total of b yo-yos, let r be red yo-yos and y be yellow yo-yos,
b = 5r + 9y, So many more yellow yo-yos than red yo-yos are r = (b -9y)/5.

Question 14.
The length of a rectangle is 4n centimeters, and it is twice as long as the width. The perimeter of the rectangle is twice as long as that of an equilateral triangle. Find the side length of the triangle in terms of n. (Chapter 10)
Answer:
Side length of the triangle in terms of n is 2n,

Explanation:
Given the length of a rectangle is 4n centimeters and it is twice as long as the width. The perimeter of the rectangle is twice as long as that of an equilateral triangle, length of rectangle = 4n and width is 4n/2 = 2n, So Perimeter p = 2(4n + 2n) = 12n, let a be the side length of the triangle so 12n : 2 = 6n, P = 3a, a = 6n : 3, a = 2n, So Side length of the triangle in terms of n is 2n.

Question 15.
The length of the minute hand of a clock is 6.5 centimeters. How far does the tip of the minute hand travel in an hour? Use 3.14 as an approximation for π. (Chapter 11)
Answer:
The minute hand travel in an hour is 40.82 centimeters,

Explanation:
Given the length of the minute hand of a clock is 6.5 centimeters. The tip of the minute hand travel in an hour is the circumference with radius as 6.5 centimeters, so it is 2π r = 2 X 3.14 X 6.5 centimeters = 40.82 centimeters.

Question 16.
The area of trapezoid ABCE is 36 square meters. Find the height of the trapezoid. (Chapter 10)

Answer:
The height of the trapezoid is 6 meters,

Explanation:
Area of trapezoid is \(\frac{1}{2}\)(a +b) X h where a and b are bases of parallel sides given a = 7 m, b = 5 m and let height be h m, therefore 36 square meters = \(\frac{1}{2}\)(7 m + 5 m) X h m, height = 2 X 36 sq mts/12 mts = 6 meters.

Find the area of the shaded region. (Chapter 10)

Question 17.

Answer:
The area of the shaded region is 56 square feet,

Explanation:
Given area consists of 2 rectangles one bigger one with length 12 ft and width 8 ft so area is 12 ft X 8 ft = 96 square feet and another small rectangle inside the bigger rectangle with a length 8 ft and width 5 ft so area is 8 ft X 5 ft = 40 square feet now the area of the shaded region is 96 square feet – 40 square feet = 56 square feet.

Question 18.

Answer:
The area of the shaded region is 172.5 square inches,

Explanation:
The shape is square inside with 2 non shaded regions are trapezoid and triangle, First area of square with side length 15 in is 15 inches X 15 inches = 225 square inches, Now area of non shaded region first trapezoid we have a as 2.5 inches, b as 7.5 inches, height 7.5 inches, so area is (2.5 inches + 7.5 inches)/2 X 7.5 inches = 37.5 square inches, Area of the non shaded triangle with base 4 inches and height 7.5 inches is 1/2 X 4 inches X 7.5 inches = 15 square inches,
Total area of non shaded region is 37.5 square inches + 15 square inches = 52.5 square inches, Now the area of the shaded region is 225 square inches – 52.5 square inches = 172.5 square inches.

Find the area of the shaded region. Use 3.14 as an approximation for π.

Question 19.

Answer:
The area of the shaded region is 113.64 square centimeters,

Explanation:
The area of the shaded region is area of triangle – area of the circle so first area of the triangle is \(\frac{1}{2}\) X base X height here base is 25 cm and height is 21.4 cm so area is \(\frac{1}{2}\) X 25 cm X 21.4 cm = 267.5 square centimeters, Area of circle is πr² we have diameter as 14 cm so radius = 14 cm/2 = 7 cm so area is 3.14 X 7 cm X 7 cm = 153.86 sq cms now area of the shaded region is 267.5 sq cms – 153.86 sq cms = 113.64 sq cms.

Question 20.

Answer:
The area of the shaded region is 75.11 square meters,

Explanation:
Given a circle and square in it and in square we have 2 small shaded triangles, First the area of the square is 10 m X 10 m = 100 square meters, The area of the circle with radius 7 m is πr² = 3.14 X 7 m X 7 m = 153.86 square meters,
Area of shaded region without square is 153.86 square meters – 100 square meters = 53.86 square meters, We have 2 triangles shaded regions one traingle area with base side 5 m and height 5 m so its area is 1/2 X 5 m X 5 m = 12.5 square meters another area of square is 1/2 X 5 m X 3.5 m = 8.75 square meters so area of 2 small shaded triangles is 12.5 sq mts + 8.75 sq mts = 21.25 square meters, Now area of the only shaded region is 53.86 square meters + 21.25 square meters = 75.11 square meters.
Solve.

Question 21.
The diagram shows the plan of a square garden. The side length of each grid square is 2 meters. (Chapters 9, 10)

a) A triangular region ABP is surrounded with a wooden fence. The shortest possible distance from point P to \(\overline{A B}\) is 8 meters, and triangle ABP is an isosceles triangle with base \(\overline{A B}\). Find the coordinates of point P.
Answer:
We have A(-8,8) and B(-8,-8) and triangle ABP is an isosceles triangle with base AB means AP = BP,
Square root of  (x – (-8))² + (y – 8)² = Square root of (x  – (-8))² + (y -(-8))² ,
-16y + 64 = 16y + 64,
32y = 0,

b) Find the area and perimeter of the garden.
Answer:
Area is 64 square meters, Perimeter of the garden is 32 meters,

Explanation:
Given the plan of a square garden with 8 mts length so area is 8 mts X 8 mts = 64 square meters and perimeter of the garden is 4 X 8 meters = 32 meters.

c) Find the area of the garden that lies outside triangle ASP in square meters.
Answer:

Solve. Use graph paper to answer the question.

Question 22.
An aspen tree is 300 centimeters tall. It grows 15 centimeters taller each month. The height of the tree, h centimeters, over t months is given by h = 300 + 15t. Copy and complete the table, Graph the relationship between t and h. Use 1 unit on the horizontal axis to represent 1 month and 1 unit on the vertical axis to represent 15 centimeters. Start your vertical axis at 300 centimeters. (Chapters 8, 9)
a)

Answer:

Explanation:
Updated the given table as shown above.

b) What is the height of the tree after 3 months?
Answer:
The height of the tree after 3 months is 345 centimeters,

Explanation:
If an aspen tree is 300 centimeters tall. It grows 15 centimeters taller each month. So after 3 months it will grow
300 centimeters + 15 centimeters + 15 centimeters + 15 centimeters = 345 centimeters.

c) Assuming the growth of the tree is constant for the next year, what is the height of the tree after 10 months?
Answer:
The height of the tree after 10 months is 450 centimeters,
Explanation:
Assuming the growth of the tree is constant for the next year, So the height of the tree after 10 months will be 300 centimeters + 10 X 15 centimeters = 300 centimeters + 150 centimeters = 450 centimeters.

d) If the tree is at least 360 centimeters tall, how many months have passed? Express your answer in the form of an inequality in terms of t, where t stands for the number of months that have passed.
Answer:
360 = 300 + 15t,

Explanation:
Given if the tree is at least 360 centimeters tall, how many months have passed so expressing my answer in the form of an inequality in terms of t, where t stands for the number of months that have passed is 360 = 300 + 15t.

e) Name the dependent and independent variables.
Answer:
Dependent is time and height and independent is 15 centimeters taller each month,

Explanation:
Given an aspen tree is 300 centimeters tall. It grows 15 centimeters taller each month. The height of the tree, h centimeters, over t months is given by h = 300 + 15t. Here dependent is time and height and independent is 15 centimeters taller each month.

Solve. Show your work.

Question 23.
A circular garden is surrounded by a cement path that is 1.5 meters wide. Find the area of the path. Use 3.14 as an approximation for n. (Chapter 11)

Answer:
The area of the path 40.035 square meters,

Explanation:
Given the radius of the circle with circular path is 5 m so its area is πr² = 3.14 X 5 m X 5 m = 78.5 square meter, now area of the circle without the path as radius is 5 m – 1.5 m = 3.5 m is 3.14 X 3.5 m X 3.5 m = 38.465 square meter, Now the area of the path only is 78.5 square meter – 38.465 square meter = 40.035 square meter.

Question 24.
Figure PRSV is a parallelogram. The length of \(\overline{V U}\) and \(\overline{U T}\) are equal, The area of parallelogram QRSW is 30 square inches. The length of \(\overline{R S}\) is 6 inches. (Chapter 10)

a) Find the height of parallelogram QRSW.
Answer:
The height of parallelogram QRSW is 5 inches,

Explanation:
Given the area of parallelogram QRSW is 30 square inches. The length of \(\overline{R S}\) is 6 inches.
As area of parallelogram is base X height we have base as 6 in. so height of the parallelogram QRSW is 30 square inches/6 inches = 5 inches.

b) Find the area of triangle PRV.
Answer:
The area of triangle PRV is 30 square inches,

Explanation:
As given the length of \(\overline{V U}\) and \(\overline{U T}\) are equal, So length of \(\overline{V U}\) = 5 inches so \(\overline{U T}\) = 5 inches the total length of VT is length of \(\overline{V U}\) + length of \(\overline{U T}\)  = 5 inches + 5 inches = 10 inches and PR = VT = 10 inches, PV = RS = 6 inches so the area of triangle PRV is 1/2 X RS X PR = 1/2 X 6 inches X 10 inches = 30 square inches.

c) If you did not know the length of \(\overline{R S}\), explain how you could find the area of triangle PRV.
Answer:
Through Parallelogram,

Explanation:
As we know the area of a triangle is half the product of any of its sides and the corresponding altitude. If a triangle and a parallelogram are on the same base and between the same parallels, then the area of the triangle is equal to half of the area of the parallelogram.

This handy Math in Focus Grade 6 Workbook Answer Key Chapter 11 Lesson 11.3 Real-World Problems: Circles detailed solutions for the textbook questions.

Math in Focus Grade 6 Course 1 B Chapter 11 Lesson 11.3 Answer Key Real-World Problems: Circles

Math in Focus Grade 6 Chapter 11 Lesson 11.3 Guided Practice Answer Key

Complete. Use 3.14 as an approximation for π.

Question 1.
The circumference of the moon is the approximate distance around a circle with radius 1,736 kilometers. Find the circumference of the moon.
a) Round your answer to the nearest 10 kilometers.
Circumference of moon = 2πr
≈ • •
= km
The circumference of the moon to the nearest 10 kilometers is kilometers.
Answer:
The circumference of the moon to the nearest 10 kilometers is 10,900 kilometers,

Explanation:
Given the radius of circle as 1,736 kilometers, So circumference of the moon is 2πr = 2 X 3.14 X 1,736 kilometers = 10,902.08 kilometer nearest 10 is 10,900 kilometers.

b) Round your answer to the nearest 1,000 kilometers.
The circumference of the moon to the nearest 1,000 kilometers is kilometers.
Answer:
The circumference of the moon to the nearest 1,000 kilometers is 11,000 kilometers,

Explanation:
We got the circumference of the moon as 10,902.08 to the nearest 1,000 kilometers rounding we get 11,000 kilometers.

Question 2.
A greeting card is made up of three semicircles. O is the center of the large semicircle. Sarah wants to decorate the distance around the card with a ribbon. How much ribbon does Sarah need? Round your answer to the nearest inch.

Length of semicircular arc AB = \(\frac{1}{2}\) • 2πr
≈ • • •
= 1 • •
= in.
Semicircular arcs AO and OB have the same length.
Total length of semicircular arcs AO and OB
= 2 • \(\frac{1}{2}\) • πd
≈ • • •
= 1 • •
= in.
Distance around the card
= length of semicircular arc AB + total length of semicircular arcs AO and OB
= +
= in.
≈ in.
Sarah needs approximately inches of ribbon
Answer:
Sarah needs approximately 49 inches of ribbon,

Explanation:
A greeting card is made up of three semicircles. O is the center of the large semicircle. Sarah wants to decorate the distance around the card with a ribbon. Ribbon does Sarah need length of semicircular arc AB = \(\frac{1}{2}\) • 2πr = πr, radius is 5.2 in. so length of semicircular arc AB = 3.14 X 5.2 in. = 16.328 inches, Total length of semicircular arcs AO and OB = 2 X \(\frac{1}{2}\) X πd as diameter is 5.2 in. + 5.2 in. = 10.4 inches, So total length of semicircular arcs AO and OB = 3.14 X 10.4 inches = 32.656 inches. Now distance around the card is length of semicircular arc AB + total length of semicircular arcs AO and OB = 16.328 inches + 32.656 inches = 48.984 inches approximately 49 inches.

Question 3.
As part of her artwork, Sally bends a length of wire into the shape shown. The shape is made up of a semicircle and a quadrant. Find the length of the wire.

Distance around the shape
= length of semicircular arc PQ + length of arc RO + RP + OQ
= + + +
= cm
The length of the wire is approximately centimeters.
Answer:
The length of the wire is approximately 81 centimeters,

Explanation:
As part of her artwork, Sally bends a length of wire into the shape shown. The shape is made up of a semicircle and a quadrant. To find the length of the wire first we calculate length of semicircular arc \(\frac{1}{2}\) X 2πr = πr as r = 12 cm so it is 3.14 X 12 cm = 37.68 cm, Length of arc RO = \(\frac{1}{4}\) X 2πr = \(\frac{1}{2}\) X πr = \(\frac{1}{2}\) X 3.14 X 12 cm = 18.84 cm, Now Distance around the shape
= length of semicircular arc PQ + length of arc RO + RP + OQ = 37.68 cm + 18.84 cm + 12 cm + 12 cm = 80.52 cm approximately 81 centimeters.

Complete. Use \(\frac{22}{7}\) as an approximation for π.

Question 4.
Judy baked a pizza and had part of it for lunch. After the meal, the shape of the remaining pizza is made up of a semicircle and a quadrant. Find the area of the remaining pizza.

Area of remaining pizza = area of quadrant + area of semicircle
= +
= cm²
The area of the remaining pizza is approximately square centimeters.
Answer:
The area of the remaining pizza is approximately 2,885 square centimeters,

Explanation:
Given Judy baked a pizza and had part of it for lunch. After the meal, the shape of the remaining pizza is made up of a semicircle and a quadrant. So to find the area of the remaining pizza first we calculate area of quadrant \(\frac{1}{4}\) X πr²  as radius is 35 cm it is \(\frac{1}{4}\) X 3.14 X 35 cm X 35 cm = 961.625 square centimeters, Second we calculate area of semicircle = \(\frac{1}{2}\) X πr² = \(\frac{1}{2}\) X 3.14 X 35 cm X 35 cm = 1,923.25 square centimeters, So area of remaining pizza = area of quadrant + area of semicircle =
961.625 square centimeters + 1,923.25 square centimeters = 2,884.875 square centimeters approximately 2,885 square centimeters.

Question 5.
A rug is made up of a quadrant and two semicircles. Find the area of the rug.

Total area of two semicircles = 2 • \(\frac{1}{2}\) • πr²
≈ • • •
= 1 • • •
= in.²
Area of figure = area of quadrant + total area of two semicircles
= +
= in.²
The area of the rug is approximately square inches.
Answer:
The area of the rug is approximately 2,770 square inches,

Explanation:
A rug is made up of a quadrant and two semicircles. To calculate the area of the rug first we calculate area of quadrant \(\frac{1}{4}\) X πr²  as radius is 42 in. \(\frac{1}{4}\) X 3.14 X 42 in. X 42 in. = 1,384.74 square inches, radius of semicircle is half the diameter 42 in/2 = 21 in. now total area of two semicircles = 2 X \(\frac{1}{2}\) X πr² = 3.14 X 21 in. X 21 in. = 1,384.74 square inches therefore area of figure = area of quadrant + total area of two semicircles = 1,384.74 square inches + 1,384.74 square inches = 2,769.48 square inches
approximately 2,770 square inches.

Question 6.
The diameter of a bicycle wheel is 60 centimeters. How far does the wheel travel when it makes 35 revolutions? Give your answer in meters.
Circumference of wheel = πd
≈ •
= cm
Distance traveled = circumference of wheel • number of revolutions
= •
= cm
= m Divide by 100 to convert to meters.
The wheel travels approximately meters.
Answer:
The wheel travels approximately 66 meters,

Explanation:
Given the diameter of a bicycle wheel is 60 centimeters. Circumference of wheel = πd = 3.14 X 60 centimeters = 188.4 centimeters, Distance traveled = circumference of wheel X number of revolutions = 188.4 centimeters X 35 revolutions = 6,594 now we convert to meters by dividing 100 so we get 6,594/100 = 65.94 meters approximately
66 meters.

Question 7.
A park is shaped like the diagram below. It is a rectangle with semicircles at the two ends. There is a running track around the park.

a) The total length of the track is 220 yards. Find the length of \(\overline{P S}\).
The track is made up of semicircular arcs PQ and SR, and sides PS and QR. Semicircular arcs PQ and SR are equal.
Total length of semicircular arcs PQ and SR
= 2 • \(\frac{1}{2}\) • πd
≈ 1 • •
= yd
The length of \(\overline{P S}\) and \(\overline{Q R}\) are equal.
Total length of track = 220
PS + QR + total length of semicircular arcs PQ and SR = 220
PS + QR + = 220
PS + QR = 220
2 • PS =
PS =
= ys
The length of \(\overline{P S}\) is approximately yards.
Answer:
The length of \(\overline{P S}\) is approximately 45 yards.

Explanation:
The total length of the track is 220 yards. The length of \(\overline{P S}\), The track is made up of semicircular arcs PQ and SR, and sides PS and QR. Semicircular arcs PQ and SR are equal. Total length of semicircular arcs PQ and SR ith diameter 35 yd is 2 X \(\frac{1}{2}\) X πd = 3.14 X 35 yd = 109.9 yards,
The length of \(\overline{P S}\) and \(\overline{Q R}\) are equal.
Total length of track = 220 yards,
PS + QR + total length of semicircular arcs PQ and SR = 220 yards,
PS + QR + 109.9 yards = 220 yards,
PS + QR = 220 yards – 109.9 yards = 90.1 yards, (PS = QR),
2 PS = 90. 1 yards,
PS = 90.1 yards/2 = 45.05 yards approximately 45 yards.

b) A jogger runs once around the track in 125 seconds. What is his average speed in yards per second?
The jogger runs yards in 125 seconds.
125 seconds →
1 second →
= yd
The average speed of the jogger is yards per second.
Answer:
The average speed of the jogger is 1.76 yards per second,

Explanation:
Given a jogger runs once around the track in 125 seconds, The jogger runs 220 yards in 125 seconds.
125 seconds = 220 yards, 1 second = 220/125 yards = 1.76 yards, So the average speed of the jogger is 1.76 yards per second.

c) A gardener is hired to water the grass in the park. Using a machine, he waters 4 square yards per second. How many minutes will he take to water the entire park? Round your answer to the nearest minute.
Radius = diameter ÷ 2
= ÷ 2
= yd
The areas of the two semicircles are equal.
Total area of two semicircles = 2 • \(\frac{1}{2}\) • πr²
= 2 • \(\frac{1}{2}\) • •
= 1 • • •
= yd²
Area of rectangle PQRS = lw
= •
= yd²
Area of park = area of rectangle PQRS + total area of two semicircles
= +
= yd²
Time taken = area of park ÷ rate of watering park
= ÷
= s
= min
≈ min Round to the nearest minute.

The gardener will take approximately minutes to water the entire park.
Answer:
The gardener will take approximately 11 minutes to water the entire park,

Explanation:
Given A gardener is hired to water the grass in the park. Using a machine, he waters 4 square yards per second. So many minutes will he take to water the entire park, Radius = diameter ÷ 2 = 35 yd ÷ 2 = 17.5 yd, The areas of the two semicircles are equal. Total area of two semicircles = 2 X \(\frac{1}{2}\) X πr² = 3.14 X 17.5 yd X 17.5 yd = 961.625 yd²,
Area of rectangle PQRS = length X width,
= 45  yd X 35 yd,
= 1,575 yd²,
Area of park = area of rectangle PQRS + total area of two semicircles
= 1,575 yd² + 961.625 yd² = 2,536.625 yd²,
Time taken = area of park ÷ rate of watering park
= 2,536.625 yd² X seconds ÷ 4 yd² ,
=  634.15625 s
= 634.15625 / 60 minutes = 10.5692 min
≈ 11 min round to the nearest minute.

Math in Focus Course 1B Practice 11.3 Answer Key

Solve. Show your work.

Question 1.
The radius of a circular pond is 8 meters. Find its area and circumference. Use 3.14 as an approximation for π.
Answer:
Area = 200.96 square meters,
Circumference = 50.24 meters,

Explanation:
Given the radius of a circular pond is 8 meters so area is πr² = 3.14 X 8 mts X 8 mts = 200.96 square meters,
Circumference = 2πr = 2 X 3.14 X 8 mts = 50.24 meters.

Question 2.
The diameter of a metal disc is 26 centimeters. Find its area and circumference. Use 3.14 as an approximation for π.
Answer:
Area = 530.66 square centimeters,
Circumference = 81.64 centimeters,

Explanation:
Given the diameter of a metal disc is 26 centimeters so radius = diameter ÷ 2 = 26 cms ÷ 2 = 13 cms, so area is πr² = 3.14 X 13 cms X 13 cms = 530.66 square centimeters, Circumference = 2πr = 2 X 3.14 X 13 cms = 81.64 centimeters.

Question 3.
The shape of a carpet is a semicircle. Use \(\frac{22}{7}\) as an approximation for π.
a) Find its area.
Answer:
Area of semicircle carpet is 76.93 square feet,

Explanation:
Area of the carpet which is semi-circle with diameter 14 ft first radius = 14 ft/2 = 7 ft and area of carpet is \(\frac{1}{2}\) X πr² = \(\frac{1}{2}\) X 3.14 X 7 ft X 7 ft = 76.93 square feet.

b) Janice wants to put a fringed border on all sides of the carpet. How many feet of fringe are needed?

Answer:
21.98 feet of fringe is needed,

Explanation:
Given Janice wants to put a fringed border on all sides of the carpet. So feet of fringe are needed \(\frac{1}{2}\) X 2πr = πr and we have diameter as 14 ft radius = diameter/2 = 14 ft/2 = 7 ft now 3.14 X 7 ft = 21.98 feet.

Question 4.
The circumference of the rim of a wheel is 301.44 centimeters. Find the diameter of the rim. Use 3.14 as an approximation for π.
Answer:
The diameter of the rim is 96 centimeters,

Explanation:
Given the circumference of the rim of a wheel is 301.44 centimeters. So it is 301.44 cms = 2πr ,
r = 301.44 cms/2 X 3.14 = 310.44 cms/6.28 = 48 cms, Now diameter = 2r = 2 X 48 cms = 96 centimeters.

Question 5.
A Japanese fan is made out of wood and cloth. The shape of the fan is made up of two overlapping quadrants. What is the area of the portion that is made of cloth? Use \(\frac{22}{7}\) as an approximation for π.

Answer:
The area of the portion that is made of cloth is 1,846.32 square centimeters,

Explanation:
Given a Japanese fan is made out of wood and cloth. The shape of the fan is made up of two overlapping quadrants.
First we calculate area of cloth and wood with radius 28 cm as we know area πr² = 3.14 X 28 cm X 28 cm = 2,461.76 square cms, Area of wood with radius 14 cms is 3.14 X 14 cms X 14 cms = 615.44 square centimeters, Therefore area of the cloth is area of cloth and wood – area of the wood = 2,461.76 square centimeters – 615.44 square centimeters = 1,846.32 square centimeters.

Question 6.
A pancake restaurant serves small silver-dollar pancakes and regular-size pancakes. Use 3.14 as an approximation for π.

a) What is the area of a small silver dollar-pancake? Round your answer to the nearest tenth of an inch.
Answer:
The area of a small silver dollar-pancake to the nearest tenth of an inch is 10 square inches,

Explanation:
The area of small silver dollar-pan cake with diameter 3.5 in. as radius = 3.5 in./2 = 1.75 in. now area = πr² = 3.14 X 1.75 in. X 1.75 in. = 9.61625 square inches to the nearest tenth is 10 square inches.

b) What is the area of a regular-size pancake? Round your answer to the nearest tenth of a square inch.
Answer:
The area of a regular-size pancake to the  nearest tenth of a square inch is 28 square inches,

Explanation:
The area of a regular-sizepan cake with diameter 6 in. as radius = 6 in./2 = 3 in. now area = πr² = 3.14 X 3 in. X 3 in. = 28.26 square inches to the nearest tenth is 28 square inches.

c) If the total price of 6 small silver-dollar pancakes is the same as the total price of 3 regular-size pancakes, which is a better deal?
Answer:
3 regular-size pancakes is a better deal,

Explanation:
6 X area of small silver dollar-pan cake = 6 X 9.61625 square inches = 57.6975 sqaure inches and 3 X area of a regular-sizepan cake = 3 X 28.26 square inches = 84.78 sqaure inches if the total price of 6 small silver-dollar pancakes is the same as the total price of 3 regular-size pancakes then the better deal is 3 regular-size pancakes as it has bigger area.

Question 7.
A park is shaped like a rectangle with a semicircle on one end, and another semicircle cut out of one side.
a) Find the distance around the park.
Answer:
The distance around the park is 480 m,

Explanation:
The semicircle on one end is equal to another semicircle which is cut out of one side, so the distance around the park is the rectangle so it is 2 X (170 m + 70 m) = 2 X 240 m = 480 m.

b) Find the area of the park. Use \(\frac{22}{7}\) as an approximation for π.

Answer: 11,900 square meters

Explanation:
As the semicircle on one end is equal to another semicircle which is cut out of one side, so it the complete rectangle so the area of rectangle is length X breadth = 170 m X 70 m = 11,900 square meters.

Question 8.
The diameter of a circular fountain in a city park is 28 feet. A sidewalk that is 3.5 feet will be built around the fountain. Use \(\frac{22}{7}\) as an approximation for π.
a) Find the area of the sidewalk.
Answer:
The area of the sidewalk is 346.185 square feet,

Explanation:
Given the diameter of a circular fountain in a city park is 28 feet. A sidewalk that is 3.5 feet will be built around the fountain so first area of the circular fountain with diameter 28 ft, radius = 28 ft/2 = 14 ft so area of the cirular fountain is πr² = 3.14 X 14 ft X 14 ft = 615.44 square feet now the area of circular fountain with the side walk is
radius = 28 ft + 7 ft /2= 35 ft/2 = 17.5 ft so area is πr² = 3.14 X 17.5 ft X 17.5 ft =
If the sidewalk is 3.5 feet means it is radius so the area = πr² = 3.14 X 3.5 feet X 3.5 feet = 961.625 square feet, So only the side walk area = area of circular fountain with side walk – area of circular fountain = 961.625 square feet – 615.44 square feet = 346.185 square feet.

b) 0.8 bag of concrete will be needed for every square foot of the new sidewalk. What is the minimum number of bags needed?

Answer:
The minimum number of bags needed are 277 bags,

Explanation:
The area of the sidewalk is 346.185 square feet and 0.8 bag of concrete will be needed for every square foot of the new sidewalk. So the minimum number of bags needed are 346.185 X 0.8 = 276.948 bags approximately 277 bags.

Question 9.
The diagram shows an athletic field with a track around it, The track is 4 feet wide. The field is a rectangle with semicircles at the two ends. Find the area of the track. Use 3.14 as an approximation for π.

Answer:
The area of the track is 1,110.4 square feet,

Explanation:
The field is a rectangle with semicircles at the two ends. First we calculate the area with out the track so we have
rectangle area as length = 84 ft – 8 ft = 76 ft and width is 36 ft so area is 76 ft X 36 ft = 2,736 square feet,
Now area of two semicircles with diameter 36 ft so radius is 36 ft/2 = 18 ft, so area is 2 X \(\frac{1}{2}\) X πr² = 3.14 X 18 ft X 18 ft = 1,017.36 square feet, Field area is 2,736 sq ft + 1,017.36 sq ft = 3,753.36 square feet,
Now field area with track area of rectangle is length 76 ft and width is 36 ft + 8 ft = 44 ft is 76 ft X 44 ft = 3,344 square feet, area of two semicircles with diameter 44 ft so radius is 44 ft/2 = 22 ft, so area is 2 X \(\frac{1}{2}\) X πr² = 3.14 X 22 ft X 22 ft = 1,519.76 square feet, Field area with track is 3,344 square feet + 1,519.76 square feet = 4,863.76 square feet. Now the area of the track is field area with track – field area = 4,863.76 square feet – 3,753.36 square feet = 1,110.4 square feet.

Question 10.
The petal of a paper flower is created by cutting along the outlines of two overlapping quadrants within a square. Use 3.14 as an approximation for π.
a) Find the distance around the shaded part.
Answer:
The distance around the shaded part is 15.7 cm,

Explanation:
Distance around the shaded part is of the 2 quadrants = 2 X circumference ÷ 4 and C = πd or 2πr as given diameter = 10 cm, So distance around the shaded part  =  2 X πd ÷ 4 = 2 X 3.14 X 10 cm ÷ 4 = 62.8 cm ÷ 4 = 15.7 cm.

b) Find the area of the shaded part.

Answer:
39.25 square centimeters,

Explanation:
The area of the 2 quardants is 2 X \(\frac{1}{4}\) X πr² as diameter is 10 cm radius is 10 cm/2 = 5 cm,
Now area is 2 X \(\frac{1}{4}\) X 3.14 X 5 cm X 5 cm = 39.25 square centimeters.

Question 11.
Wheels A and B are placed side by side on a straight road. The diameter of wheel A is 56 inches. The diameter of wheel B is 35 inches. Suppose each wheel makes 15 revolutions. Find the distance between the wheels after they have made these 15 revolutions.
Use \(\frac{22}{7}\) as an approximation for π.

Answer:
The distance between the wheels after they have made these 15 revolutions is 494.55 inches,

Explanation:
Given wheels A and B are placed side by side on a straight road. The diameter of wheel A is 56 inches. The diameter of wheel B is 35 inches. Radius of wheel A is diameter of wheel A/2 = 56 inches/2 = 28 inches, Radius of wheel B is diameter of wheel B/2 = 35 inches/2 = 17.5 inches, Circumference of wheel A = πr = 3.14 X 28 inches = 87.92 inches,
Circumference of wheel B = πr = 3.14 X 17.5 inches = 54.95 inches, If Wheel A makes 15 revolutions the distance is
87.92 inches X 15 = 1,318.8 inches and If wheel B makes 15 revolutions the distance is 54.95 inches X 15 = 824.25 inches so the distance between the wheels after they have made these 15 revolutions is distance of 15 revolutions of wheel A – distance of 15 revolutions of wheel B = 1,318.8 inches – 824.25 inches = 494.55 inches.

Question 12.
Nine identical circles are cut from a square sheet of paper whose sides are 36 centimeters long. If the circles are as large as possible, what is the area of the paper that is left after all the circles are cut out? Use 3.14 as an approximation for π.
Answer:
The area of the paper that is left after all the circles are cut out is 278.64 square centimeters,

Explanation:
Given nine identical circles are cut from a square sheet of paper whose sides are 36 centimeters long. If the circles are as large as possible, So the area of the paper that is left after all the circles are cut out is area of the square – area of 9 circles, The area of the square is 36 cms X 36 cms = 1,296 square cms, diameter = 38 cms/3 approximately 12 cms, radius is 12 cms / 2 cms = 6 cms, Area of 9 circles is 9 X πr² = 9 X 3.14 X 6 cms X 6 cms = 1,017.36 square centimeters, therefore the area of the paper that is left after all the circles are cut out = 1,296 sq cms – 1,017.36 sq cms = 278.64 square centimeters.

Question 13.
A designer drew an icon as shown below. O is the center of the circle, and AB is a diameter. Two semicircles are drawn in the circle. If \(\overline{A B}\) is 28 millimeters, find the area of the shaded part. Use \(\frac{22}{7}\) as an approximation for π.

Answer:
The area of the shaded part is 307.72 square millimeters,

Explanation:
Given a designer drew an icon as shown below. O is the center of the circle, and AB is a diameter. Two semicircles are drawn in the circle. If \(\overline{A B}\) is 28 millimeters, if we see the design the shaded region is half of the circle so the area of the shaded part as diameter is 28 mm its radius is 28 mm/2 = 14 mm, Now area of shaded region = \(\frac{1}{2}\) X πr² = \(\frac{1}{2}\) X 3.14 X 14 mm X 14 mm = 307.72 square millimeters.

Use graph paper. Solve.

Question 14.
Mary wants to draw the plan of a circular park on graph paper. The coordinates of the center of the park are A (3, 4). The circle has a radius of 3 units.
a) Use a compass and draw the plan of the circular park on graph paper.
Answer:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Explanation:
Used a compass and drawn the plan of the circular park on graph paper as shown above.

b) Assume that the y-axis points north and south. A barbecue pit is located at the northernmost part of the park. Plot and label the location of the barbecue pit as point B. Give the coordinates of point B.
Answer:

c) Connect points A, B, and the origin to form a triangle. Find the area of the triangle.
Answer:

Question 15.
A wire is bent to make the shape below. The shape is made up of four identical circles. Each circle intersects two other circles. The four circles meet at a common point T, which is the center of square PQRS. Use \(\frac{22}{7}\) as an approximation for π.
a) Find the length of the wire.
Answer:
The length of the wire is 112 cm,

Explanation:
Given to find the length of the wire it is perimeter of the square and length of square is 28 cm therfore length of the wire is 4 X 28 cm = 112 cms.

b) Find the area of the whole shape.

Answer:
The area of the whole space is 2,014.88 square centimeters,

Explanation:
Given a wire is bent to make the shape below. The shape is made up of four identical circles. Each circle intersects two other circles. The four circles meet at a common point T, which is the center of square PQRS. So the area of the whole shape is (4 X \(\frac{1}{2}\) the area of circle) + area of the square, radius = diameter/2 = 28 cm/2 = 14 cm, So [4 X (\(\frac{1}{2}\) X 3.14 X 14 cm X 14 cm)] + (28 cm X 28 cm) = (2 X 615.44 square centimeter) +  784 square centimeters = 1,230.88n square centimeters + 784 square centimeters = 2,014.88 square centimeters.

Brain @ Work

Question 1.
The figure shows two identical overlapping quadrants. Find the distance around the shaded part. Use 3.14 as an approximation for π. Round your answer to the nearest tenth of a centimeter.

Answer:
The distance around the shaded part is 47.1 cm,

Explanation:
Distance around the shaded part is of the 2 quadrants = 2 X circumference ÷ 4 and C = πd or 2πr as given radius = 15 cm, So distance around the shaded part  =  2 X 2 X 3.14 X 15  cms ÷ 4 = 3.14 X 15 cm = 47.1 cm.

Question 2.
A cushion cover design is created from a circle of radius 7 inches, and 4 quadrants. Find the total area of the shaded parts of the design. Use \(\frac{22}{7}\) as an approximation for π.

Answer:
The total area of the shaded parts of the design is 42.14 square inches,

Explanation:
Given a cushion cover design is created from a circle of radius 7 inches, and 4 quadrants. The total area of the shaded parts of the design is area of square – 4 X area of the quadrant = 14 in X 14 in – 4 X \(\frac{1}{4}\) X 3.14 X 7 in X 7 in = 196 square inches – 153.86 square inches = 42.14 square inches.

Question 3.
Two identical wheels are placed along a straight path so that their centers are 9.31 meters apart. The radius of each wheel is 3.5 centimeters. They are pushed towards each other at the same time, each making one revolution per second. How long does it take for them to knock into each other? Use \(\frac{22}{7}\) as an approximation for π.

Answer:
Long does it will take for them to knock into each other is 4.0926 meters,

Explanation:
Given two identical wheels are placed along a straight path so that their centers are 9.31 meters apart. The radius of each wheel is 3.5 centimeters. Circumfrence of the one wheel is 2 X 3.14 X 3.5 cms = 21.98 cms converting into meters 21.98 X 0.01 m = 0.2198 meters so 2 wheels means 0.2198 m X 2 =  0.4396 m their centers are apart of 9.31 meters so the distance for the 2 wheels to knock into each other is 9.31 meters X 0.4396 = 4.0926 meters.

Question 4.
A stage prop is made up of a semicircle and a quadrant. Its area is 924 square inches, Find the value of x. Use \(\frac{22}{7}\) as an approximation for π.

Answer:
The value of x is 9.903 inches,

Explanation:
Given a stage prop is made up of a semicircle and a quadrant. Its area is 924 square inches with radius 2x in. So 924 sqaure inches = area of the quadrant + area of the semicircle,
924 sq in = \(\frac{1}{4}\) X 3.14 X 2x in X 2x in + \(\frac{1}{2}\) X 3.14 X 2x in X 2x in,
924 sq in = 3.14x² sq in + 6.28x² sq in,
924 sq in =9.42x² sq in,
x² = 924 sq in/9.42 sq in = 98.089 sq in, therefore x = square root of 98.089 sq in = 9.903 inches.

This handy Math in Focus Grade 6 Workbook Answer Key Chapter 11 Lesson 11.2 Area of a Circle detailed solutions for the textbook questions.

Math in Focus Grade 6 Course 1 B Chapter 11 Lesson 11.2 Answer Key Area of a Circle

Math in Focus Grade 6 Chapter 11 Lesson 11.2 Guided Practice Answer Key

Complete. Use 3.14 as an approximation for π.

Question 1.
Find the area of a circle that has a radius of 18 centimeters.
Area = πr²
≈ •
= •
= cm²
The area of the circle is approximately square centimeters
Answer:
The area of the circle is approximately 1,017 square centimeters,

Explanation:
Given radius as 18 centimeters as the area of the circle is πr², So area = 3.14 X 18 cms X 18 cms = 1,017.36 cm² approximately 1,017 square centimeters.

Question 2.
Find the area of a circle that has a radius of 15 inches.
Area = πr²
≈ •
= •
= in.²
The area of the circle is approximately square inches.
Answer:
The area of the circle is approximately 707 square inches,

Explanation:
Given radius as 15 inches as the area of the circle is πr², So area = 3.14 X 15 in X 15 in =  706.5 in² approximately 707 square inches.

Question 3.
Find the area of a circle that has a diameter of 26 centimeters.
Radius = diameter ÷ 2
= ÷
= cm
Area = πr²
≈ • •
= •
= cm²
The area of the circle is approximately square centimeters.
Answer:
The area of the circle is approximately 531 square centimeters,

Explanation:
Given diameter as 26 cms so radius = diameter ÷ 2 = 26 cms ÷ 2 = 13 cms as the area of the circle is πr², So area = 3.14 X 13 cms X 13 cms = 530.66 cm² approximately 531 square centimeters.

Complete. Use \(\frac{22}{7}\) as an approximation for π.

Question 4.
Find the area of a quadrant.

The area of a quadrant is approximately square feet.
Answer:
The area of the quadrant is approximately 154 square feet,

Explanation:
Given radius is 14 ft as area of quardant is \(\frac{1}{4}\) X area of circle = \(\frac{1}{4}\) X πr² = \(\frac{1}{4}\) X 3.14 X 14 feet X 14 feet = 153.86 square feet approximately 154 square feet.

Question 5.
The diameter of a circle is 42 inches. Find the area of a quadrant.

Radius = diameter ÷ 2
= ÷ 2
= in.
Area of quadrant = • area of circle
= • πr²
≈ • •
= • • •
= in.²
The area of a quadrant is approximately square inches.
Answer:
The area of the quadrant is approximately 346 square inches,

Explanation:
Given diameter as 42 inches so radius = 42 inches/2 = 21 inches as area of quardant is \(\frac{1}{4}\) X area of circle = \(\frac{1}{4}\) X πr² = \(\frac{1}{4}\) X 3.14 X 21 inches X 21 inches = 346.185 square inches approximately 346 square inches.

Math in Focus Course 1B Practice 11.2 Answer Key

Find the area of each circle. Use 3.14 as an approximation for π.

Question 1.

Answer:
The area of the circle is 314 square centimeters,
Explanation:

Explanation:
Given radius as 10 cm as the area of the circle is πr², So area = 3.14 X 10 cm X 10 cm = 314 cm² .

Question 2.

Answer:
The area of the circle is approximately 1,963 square inches,

Explanation:
Given diameter as 50 inches so radius = diameter ÷ 2 = 50 in ÷ 2 = 25 in as the area of the circle is πr², So area = 3.14 X 25 in X 25 in =1962.5 inches² approximately 1,963 square inches.

Find the area of each semicircle. Use \(\frac{22}{7}\) as an approximation for π.

Question 3.

Answer:
The area of semicircle is 307.72 square feet,

Explanation:
Given diameter as 28 ft  so radius = diameter ÷ 2 = 28 ft ÷ 2 = 14 ft as the area of the circle is πr²,  So area = 3.14 X 14 ft X 14 ft = 615.44 square feet, The area of semicircle is half the area of the circle so it is \(\frac{1}{2}\) X
615.44 square feet = 307.72 square feet.

Question 4.

Answer:
The area of semicircle is 76.93 square meter,

Explanation:
Given radius as 7 m the area of the circle is πr²,  So area = 3.14 X 7 m X 7 m = 153.86 square meter, The area of semicircle is half the area of the circle so it is \(\frac{1}{2}\) X
153.86 square meter = 76.93 square meter.

Find the area of each quadrant to the nearest tenth. Use 3.14 as an approximation for π.

Question 5.

Answer:
The area of quardant is 113.04 square inches,

Explanation:
Given radius as 12 inches the area of the circle is πr²,  So area = 3.14 X 12 in. X 12 in. = 452.16 square inches, The area of quardant is 1/4 of the area of the circle so it is \(\frac{1}{4}\) X 452.16 square inches = 113.04 square inches.

Question 6.

Answer:
The area of quardant is 283.385 square meters,

Explanation:
Given radius as 19 meters the area of the circle is πr²,  So area = 3.14 X 19 m X 19 m = 1,133.54 square meters, The area of quardant is 1/4 of the area of the circle so it is \(\frac{1}{4}\) X 1,133.54 square meters = 283.385 square meters.

Solve. Show your work.

Question 7.
A circular pendant has a diameter of 7 centimeters. Find its area. Use \(\frac{22}{7}\) as an approximation for π.
Answer:
Area = 38.465 square centimeters,

Explanation:
Given a circular pendant has a diameter of 7 centimeters. So radius = diameter ÷ 2 = 7 cms ÷ 2 = 3.5 cms, As area of circle = πr², So area = 3.14 X 3.5 cms X 3.5 cms = 38.465 square centimeters.

Question 8.
The shape of the stage of a lecture theater is a semicircle. Find the area of the stage. Use 3.14 as an approximation for π.

Answer:
The area of the stage is 226.08 square meter,

Explanation:
Given shape of the stage of a lecture theater is a semicircle with radius 12 m, First area of circle is πr², So area = 3.14 X 12 m X 12 m = 452.16 square meter, Now area of semicircle is \(\frac{1}{2}\) X 452.16 sq mt = 226.08 square meter.

Question 9.
The shape of a balcony floor is a quadrant. Find the area of the balcony floor. Use 3.14 as an approximation for π.

Answer:
The area of the balcony floor is 113.04 square feet,

Explanation:
Given shape of a balcony floor is a quadrant with radius 12 ft, First area of circle is πr², So area = 3.14 X 12 ft X 12 ft = 452.16 square feet, Now area of quadrant is \(\frac{1}{4}\) X 452.16 square feet = 113.04 square feet.

Question 10.
The cost of an 8-inch pizza is $4. The cost of a 16-inch pizza is $13. Use 3.14 as an approximation for π.
a) How much greater is the area of the 16-inch pizza than the area of the 8-inch pizza?
Answer:
Greater is the area of the 16-inch pizza than the area of the 8-inch pizza is 150.72 square inches,

Explanation:
Area of 8 inch pizza radius is 8 inch ÷ 2 = 4 inch, Area of 8 inch pizza = 3.14 X 4 inch X 4 inch = 50.24 square inch,
Area of 16 inch pizza is 16 inch ÷ 2 = 8 inch, Area of 16 inch pizza = 3.14 X 8 inch X 8 inch = 200.96 square inch,
Now the area of the 16-inch pizza than the area of the 8-inch pizza is 200.96 square inch – 50.24 square inch = 150.72 square inches.

b) Which is a better deal? Explain your reasoning.
Answer:
8 inch pizza,

Explanation:
Cost of 16 inch pizza is $13, So cost of 1 inch pizza is $13/16 = $0.8125,
Cost of 8 inch pizza is $4, So cost of 1 inch pizza is $4/8 = $0.5 therefore it is better to buy 8 inch pizza as in 1 inch it cost $0.5 than 16 inch pizza in which 1 inch is $0.8125.

Question 11.
Four identical drinking glasses each have a radius of 5 centimeters. The glasses are arranged so that they touch each other as shown in the figure below. Find the area of the green portion. Use 3.14 as an approximation for π.

Answer:
The area of the green portion is 21.5 square centimeters,

Explanation:
Given to find the area of green portion so it is the area of square minus area of 4 quadrants so ((the area of square – 4(\(\frac{1}{4}\) X πr²)) = 10 cms x 10 cms – 4(\(\frac{1}{4}\) X 3.14 X 5 cms X 5 cms) = 100 square cms – 3.14 X 25 square cms = 100 square cms – 78.5 square cms =  21.5 square centimeters.

Question 12.
The figure is made up of trapezoid ABCD and a semicircle. The height of trapezoid ABCD is \(\frac{5}{6}\) the length of \(\overline{B C}\). Find the area of the figure. Use \(\frac{22}{7}\) as an approximation for π.

Answer:
Total area of trapezoid ABCD and a semicircle is 18,235.7925 square meters,

Explanation:
The area of trapezoid is 1/2 X (base1 + base2) X height, where base1 and base2 are lengths of parallel sides, we have height as \(\frac{5}{6}\) X 124.8 m = 104 m, Now area of trapezoid is 1/2 X (124.8 m + 119 m) X 104 m = 1/2 X 243.8 m X 104 m =12,677.6 square meters. First we calculate area of circle as diameter is 119 m so radius = 119 m/2 = 59.5 m, Area = 3.14 X 59.5 m x 59.5 m = 11,116.385 square meter, Now area of semicircle is half the area of circle = 11,116.385 square mt/2 = 5,558.1925 square meter, So total area of trapezoid ABCD and a semicircle is
area trapezoid + area of semicircle = 12,677.6 square meters + 5,558.1925 square meters = 18,235.7925 square meters.