Math in Focus

Go through the Math in Focus Grade K Workbook Answer Key Chapter 4 Counting and Numbers 0 to 10 to finish your assignments.

Math in Focus Kindergarten Chapter 4 Answer Key Counting and Numbers 0 to 10

Lesson 1 Composing and Decomposing 5

Count, write, and circle.

Question 1.

Answer:

Explanation:
There are 2 red squares and 2 green squares, 4 in all. So, i circled 4.

Question 2.

Answer:

Explanation:
There are 3 red squares and 2 green squares, 4 in all. So, i circled 5.

Color, count and write.

Question 1.

4 is __________ and ___________.
Answer:

Explanation:
I colored and wrote 2 blue squares and 2 yellow squares.

Question 2.

5 is __________ and ___________.
Answer:

Explanation:
I colored and wrote 2 blue squares and 3 yellow squares.

Question 3.

5 is __________ and ___________.
Answer:

Explanation:
I colored and wrote 2 blue squares and 2 yellow squares.

Lesson 2 Counting and Ordering Up to 10

Are there enough? Color.

Answer:

Explanation:
There are 4 bananas, 4 strawberries and 4 pears.
The other fruits are not enough as they are less than four each.

Draw enough . Count and write.

Answer:

Explanation:
I drew 5 apples and there are 5 ants and 5 apples.

Count and write.

Question 1.

Answer:

Explanation:
I counted and wrote the number 6 as there are 6 ants.

Question 2.

Answer:

Explanation:
I counted and wrote the number 5 as there are 5 ants.

Question 3.

Answer:

Explanation:
I counted and wrote the number 9 as there are 9 ants.

Question 4.

Answer:

Explanation:
I counted and wrote the number 5 as there are 5 ants.

Draw one more. Count and write.

Question 1.

Answer:

Explanation:
I drew 1 more apple and i counted and wrote the number 2 as there are 2 apples.

Question 2.

Answer:

Explanation:
I drew 1 more apple and i counted and wrote the number 3 as there are 3 apples.

Question 3.

Answer:

Explanation:
I drew 1 more apple and i counted and wrote the number 4 as there are 4 apples.

Question 4.

Answer:

Explanation:
I drew 1 more apple and i counted and wrote the number 5 as there are 5 apples.

Count and write.

Question 1.

Answer:

Explanation:
I counted and wrote the number 3 as there are 3 turnips.

Question 2.

Answer:

Explanation:
I counted and wrote the number 2 as there are 2 capsicums.

Question 3.

Answer:

Explanation:
I counted and wrote the number 5 as there are 5 green peas.

Question 4.

Answer:

Explanation:
I counted and wrote the number 6 as there are 6 chillies.

Draw, count and write.

Question 1.

Answer:

Explanation:
I drew 1 more carrot and i counted and wrote the number 2.

Question 2.

Answer:

Explanation:
I drew 3 more tomatoes and i counted and wrote the number 4.

Question 3.

Answer:

Explanation:
I drew 3 more potatoes and i counted and wrote the number 4.

Lesson 3 Using Your Fingers and Toes to Count On

Count and write.

Question 1.

Answer:

Explanation:
I counted and wrote 0 and 1 is 1.

Question 2.

Answer:

Explanation:
I counted and wrote 1 and 1 is 2.

Question 3.

Answer:

Explanation:
I counted and wrote 2 and 1 is 3.

Question 4.

Answer:

Explanation:
I counted and wrote 3 an d1 is 4.

Question 5.

Answer:

Explanation:
I counted and wrote 4 and 1 is 5.

Lesson 4 Same Number and More

Look and talk.

Answer:

How many more? Count and write.

Answer:

Explanation:
I counted and wrote the numbers how many more are needed for 5 people.

Count and write.

Question 1.

__________ more flowers are needed.
Answer:
2 more flowers are needed.

Question 2.

__________ more flowers are needed.
Answer:
3 more flowers are needed.

Question 3.

__________ more flowers are needed.
Answer:
4 more flowers are needed.

Lesson 5 Fewer Than

Circle.

Which group has fewer than 3?

Answer:

Explanation:
I circled the group that has fewer than 3.

Which group has fewer than 5?

Answer:

Explanation:
I circled the group that has fewer than 5.

Which group has fewer than 7?

Answer:

Explanation:
I circled the group that has fewer than 7.

Which group has fewer than 9?

Answer:

Explanation:
I circled the group that has fewer than 9.

Lesson 6 How Many in All?

Draw one more. How many are there in all?

Question 1.

Answer:

Explanation:
I drew 1 more orange and i counted and wrote the number 5.

Question 2.

Answer:

Explanation:
I drew 1 more ball and i counted and wrote the number 3.

Question 3.

Answer:

Explanation:
I drew 1 more balloon and i counted and wrote the number 4.

Circle, count and write.

Question 1.
Vicki eats 2 grapes. Circle the grapes that are left behind.

___________ grapes are left behind.
Answer:

Explanation:
4 grapes are left behind. So, i circled 4 grapes.

Question 2.
2 birds fly away. Circle the birds that stay behind.

___________ birds stay behind.
Answer:

Explanation:
8 birds stay behind. So, i circled 8 birds.

Question 3.
4 horses trot away. Circle the horses that stay behind.

____________ horses stay behind.
Answer:

Explanation:
3 horses stay behind. So, i circled 3 horses.

Practice the problems of Math in Focus Grade 6 Workbook Answer Key Chapter 7 Review Test to score better marks in the exam.

Math in Focus Grade 6 Course 1 A Chapter 7 Review Test Answer Key

Concepts and Skills

Write an algebraic expression for each of the following.

Question 1.
A number that is 5 more than twice x
Answer:
2x + 5
Explanation:
Let’s take a number as variable x.
A number that is 5 more than twice x
2x+ 5.
A number that is 5 more than twice x
is written as 2x+5

Question 2.
The total cost, in dollars, of 4 pencils and 5 pens if each pencil costs w cents and each pen costs 2w cents
Answer:
4w + 10w = 14w
Explanation:
The total cost, in dollars, of 4 pencils and 5 pens
given, each pencil costs w cents and each pen costs 2w cents
pencils = w cents
4 pencils = 4w cents
pen = 2w cents
5 pens = 10w cents

Question 3.
The length of a side of a square whose perimeter is r units
Answer:
\(\frac{r}{4}\)
Explanation:
perimeter of a square = side x 4
side = perimeter / 4
perimeter is given as r
side = r/4

Question 4.
The perimeter of a rectangle whose sides are of lengths (3z + 2) units and (2z + 3) units
Answer:
(10z + 10) units
Explanation:
The perimeter of a rectangle whose sides are of lengths (3z + 2) units and (2z + 3) units
perimeter of a rectangle = 2 [(3z + 2) + (2z + 3)] units
=2 [ 5z + 5]
= 10z + 10

Evaluate each expression for the given value of the variable.

Question 5.
3(x + 4) – \(\frac{x}{2}\) when x = 2
Answer:
17
Explanation:
3(x + 4) – \(\frac{x}{2}\) when x = 2
= 3(2 + 4) – \(\frac{2}{2}\)
= [(3 x 6) – 1]
= 18 – 1 = 17

Question 6.
\(\frac{5 p+9}{2}\) + \(\frac{2 p+5}{3}\) when p = 5
Answer:
22
Explanation:
\(\frac{5 p+9}{2}\) + \(\frac{2 p+5}{3}\) when p = 5
= \(\frac{5 x 5 + 9}{2}\) + \(\frac{2 x 5 + 5}{3}\)
= \(\frac{25+9}{2}\) + \(\frac{10+5}{3}\)
= \(\frac{34}{2}\) + \(\frac{15}{3}\)
= \(\frac{(3×34)+(2×15)}{6}\)
= \(\frac{102+30}{6}\)
= \(\frac{132}{6}\)
= 22

Simplify each expression.

Question 7.
24k + 11 – 5k – 4
Answer:
19k + 7
Explanation:
24k + 11 – 5k – 4
= 19k + 7

Question 8.
10 + 13h – 6 – 4h + 9 + 12h
Answer:
13 + 21h
Explanation:
10 + 13h – 6 – 4h + 9 + 12h
= 10 – 6 + 9 + 13h – 4h + 12h
= 4 + 9 + 9h + 12h
= 13 + 21h

Expand each expression.

Question 9.
5(m + 3) + 2(m + 8)
Answer:
7m + 31
Explanation:
5(m + 3) + 2(m + 8)
= (5m + 15) + (2m + 16)
= 7m + 31

Question 10.
9(x + 2) + 4(5 + x)
Answer:
13x + 38
Explanation:
9(x + 2) + 4(5 + x)
= 9x + 18 + 20 + 4x
= 13x + 38

Factor each expression.

Question 11.
5a – 25
Answer:
5(a – 5)
Explanation:
5a – 25
Take 5 as a common factor
5(a – 5)

Question 12.
28 – 7x
Answer:
7(4 – x)
Explanation:
28 – 7x
Take 7 as a common factor
7(4 – x)

Question 13.
12z + 28 – 7z – 3
Answer:
5(z + 5)
Explanation:
12z + 28 – 7z – 3
= 5z – 25
= 5(z + 5)

State whether each pair of expressions are equivalent.

Question 14.
3(x + 5) and 5(x + 3)
Answer:
No, pair of expression is not equivalent.
Explanation:
3(x + 5) and 5(x + 3)
3x + 15 = 5x + 15

Question 15.
6y – 26 and 2(3y – 13)
Answer:
Yes, pair of expression equivalent.
Explanation:
6y – 26 and 2(3y – 13)
6y – 26 = 6y – 26

Question 16.
18 – 12p and 3(5 + 6p) + 3(2p + 1)
Answer:
No, pair of expression is not equivalent.
Explanation:
18 – 12p and 3(5 + 6p) + 3(2p + 1)
18 – 12p = 15 + 18p + 6p + 3
18 – 12p = 18 + 24p

Question 17.
15 – 5q and 5(q – 3)
Answer:
No, pair of expression is not equivalent.
Explanation:
15 – 5q and 5(q – 3)
15 – 5q = 5q – 15

Problem Solving

Solve. Show your work.

Question 18.
Juan is g years old and Eva is 2 years younger than Juan.
a) Find the sum of their ages in terms of g.
Answer:
(2g – 2) years
Explanation:
Juan is g years old
Eva is 2 years younger than Juan = (g – 2)
the sum of their ages in terms of g,
g + (g – 2)
So, (2g – 2) years

b) Find the sum of their ages in g years’ time, in terms of g.
Answer:
(4g – 2) years
Explanation:
Juan is g years old
Eva is 2 years younger than Juan = (g – 2)
the sum of their ages in g years’ time, in terms of g.

Question 19.
Parker bought 4 times as many marbles as Molly. Cole bought 5 fewer marbles than Parker. If Molly bought p marbles, how many marbles did Cole buy?
Answer:
(4p – 5) marbles.
Explanation:
Number of marbles Molly bought = p
Parker bought 4 times as many marbles as Molly = 4p
Cole bought 5 fewer marbles than Parker = (4p – 5)
Number of marbles bought by Cole = (4p  – 50

Question 20.
Andrea baked h muffins. Violet baked 8 fewer muffins than Andrea. Find the average number of muffins baked by both girls, in terms of h.
Answer:
(h – 4) muffins
Explanation:
Andrea baked h muffins.
Violet baked 8 fewer muffins than Andrea = (h – 8)
Average number of muffins baked by both girls, in terms of h
h + (h – 8) ÷ 2
= (2h – 8) ÷  2
= (h – 4)

Question 21.
A square garden has a side length that is 3 meters shorter than the length of a rectangular garden. Find the perimeter of the rectangular garden in terms of y.

Answer:
(8y + 8) meters
Explanation:
A square garden has a side length that is 3 meters shorter than the length of a rectangular garden.
side of square = perimeter/4
= 8y/4 = 2y
3 meters shorter than the length of a rectangular garden
= 2y + 3
Perimeter of a rectangle garden = 2 [ length + width]
Perimeter of a rectangle garden = 2 [ 2y + 3  + 2y + 1]
=2[4y+4]
=8y + 8

Question 22.
Mrs. Roberts sewed m shirts using 2 yards of cloth for each shirt. She also sewed (m + 2) dresses, using 5 yards of cloth for each dress.
a) How much cloth did she use altogether? Give your answer in terms of m.
Answer:
(7m + 10) yards
Explanation:
Mrs. Roberts sewed m shirts using 2 yards of cloth for each shirt.
She also sewed (m + 2) dresses, using 5 yards of cloth for each dress.
2m + 5 x (m+2) =
= 2m + 5m + 10
= 7m + 10 yards cloth used
b) If m = 7, find how much more cloth she used to sew the dresses than the shirts.
Answer:
31 yards
Explanation:
m = 7
5(m+2) – 2m
=5(7 + 2 ) – 2 x 7
= 5 x 9 – 14
= 45 – 14
= 31

Question on 23.
A glass jug can hold (6p + 8) quarts more water than a plastic container. 2 glass jugs and 2 plastic containers contain 56p quarts of water in all.
a) How much water can the plastic container hold? Give your answer in terms of p.
Answer:
(11p – 4) quarts
Explanation:
Let X = quarts the plastic container can hold
Then the jug can hold (6p+8) + X
2 jugs and 2 plastic containers = 56p quarts  ==>
2(6p + 8 + X) + 2X = 56p
12p + 16 + 2X + 2X = 56p
4X = 44p – 16
X = 11p – 4
The plastic container can hold 11p – 4 quarts of water

b) If p = 3, find how much water can 1 glass jug and 1 plastic container hold in all.
Answer:
84 quarts
Explanation:
A glass jug can hold (6p + 8) quarts

Question 24.
Mr. Lee can paint 20 chairs in t hours. He uses 3 liters of paint for every 12 chairs that he painted.
a) Find, in terms of t, the number of chairs that he can paint in 3 hours.
Answer:
\(\frac{60}{t}\) chairs
Explanation:
Mr. Lee can paint 20 chairs in t hours
let X number of chairs that he can paint in 3 hours.
20 => t
X => 3
3 x 20 = X x t
Xt = 60
X = \(\frac{60}{t}\) chairs

b) Find, in terms of t, the time taken by Mr. Lee to paint 7 chairs.
Answer:
\(\frac{7}{20}\) t hours
Explanation:
20 => t
7 => X
X x 20 = 7 x t
X = \(\frac{7}{20}\) t

c) If t = 4, find the amount of paint Mr. Lee has used after painting for 4 hours.
Answer:
5 liters
Explanation:
He uses 3 liters of paint for every 12 chairs that he painted.
Mr. Lee can paint 20 chairs in t hours
20 => t
if t  = 4
Mr. Lee can paint 20 chairs in 4 hours
He uses 3 liters of paint for every 12 chairs that he painted.
x ltrs for 20 chairs
3 => 12
x => 20
3 x 20 = x x 12
x = 60 / 12
x = 5 ltrs

Practice the problems of Math in Focus Grade 6 Workbook Answer Key Chapter 7 Lesson 7.3 Simplifying Algebraic Expressions to score better marks in the exam.

Math in Focus Grade 6 Course 1 A Chapter 7 Lesson 7.3 Answer Key Simplifying Algebraic Expressions

Math in Focus Grade 6 Chapter 7 Lesson 7.3 Guided Practice Answer Key

Simplify each expression. Then state the coefficient of the variable In the expression.

Question 1.
x + x + x + x + x
Answer:
5x
Explanation:
x is the variable or constant,
add all the variables, we get 5x.
So, 5 is the coefficient of the variable x.

Question 2.
y + y + 6
Answer:
2y + 6
Explanation:
y is the variable, add all the variables
y + y + 6
2y + 6
So, 25 is the coefficient of the variable y

Question 3.
m + m + m + 5 + 4
Answer:
3m + 9
Explanation:
3 is the coefficient of the variable m,
add of the variables.
m + m + m + 5 + 4
3m+9

Question 4.
n + n + n + n + n + n + 12 – 8
Answer:
6n + 4
Explanation:
n is the variable, add all the variables at first.
n + n + n + n + n + n + 12 – 8
6n + 4
6 is the coefficient of the variable 2

Solve.

Question 5.
A square has sides of length x centimeters. Find the perimeter of the square in terms of x.

The perimeter of the square is centimeters.
Answer:
4x
Explanation:
x + x + x + x = 4x
The perimeter of the square is 4x centimeters.

Question 6.
The figure shows a trapezoid. The length of each side is given as shown. Find the perimeter of the trapezoid in terms of w.

10cm
The perimeter of the trapezoid is centimeters.
Answer:
3w + 10 cm
Explanation:
The perimeter of the trapezoid is 3w + 10 centimeters.

Hands-On Activity

Materials:

• paper
• ruler
• scissors

RECOGNIZE SIMPLIFIED EXPRESSIONS ARE EQ UIVALENT)

Work in pairs.

STEP 1: Make the following set of paper strips.
Let the length of the shortest strip be m units. Make and label 5 such strips.

Make and label 4 more strips of lengths 2m units, 3m units, 4m units, and 5m units.

STEP 2: Take one of the longer strips and place it horizontally.

STEP 3: Ask your partner to use the pieces of the shortest strips to match the length of the chosen strip in STEP 2.
Example

STEP 4: Write an algebraic expression to describe the number of short strips used, and simplify it. For example in STEP 3, write m + m + m = 3m.

STEP 5: Repeat the activity with other lengths of strips.
Activity 1:

m + m + m + m = 4m
Activity 2:

m + m + m + m + m = 5m

Math Journal How do the lengths of the strips show that the expressions are equivalent?
Answer:
We use the fraction strips to show differences between different fractions.
Explanation:
For example;
We start with halves and quarters.
Ask each student to pull out a 1/2 foot strip.
Now ask them to clear their desks of everything but the 1/4 pieces and the one 1/2 strip.
Then ask them to put down enough quarter pieces to make the same length as the half piece.
We can show the lengths of the strips are equivalent with expressions.

Complete.

Question 7.
Simplify x + 8x.
x + 8x =

Answer:
9x
Explanation:
We use the fraction strips to show differences between different fractions.

Simplify each expression.

Question 8.
3r + 2r
Answer: 5r
Explanation:
r is the variable in the given expression
3r + 2r
= r + r + r + r + r
= 5r
5 is the coefficient of the variable r.

Question 9.
5y + 6y
Answer:
11y
Explanation:
y is the variable in the given expression.
y + y + y + y + y + y +y + y + y + y + y = 11y
11 is the coefficient of the variable y.

State whether each pair of expressions are equivalent.

Question 10.
3a and a + a + a
Answer:
YES, pair of expressions are equivalent.
Explanation:
3a  can be expanded as a + a + a = 3a
a + a + a =  3a
So, the given statement is correct.

Question 11.
2h + 2h and 4h
Answer:
YES, pair of expressions are equivalent.
Explanation:
2h can be expanded as h + h
2h + 2h = 4h
h + h + h + h = 4h

Question 12.
2k + 5 and (k + k) • 5
Answer:
NO, pair of expressions are not equivalent.
Explanation:
2k + 5 and (k + k) • 5
2k + 5 is not equal to 2k.5
2k + 5 is not equal to 10k

Question 13.
6z + 4z and 10 + 2z
Answer:
NO, pair of expressions are notequivalent.
Explanation:
6z + 4z = 10z
10 + 2z is another expression
10z not equal to 10 + 2z

Question 14.
1p + 3p and 13p
Answer:
NO, pair of expressions are not equivalent.
Explanation:
1p + 3p = 4p
4p is not equal to 13p

Question 15.
3n + 2 + 4n and 2 + 7n
Answer:
YES, pair of expressions are equivalent.
Explanation:
both are equal
3n + 2 + 4n
= 3n + 4n + 2
= 7n + 2

Complete

Question 16.
Simplify 4s – s

4s – s =
Answer:
3s
Explanation:
We use the fraction strips to show differences between different fractions.

Simplify each expression.

Question 17.
12z – 7z
Answer:
5z
Explanation:
Subtract 12 from 7
12z – 7z = 5z

Question 18.
3p – 3p
Answer:
0
Explanation:
Subtracting 3p from 39 we get 0
3p – 3p = 0

State whether each pair of expressions are equivalent.

Question 19.
f – 6 and 6 – f
Answer:
NO, pair of expressions are not equivalent.
Explanation:
f- 6 is positive if f > 6
6-f is negative for the above f value.

Question 20.
5c – 5c and a – a
Answer:
YES, pair of expressions are equivalent.
Explanation:
5c – 5c = 0
a – a = 0
both are correct.

Simplify each expression.

Question 21.
(j + 3j) + 2j = + 2j
=
Answer:
6j
Explanation:
(j + 3j) + 2j
= 4j + 2j
= 6j

Question 22.
4j + 5j + 2j
Answer:
11j
Explanation:
4j + 5j + 2j
= 9j + 2j
= 11j

Question 23.
9t – 3t – 4t
Answer:
2t
Explanation:
9t – 3t – 4t
= 6t – 4t
= 2t

Question 24.
5t – t – 4t
Answer:
0
Explanation:
5t – t – 4t
= 4t – 4t = 0

Question 25.
8w – 6w + 3w
Answer:
5w
Explanation:
8w – 6w + 3w
= 2w + 3w
= 5w

Question 26.
7w + 2w – 6w
Answer:
3w
Explanation:
7w + 2w – 6w
= 9w – 6w = 3w

Complete.

Question 27.
The figure shows a quadrilateral. Find the perimeter of the quadrilateral.

6x + 6 + 2x + 2 = 6x + 2x + 6 + 2
= +
The perimeter of the quadrilateral is units.
Answer:
8x + 8 units
Explanation:
6x + 6 + 2x + 2
= 6x + 2x + 6 + 2
= 8x + 8 units
The perimeter of the quadrilateral is 8x + 8 units.

Simplify each expression.

Question 28.
4x – 3 + 3x
Answer:
7x – 3
Explanation:
4x – 3 + 3x
= 4x + 3x – 3
= 7x – 3

Question 29.
5y + 4 = 2y
Answer:
3y + 4 =0
Explanation:
5y + 4 = 2y
5y – 2y + 4
3y + 4 = 0

Question 30.
8y – 7 – 4y
Answer:
4y – 7
Explanation:
8y – 7 – 4y
subtract the variables first,
= 4y – 7

Question 31.
7z + 9 – 2z – 2
Answer:
5z + 7
Explanation:
7z + 9 – 2z – 2
Bring all the variables and coefficients to one side.
= 7z – 2z +9 – 2
= 5z + 7

Question 32.
5 + 11 z – 4 + 6z
Answer:
1 + 17z
Explanation:
5 + 11 z – 4 + 6z
Bring all the variables and coefficients to one side.
= 5 – 4 + 11 z + 6z
= 1 + 17z

Question 33.
8g + 10 – 3g + 7
Answer:
5g + 17
Explanation:
8g + 10 – 3g + 7
Bring all the variables and coefficients to one side.
=8g – 3g + 10 + 7
=5g + 17

Question 34.
12 + 6g – 5 – 4g
Answer:
7 – 2g
Explanation:
12 + 6g – 5 – 4g
Bring all the variables and coefficients to one side.
=12 – 5 + 6g – 4g
=7 – 2g

Question 35.
27 + 3r – 9 + 15r
Answer:
18 +18r
Explanation:
27 + 3r – 9 + 15r
Bring all the variables and coefficients to one side.
=27 – 9 +3r + 15r
=18 +18r

Math in Focus Course 1A Practice 7.3 Answer Key

Simplify each expression. Then state the coefficient of the variable in each expression.

Question 1.
u + u + u + u
Answer:
4u
Explanation:
Add all the variable to get the sum
u + u + u + u = 4u

Question 2.
v + v + 5 – 2
Answer:
2v + 3
Explanation:
v + v + 5 – 2
Add all the variable at first,
then subtract the coefficient.
Finally, write an algebraic expression.
= 2v + 5 – 2
= 2v + 3

Question 3.
w + w + w + w + w + w + 15 – 7
Answer:
6w + 8
Explanation:
Add all the variable at first,
then subtract the coefficient.
Finally, write an algebraic expression.
w + w + w + w + w + w + 15 – 7
= 6w + 15 – 7
= 6w + 8

Simplify each expression.

Question 4.
3p + p
Answer:
4p
Explanation:
3p + p
Expand the variables at first, then add.
= p + p + p + p
= 4p

Question 5.
4p + 5p
Answer:
9p
Explanation:
4p + 5p
Expand the variables at first, then add.
p + p + p + p + p + p + p + p + p
= 9p

Question 6.
7p – 2p
Answer:
5p
Explanation:
7p -2 p
Expand the variables at first, then subtract.
= (p + p + p + p + p + p + p) – (p + p)
= 5p

Question 7.
3p – 2p + 5p
Answer:
6p
Explanation:
3p – 2p + 5p
3p – 2p = 1p
1p + 5p = 6p

Question 8.
2p + 3p + 4p + 5p – 6p – 7p
Answer:
p
Explanation:
2p + 3p + 4p + 5p – 6p – 7p
14p – 13p = p

State whether each pair of expressions are equivalent.

Question 9.
5x and x + x + 3x
Answer:
YES, pair of expressions are equivalent.
Explanation:
x + x + 3x = 5x
So, 5x and x + x + 3x are equivalent.

Question 10.
4y + 2y + y and 5y + y
Answer:
NO, pair of expressions are not equivalent.
Explanation:
4y + 2y + y and 5y + y
4y+2y+y = 7y
5y+y=6y
7y is not equal to 6y

Question 11.
2z + 5 and z + 8 + z – 3
Answer:
YES, pair of expressions are equivalent.
Explanation:
2z + 5 and z + 8 + z – 3
2z + 5
z + 8 + z – 3 = 2z+5
both are equal

Question 12.
2w – 5 and 5 – 2w
Answer:
No, pair of expressions are not equivalent.
Explanation:
2w – 5 and 5 – 2w
2w – 5, here coefficient is negative and variable is positive.
-2w + 5, here coefficient is positive and variable is negative.

Question 13.
11u – 4u and 11 – 4 + u
Answer:
No, pair of expressions are not equivalent.
Explanation:
11u – 4u and 11 – 4 + u
11u – 4u = 7u
11 – 4 + 4 = 7 + u
7 + u is not 7u

Question 14.
3v + v and \(\frac{12v}{3}\)
Answer:
Yes, pair of expressions are equivalent.
Explanation:
3v + v and \(\frac{12v}{3}\)
3v + v = 4v
\(\frac{12v}{3}\) = 4v
4v is equal \(\frac{12v}{3}\)

Simplify each expression.

Question 15.
3x + 5 + 4x + 6
Answer:
7x + 11
Explanation:
3x + 5 + 4x + 6
Add all the variables and coefficients.
= 3x + 4x + 5 + 6
= 7x + 11

Question 16.
3x + 2x + 3x + 2
Answer:
8x + 2
Explanation:
3x + 2x + 3x + 2
Add all the variables at first,
= 8x + 2

Question 17.
17 + 4w – 12 – w
Answer:
5 + 3w
Explanation:
17 + 4w – 12 – w
= 17 -12 + 4w – w
= 5 + 3w

Question 18.
9 + 5u + 6u – 7 – 8u + 4
Answer:
6 + 3u
Explanation:
9 + 5u + 6u – 7 – 8u + 4
= 9 – 7 + 4 + 5u + 6u – 8u
= 6 + 3u

Solve.

Question 19.
A book has a length of (b + 2) inches and a width of b inches. Write a simplified expression for the perimeter of the book.

Answer:
4b + 4 inches
Explanation:
A book has a length of (b + 2) inches and a width of b inches.
Perimeter = 2(length + width)
= b + 2 + b + b + 2 + b
Add all the variables and coefficients to write an equation.
= 4b + 4 inches is the perimeter of the book.

Question 20.
The figure shows a quadrilateral. The length of each side is given as shown. Find the perimeter of the quadrilateral in terms of z.

Answer:
3z + 15 cm
Explanation:
The length of each side is given as shown.
The perimeter of the quadrilateral in terms of z.
p = z + 8 + z + 3 + z + 4
= 3z + 15 cm

Question 21.
Anne is currently h years old. Bill is currently 2h years old and Charles is currently 8 years old. Find an expression for each person’s age after h years. Then find an expression for the sum of their ages after h years.
Answer:
8 + h years
Explanation:
Anne h years
Bill is 2h years
Charles is 8 years
an expression for each person’s age after h years
Anne h + h = 2h years
Bill is 2h + h = 3h years
Charles is 8 + h years

Question 22.
There are 18 boys in a class. There are w fewer boys than girls. How many students are there in the class?
Answer:
36 – w
Explanation:
Boys  = 18
Girls = 18 – w
Total number of students = 18 + 18 – w
= 36 – w

Question 23.
A rectangular garden has a length of (y + 2) yards and a width of (4y – 1) yards. Find the perimeter of the garden in terms of y.
Answer:
10y + 2
Explanation:
A rectangular garden has a length of (y + 2) yards and a width of (4y – 1) yards.
The perimeter of the garden in terms of y.
p = y + 2 + 4y – 1 + y + 2 + 4y – 1
p = 10y + 2
the perimeter of the garden in terms of y is = 10y + 2

Question 24.
Kayla had 64b dollars. She gave \(\frac{1}{8}\) of it to Luke and spent $45. How much money did Kayla have left? Express your answer in terms of b.
Answer:
\(\frac{520b -360} {8}\)
Explanation:
Kayla had 64b dollars.
She gave \(\frac{1}{8}\) of it to Luke and spent $45.
64b- \(\frac{1}{8}\) -45
= \(\frac{64b x  8 + 8b -45 x 8} {8}\)
= \(\frac{520b – 360} {8}\)

Question 25.
A rectangle has a length of (2m + 1) units and a width of (10 – m) units. A square has sides of length \(\frac{2 m+1}{2}\) units.
a) Find the perimeter of the rectangle.
Answer:
2m + 22
Explanation:
length of (2m + 1) units and a width of (10 – m) units
the perimeter of the rectangle is
= 2m + 1 + 10 – m + 2m + 1 + 10 – m
= 2m + 22

b) Find the perimeter of the square.
Answer:
4 x  \(\frac{2 m+1}{2}\) units
Explanation:
length \(\frac{2 m+1}{2}\) units
Perimeter of a square =4 x  \(\frac{2 m+1}{2}\) units

c) Find the sum of the perimeters of the two figures if m = 6.
Answer:
26 units
Explanation:
the perimeters of the two figures if m = 6
Perimeter of a square = 4 x  \(\frac{2 m+1}{2}\) units
= 4 x  \(\frac{2 x 6+1}{2}\) units
= 4 x  \(\frac{13}{2}\) units
= 26 units
the perimeter of the rectangle is = 2m + 22 units
=2 x 9 + 22
= 18 + 22
= 40 units

d) If m – 6, the perimeter of the rectangle is greater than the perimeter of the square. Find how many units greater the rectangle’s perimeter is than the square’s perimeter.
Answer:
40 units
Explanation:
the perimeters of the two figures if m = 6
Perimeter of a square = 26 units
the perimeter of the rectangle is = 40 units

Question 26.
Math Journal Rita simplified the expression 10w – 5w + 2w in this way:

Is Rita’s answer correct? If not, explain why it is incorrect.
Answer:
Is Rita’s answer Wrong
Explanation:
In the above math journal of Rita,
she wrote the express as -5w + 2w = – 7
where as the correct express is 10w – 5w = 5w
5w + 2w = 7w

Go through the Math in Focus Grade 6 Workbook Answer Key Chapter 4 Lesson 4.1 Comparing Two Quantities to finish your assignments.

Math in Focus Grade 6 Course 1 A Chapter 4 Lesson 4.1 Answer Key Comparing Two Quantities

Math in Focus Grade 6 Chapter 4 Lesson 4.1 Guided Practice Answer Key

Complete.

Question 1.
The ratio of the number of CDs to the number of CD sleeves is : .

Answer:
3 : 5
Explanation:
The number of CDs = 3
The number of sleeves = 5
the ratio of CDs = 3:5

Question 2.
Mrs. Carter buys 4 bags of apples and 7 bags of oranges. Each bag has an equal number of fruits.

The ratio of the number of apples to the number of oranges is : .
Answer:
4 : 7
Explanation:
The number of apples = 4
The number of oranges = 7
The ratio of the number of apples to the number of oranges is 4:7

Question 3.
Desiree has a cat that weighs 12 pounds and a dog that weighs 13 pounds.

a) The ratio of the weight of the cat to the weight of the dog is : .
Answer:
12 : 13
Explanation:
The weight of cat = 12lb
The weight of dog = 13lb
The ratio of the weight of the cat to the weight of the dog is =12:13

b) The ratio of the weight of the dog to the weight of the cat is : .
Answer:
13 : 12
Explanation:
The weight of cat = 12lb
The weight of dog = 13lb
The ratio of the weight of the dog to the weight of the cat is = 13:12

State whether each of the following can be expressed as a ratio.

Question 4.
6 kg and 84 kg
Answer:
yes, expressed as 1 : 14
Explanation:
The ratio of 6 kg and 84kg is expressed as
= 6 : 84
=1 : 14

Question 5.
72 ft and 2 yd
Answer:
yes, expressed as 1 : 12
Explanation:
The ratio of  72ft and 2yds
To express in ration we should convert the both the numbers in one unit( Either in feet of yards) as shown below
(2yd = 6ft)
= 6 : 72
= 1 : 12

Question 6.
3 oz and 30 in.
Answer:
yes, expressed as 1 : 6
Explanation:
The ratio of 3 oz and 30 in.
3 oz = 5in
3 oz = 5.2 in
= 5 : 30
= 1:6

Complete.

Question 7.
13 cm : 2 m = 13 cm : cm Think: 1 m = 100 cm, so 2 m = cm.
= : .
Answer:
13 : 200
Explanation:
13 cm : 2 m = 13 cm : 200 cm
Think: 1 m = 100 cm,
so 2 m = 200 cm.
= 13 : 200.

Question 8.
5 kg : 13 g = g : g Think: 1 kg = 1,000 g, so 5 kg = g.
= : .
Answer:
5000 : 13
Explanation:
5 kg : 13 g = 5000 g : 13 g
Think: 1 kg = 1,000 g,
so 5 kg = 5000 g.
5000 : 13

Question 9.
9 mL : 7 L = mL : mL Think: 1 L = 1,000 mL, so 7 L = mL.
= : .
Answer:
9 : 7000
Explanation:
9 mL : 7 L = 9 mL : 7000 mL
Think: 1 L = 1,000 mL,
so 7 L = 7000 mL.
= 9 : 7000

Question 10.
John keeps 17 angelfish and 24 guppies.

a) Find the ratio of the number of angelfish to the number of guppies.
The ratio of the number of angelfish to the number of guppies is : .
Answer:
17 : 24
Explanation:
John keeps 17 angelfish and 24 guppies

The ratio of the number of angelfish to the number of guppies is 17 : 24

b) Find the ratio of the number of guppies to the total number of fish.
Total number offish = +
=
The ratio of the number of guppies to the total number of fish is 24 : 41.
Answer:
24 : 58
Explanation:
John keeps 17 angelfish and 24 guppies
Total number offish = 17 + 41 = 58
The ratio of the number of guppies to the total number of fish is 24 : 58.

Question 11.
Alex has a string 72 centimeters long. He cuts it into two pieces. The length of the shorter piece is 31 centimeters. What is the ratio of the length of the longer piece to the total length of the string?

72 – =
The length of the longer piece is centimeters,
cm : cm = :
The ratio of the length of the longer piece to the total length of the string is : .
Answer:
41 : 72
Explanation:

72 – 31 = 41
The length of the longer piece is 41 centimeters,
31 cm : 41 cm = 31 : 41
The ratio of the length of the longer piece to the total length of the string is 41 : 72.

Question 12.
The total amount of money Pamela and Rachel saved in a week was $45. Pamela saved $13.
a) Find the amount of money Rachel saved.
Answer:
$32
Explanation:
The total amount of money Pamela and Rachel saved in a week was $45
Pamela saved $13.
$45 – $13 = $32
the amount of money Rachel saved = $32

b) Find the ratio of the total amount of money Pamela and Rachel saved to the amount of money Rachel saved.
Answer:
4 : 9
Explanation:
The total amount of money Pamela and Rachel saved in a week was $45
The amount of money Rachel saved = $32
The ration of the total amount of money Pamela and Rachel saved to the amount of money Rachel saved is 32 : 72
4 : 9  (8 x 4 = 32  and  8 x 9 = 72)

Complete. You may draw a model to help you.

Question 13.
The ratio of the number of white parrots to the number of green parrots is 7 : 8. Write a ratio to describe the number of green parrots to the total number of parrots.
Total number of parrots = + = units
The ratio of the number of green parrots to the total number of parrots is : or \(\frac{?}{?}\).
Answer:
8 : 15
Explanation:
The ratio of the number of white parrots to the number of green parrots is 7 : 8.
Total number of parrots = 7 + 8 = 15 units
The ratio of the number of green parrots to the total number of parrots is 8 : 15 or
\(\frac{8}{15}\).

Solve. You may draw a model to help you.

Question 14.
The ratio of the number of male teachers to the number of female teachers in a school is 5 : 9.
a) What fraction of the teachers are females?
Answer:
\(\frac{9}{14}\).

b) What fraction of the teachers are males?
Answer:
\(\frac{5}{14}\).
Explanation:
The ratio of the number of male teachers to the number of female teachers in a school is 5 : 9.
Total teachers = 5 + 9 = 14
fraction of male teachers = \(\frac{9}{14}\).

fraction of female teachers = \(\frac{5}{14}\)

Complete.

Question 15.
The height of a coconut tree is 48 feet and the height of a papaya tree is 12 feet.
a) How many times the height of the papaya tree is the height of the coconut tree?

The height of the coconut tree is times the height of the papaya tree.
Answer:
4ft
Explanation:
The height of a coconut tree is 48 feet and the height of a papaya tree is 12 feet.
The height of the coconut tree is 4 times the height of the papaya tree.

b) How many times the height of the coconut tree is the height of the papaya tree?

The height of the papaya tree is times the height of the coconut tree.
Answer:
\(\frac{1}{4}\).
Explanation:
The height of a coconut tree is 48 feet and the height of a papaya tree is 12 feet.
The height of the papaya tree is 1/4 times the height of the coconut tree.

Math in Focus Course 1A Practice 4.1 Answer Key

Write two ratios to compare the quantities.

Question 1.

Answer:
9 : 5
Explanation:
there are 9 sun flowers
there are 5 flower vases
the ratio of flowers and vases are 9:5

Question 2.
Catherine has 23 video game disks and Dylan has 37 video game disks.

Answer:
23 : 37
Explanation:
Catherine has 23 video game disks and Dylan has 37 video game disks.
Ratio of the both video games = 23 : 37

Question 3.
In a school, there are 8 classes in the sixth grade and 7 classes in the seventh grade. Each class has an equal number of students.

Answer:
8 : 7
Explanation:
In a school, there are 8 classes in the sixth grade and 7 classes in the seventh grade.
Each class has an equal number of students
Ratio of the students = 8 : 7

State whether each of the following can be expressed as a ratio.

Question 4.
6 cm and 60 g
Answer:
No, it can’t be expressed in ratio.
Explanation:
Both the quantities must be of same units,
here both are different in units, which can not be compared.

Question 5.
54 kg and 54 m
Answer:
No, it can’t be expressed in ratio.
Explanation:
Both the quantities must be of same units,
here both are different in units, which can not be compared.

Question 6.
12 g and 45 kg
Answer:
Yes, it can be expressed in ratio.
Explanation:
Both the quantities must be of same units.
45Kg = 45 x 1000 = 45000g
1kg = 1000g
12 : 45000

Question 7.
87 ft and 93 yd
Answer:
Yes, it can be expressed in ratio.
Explanation:
Both the quantities must be of same units.
1yd = 3ft
93 x 3  = 279 ft
87 ft : 279 ft
29 : 93

Solve. Show your work.

Question 8.
There are 123 students in a drama club. 65 of them are girls.
a) What is the ratio of the number of boys to the number of girls?
Answer:
58 : 65
Explanation:
There are 123 students in a drama club.
65 of them are girls.
number of boys 123 – 65 =58
the ratio of the number of boys to the number of girls = 58 : 65

b) What is the ratio of the number of boys to the total number of students?
Answer:
58 : 123
Explanation:
There are 123 students in a drama club.
65 of them are girls.
the number of boys 123 – 65 =58
the ratio of the number of boys to the total number of students = 58 : 123

Question 9.
There were 37 clarinet players and 150 trumpet players who tried out for a music competition.

a) Find the ratio of the number of clarinet players to the number of trumpet players.
Answer:
37 : 150
Explanation:
There were 37 clarinet players and 150 trumpet players who tried out for a music competition.
The ratio of the number of clarinet players to the number of trumpet players = 37 : 150

b) Find the ratio of the number of trumpet players to the total number of players.
Answer:
150 : 187
Explanation:
There were 37 clarinet players and 150 trumpet players who tried out for a music competition.
the total number of players = 37 + 150 = 187
the ratio of the number of trumpet players to the total number of players = 150 : 187

Question 10.
CD Carl had $43 when he entered the museum gift shop. After spending some money, he had $18 left.
a) Find the amount of money Carl spent.
Answer:
$25
Explanation:
CD Carl had $43
he had $18 left.
He spend money 43 – 18 = 25

b) Find the ratio of the amount of money Carl spent to the amount of money he had when he entered the gift shop.
Answer:
25 : 43
Explanation:
CD Carl had $43
he spend $25.
the ratio of the amount of money Carl spent to the amount of money he had 25 : 43

Question 11.
Kelvin’s monthly allowance is $42 and Mike’s monthly allowance is $63. How many times Mike’s monthly allowance is Kelvin’s monthly allowance?
Answer:
\(\frac{3}{2}\).
Explanation:
Kelvin’s monthly allowance is $42
Mike’s monthly allowance is $63
\(\frac{63}{42}\) i.e., { 21 x 2 = 42 & 21 x 3 = 63}

Question 12.
The ratio of the weight of vegetables sold to the weight of fruits sold is 45 : 144.
a) How many times the weight of vegetables sold is the weight of fruits sold?
Answer:
\(\frac{45}{144}\)

b) What fraction of the total weight of vegetables and fruits sold is the weight of vegetables sold?
Answer:
\(\frac{144}{189}\)
Explanation:
The ratio of the weight of vegetables sold to the weight of fruits sold is 45 : 144.
Total vegetables and fruits = 144 + 45 = 189
number of times the weight of vegetables sold is the weight of fruits sold
= \(\frac{45}{144}\)
fraction of the total weight of vegetables and fruits sold is the weight of vegetables sold
= \(\frac{144}{189}\)

Question 13.
The ratio of the length to the width of a rectangle is 5 : 2.
a) Express the difference between the length and the width of the rectangle as a fraction of the length of the rectangle.
Answer:
The difference is 3
Explanation:
The ratio of the length to the width of a rectangle is 5 : 2.
difference between width and length
= 5 – 2 = 3
b) Express the width of the rectangle as a fraction of the perimeter of the rectangle.
Answer:
\(\frac{1}{7}\)
Explanation:
The ratio of the length to the width of a rectangle is 5 : 2.
difference between width and length
= 5 – 2 = 3
perimeter = \(\frac{2}{14}\)
\(\frac{1}{7}\)

Math Journal Describe a situation that each ratio could represent.

Question 14.
5 : 16
Answer:
5 : 16
Number of base ball players : Number of foot ball players
Explanation:
The ratio of Baseball players and Football players are 5 : 16

Question 15.
98 : 3
Answer:
98 : 3
Number of students present : Number of students absent
Explanation:
The ratio of students present in a class are,
98 : 3

Question 16.
1,000 : 1
Answer:
Scale for maps
1000 m : 1 km
Explanation:
The ratio of scales on maps are represented as meters and kilometers.

Go through the Math in Focus Grade 6 Workbook Answer Key Chapter 4 Review Test to finish your assignments.

Math in Focus Grade 6 Course 1 A Chapter 4 Review Test Answer Key

Concepts and Skills

Write each ratio in simplest form.

Question 1.
8 : 24
Answer:
1 : 3
Explanation:
The simplest form of 8 : 24 is 1 : 3
Divide each side by the highest common factor.
That is the highest number which divides evenly into both numbers.
G.C.M for 8 and 24 is 8

Question 2.
6 : 20
Answer:
3 : 10
Explanation:
The simplest form of 6 : 20 is 3 : 10
Divide each side by the highest common factor.
That is the highest number which divides evenly into both numbers.
G.C.M for 6 and 20 is 2

Question 3.
14 : 49
Answer:
2 : 7
Explanation:
The simplest form of 14 : 49 is 2 : 7
Divide each side by the highest common factor.
That is the highest number which divides evenly into both numbers.
G.C.M for 14 and 49 is 7

Question 4.
27 : 72
Answer:
3 : 8
Explanation:
The simplest form of 27 : 72 is 3 : 8
Divide each side by the highest common factor.
That is the highest number which divides evenly into both numbers.
G.C.M for 27 and 72 is 9

Question 5.
14 : 49
Answer:
2 : 7
Explanation:
The simplest form of 14 : 49 is 2 : 7
Divide each side by the highest common factor.
That is the highest number which divides evenly into both numbers.
G.C.M for 14 and 49 is 7

Question 6.
45 : 30
Answer:
3 : 2
Explanation:
The simplest form of 45 : 30 is 3 : 2
Divide each side by the highest common factor.
That is the highest number which divides evenly into both numbers.
G.C.M for 45 and 30 is 15

Question 7.
27 : 72
Answer:
3 : 8
Explanation:
The simplest form of 27 : 72 is 3 : 8
Divide each side by the highest common factor.
That is the highest number which divides evenly into both numbers.
G.C.M for 27 and 72 is 9

Question 8.
64 : 56
Answer:
8 : 7
Explanation:
The simplest form of 64 : 56 is 8 : 7
Divide each side by the highest common factor.
That is the highest number which divides evenly into both numbers.
G.C.M for 64 and 56 is 8

Find the missing term in each pair of equivalent ratios.

Question 9.
1 : 3 = 6 :
Answer: 18
Explanation:
product of extremes = product of means
a:b = c:d
\(\frac{a}{b}\) = \(\frac{c}{d}\)
ad = bc
a = 1, b = 3, c = 6, d = x
1 x x = 3 x 6
x = 18

Question 10.
4 : 7 = : 21
Answer: 12
Explanation:
product of extremes = product of means
a:b = c:d
\(\frac{a}{b}\) = \(\frac{c}{d}\)
ad = bc
a = 4, b = 7, c = x, d = 21
4 x 21 = 7 x x
84 = 7x
x = \(\frac{84}{7}\)
x = 12

Question 11.
25 : 15 = : 3
Answer: 5
Explanation:
product of extremes = product of means
a:b = c:d
\(\frac{a}{b}\) = \(\frac{c}{d}\)
ad = bc
a = 25, b = 15, c = x, d = 3
25 x 3 = 15 x x
15x = 75
x = \(\frac{75}{15}\)
x = 5
Question 12.
54 : 36 = 18 :
Answer: 12
Explanation:
product of extremes = product of means
a:b = c:d
\(\frac{a}{b}\) = \(\frac{c}{d}\)
ad = bc
a = 54, b = 36, c = 18, d = x
54 x x = 36 x 18
54x = 648
x = \(\frac{648}{54}\)
x = 12

Question 13.
4 : = 20 : 25
Answer: 5
Explanation:
product of extremes = product of means
a:b = c:d
\(\frac{a}{b}\) = \(\frac{c}{d}\)
ad = bc
a = 4, b = x, c = 20, d = 25
4 x 25 = x x 20
100 = 20x
x = \(\frac{100}{20}\)
x = 5

Question 14.
: 9 = 48 : 72
Answer: 6
Explanation:
product of extremes = product of means
a:b = c:d
\(\frac{a}{b}\) = \(\frac{c}{d}\)
ad = bc
a = x, b = 9, c = 48, d = 72
x x 72 = 9 x 48
72x = 432
x = \(\frac{432}{72}\)
x = 6

Question 15.
28 : = 4 : 6
Answer: 42
Explanation:
product of extremes = product of means
a:b = c:d
\(\frac{a}{b}\) = \(\frac{c}{d}\)
ad = bc
a = 28, b = x, c = 4, d = 6
28 x 6 = x x 4
168 = 4x
x = \(\frac{168}{4}\)
x = 42

Question 16.
: 36 = 21 : 12
Answer: 63
Explanation:
product of extremes = product of means
a:b = c:d
\(\frac{a}{b}\) = \(\frac{c}{d}\)
ad = bc
a = x, b = 36, c = 21, d = 12
x x 12 = 36 x 21
12x = 756
x = \(\frac{756}{12}\)
x = 63

Problem Solving

Solve. Show your work.

Question 17.
The city has 28 fire engines and 36 fire trucks. Find the ratio of the number of fire engines to the number of fire trucks in simplest form.
Answer:
7 : 9
Explanation:
The city has 28 fire engines and 36 fire trucks.
Ratio is 28 : 36
The ratio of the number of fire engines to the number of fire trucks in simplest form.
7 : 9

Question 18.
Of the 80 students who signed up for after school clubs, 16 students signed up for the art club, and the rest signed up for other clubs. Find the ratio of the number of students who signed up for the art club to the number of students who signed up for other dubs. Give your answer in simplest form.
Answer:
1 : 4
Explanation:
80 students who signed up for after school clubs,
16 students signed up for the art club,
the rest signed up for other clubs = 80 – 16 = 64
The ratio of the number of students who signed up for the art club to the number of students who signed up for other dubs are 16 : 64
simplest form of ratio is 1 : 4

Question 19.
On Saturday, Alison used her cell phone for 36 minutes. On Sunday, she used her cell phone for 18 minutes more than on Saturday. Find the ratio of the number of minutes Alison used on Saturday to the total number of minutes on Saturday and Sunday. Give your answer in simplest form.
Answer:
2 : 5
Explanation:
On Saturday 36 min
On Sunday 36 + 18 = 54 min
the number of minutes Alison used on Saturday to the total number of minutes on Saturday and Sunday is
36 + 54 =90 min
the ratio is 36 : 90 = 2 : 5

Question 20.
Daniel is 12 years old. Elliot is 15 years older than Daniel. Frank is 3 years younger than Elliot. Find the ratio of Daniel’s age to Frank’s age. Give your answer in simplest form.
Answer:
1: 2
Explanation:
Daniel’s age is 12 years
Elliots age is 15 + 12 = 27 years
Franks age is 27 -3 = 24 years
the ratio of Daniel’s age to Frank’s age
12 : 24
1 : 2

Question 21.
The ratio of the number of left-handed batters to the number of right-handed batters is 5 : 8. There are 45 left-handed batters.
a) How many right-handed batters are there?
Answer: 72
Explanation:
45 : 5  = x : 8
x =(45 x 8) /5 = 72

b) Find the ratio of the number of left-handed batters to the total number of batters. Give your answer in simplest form.
Answer:
5 : 13
Explanation:
45 + 72 = 117 total number of batters
45 : 117
5 : 13

Question 22.
Mrs. Johnson gave a sum of money to her son and daughter in the ratio 5 : 6. Her daughter received $2,400. How much did Mrs. Johnson give away in all?
Answer:
$2000
Explanation:
son and daughter in the ratio 5 : 6
6 : x = 5 x 2400
x = (5x 2400)/6 = 2000

Question 23.
The ratio of the number of boys to the number of girls in a school is 5 : 7. If there are 600 students in the school, how many girls are there?
Answer:
350 girls
Explanation:
ratio of boys and girls = 5 + 7 = 12
total students = 600
number of girls =7 x 600/12 = 50 x 7 = 350

Question 24.
The murals in a school are painted by its grade 6 and grade 7 students. The number of mural painters from grade 6 and the number of mural painters from grade 7 is the same for each mural in the school. Copy and complete the table.

Answer:

Explanation:
The number of mural painters from grade 6 and the number of mural painters from grade 7 is the same for each mural in the school
Each time the number of Murals in Garde 6 increased with 5 multiple.
At the same time in Grade 7 also increased with 7 multiple.

Question 25.
A sum of money was shared among Aaron, Ben, and Charles in the ratio 2 : 5 : 7. If Charles’s share was $1,180 more than Aaron’s share, what was the original sum of money shared?
Answer:
Explanation:
Aaron, Ben, and Charles in the ratio 2 : 5 : 7
2+5+7= 14

Aaron: 2x
Ben: 5x
Charles:7x

7x = 2x + 1180
5x = 1180
x = 236

2x + 5x + 7x = 14x
14x = 14*236 = $3304

Question 26.
The ratio of the number of beads Karen had to the number of beads Patricia had was 2 : 5. After Patricia bought another 75 beads, the ratio became 4 : 15. How many beads did each girl have at first?
Answer:
36 : 90 beads
Explanation
2:5
2x : 5x
2x : 5x+75 = 4:15
2x X 15 = 4 ( 5x + 75)
30x = 20x + 180
30x-20x = 180
10x=180
x=180/10 = 18
2x:5x
18 x2 : 5×18
36 : 90

Question 27.
Mr. Young had some bottles of apple juice and orange juice. The ratio of the number of bottles of apple juice to the number of bottles of orange juice was 3 : 2. After he sold 64 bottles of apple juice, the ratio became 1 : 6. How many bottles of apple juice and orange juice did Mr. Young have altogether in the end?
Answer:
56 bottles
Explanation:
A : O = 3 : 2
Apple  = 3 x 3 = 9
Orange = 2 x 3 = 6 =>6
64/8 = 8
8 x 6 = 48
48 + 8 = 56 bottles left out juice

Go through the Math in Focus Grade 6 Workbook Answer Key Chapter 4 Ratio to finish your assignments.

Math in Focus Grade 6 Course 1 A Chapter 4 Answer Key Ratio

Math in Focus Grade 6 Chapter 4 Quick Check Answer Key

Express each fraction as two equivalent fractions using division.

Question 1.
\(\frac{3}{4}\)

Answer:
\(\frac{6}{8}\) and \(\frac{9}{12}\)

Explanation:
To find equivalent fractions,
just multiply the numerator and denominator of that reduced fraction (3/4) by any integer number, i.e., multiply by 2, 3, 4 and so on.

\(\frac{3×2}{4×2}\) = \(\frac{6}{8}\)

\(\frac{3×3}{4×3}\) = \(\frac{9}{12}\)

Question 2.
\(\frac{7}{9}\)

Answer:
\(\frac{14}{18}\) and \(\frac{21}{27}\)

Explanation:
To find equivalent fractions,
just multiply the numerator and denominator of that reduced fraction (7/9) by any integer number, i.e., multiply by 2, 3, 10, 30 and so on.

\(\frac{7×2}{9×2}\) = \(\frac{14}{18}\)

\(\frac{7×3}{9×3}\) = \(\frac{21}{27}\)

Question 3.
\(\frac{6}{11}\)

Answer:
\(\frac{12}{22}\) and \(\frac{18}{33}\)

Explanation:
To find equivalent fractions,
just multiply the numerator and denominator of that reduced fraction (6/11) by any integer number, i.e., multiply by 2, 3, 10, 30 and so on.

\(\frac{6×2}{11×2}\) = \(\frac{12}{22}\)

\(\frac{6×3}{11×3}\) = \(\frac{18}{33}\)

Question 4.
\(\frac{16}{56}\)

Answer:
\(\frac{4}{14}\) and \(\frac{2}{7}\)

Explanation:
To find equivalent fractions,
Divide the numerator and denominator to reduce fraction (16/56) by 4 and 2.

\(\frac{16}{56}\) ÷ 4

= \(\frac{4}{14}\) ÷ 2

= \(\frac{2}{7}\)

Question 5.
\(\frac{21}{63}\)

Answer:
\(\frac{1}{3}\) and \(\frac{2}{6}\)

Explanation:
To find equivalent fractions,
Divide the numerator and denominator to reduce fraction \(\frac{21}{63}\) by 21.

\(\frac{21}{63}\) ÷ 1

= \(\frac{1}{3}\)

Multiply with integer 2
\(\frac{1×2}{3×2}\)

= \(\frac{2}{6}\)

Question 6.
\(\frac{35}{140}\)

Answer:
\(\frac{1}{4}\) and \(\frac{2}{8}\)

Explanation:
To find equivalent fractions,
Divide the numerator and denominator to reduce fraction \(\frac{35}{140}\) by 35.

\(\frac{35}{140}\) ÷ 35

= \(\frac{1}{4}\)

Multiply with integer 2
\(\frac{1×2}{4×2}\)

= \(\frac{2}{8}\)

Find the unknown numerator or denominator in each pair of equivalent fractions.

Question 7.
\(\frac{3}{8}\) = \(\frac{?}{56}\)

Answer: 21
Explanation:
Use cross multiplication to calculate the unknown variable x or fraction.

\(\frac{3}{8}\) = \(\frac{?}{56}\)

\(\frac{3×56}{8}\) = \(\frac{168}{8}\) = 21

Question 8.
\(\frac{7}{9}\) = \(\frac{21}{?}\)

Answer: 27
Explanation:
Use cross multiplication to calculate the unknown variable x or fraction.

\(\frac{7}{9}\) = \(\frac{21}{?}\)

\(\frac{9×21}{7}\) = \(\frac{189}{7}\) = 27

Question 9.
\(\frac{?}{11}\) = \(\frac{30}{55}\)

Answer: 6
Explanation:
Use cross multiplication to calculate the unknown variable x or fraction.

\(\frac{?}{11}\) = \(\frac{30}{55}\)

\(\frac{30×11}{55}\) = \(\frac{330}{55}\) = 6

Question 10.
\(\frac{6}{?}\) = \(\frac{42}{84}\)

Answer: 12
Explanation:
Use cross multiplication to calculate the unknown variable x or fraction.

\(\frac{6}{?}\) = \(\frac{42}{84}\)

\(\frac{6×84}{42}\) = \(\frac{504}{42}\) = 12

Writing fractions in simplest form

Question 11.
\(\frac{5}{45}\)

Answer:
\(\frac{1}{9}\)

Explanation:
If the numerator and denominator can be divided by the same number,
which is called a common factor a simple form of fraction.
See if at least one number in the fraction is a prime number.

Question 12.
\(\frac{18}{63}\)

Answer:
\(\frac{2}{7}\)

Explanation:
If the numerator and denominator can be divided by the same number,
which is called a common factor a simple form of fraction.
See if at least one number in the fraction is a prime number.

Question 13.
\(\frac{22}{55}\)

Answer:
\(\frac{2}{5}\)

Explanation:
If the numerator and denominator can be divided by the same number,
which is called a common factor a simple form of fraction.
See if at least one number in the fraction is a prime number.

Find the unknown measurement.

Question 14.
cm = 4 m
Answer:
400 cm
Explanation:
1m = 100 cm
4 m = 4 x 100
= 400 cm

Question 15.
9.8 kg = g
Answer:
9800 g
Explanation:
1 kg = 1000 g
9.8 kg = 1000 x 9.8
= 9800 g

Question 16.
6 ft = yd
Answer:
2 yards
Explanation:
1 feet = 0.33 yards
6 feet = 0.333 x 6
= 2 yards

Question 17.
10 L = mL
Answer:
10000 mL
Explanation:
1 L = 1000 mL
10 L = 1000 x 10
= 10000 mL

Question 18.
yd = 72 in.
Answer:
2 yards
Explanation:
1 yd = 36 in
72 in = 72/36 = 2

Question 19.
5 lb = oz
Answer:
80 oz
Explanation:
1 lb = 16 oz
5 lb = 16 x 5 = 80

Find the values of P and Q.

Question 20.

Answer:
P = 24 and Q = 6
Explanation:
Each block  is of 6 units, as shown below picture

P is of four blocks 6 x 4 = 24 and
Q is 6

Question 21.

Answer:
P = 12 and Q = 16
Explanation:

Each bock is of 4 units
P is 4 x 3 =12 and Q is 4 x 4 = 16

Go through the Math in Focus Grade 6 Workbook Answer Key Chapter 5 Rates to finish your assignments.

Math in Focus Grade 6 Course 1 A Chapter 5 Answer Key Rates

Math in Focus Grade 6 Chapter 5 Quick Check Answer Key

Multiply

Question 1.
268 × 13
Answer:
3,484
Explanation:

Question 2.
54 × 471
Answer:
25,434
Explanation:

Question 3.
532 × 48
Answer:
25,536
Explanation:

Question 4.
75 × 698
Answer:
52,350
Explanation:

Find each product. Express the product in simplest form.

Question 5.
4 × \(\frac{5}{32}\)
Answer:
\(\frac{5}{8}\)
Explanation:
The first step when multiplying fractions is to multiply the two numerators.
The second step is to multiply the two denominators.
Finally, simplify the new fractions.
= 4 × \(\frac{5}{32}\)
= 4 x 5 ÷ 32
= \(\frac{20}{32}\)
= \(\frac{5}{8}\)

Question 6.
\(\frac{7}{12}\) × 36
Answer: 21
Explanation:
The first step when multiplying fractions is to multiply the two numerators.
The second step is to multiply the two denominators.
Finally, simplify the new fractions.
\(\frac{7}{12}\) x 36
= \(\frac{(7)(36)}{12}\)
= \(\frac{252}{12}\) = 21

Question 7.
3\(\frac{2}{7}\) × 5
Answer:
\(\frac{115}{7}\)
Explanation:
The first step when multiplying fractions is to multiply the two numerators.
The second step is to multiply the two denominators.
Finally, simplify the new fractions.
3\(\frac{2}{7}\) x 5
= \(\frac{23}{7}\) x 5
= \(\frac{115}{7}\)

Question 8.
9\(\frac{1}{2}\) × 8
Answer: 76
Explanation:
The first step when multiplying fractions is to multiply the two numerators.
The second step is to multiply the two denominators.
Finally, simplify the new fractions.
9\(\frac{1}{2}\) x 8
= \(\frac{19}{2}\) x 8
= \(\frac{152}{2}\) = 76

Find each product. Express the product in simplest form.

Question 9.
\(\frac{2}{7}\) × \(\frac{63}{84}\)
Answer:
\(\frac{3}{14}\)
Explanation:
The first step when multiplying fractions is to multiply the two numerators.
The second step is to multiply the two denominators.
Finally, simplify the new fractions.
\(\frac{2}{7}\) x \(\frac{63}{84}\)
= \(\frac{(2)(63)}{(7)(84)}\)
= \(\frac{126}{588}\)
= \(\frac{3}{14}\)

Question 10.
\(\frac{11}{18}\) × \(\frac{3}{44}\)
Answer: 24
Explanation:
The first step when multiplying fractions is to multiply the two numerators.
The second step is to multiply the two denominators.
Finally, simplify the new fractions.
\(\frac{11}{18}\) x \(\frac{3}{44}\)
= \(\frac{11 X3}{18 x 44}\)
= \(\frac{33}{792}\) = 24

Find each quotient. Express the quotient in simplest form.

Question 11.
\(\frac{6}{7}\) ÷ 30
Answer:
\(\frac{1}{35}\)
Explanation:
Dividing fractions is equal to the multiplication of a fraction by the reciprocal of another fraction.
A fraction has a numerator and a denominator.
When we divide one fraction by another, we almost multiply the fractions.
\(\frac{6}{7}\) ÷ 30
= \(\frac{6}{7}\) ÷ \(\frac{30}{1}\)
= \(\frac{6}{7 X 30}\)
= \(\frac{6}{210}\)
= \(\frac{2}{70}\)
= \(\frac{1}{35}\)

Question 12.
72 ÷ \(\frac{9}{10}\)
Answer:
\(\frac{1}{80}\)
Explanation:
Dividing fractions is equal to the multiplication of a fraction by the reciprocal of another fraction.
A fraction has a numerator and a denominator.
When we divide one fraction by another, we almost multiply the fractions.
\(\frac{9}{10}\) ÷ 72
= \(\frac{9}{10}\) ÷ \(\frac{72}{1}\)
= \(\frac{9}{10 X 72}\)
= \(\frac{9}{720}\)
= \(\frac{1}{80}\)

Question 13.
\(\frac{7}{9}\) ÷ 49
Answer:
\(\frac{1}{63}\)
Explanation:
Dividing fractions is equal to the multiplication of a fraction by the reciprocal of another fraction.
A fraction has a numerator and a denominator.
When we divide one fraction by another, we almost multiply the fractions.
\(\frac{7}{9}\) ÷ 49
= \(\frac{7}{9}\) ÷ \(\frac{49}{1}\)
= \(\frac{7}{9 X 49}\)
= \(\frac{7}{441}\)
= \(\frac{1}{63}\)

Question 14.
56 ÷ \(\frac{8}{11}\)
Answer:
\(\frac{1}{77}\)
Explanation:
Dividing fractions is equal to the multiplication of a fraction by the reciprocal of another fraction.
A fraction has a numerator and a denominator.
When we divide one fraction by another, we almost multiply the fractions.
\(\frac{8}{11}\) ÷ 56
= \(\frac{8}{11}\) ÷ \(\frac{56}{1}\)
= \(\frac{8}{11 X 56}\)
= \(\frac{8}{616}\)
= \(\frac{1}{77}\)

Question 15.
\(\frac{4}{9}\) ÷ \(\frac{36}{135}\)
Answer:
\(\frac{5}{3}\)
Explanation:
Dividing fractions is equal to the multiplication of a fraction by the reciprocal of another fraction.
A fraction has a numerator and a denominator.
When we divide one fraction by another, we almost multiply the fractions.
= \(\frac{4}{9}\) ÷ \(\frac{36}{135}\)
= \(\frac{4}{9}\) x \(\frac{135}{36}\)
= \(\frac{4 X 135}{9 X 36}\)
= \(\frac{540}{324}\)
= \(\frac{5}{3}\)

Question 16.
\(\frac{77}{92}\) ÷ \(\frac{11}{42}\)
Answer:
\(\frac{5467}{56}\) OR 97\(\frac{35}{56}\)
Explanation:
Dividing fractions is equal to the multiplication of a fraction by the reciprocal of another fraction.
A fraction has a numerator and a denominator.
When we divide one fraction by another, we almost multiply the fractions.
= \(\frac{77}{92}\) ÷ \(\frac{11}{142}\)
= \(\frac{77}{92}\) x \(\frac{142}{11}\)
= \(\frac{77 X 142}{92 X 11}\)
= \(\frac{10,934}{1,012}\)
= \(\frac{5467}{56}\)

Find the value of each set.

Question 17.
If 7 units represent 98 liters, find the value of 15 units.
Answer:
210 liters
Explanation:
First, find out the value of 1 unit,
98 ÷ 7 = 14 liters.
Then, multiply 14 by 15 which is 210 liters.

Question 18.
If 13 units represent 143 square meters, find the value of 24 units.
Answer:
264 square meters
Explanation:
First, find out the value of 1 unit,
143 ÷ 13 = 11 liters.
Then, multiply 11 by 24 which is 264 liters.

Express each ratio in simplest form.

Question 19.
4 km : 370 m
Answer:
400m : 37m
Explanation:
The simplest form of the ratio is when the numbers are expressed as natural numbers with no common factors.
convert km in m
1km = 1000m
4 km : 370 m
4000m : 370
The common factor in the above equation is 2, 5.

Question 20.
66 L : 120 mL
Answer:
550mL : 1mL
Explanation:
The simplest form of the ratio is when the numbers are expressed as natural numbers with no common factors.
Convert liter into milliliters
1 L = 1000mL
66 L : 120 mL
66000 : 120mL
550mL : 1mL
The common factor in the above equation is 2, 3, 5.

Question 21.
15 in. : 5 ft
Answer:
12ft : 1ft
Explanation:
The simplest form of the ratio is when the numbers are expressed as natural numbers with no common factors.
Convert inches to feet
1 feet = 12 inches
15 in. : 5 ft
= 15 x 12 : 5
= 60 : 5
= 12 : 1
The common factor in the above equation is 5.

Question 22.
270 qt: 105 gal
Answer:
Explanation:
The simplest form of the ratio is when the numbers are expressed as natural numbers with no common factors.
Convert quarter to gallon
1gal = 4 qt
270 qt : 105 gal
= 270 : 105 x 4
= 270 : 420
= 9 : 14
The common factor in the above equation is 30.

Find two ratios equivalent to each ratio.

Question 23.
4 : 9
Answer:
8 : 18
Explanation:
Equivalent ratios are ratios that make the same comparison of numbers.
Two ratios are equivalent if one can be expressed as a multiple of the other.
multiples of 4 (4, 8, 12, 16, 20, 24, 28, . . . ) and 9 (9, 18, 27, 36, . . . . )

Question 24.
5 : 13
Answer:
10 : 26
Explanation:
Equivalent ratios are ratios that make the same comparison of numbers.
Two ratios are equivalent if one can be expressed as a multiple of the other.
multiples of 5 (5, 10, 15, 20, . . . ) and 13 (13, 26, 39, 52, 65, 78, 91, . . . . )

Go through the Math in Focus Grade 6 Workbook Answer Key Chapter 3 Lesson 3.4 Real-World Problems: Fractions and Decimals to finish your assignments.

Math in Focus Grade 6 Course 1 A Chapter 3 Lesson 3.4 Answer Key Real-World Problems: Fractions and Decimals

Math in Focus Grade 6 Chapter 3 Lesson 3.4 Guided Practice Answer Key

Solve.

Question 1.
The cost of carpeting a square yard is $8.60. How carpet 9.7 square yards?

9.7 × $ = $
It costs $ to carpet 9.7 square yards.
Answer:
9.7 × $8.60 = $83.42
It costs $83.42 to carpet 9.7 square yards.
Explanation:
The cost of carpeting a square yard is $8.60.
Cost of carpet 9.7 square yards
9.7 × $8.60 = $83.42
It costs $83.42 to carpet 9.7 square yards.

Question 2.
A roll of cloth 12 meters long is cut into smaller pieces of the same size. Each piece is 0.75 meter long. How many small pieces of cloth can be cut from the 12-meter roll?

12 =
small pieces of cloth can be cut from the 12-meter roll.
Answer:
12 ÷ 0.75 = 16
16 small pieces of cloth can be cut from the 12-meter roll.
Explanation:
A roll of cloth 12 meters long is cut into smaller pieces of the same size.
Each piece is 0.75 meter long.
Number of small pieces of cloth can be cut from the 12-meter roll
12 ÷ 0.75 = 16

Question 3.
Rosie buys 2.24 pounds of sliced ham to make sandwiches. She uses 0.16 pound for a sandwich. How many sandwiches can Rosie make?
÷ =
Rosie can make sandwiches.
Answer:
2.24÷ 0.16 = 14
Rosie can make 14 sandwiches.
Explanation:
Rosie buys 2.24 pounds of sliced ham to make sandwiches.
She uses 0.16 pound for a sandwich.
Number of sandwiches can Rosie make
2.24 ÷ 0.16 = 14

Question 4.
Bryce has $12.75. She wants to buy gifts for friends at a souvenir shop. If each souvenir costs $0.85, how many souvenirs can Bryce buy?
$ $ =
Bryce can buy souvenirs.
Answer:
$12.75 ÷ $0.85 = 15
Bryce can buy 15 souvenirs.
Explanation:
Bryce has $12.75.
She wants to buy gifts for friends at a souvenir shop.
Each souvenir costs $0.85,
Number of souvenirs can Bryce buy
$12.75 ÷ $0.85 = 15

Question 5.
A paper artist uses 18 paper rectangles for a collage. He uses 6 of them for the border of the artwork. He then cuts the remaining paper rectangles into equal strips, each \(\frac{1}{4}\) of a paper rectangle. How many strips of paper does he cut?

Answer:
48 strips of paper
Explanation:

A paper artist uses 18 paper rectangles for a collage.
He uses 6 of them for the border of the artwork.
He then cuts the remaining paper rectangles into equal strips, each \(\frac{1}{4}\) of a paper rectangle
Total strips of paper he cut
18 – 6 = 12
12 ÷ \(\frac{1}{4}\)
=48

Question 6.
Sophie buys a roll of string that is 20 meters long. She uses 3 meters of string to tie a parcel. She then cuts the remaining string into equal pieces, each \(\frac{1}{2}\) meter long. How many pieces does Sophie cut?
÷ \(\frac{?}{?}\) = ×
=
Sophie cuts pieces.
Answer:
Sophie cuts 34 pieces.
Explanation:
Sophie buys a roll of string that is 20 meters long.
She uses 3 meters of string to tie a parcel.
20 – 3 = 17 mtrs
the roll of string after use 3 mtrs is 17 mtrs
17 ÷ \(\frac{1}{2}\) = 17 × 2
= 34
Sophie cuts 34 pieces.

Question 7.
Andrea buys 75 cups of cranberry-apple juice for a party. She uses \(\frac{2}{5}\) of the juice to make punch. She then uses the remaining juice to pour single servings that are \(\frac{5}{6}\) cup each. How many single servings does Andrea pour?



The remaining juice is poured into single servings that are \(\frac{5}{6}\) cup each.
÷ \(\frac{?}{?}\) = × \(\frac{?}{?}\)
=
Andrea pours single servings.
Answer:
The remaining juice is poured into single servings that are \(\frac{5}{6}\) cup each.

45 ÷ \(\frac{5}{6}\) = 45 × \(\frac{6}{5}\)
= 54
Andrea pours 54 single servings.
Explanation:
Andrea buys 75 cups of cranberry-apple juice for a party.
She uses \(\frac{2}{5}\) of the juice to make punch.
She then uses the remaining juice to pour single servings that are \(\frac{5}{6}\) cup each.
Number of single servings does Andrea pour

Andrea pours 54 single servings.

Question 8.
Meredith bakes 5 pumpkin pies. She cuts the pies into quarters and distributes the pieces equally among her neighbors. Each neighbor receives \(\frac{3}{4}\) of a pie.
a) How many neighbors does Meredith distribute the pies to?
5 ÷ \(\frac{3}{4}\) = × \(\frac{?}{?}\)
=
Meredith distributes the pies to neighbors.

How many neighbors does Meredith distribute the pies to?
Answer:
6\(\frac{2}{3}\)
Explanation:
5 ÷ \(\frac{3}{4}\) = 5 × \(\frac{4}{3}\)

= \(\frac{20}{3}\)

= 6\(\frac{2}{3}\)

Meredith distributes the pies to 6 neighbors.

b) What fraction of a pie is left?
× \(\frac{3}{4}\) =
5 – =
of a pie is left.
Answer:
\(\frac{1}{2}\) pie is left over.
Explanation:
\(\frac{1}{2}\) ÷ 6

= \(\frac{1}{2}\) x \(\frac{1}{6}\)

= \(\frac{1}{12}\)

c) What is the total amount of pie each neighbor will receive if the remainder is divided evenly and distributed to each neighbor?
5 ÷ =
Each neighbor will receive of a pie in total.
Answer:
Each neighbor will receive \(\frac{5}{6}\) of each pie.
Explanation:
Meredith bakes 5 pumpkin pies.
She cuts the pies into quarters and distributes the pieces equally among her neighbors.\(\frac{3}{4}\) + \(\frac{1}{12}\)

= \(\frac{9}{12}\) + \(\frac{1}{12}\)

= \(\frac{10}{12}\)

= \(\frac{5}{6}\)

Question 9.
Some bottles containing \(\frac{2}{5}\) gallon of water each are used to fill a 7-gallon container. How many of these bottles are needed to fill the container with water to its brim?

7 ÷ \(\frac{2}{5}\) = × \(\frac{?}{?}\)
=
bottles are needed to fill the container with water to its brim.
Answer:
7 ÷ \(\frac{2}{5}\) = 7 × \(\frac{5}{2}\)

= \(\frac{35}{2}\)

17 \(\frac{1}{2}\) bottles are needed to fill the container with water to its brim.

Explanation:
Bottles containing \(\frac{2}{5}\) gallon of water used to fill a 7-gallon container.
Number of bottles are needed to fill the container with water to its brim
7 ÷ \(\frac{2}{5}\) = 7 × \(\frac{5}{2}\)

= \(\frac{35}{2}\)

= 17 \(\frac{1}{2}\)

Question 10.
Lilian has a part-time job. Each month, she spends \(\frac{1}{3}\) of her earnings on clothes, saves \(\frac{3}{8}\) of the remainder, and spends the rest of her earnings on food.
a) What fraction of her earnings does she spend on food?
b) If she earns $540, how much does she spend on food each month?
Method 1
a)

The model shows that:
She spends \(\frac{?}{?}\) of her earnings on food.
Answer:
She spends \(\frac{5}{8}\) of her earnings on food.

b) units → $540
1 unit → $ ÷ = $
units → × $ = $
She spends $ on food each month.
Answer:
She spends $225 on food each month.
Explanation:
12 units → $540
1 unit → $540 ÷ 12 = $45
5 units → 5 × $45 = $225
She spends $225 on food each month.
Method 2
a)

Answer:
She spends \(\frac{5}{8}\)
Explanation:

b) \(\frac{?}{?}\) × $540 = $
She spends $ on food each month.
Answer:
$225
Explanation:
540 x (1/3) = 180 for cloths
540 – 180 = 360
\(\frac{5}{8}\) × $360 = $225
She spends $225 on food each month.

Question 11.
\(\frac{2}{3}\) of a square is colored green. Asha cuts this green part into a number of pieces so that each piece is \(\frac{1}{9}\) of the whole square.
a) Find the number of pieces Asha has.
\(\frac{?}{?}\) ÷ \(\frac{?}{?}\) = \(\frac{?}{?}\) ×
=
Asha has pieces.
Answer:
Asha has 6 pieces.
Explanation:
\(\frac{2}{3}\) ÷ \(\frac{1}{9}\) = \(\frac{2}{3}\) × 9
= 6
Asha has 6 pieces.

b) If the area of the square is 45 square inches, what is the area of each piece?
Area of green part = \(\frac{?}{?}\) × 45 = in.²
Area of each piece = ÷ = in.²
The area of each piece is square inches.
Answer:
The area of each piece is 5 square inches.
Explanation:
Area of green part = \(\frac{1}{9}\) × 45 = 5 in.²
Area of each piece = 45 ÷ 9 = 5 in.²
The area of each piece is 5 square inches.

Question 12.
\(\frac{3}{5}\) of the students in a class were boys. The teacher divided the boys equally into groups so that each group of boys had \(\frac{1}{10}\) of the number of students in the class. The teacher then divided the girls equally into groups such that each group of girls had \(\frac{1}{5}\) of the number of students in the class.
a) Find the number of groups of boys and the number of groups of girls.

There were groups of boys and groups of girls.
Answer:
There were 6 groups of boys and 2 groups of girls.
Explanation:

b) If there were 16 girls in the class, how many boys were there ¡n each group?
\(\frac{2}{5}\) of the class → 16 students
\(\frac{1}{5}\) of the class → students
\(\frac{3}{5}\) of the class → students
÷ 6 =
There were boys in each group.
Answer:
There were 4 boys in each group.
Explanation:
\(\frac{2}{5}\) of the class → 16 students
\(\frac{1}{5}\) of the class → 8 students
\(\frac{3}{5}\) of the class → 24 students
24 ÷ 6 =
There were 4 boys in each group.

Math in Focus Course 1A Practice 3.4 Answer Key

Solve. Show your work.

Question 1.
An ounce of pine nuts costs $1.40. If Ellen buys 2.5 ounces of pine nuts, how much will she have to pay?
Answer:
$3.50
Explanation:
One ounce cost $1.40,
she needed 2 and a half ounces
so, 1.40 x 2.5 = $3.50

Question 2.
25 pints of apple juice are poured into \(\frac{1}{2}\) pint bottles. How many bottles can be filled with apple juice?
Answer:
50 bottles can be filled with apple juice
Explanation:
25 pints of apple juice are poured into \(\frac{1}{2}\) pint bottles.
25 ÷ \(\frac{1}{2}\)
= 25 x 2 = 50 bottles

Question 3.
40 pounds of sugar are repackaged into packets of \(\frac{1}{16}\) pound each. How many packets of sugar are there?
Answer:
640 packets of sugar
Explanation:
40 pounds of sugar are repackaged into packets of \(\frac{1}{16}\) pound each.
Total packets of sugar 40 ÷ \(\frac{1}{16}\)
= 40 x 16 = 640

Question 4.
Tom used \(\frac{5}{8}\) yard of ribbon to tie weights on the tail of his kite. He cut the length of ribbon into equal pieces that were \(\frac{1}{12}\) yard long. How many pieces, each \(\frac{1}{12}\) yard long, did Tom cut from the \(\frac{5}{8}\) yard ribbon?

Answer:
7 \(\frac{2}{4}\) pieces
Explanation:
Tom used \(\frac{5}{8}\) yard of ribbon to tie weights on the tail of his kite.
He cut the length of ribbon into equal pieces that were \(\frac{1}{12}\) yard long.
Number of pieces = \(\frac{5}{8}\)÷ \(\frac{1}{12}\)
= 5 x \(\frac{12}{8}\)
= \(\frac{60}{8}\)
= 7 \(\frac{2}{4}\)

Question 5.
A carpenter has a 6-foot long board. He wants to cut the board into pieces that are \(\frac{4}{5}\) foot long.
a) How many pieces of length \(\frac{4}{5}\) foot can the carpenter cut from the board?
Answer:
7 \(\frac{2}{4}\) pieces
Explanation:
A carpenter has a 6-foot long board.
He wants to cut the board into pieces that are \(\frac{4}{5}\) foot long.
Number of pieces he get from the length \(\frac{4}{5}\)
6 ÷ \(\frac{4}{5}\)
= \(\frac{30}{4}\)
= 7 \(\frac{2}{4}\)

b) What length of the original board will be left after the carpenter has cut all the pieces that are \(\frac{4}{5}\) foot long?
Answer:
\(\frac{2}{4}\) pieces
Explanation:
A carpenter has a 6-foot long board.
He wants to cut the board into pieces that are \(\frac{4}{5}\) foot long.
Number of pieces he get from the length \(\frac{4}{5}\)
6 ÷ \(\frac{4}{5}\)
= \(\frac{30}{4}\)
= 7 \(\frac{2}{4}\)
Carpenter will cut 7 equal pieces of \(\frac{4}{5}\)
\(\frac{2}{4}\) is remaining.

Question 6.
A candle maker has 4\(\frac{1}{2}\) pounds of clear wax. He wants to cut the wax into pieces that are \(\frac{2}{3}\) pound each.
a) How many \(\frac{2}{3}\)-pound pieces can he divide the wax into?
Answer:
6 pieces

b) How much wax is left over?
Answer:
\(\frac{3}{4}\)
Explanation:
A candle maker has 4\(\frac{1}{2}\) pounds of clear wax.
He wants to cut the wax into pieces that are \(\frac{2}{3}\) pound each.
4\(\frac{1}{2}\) ÷ \(\frac{2}{3}\)
= \(\frac{9}{2}\) ÷ \(\frac{2}{3}\)
= \(\frac{9}{3}\) x \(\frac{2}{2}\)
= \(\frac{27}{4}\)
= 6 \(\frac{3}{4}\)

Question 7.
A roll of ribbon was 9 meters long. Kevin cut 8 pieces of ribbon, each of length 0.8 meter, to tie some presents. He then cut the remaining ribbon into some pieces, each of length 0.4 meter.
a) How many pieces of ribbon, each 0.4 meter in length, did Kevin have?
Answer:
6 pieces of 0.4 length

b) What was the length of ribbon left over?
Answer:
\(\frac{2}{4}\)
Explanation:
A roll of ribbon was 9 meters long.
Kevin cut 8 pieces of ribbon, each of length 0.8 meter, to tie some presents.
8 x 0.8 = 6.4 meters are used out of 6
So, 6 – 6.4 = 2.6
He then cut the remaining ribbon into some pieces, each of length 0.4 meter.
2.6 ÷ 0.4 = 6\(\frac{2}{4}\)

Question 8.
Mike has a large tropical fish collection. He gives \(\frac{2}{3}\) of his fish to a local high school. Then he gives \(\frac{2}{5}\) of the remaining fish to an elementary school. In the end, he has 30 fish left. How many fish did Mike have at first?
Answer:
150 fishes at first
Explanation:
Let, number of fishes = x
He gives \(\frac{2}{3}\) of his fish to a local high school.
That means \(\frac{1}{3}\) of his fish left.
Then he gives \(\frac{2}{5}\) of the remaining fish to an elementary school.
\(\frac{2}{3}\) x \(\frac{1}{3}\) x X = 30
\(\frac{3}{15}\)x X = 30
\(\frac{1}{5}\) x X = 30
X = 30 x 5 = 150

Question 9.
A costume designer has 40 yards of red fabric to make costumes for a musical in which 8 performers will wear red dresses and 14 people will wear red scarves. The costume maker uses 3\(\frac{1}{2}\) yards for each dress, and \(\frac{3}{4}\) yard for a scarf.
a) After making the dresses and scarves, the costume designer uses the leftover fabric to make some sashes for the dresses. If each sash uses \(\frac{1}{4}\) yard of fabric, how many sashes can be made?
Answer:
6 sashes

b) The costume designer decides to make the sashes a little smaller, so that each of the 8 dresses can have a sash. What fraction of a yard of fabric should the costume maker use to make each sash?
Answer:
\(\frac{1}{4}\) yard of fabric
Explanation:
Total fabric used for dresses= 8×3.5= 28 yards
Total fabric used for scarfs=14×0.75= 10.5 yards
Remaining fabric=40-28-10.5= 1.5 yards
Sashes made=1.5/0.25= 6 sashes.

Question 10.
Jack read \(\frac{1}{6}\) of a book on Monday and another \(\frac{1}{3}\) of it on Tuesday. He took another 4 days to finish reading the book. He read the same number of pages on each of the 4 days.
a) What fraction of the book did he read on each of the 4 days?
Answer:
Jack read \(\frac{1}{4}\)

b) If he read 40 pages on each of these 4 days, find the number of pages in the book.
Answer:
160 pages
Explanation:
Jack read \(\frac{1}{6}\) of a book on Monday and
another \(\frac{1}{3}\) of it on Tuesday
Total pages on both days \(\frac{1}{6}\) + \(\frac{1}{3}\)
= \(\frac{3}{6}\) = \(\frac{1}{2}\)
If he read 40 pages on each of these 4 days
Total number of pages in a book
40 x 4 = 160
fraction of the book he read on each of the 4 days
\(\frac{40}{160}\)
= \(\frac{1}{4}\)

Question 11.
The school librarian has $100 to spend on some books for the school. She wants to order many copies of the same book so an entire classroom can read the book. Each copy costs $3.95, Shipping for the books will be $6.95.
a) How many copies can the librarian order?
Answer:
23 books
Explanation:
100 – 6.95 = 93.05
93.05/3.95 =23.5
23.5 is rounded to 23 books

b) Describe how you can use estimation to decide if your answer to part a) is reasonable.
Answer:
23 books at cost of $4 is reasonable.
Explanation
25 book at cost $4 = $100
estimated cost of each book is $4
23 book cost is $92
$8 is for Shipping chargers
23 books  at $4 is reasonable

Question 12.
Jason cycled for 3 hours. He cycled 7\(\frac{1}{2}\) miles each hour for the first two hours and the distance he cycled for the third hour was \(\frac{1}{4}\) of the total distance he cycled in 3 hours. What was the total distance Jason cycled in 3 hours?
Answer:
8\(\frac{2}{4}\) miles
Explanation:
Jason cycled 7\(\frac{1}{2}\) miles each hour for the first two hours and
the distance he cycled for the third hour was \(\frac{1}{4}\)
The total distance he cycled in 3 hours,
7\(\frac{1}{2}\) + \(\frac{1}{4}\)
= \(\frac{15}{4}\) + \(\frac{1}{4}\)
= \(\frac{34}{4}\)
= 8\(\frac{2}{4}\)

Question 13.
The length of a field was 20 yards. Mr. Matsumoto planted a row of peas every \(\frac{3}{4}\) yard.
a) How many rows of peas did Mr. Matsumoto plant?
Answer:
15 rows planted

b) What was the remaining length of field?
Answer:
5 yards
Explanation:
The length of a field was 20 yards.
Mr. Matsumoto planted a row of peas every \(\frac{3}{4}\) yard
20x \(\frac{3}{4}\) yard
= \(\frac{60}{4}\) = 15
Remaining length of the field 20 – 15 = 5 yards

Question 14.
A sign in an elevator says the elevator can lift up to 450 kilograms. John has 10 boxes that weigh 13.75 kilograms each, and a number of additional boxes that weigh 15.5 kilograms each. If he puts the 10 boxes on the elevator, how many of the additional boxes can be lifted in the same load?
Answer:
20 additional boxes
Explanation:
Weight of each box = 13.75 kg
Weight of 10 boxes = 13.75 x 10 = 137.5 kg
Space left in the box can be adjusted = 450 – 137.5 = 312.5 kg
Weight of additional boxes = 15.5 kg
Additional boxes can be lifted in the same load = 312.5 + 15.5 = 20.1 kg

Question 15.
Ken had a number of colored marbles in a bag. \(\frac{1}{4}\) of the marbles were red, \(\frac{2}{3}\) of the remaining marbles were blue, and the rest were yellow. Given that there were 120 red and yellow marbles altogether, how many marbles were there in the bag?

Answer:
240 marbles
Explanation
\(\frac{1}{4}\) = 0.25 of the marbles were red,
\(\frac{2}{3}\) x 0.75 = 0.5 of the remaining marbles were blue,
rest are yellow
1 – [0.25 + 0.5] = 0.25 [yellow]
0.25 + 0.25 = 0.5
RED + YELLOW = 120
total
RED + YELLOW+ BLUE = 120 + 120 = 240

Question 16.
Rachel used \(\frac{3}{8}\) of her money to buy some blouses and \(\frac{2}{5}\) of the remainder to buy 2 pairs of pants. A pair of pants costs 3 times as much as a blouse. How many blouses did she buy?
Answer:
9 blouses
Explanation:
Rachel used \(\frac{3}{8}\) of her money to buy some blouses,
If total cloth is 8 parts, out of which 3 parts are used.
So, \(\frac{5}{8}\) left.
She used \(\frac{2}{5}\) of the remainder to buy 2 pairs of pants.
\(\frac{2}{5}\) x \(\frac{5}{8}\)

= \(\frac{10}{40}\)
= \(\frac{1}{4}\) used for 2 pairs of pants \(\frac{1}{8}\) each.
A pair of pants costs 3 times as much as a blouse.
So, each blouse cost \(\frac{1}{24}\)
Number of blouses she buy are
\(\frac{3}{8}\) x \(\frac{24}{1}\)
= \(\frac{72}{8}\)
= 9

Question 17.
Sheila went shopping and spent $120 on a coat. She then used \(\frac{2}{3}\) of the remaining money to buy a dress. She was left with \(\frac{1}{5}\) of her original amount of money. How much did Sheila have at first?
Answer:
let Sheila has  $M first
Explanation:
Let M = original amount of money
M – 120 – [(2/5)(M-120)] = M/5
M – 120 – 2M/5 + 48 = M/5
M – 2M/5 – M/5 = 120 – 48
5M/5 – 2M/5 – M/5 = 72
2M/5 = 72
M = 72(5/2)
M = 36(5)
M = $180
Sheila started with $180
180 – 120 = 60
60 – (2/5)(60) = 36
180(1/5) = 36
She has $36 left and that is indeed 1/5 of the original $180.

Brain @ Work

Question 1.
Alex, Beth and Carol share a sum of money. Alex receives 0.7 of the sum of money. Beth and Carol receive the rest of the money. If Beth receives \(\frac{5}{12}\) of the money shared by both her and Carol, and Carol receives $847, how much money does Alex get?
Answer:
Alex’s share = $3,388
Explanation:
Let the total money be X ,
Alex = 0.70 or \(\frac{7}{10}\) of total amount

Share of Beth + Carol = 1 – share of Alex
= 1 – \(\frac{7}{10}\)
=\(\frac{3}{10}\)
Beth and Carol = \(\frac{3}{10}\) of total amount
Beth = \(\frac{5}{12}\) of the \(\frac{3}{10}\) of total amount
Carol = \(\frac{7}{12}\) of the \(\frac{3}{10}\) of total amount

Since Carol receives $847,
Share of Beth + Carol = share of Carol ÷ \(\frac{7}{12}\)
= 847 ÷ \(\frac{7}{12}\)
= 847 x \(\frac{12}{7}\)
= $1425

Total amount = Alex +Beth + Carol
we know,
Beth + Carol = \(\frac{3}{10}\)  x total amount = $1425
total amount = 1452 ÷ \(\frac{3}{10}\)
= $1,452 x \(\frac{10}{3}\)
= 4,840

Total amount = Alex + Beth + Carol
4840 = Alex + 1452
Alex = 4840  – 1452
Alex = 3388
Alex’s share = $3,388 

Go through the Math in Focus Grade 6 Workbook Answer Key Chapter 2 Review Test to finish your assignments.

Math in Focus Grade 6 Course 1 A Chapter 2 Review Test Answer Key

Concepts and Skills

Write the opposite of each number.

Question 1.
-47
Answer:
The opposite number of -47 is 47.

Question 2.
56
Answer:
The opposite number of 56 is -56.

Question 3.
-78
Answer:
The opposite number of -78 is 78.

Draw a horizontal number line to represent each set of numbers.

Question 4.
-41, -37, -34, -30, -28, -25, -22
Answer:

The given numbers -41, -37, -34, -30, -28, -25, -22 are represented in the above horizontal number line. On a horizontal number line, the numbers become greater as you move to the right, and less as you move to the left.

Question 5.
-133, -129, -126, -122, -119
Answer:

The given numbers -133, -129, -126, -122, -119 are represented in the above horizontal number line. On a horizontal number line, the numbers become greater as you move to the right, and less as you move to the left.

Draw a vertical number line to represent each set of numbers.

Question 6.
-8, -6, -2, 1, 3, 4
Answer:

The given numbers -8, -6, -2, 1, 3, 4 are represented in the above vertical number line. On a vertical number line, the numbers become greater as you move up, and lesser as you move down.

Question 7.
Odd numbers greater than -40 but less than -28.
Answer:
Odd numbers greater than -40 but less than -28 are -39, -37, -35, -33, -31, -29.

The odd numbers greater than -40 but less than -28 are -39, -37, -35, -33, -31, -29 are represented in the above vertical number line. On a vertical number line, the numbers become greater as you move up, and lesser as you move down.

Write a positive or negative number to represent each situation.

Question 8.
A deposit of $94
Answer:
$94
Explanation:
The given situation is a deposit of $94. The situation is represented with positive number $94.

Question 9.
181°F below zero.
Answer:
-181°F
Explanation:
The given situation is 181°F below zero. Below represents negative number and above represents positive number. So the situation is represented with negative number -181°F.

Question 10.
The plane’s altitude is 23,920 feet
Answer:
23,920 feet
Explanation:
The given situation is the plane’s altitude is 23,920 feet. The situation is represented with positive number 23,920 feet.

Question 11.
The elevation of a sunken ship that is 11 meters beneath the ocean’s surface
Answer:
-11 meters
Explanation:
The given situation is the elevation of a sunken ship that is 11 meters beneath the ocean’s surface. Here the word beneath represents negative sign. So the situation is represented with negative number -11 meters.

Question 12.
A gain of 35 yards
Answer:
35 yards
Explanation:
The given situation is a gain of 35 yards. The word gain represents a positive sign. The situation is represented with positive number 35 yards.

Copy and complete each Inequality using > or <.

Question 13.
-14  -18
Answer:
-14 -18
Explanation:
The given numbers are -14 and -18. Here we have to compare both the numbers. By comparing the above given numbers the number -14 is greater than -18.

Question 14.
17  -11
Answer:
17 -11
Explanation:
The given numbers are 17 and -11. Here we have to compare both the numbers. By comparing the above given numbers the number 17 is greater than -11.

Question 15.
-34  23
Answer:
-34  23
Explanation:
The given numbers are -34 and 23. Here we have to compare both the numbers. By comparing the above given numbers the number -34 is less than 23.

Question 16.
-157  -145
Answer:
-157 -145
Explanation:
The given numbers are -157 and -145. Here we have to compare both the numbers. By comparing the above given numbers the number -157 is less than -145.

Order the numbers in each set.

Question 17.
Order the numbers from greatest to least:
15, -14, 7, 2, -5, -6, -9
Answer:
The number from greatest to least: 15, 7, 2, -5, -6, -9, -14.

Question 18.
Order the numbers from least to greatest: 112, -140, -50, 51, -122, 175, -182
Answer:
The numbers from least to greatest: -182, -140, -122, -50, 51, 112, 175.

Write an Inequality for each of the following statements using > or <.

Question 19.
-112°C is warmer than -143°C.
Answer:
-112°C > -143°C
Explanation:
The given situation is -112°C is warmer than -143°C. Here we have to compare both the temperatures. After comparing the temperatures -112°C is greater than -143°C.

Question 20.
The lowest recorded temperature yesterday was -4°C, which is colder than today’s lowest recorded temperature of 4°C.
Answer:
-4°C < 4°C
Explanation:
After comparing the above given temperatures and situation -4°C is less than 4°C.

Write the absolute value of each number.

Question 21.
|79|
Answer:
The absolute value of the number |79| is 79.

Question 22.
|-88|
Answer:
The absolute value of the number |-88| is 88.

Question 23.
|-102|
Answer:
The absolute value of the number |-102| is 102.

Copy and complete each inequality using > or <.

Question 24.
|-65|  |-57|
Answer:
|-65| |-57|
Explanation:
The absolute value of the number |-65| is 65.
The absolute value of the number |-57| is 57.
So, the number |-65| is greater than |-57|.

Question 25.
|111|  |-124|
Answer:
|111|  |-124|
Explanation:
The absolute value of the number |111| is 111.
The absolute value of the number |-124| is 124.
So, the number |111| is less than |-124|.

Question 26.
|-153|  |135|
Answer:
|-153|  |135|
Explanation:
The absolute value of the number |-153| is 153.
The absolute value of the number |135| is 135.
So, the number |-153| is greater than |135|.

Question 27.
|-209|  |-278|
Answer:
|-209|  |-278|
Explanation:
The absolute value of the number |-209| is 209.
The absolute value of the number |-278| is 278.
So, the number |-209| is less than |-278|.

Problem Solving

Answer the questions.

Question 28.
The Afar Depression is a land formation in Africa. At one location in the Afar Depression, the elevation is -75 meters. At another location, the elevation is -125 meters. Write an inequality to compare the elevations. Which elevation is farther from sea level?
Answer:
-75 meters > -125 meters
The elevation -125 meters is farther from sea level.
Explanation:
At one location in the Afar Depression, the elevation is -75 meters.
At another location, the elevation is -125 meters.
After comparing the above elevations -75 meters is greater than -125 meters.

Question 29.
The table shows temperature readings taken at the same location at three different times.

a) At what time was the location the coldest?
Answer:
At 12:30 A.M. the location is coldest.

b) Between 12:30 A.M. and 8:30 A.M., the temperature was always rising. Between what two times shown in the table did the temperature reach 0°C?
Answer:
Between 4:30 A.M. and 8:30 A.M. the temperature reaches to 0°C as we can observe in the above table.

Question 30.
Clarence owes his brother Joe $240, and his best friend Tristan $166. His sister Chloe owes Clarence $275, and his friend Luke owes Clarence $150.
a) Clarence writes the number -240 to represent the amount he owes his brother Joe. What numbers should Clarence use to represent the other amounts given above?
Answer:
-166 to represent the amount Clarence owes to his best friend Tristan.
275 to represent the amount Chloe  owes to Clarence.
150 to represent the amount Luke owes to Clarence.

b) Who owes the most money?
Answer:
Clarence owes the most money($240 + $166 = $406).

c) How much does Clarence owe in total?
Answer:
Clarence owes $406 in total.

d) Which is greater, the amount of money Clarence owes, or the amount of money that people owe him?
Answer:
The amount of money that people owe to Clarence is greater.
Explanation:
The total amount that people owes to Clarence is $275 + $150 = $475
The total amount that Clarence owes to others is $240 + $166 = $406

Go through the Math in Focus Grade 6 Workbook Answer Key Chapter 2 Lesson 2.2 Absolute Value to finish your assignments.

Math in Focus Grade 6 Course 1 A Chapter 2 Lesson 2.2 Answer Key Absolute Value

Math in Focus Grade 6 Chapter 2 Lesson 2.2 Guided Practice Answer Key

Use the number line to find the absolute value of each of the following numbers.

Question 1.
|-10|
Answer:

The absolute value of the number |-10| is 10.
-10 is 10 units from 0.

Question 2.
|3|
Answer:

The absolute value of the number |3| is 3.
3 is 3 units from 0.

Question 3.
|-8|
Answer:

The absolute value of the number |-8| is 8.
-8 is 8 units from 0.

Question 4.
|1|
Answer:

The absolute value of the number |1| is 1.
1 is 1 units from 0.

Question 5.
|-7|
Answer:

The absolute value of the number |-7| is 7.
-7 is 7 units from 0.

Question 6.
|0|
Answer:

The absolute value of the number |0| is 0.

Write the absolute value of each number.

Question 7.
|-23|
Answer:
The absolute value of the number |-23| is 23.

Question 8.
|41|
Answer:
The absolute value of the number |41| is 41.

Question 9.
|-38|
Answer:
The absolute value of the number |-38| is 38.

Question 10.
|114|
Answer:
The absolute value of the number |114| is 114.

Question 11.
|-132|
Answer:
The absolute value of the number |-132| is 132.

Question 12.
|506|
Answer:
The absolute value of the number |506| is 506.

Use absolute values to interpret real-world situations.

a) The figure shows a section from Keith’s bank account statement.

As of May 31, Keith had $280 ¡n his bank account. On June 24, after he withdrew $490, he had -$170 in his bank account.
|-170| = 170
This means that Keith had overdrawn $170.

b) A dog is standing on a cliff, 35 feet above sea level. A dolphin is swimming 6 feet below sea level. An octopus is moving along the seabed, 40 feet below sea level.
You can use positive and negative numbers to show the elevation of the animals relative to sea level.

Dog’s elevation = 35 ft
Dolphin’s elevation = -6 ft
Octopus’s elevation = -40 ft
To decide which animal ¡s farthest from sea level, you do not need to think about whether the animals are above or below sea level. You can use absolute values to compare distances.
Distance of dog from sea level = |35|
= 35 ft
Distance of dolphin from sea level = |-6|
= 6 ft
Distance of octopus from sea level = |-40|
= 40 ft
The octopus is farthest from sea level.

Answer the questions.

Question 13.
Joe owes his sister Lisa $12, and his sister Kelly $18. His brother David owes Joe $20.

a) Joe writes the number -12 to represent the amount he owes Lisa. What numbers should Joe use to represent the other amounts given above?
Answer:
-18 to represent the amount he owes Kelly.
20 to represent the amount his brother David owes Joe.

b) Which person owes the most money?
Answer:
Joe owes the most money($12 + $18 = $30)

Question 14.
At a parking garage, you can park underground or above ground. The lowest part of the underground parking is 40 feet below ground level. The highest part of the parking garage is 20 feet above ground level. The limousine parking area is 23 feet below ground level.
a) Use positive and negative numbers to represent the locations, with respect to ground level, of the three different parts of the parking
garage.
Answer:
The lowest part of the underground parking is 40 feet below ground level which represents -40 feet.
The highest part of the parking garage is 20 feet above ground level which represents 20 feet.
The limousine parking area is 23 feet below ground level which represents -23 feet.

b) Which part of the parking garage is closest to ground level?
Answer:
The highest part of the parking garage is 20 feet above ground level which is closest among all levels of parking.

Math in Focus Course 1A Practice 2.2 Answer Key

Use the number line to find the absolute value of each of the following numbers.

Question 1.
|-11|
Answer:

The absolute value of the number |-11| is 11.
-11 is 11 units from 0.

Question 2.
|4|
Answer:

The absolute value of the number |4| is 4.
4 is 4 units from 0.

Question 3.
|-6|
Answer:

The absolute value of the number |-6| is 8.
-6 is 6 units from 0.

Write the absolute value of each number.

Question 4.
|35|
Answer:
The absolute value of the number |35| is 35.

Question 5.
|-46|
Answer:
The absolute value of the number |-46| is 46.

Question 6.
|-77|
Answer:
The absolute value of the number |-77| is 77.

Copy and complete each inequality using > or <.

Question 7.
|-26| |30|
Answer:
|-26|  |30|
Explanation:
The absolute value of the number |-26| is 26.
The absolute value of the number |30| is 30.
So, the number |-26| is less than |30|.

Question 8.
|-92| |-114|
Answer:
|-92|  |-114|
Explanation:
The absolute value of the number |-92| is 92.
The absolute value of the number |-114| is 114.
So, the number |-92| is less than |-114|.

Question 9.
|511| |-500|
Answer:
|511|  |-500|
Explanation:
The absolute value of the number |511| is 511.
The absolute value of the number |-500| is 500.
So, the number |511| is greater than |-500|.

Question 10.
|-707| |-628|
Answer:
|-707|  |-628|
Explanation:
The absolute value of the number |-707| is 707.
The absolute value of the number |-628| is 628.
So, the number |-707| is greater than |-628|.

Answer the questions.

Question 11.
Two numbers have an absolute value of 16. Which of the two numbers is greater than 12?
Answer:
The absolute value of -16 is 16 and the absolute value of 16 is 16. So we know that 16 is greater than 12 but -16 is not.

Question 12.
Math Journal Jesse graphed a point to represent the absolute value of a number on a number line. If the original number is less than -10, describe all the possible values for the point Jesse graphed on the number line. Explain
your thinking.
Answer:
The possible values for the point Jesse graphed on the number line will be the values which are less than or equal to -11
Explanation:
Jessie pointed a number less than -10 on the number line. So the possible values will start from -11 and will decrease further.

Question 13.
The table shows a monthly bank account statement for the period March to July.

a) For which months is the account overdrawn?
Answer:
The account overdrawn in March, April and July months.

b) How much was the bank owed in March?
Answer:
The bank owed $450 in March.

c) In which month was the account overdrawn by the greatest amount?
Answer:
In March month the account overdrawn by the greatest amount with $450.

d) In which month was the account overdrawn by the least amount?
Answer:
In April month the account overdrawn by the least amount with $180.

e) How much was the bank owed ¡n total?
Answer:
$450 + $180 + $240 = $870
In total the bank owed is $870.

Question 14.
The table shows some locations with their elevations.

a) Which location is the closest to sea level?
Answer:
The location closest to sea level is Laguna Salada.

b) Which locations are within 200 feet of sea level?
Answer:
The locations within 200 feet of sea level are Salton City and Laguna Salada.

c) How much farther from sea level is Desert Shores than Salton City?
Answer:
-75.4 farther from sea level is Desert Shores than Salton City.

d) Write the locations ¡r, order from the location that is farthest from sea level to the location that is closest to sea level.
Answer:
The location that is farthest from sea level to the location that is closest to sea level are Bombay Beach, Desert Shores, Salton City, Laguna Salada.

Question 15.
The table shows the average surface temperature of some planets.

a) Which planet has the highest average surface temperature?
Answer:
Earth has the highest average surface temperature.

b) Which planet has the lowest average surface temperature?
Answer:
Uranus has the lowest average surface temperature.

c) On Earth, the boiling temperature of water at sea level is 100°C. Which planet has an average surface temperature that is closest to this temperature?
Answer:
Earth has an average surface temperature that is closet to the temperature 100°C.

d) Order the temperatures from lowest to highest.
Answer:
The temperature from lowest to highest are -218, -108, -53, 14.

Brain @ Work

Question 1.
You can interpret a negative sign in front of a number as meaning “the opposite of.” So, -3 means the opposite of 3.

a) What number is -(-3) the opposite of?
Answer:
The opposite number of -(-3) is -3.

b) What number is -(-3) equal to?
Answer:
The number -(-3) is equal to 3.

Question 2.
On a certain day, the maximum recorded temperature was 15°C and the minimum recorded temperature was -8°C. How many degrees Celsius was the difference between the recorded maximum and recorded minimum temperatures?
Answer:
The maximum recorded temperature was 15°C.
The minimum recorded temperature was -8°C.
The difference between the recorded maximum temperature and recorded minimum temperature is 15°C + 8°C = 23°C
Explanation:
On a number line 15°C is 15 units after 0.
-8°C on a number line 8 units before 0.
So, the total difference is 15°C + 8°C = 23°C