Math in Focus

Practice the problems of Math in Focus Grade 6 Workbook Answer Key Chapter 9 Review Test to score better marks in the exam.

Math in Focus Grade 6 Course 1 B Chapter 9 Review Test Answer Key

Concepts and Skills

Use the coordinate plane below.

Question 1.
Give the coordinates of points A, B, C, D, and E.

Answer:
Point A:
The first coordinate represents distance along negative x axis which is 4 and the second coordinate represents distance along negative y axis which is 3.
So, the point A will be (-4,-3)
Point B:
The first coordinate represents distance along x axis which is 0 and the second coordinate represents distance along negative y axis which is 6.
So, the point B will be (0,-6)
Point C:
The first coordinate represents distance along x axis which is 2 and the second coordinate represents distance along negative y axis which is 4.
So, the point C will be (2,-4)
Point D:
The first coordinate represents distance along x axis which is 6 and the second coordinate represents distance along y axis which is 3.
So, the point D will be (6,3)
Point E:
The first coordinate represents distance along the negative x axis which is 2 and the second coordinate represents distance along y axis which is 3.
So, the point E will be (-2,3)

Use graph paper. Plot the points on a coordinate plane. In which quadrant is each point located?

Question 2.
A (3, 5), B (-2, 0), C (7, -2), D (0, -5), and E (-3, -8)
Answer:
Point A (3, 5): Since both the x and y coordinate are positive, the point A will be located in quadrant I.
Point B (-2, 0): Since the x coordinate is negative and y coordinate is origin, the point B will not lie in any quadrant. It will be located on x axis.
Point C (7, -2): Since the x coordinate is positive and y coordinate is negative, the point C will lie in quadrant IV.
Point D (0, -5): Since the x coordinate is zero and y coordinate is negative, the point D will not lie in any quadrant. It will be located on y axis.
Point E (-3, -8): Since both the x and y coordinate are negative, the point A will be located in quadrant III.

Use graph paper. Points A and B are reflections of each other about the x-axis. Give the coordinates of point B if the coordinates of point A are the following:

Question 3.
(3, 6)
Answer:
Given that the points A and B are reflections of each other about the x-axis.
Given that B is reflection of point A in the x-axis and point A is (3, 6).
If a point is reflected across the x-axis, The x-coordinate remains the same, but the y-coordinate is taken to be the additive inverse.
Thus, x coordinate will be 3 and y coordinate will be -6.
Therefore, point B will be (3,-6).

Question 4.
(-6, 2)
Answer:
Given that B is reflection of point A in the x-axis and point A is (-6, 2).
If a point is reflected across the x-axis, The x-coordinate remains the same, but the y-coordinate is taken to be the additive inverse.
Thus, x coordinate will be 3 and y coordinate will be -2.
Therefore, point B will be (-6,-2).

Question 5.
(5, -4)
Answer:
Given that B is reflection of point A in the x-axis and point A is (5, -4).
If a point is reflected across the x-axis, The x-coordinate remains the same, but the y-coordinate is taken to be the additive inverse.
Thus, x coordinate will be 5 and y coordinate will be 4.
Therefore, point B will be (5,4).

Question 6.
(-3, -5)
Answer:
Given that B is reflection of point A in the x-axis and point A is (-3, -5).
If a point is reflected across the x-axis, The x-coordinate remains the same, but the y-coordinate is taken to be the additive inverse.
Thus, x coordinate will be -3 and y coordinate will be 5.
Therefore, point B will be (-3,5).

Use graph paper. Points C and D are reflections of each other about the y-axis. Give the coordinates of point D if the coordinates of point C are the following:

Question 7.
(3, 6)
Answer:
Given that the points C and D are reflections of each other about the y-axis.
Given that C is reflection of point D in the y-axis and point C is (3, 6).
If a point is reflected across the y-axis, The x-coordinate is taken to be the additive inverse and the y-coordinate remains the same.
Thus, x cordinate will be -3 and y coordinate will be 6.
Therefore, point D will be (-3,6).

Question 8.
(-6, 2)
Answer:
Given that B is reflection of point A in the y-axis and point C is (-6, 2).
If a point is reflected across the y-axis, The x-coordinate is taken to be the additive inverse and the y-coordinate remains the same.
Thus, x cordinate will be 6 and y coordinate will be 2.
Therefore, point D will be (6,2).

Question 9.
(5, -4)
Answer:
Given that B is reflection of point A in the y-axis and point C is (5, -4).
If a point is reflected across the y-axis, The x-coordinate is taken to be the additive inverse and the y-coordinate remains the same.
Thus, x cordinate will be -5 and y coordinate will be -4.
Therefore, point D will be (-5,-4).

Question 10.
(-3, -5)
Answer:
Given that B is reflection of point A in the y-axis and point C is (-3, -5).
If a point is reflected across the y-axis, The x-coordinate is taken to be the additive inverse and the y-coordinate remains the same.
Thus, x cordinate will be 3 and y coordinate will be -5.
Therefore, point D will be (3,-5).

Use graph paper. For each exercise, plot the given points on a coordinate plane. Then connect the points in order with line segments to form a closed figure. Name each figure formed.

Question 11.
A (2, -4), B (2, 4), C (-6, 4), and D (-6, -4)
Answer:
Plot the points A,B,C and D on a coordinate plane. Connect the points in order with line segments to form a closed figure.
The formed figure has all equal sides. Therefore, it can be a square.

Question 12.
E (0, -2), F (-3, 1), G(-5, -1), and H (-2, -4)
Answer:
Plot the points E,F,G and H on a coordinate plane. Connect the points in order with line segments to form a closed figure.
The formed figure has all four sides different. Therefore, it can be a quadrilateral.

Question 13.
J (0, 1), K(1, 4), and L(-4, 3)
Answer:
Plot the points J,K and L on a coordinate plane. Connect the points in order with line segments to form a closed figure.
The formed figure has all three sides different. Therefore, it can be a triangle.

Question 14.
M (6, 5), N (3, 5), P (3, -3), and Q (6, -3)
Answer:
Plot the points M,N,P and Q on a coordinate plane. Connect the points in order with line segments to form a closed figure.
The formed figure has opposite sides equal. Therefore, it can be a rectangle.

Question 15.
A (6, -3), B (4, 2), C (-1, 2), and D (0, -3)
Answer:
Plot the points A,B,C and D on a coordinate plane. Connect the points in order with line segments to form a closed figure.
The formed figure has a pair of opposite sides parallel and equal. Therefore, it can be a parallelogram.

Question 16.
E (-1, 6), F (-3, 3), G (3, 3), and H (5, 6)
Answer:
Plot the points E,F,G and H on a coordinate plane. Connect the points in order with line segments to form a closed figure.
The formed figure has all different sides. Therefore, it can be a quadrilateral.

Question 17.
J(6, 1), K(8, -2), L (2, -2), and M (0, 1)
Answer:
Plot the points J,K,L and M on a coordinate plane. Connect the points in order with line segments to form a closed figure.
The formed figure has a pair of opposite sides parallel and equal. Therefore, it can be a parallelogram.

Question 18.
P (2, 7), Q (-1, 7), R (-5, 4), and S (4, 4)
Answer:
Plot the points P,Q,R and S on a coordinate plane. Connect the points in order with line segments to form a closed figure.
The formed figure has a pair of opposite sides parallel and equal. Therefore, it can be a trapezium.

Question 19.
T (-2, 1), U (-6, 1), V(-6, -3), and W(-2, -3)
Answer:
Plot the points T,U,V and W on a coordinate plane. Connect the points in order with line segments to form a closed figure.
The formed figure has opposite sides equal. Therefore, it can be a rectangle.

Use graph paper. Plot the points on a coordinate plane and answer the question.

Question 20.
a) Plot points A (1, -1) and B (7, -1) on a coordinate plane. Connect the two points to form a line segment.
Answer:
Point A:
The first coordinate represents distance along positive x axis which is 1 and the second coordinate represents distance along negative y axis which is 1.
So, the point A will be (1,-1)
Point B:
The first coordinate represents distance along positive x axis which is 7 and the second coordinate represents distance along negative y axis which is -1.
So, the point A will be (7,-1)

b) Point C lies above \(\overline{\mathrm{AB}}\), and is 2 units away from the x-axis. If triangle ABC is an isosceles triangle with base \(\overline{\mathrm{AB}}\), find the coordinates of point C.
Answer:
Point C lies above \(\overline{\mathrm{AB}}\), and is 2 units away from the x-axis. If triangle ABC is an isosceles triangle with base \(\overline{\mathrm{AB}}\)
Since it is an isosceles triangle, two sides will be equal. When the line segment is joined through midpoint, the formed line segments will be equal in length.
Midpoint of AB line segment is 3 units, so midpoint will be (4,2).
Thus, coordinates of point C will be (4,2)

c) Points D and E lie below \(\overline{\mathrm{AB}}\) such that ABDE is a rectangle. If BD is 5 units, find the coordinates of points D and E.
Answer:
Given that BD is 5 units.
Points D and E lie below \(\overline{\mathrm{AB}}\) such that ABDE is a rectangle.
Start from point B and move 5 units above and mark it. Thus, the required point D will be formed.
Since it is a rectangle, opposite sides will be equal. Therefore, AB will be equal to DE, which is 6 units.
Start from point D and move 6 units towrads left and mark it is as E.
Thus, the formed D point will be (7,4) and E will be (1,4).

Use graph paper. Plot each pair of points on a coordinate plane. Connect the points to form a line segment and find its length.

Question 21.
A (-1,0) and B (8, 0)
Answer:
To plot point A (-1,0):
Here, the x-coordinate of A is -1 and the y-coordinate is 0. Start at the Origin. As the x coordinate is negative, move 1 units on the negative x-axis and point it.
Thus, the required point (-1,0) is marked.
To plot point B (8, 0):
Here, the x-coordinate of A is 8 and the y-coordinate is 0. Start at the Origin. As the x coordinate is positive, move 8 units along the positive x-axis and point it.
Thus, the required point B (8, 0) is marked.

After connecting the points to form a line segment, the length will be 9 units.

Question 22.
C (-2, 4) and D (6, 4)
Answer:
To plot point C (-2, 4):
Here, the x-coordinate is -2 and the y-coordinate is 4. Start at the Origin. As the x coordinate is negative, move 2 units on the negative x-axis and 4 units on the positive y -axis.
Thus, the required point C (-2,4) is marked.
To plot point D (6, 4):
Here, the x-coordinate is 6 and the y-coordinate is 4. Start at the Origin. As the x coordinate is positive, move 4 units along the positive x-axis and 4 units along positive y -axis.
Thus, the required point D (6, 4) is marked.

After connecting the points to form a line segment, the length will be 8 units.

Question 23.
E (-6, -2) and F(-6, -6)
Answer:
To plot point E (-6, -2):
Here, the x-coordinate is 6 and the y-coordinate is-2. Start at the Origin. As the x coordinate is negative, move 6 units on the negative x-axis and 2 units on the negative y -axis.
Thus, the required point C (-2,4) is marked.
To plot point F(-6, -6):
Here, the x-coordinate is -6 and the y-coordinate is -6. Start at the Origin. As the x coordinate is negative, move 6 units along the negative x-axis and 6 units along negative y -axis.
Thus, the required point D (-6, -6) is marked.

After connecting the points to form a line segment, the length will be 4 units.

Question 24.
G (-5, -4) and H (2, -4)
Answer:
To plot point G (-5, -4):
Here, the x-coordinate is -5 and the y-coordinate is-4. Start at the Origin. As the x coordinate is negative, move 5 units on the negative x-axis and 4 units on the negative y -axis.
Thus, the required point G (-5, -4) is marked.
To plot point H (2, -4):
Here, the x-coordinate is 2 and the y-coordinate is -4. Start at the Origin. As the x coordinate is positive, move 4 units along the positive x-axis and 4 units along negative y -axis.
Thus, the required point H (2, -4) is marked.

After connecting the points to form a line segment, the length will be 7 units.

Question 25.
J(0, -3) and K(0, -8)
Answer:
To plot point J(0, -3):
Here, the x-coordinate is 0 and the y-coordinate is-3. Start at the Origin. As the x coordinate is origin, move 3 units on the negative y -axis.
Thus, the required point J(0, -3) is marked.
To plot point K(0, -8):
Here, the x-coordinate is 0 and the y-coordinate is-8. Start at the Origin. As the x coordinate is origin, move 8 units on the negative y -axis.
Thus, the required point K(0, -8) is marked.

After connecting the points to form a line segment, the length will be 5 units.

Question 26.
M (5, 2) and N (5, -5)
Answer:
To plot point M (5, 2):
Here, the x-coordinate is 5 and the y-coordinate is 2. Start at the Origin. As the x and y coordinates are positive , move 5 units on the positive x-axis and 2 units on the positive y -axis.
Thus, the required point M (5, 2) is marked.
To plot point N (5, -5):
Here, the x-coordinate is 5 and the y-coordinate is 5. Start at the Origin. Move 5 units on the positive x-axis and 2 units on the negative y -axis.
Thus, the required point N (5, -5) is marked.

After connecting the points to form a line segment, the length will be 7 units.

Problem Solving

The diagram shows the plan of a room. The side length of each grid square is 10 feet. Use the diagram to answer questions 27 to 28.

Question 27.
The eight corners of the room are labeled points A to H. Give the coordinates of each of these corners.
Answer:
The first coordinate represents distance along negative x axis which is 40 and the second coordinate represents distance along y axis which is 100.
The coordinates of A will be (-40,100)
Similarly,The coordinates of B will be (-40,20)
The coordinates of C will be (-60,20)
The coordinates of D will be (-60,-40)
The coordinates of E will be (60,40)
The coordinates of F will be (60,20)
The coordinates of G will be (40,20)
The coordinates of H will be (40,100)

Question 28.
The entrance of the room is situated along \(\overline{\mathrm{AH}}\). What is the shortest possible distance in feet between the entrance and \(\overline{\mathrm{DE}}\) of the room?
Answer:
The entrance of the room is situated along \(\overline{\mathrm{AH}}\).
The shortest distance between the the entrance and \(\overline{\mathrm{DE}}\) of the room will be the straight route from AH to DE.
The distance between AH to DE is 14 square grids.
Each square grid equals to 10 feet.
Thus, the distance between AH to DE will be 14×10 = 140 ft.

Question 29.
Diana walks across the room from point B to point G, and then walks from point G to point H. Find the total distance, in feet, that Diana walks.
Answer:
Given that Diana walks across the room from point B to point G, and then walks from point G to point H.
Distance between B and G will be 8 square grids, 8×10=80 ft.
Distance between G to H will be 8 square grids, 8×10=80 ft.
The total distance will be BG+GH = 80+80 = 160 ft

Question 30.
Calculate the floor area of the room in square feet.
Answer:
As the complete floor area is in irregular shape, we will find the area in parts.
To calculate the floor area, we will find the area of rectangle AHGB and area of CFED
Area of rectangle AHGB:
Length of AB will be 8 square grids, 8×10=80 ft.
Length of AH will be 8 square grids, 8×10=80 ft.
Area of the rectangle = length×width
= 80×80
= 6400 sq.ft
Area of rectangle CFED:
Length of CD will be 6 square grids, 6×10=60 ft.
Length of DE will be 12 square grids, 12×10=120 ft
Area of the rectangle = length×width
= 60×120
= 7200 sq.ft
Thus, the total area of floor will be 6400+7200 = 13600 sq.ft

Use graph paper. Solve.

Question 31.
An athlete took part in a race. The distance the athlete ran, v meters, after t minutes, is given by v = 300t. Graph the relationship between t and v. Use 2 units on the horizontal axis to represent 1 minute and 1 unit on the vertical axis to represent 150 meters.

a) What type of graph is it?
Answer:
It is a straight line graph.This is also called a linear graph.

b) What is the distance the athlete ran in 3.5 minutes?
Answer:
An athlete took part in a race. The distance the athlete ran, v meters, after t minutes, is given by v = 300t.
Here, t=3.5 min
v = 300t
v = 300×3.5
= 1050 m
Thus the distance the athlete ran in 3.5 min will be 1050 m.

c) What is the average speed of the athlete?
Answer:
The average speed of the athlete will be,
Average speed = \(\frac{total distance}{total time}\)
Total distance will be 0+300+600+900+1200 = 3000 m
Total time will be 0+1+2+3+4 = 10 min
= \(\frac{3000}{10}\)
= 300 meters per min

d) Assuming the athlete runs at a constant speed, what is the distance she will run in 8 minutes?
Answer:
The distance the athlete ran, v meters, after t minutes, is given by v = 300t.
The distance she will run in 8 min will be, 300×8=2400 m

e) Name the dependent and independent variables.
Answer:
From the graph, we can say v is the dependent variable and t is the independent variable.

Question 32.
A truck uses 1 gallon of diesel for every 12 miles traveled, The amount of diesel left in the gas tank, r gallons, after traveling s miles is given by s = 300 – 12r. Copy and complete the table. Graph the relationship between r and s. Use 1 unit on the horizontal axis to represent 1 gallon and 1 unit on the vertical axis to represent 12 miles. Start your horizontal axis at 17 gallons.
a)

Answer:
A truck uses 1 gallon of diesel for every 12 miles traveled, The amount of diesel left in the gas tank, r gallons, after traveling s miles is given by s = 300 – 12r.
Here s=24,
24 = 300 – 12r
24+12r = 300-12r+12r
24+12r = 300
24+12r-24 = 300-24
12r = 276
r = 276÷12 = 23 gallons
Here r=17,
s = 300 – 12×17
= 300-204
= 96 miles

b) How many gallons of diesel are left after the truck has traveled 60 miles?
Answer:
If 60 miles is travelled,
s = 300 – 12r
60 = 300 – 12r
60 + 12r = 300 – 12r + 12r
60 + 12r = 300
60 + 12r – 60 = 300-60
12r = 240
12r÷12 = 240÷12
r = 20
Thus, 20 gallons of diesel are left after the truck has traveled 60 miles

c) After the truck has traveled for 72 miles, how much farther can the truck travel before it runs out of diesel?
Answer:
After 72 miles, 19 gallons of gas were left.
Distance = 19 × mileage
72 =  19×mileage
72÷19 = (19×mileage)÷19
3.7 = mileage
The car can travel another 3.7 miles

d) If the truck travels more than 48 miles, how much diesel is left in the gas tank? Express your answer in the form of an inequality in terms of r, where r stands for the amount of diesel left.
Answer:
If the truck travels more than 48 miles, r ≥ 21, where r stands for the amount of diesel left.

Practice the problems of Math in Focus Grade 6 Workbook Answer Key Chapter 10 Area of Polygons to score better marks in the exam.

Math in Focus Grade 6 Course 1 B Chapter 10 Answer Key Area of Polygons

Math in Focus Grade 6 Chapter 10 Quick Check Answer Key

Solve.

Question 1.
The length of a rectangle is 15 meters and its width is 8 meters. Find the area of the rectangle.
Answer:
Given that the length of a rectangle is 15 meters and its width is 8 meters.
Formula to find the area of rectangle will be,
= length × width
Here length = 15m and width = 8m
= 15×8
= 120 sq.m

Solve.

Question 2.
A side length of a square is 10 centimeters. Find the area of the square.
Answer:
Given that a side length of a square is 10 cm.
Formula to find the area of a square
= length×length
= 10×10
= 100sq.cm
Thus, the area of square will be 100sq.cm

Name each figure and identify the pairs of parallel lines.

Question 3.

Answer:
Given figure RSTU is similar to rhombus.
In rhombus opposite sides are parallel to each other.
Here, line segment RU is parallel to ST and RS is parallel to UT.
Therefore, RS and UT, RU and ST forms the pairs of parallel lines.

Question 4.

Answer:
Given figure EFGH is similar to parallelogram.
In parallelogram opposite sides are parallel to each other.
Here, line segment EH is parallel to FG and EF is parallel to HG.
Therefore, EH and FG, EF and HG forms the pairs of parallel lines.

Question 5.

Answer:
Given figure has two opposite sides of different lengths which is similar to the properties of trapezium.
Therefore, we can assume the given figure to be  a trapezium.
In trapezium, a pair of sides will be parallel to each other.
Here, in the given figure JM is parallel to KL, thus forming a pair of parallel lines.

Practice the problems of Math in Focus Grade 6 Workbook Answer Key Chapter 10 Lesson 10.1 Area of Triangles to score better marks in the exam.

Math in Focus Grade 6 Course 1 B Chapter 10 Lesson 10.1 Answer Key Area of Triangles

Math in Focus Grade 6 Chapter 10 Lesson 10.1 Guided Practice Answer Key

Hands-On Activity

Materials:

• scissors
• graph paper

PROVE THE FORMULA FOR FINDING THE AREA OF A TRIANGLE

Work in pairs.

Triangle PQR is an acute triangle. \(\overline{\mathrm{Q} R}\) is the base and PX is the height.

Step 1.
Draw triangle POR on a piece of graph paper as shown. Then draw and label rectangle AQRD,
Example

Step 2.
Cut up triangle PQR into smaller triangles. Rearrange the triangles to form rectangle EQRF, as shown below.
Example

The orange, blue, and yellow figures form a rectangle.

Step 3.
Find the area of rectangle EQRF. How does its area compare to the area of rectangle AQRD?
Answer:
Area of rectangle EQRF:
Figure EQRF is occupying 8 units wide and 2 units high.
Area of rectangle = length × width
= 8×2
= 16 sq.units.
Area of rectangle AQRD:
Figure AQRD is occupying 8 units wide and 4 units high.
Area of rectangle = length × width
= 8×4
= 32 sq.units.
By observing the area of EQRF and AQRD, we can say that the area of ANPD is twice the area of AQRD.

How does the area of triangle PQR compare to the area of rectangle EQRF?
Answer:
Area of triangle MNP:
Let the base of the triangle be PQR , which is occupying 8 horizontal squares. Thus, the base will be 8 units.
The perpendicular segment between the base and the opposite vertex will be the height of the triangle which is 4 units.
Area of triangle = \(\frac{1}{2}\)×b×h
= \(\frac{1}{2}\)×8×4
= \(\frac{1}{2}\)×32
= 16 sq.units
Area of rectangle EQRF:
Figure EQRF is occupying 8 units wide and 2 units high.
Area of rectangle = length × width
= 8×2
= 16 sq.units.
By observing the area of triangle MNP and rectangle ENPF, we can say that both the areas are same.

How does the area of triangle PQR compare to the area of rectangle AQRD?
Answer:
Area of triangle PQR is 16 sq.units.
Area of rectangle AQRD is 32 sq.units
By observing the area of triangle PQR and the area of rectangle AQRD, we can say that the area of AQRD is twice the area of PQR.

Triangle MNP is an obtuse triangle. \(\overline{N P}\) is the base and MF is the height.

Step 1.
Draw triangle MNP on a piece of graph paper as shown. Then draw and label rectangle ANPD.
Example

Step 2.
Cut up triangle MNP into smaller triangles. Rearrange the triangles to form rectangle ENPF, as shown below.

Step 3.
Find the area of rectangle ENPF. How does its area compare to the area of rectangle ANPD?
Answer:
Area of rectangle ENPF:
Figure ENPF is occupying 3 units wide and 3 units high. Since all sides are equal, it is a square.
Area of square = length × length
= 3×3
= 9 sq.units.
Area of rectangle ANPD:
Figure ANPD is occupying 3 units wide and 6 units high, forming a rectangle.
Area of rectangle = length×width
= 3×6
= 18 sq.units
By observing the area of ENPF and ANPD, we can say that the area of ANPD is twice the area of ENPF.

How does the area of triangle MNP compare to the area of rectangle ENPF?
Answer:
Area of triangle MNP:
Let the base of the triangle be NP, which is occupying 3 horizontal squares. Thus, the base will be 3 units.
The perpendicular segment between the base and the opposite vertex will be the height of the triangle which is 6 units.
Area of triangle = \(\frac{1}{2}\)×b×h
= \(\frac{1}{2}\)×3×6, multiplying 1×3×6 will give 18 and denominator remains same
= \(\frac{18}{2}\)
= 9 sq.units
By observing the area of triangle MNP and rectangle ENPF, we can say that both the areas are same.

How does the area of triangle MNP compare to the area of rectangle ANPD?
Answer:
Area of triangle MNP is 9 sq.units.
Area of rectangle ANPD is 18 sq.units
By observing the area of triangle MNP and the area of rectangle ANPD, we can say that the area of ANPD is twice the area of MNP.

Complete to find the base, height, and area of each triangle. Each square measures 1 unit by 1 unit.

Question 1.

Base = units
Height = units
Area = \(\frac{1}{2}\)bh
= ∙ ∙
= units²
Answer:
Given that:
The above given triangle is drawn on a graph in which each square measures 1 unit.
Let the horizontal side be the base of triangle and the perpendicular segment between the base and the opposite vertex will be the height of the triangle.
The triangle drawn is occupying 5 horizontal square boxes which is base and 8 vertical square boxes which is height.
Thus the base length of the triangle will be 5 units and height will be 8 units.
Base = 5 units, Height = 8 units
Formula:
Area of triangle = \(\frac{1}{2}\)×b×h
= \(\frac{1}{2}\)×5×8, In numerator, multiplying 1×5×8 will give 40 and denominator remains same
= \(\frac{40}{2}\), 40 divided by 2 gives 20
Therefore, area of the given triangle will be 20 sq.units

Question 2.

Base = units
Height = units
Area = \(\frac{1}{2}\)bh
= ∙ ∙
= units²
Answer:
The above given triangle is drawn on a graph in which each square measures 1 unit.
Let the vertical side be the base of triangle and the perpendicular segment between the base and the opposite vertex will be the height of the triangle which lies on the horizontal axis.
The vertical boxes are occupying 3 square boxes and horizontally they are occupying 4 boxes.
Thus the base length of the triangle will be 3 units and height will be 4 units.
Base = 3 units, Height = 4 units
Formula:
Area of triangle = \(\frac{1}{2}\)×b×h
= \(\frac{1}{2}\)×3×4, In numerator, multiplying 1×3×4 will give 12 and denominator remains same
= \(\frac{12}{2}\), 12 divided by 2 gives 6
Therefore, area of the given triangle will be 6 sq.units

Question 3.

Base = cm
Height = cm
Area of triangle ABC
= \(\frac{1}{2}\)bh
= ∙ ∙
= cm²
Answer:
From the given figure, the height of the triangle is 1.8 cm and base length is 2.1 cm.
Base = 2.1cm
Height = 1.8cm
Area of triangle ABC = \(\frac{1}{2}\)×b×h
= \(\frac{1}{2}\)×2.1×1.8, multiplying 1×2.1×1.8 gives 3.78
= \(\frac{3.78}{2}\)
= 1.89 sq.cm

Question 4.

Base = ft
Height = ft
Area of triangle PQR
= \(\frac{1}{2}\)bh
= ∙ ∙
= ft²
Answer:
From the given figure, the height of the triangle is 2.7 ft and base length is 3.4 ft.
Base = 2.7 ft
Height = 3.4 ft
Area of triangle PQR
= \(\frac{1}{2}\)×b×h
= \(\frac{1}{2}\)×3.4×2.7
= \(\frac{1}{2}\)×9.18
= 4.59 sq.ft

Complete to find the height of triangle JKL.

Question 5.
The area of triangle JKL is 35 square meters. Find the height of triangle JKL.

Answer:
Given that the area of triangle JKL is 35 sq.m. The base of the triangle is 7m.
Area of triangle JKL = \(\frac{1}{2}\)×b×h
35 = \(\frac{1}{2}\)×b×h
35 = 0.5×7×h
35 = 3.5×h
35÷3.5 = (3.5×h)÷3.5
10 = h
The height of triangle JKL is 10 m.

Complete to find the base of each triangle.

Question 6.
The area of triangle LMN is 36 square inches. Find the base of triangle LMN.

Area of triangle LMN = \(\frac{1}{2}\)bh
= ∙ b ∙
= ∙ ∙ b
= ∙ b
÷ = b ÷
= b
The base of triangle LMN is inches.
Answer:
Given that the area of triangle LMN is 36 sq.in and the height of the triangle is 8 in.
36 = \(\frac{1}{2}\)×b×h
36 = \(\frac{1}{2}\)×b×8
36 = 4 × b
36 ÷ 4 = (4 ×b) ÷ 4
9 = b
The base of triangle LMN is 9 inches.

Question 7.
The area of triangle PQR1s 19.2 square centimeters. Find the base of triangle PQR.

Area of triangle PQR = \(\frac{1}{2}\)bh
= ∙ b ∙
= ∙ ∙ b
= ∙ b
÷ = b ÷
= b
The base of triangle PQR is inches.
Answer:
Given that the area of triangle PQR is 19.2 sq.cm and the height of the triangle is 9.6 cm.
Area of triangle PQR = \(\frac{1}{2}\)×b×h
19.2 = \(\frac{1}{2}\)×b×9.6
19.2 = 4.8 × b
19.2÷4.8 = (4.8×b)÷4.8
4 = b
The base of triangle PQR is 4 inches.

Math in Focus Course 1B Practice 10.1 Answer Key

Identify a base and a height of each triangle.

Question 1.

Answer:
Let the side BC be the base of triangle and the perpendicular segment between the base and the opposite vertex will be the height of the triangle,which is AB.
Therefore, the base of triangle is BC and height is AB.

Question 2.

Answer:
Let the horizontal side QR be the base of triangle and the perpendicular segment between the base and the opposite vertex will be the height of the triangle,which is PR.
Therefore, the base of triangle is QR and height is PR.

Copy each triangle. Label a base with the letter b and a height with the letter h.

Question 3.

Answer:

Explanation:
Let the side AB be the base of triangle and the perpendicular segment between the base and the opposite vertex will be the height of the triangle,which is AC.
Therefore, the side AB can be marked as b and the side AC as h.

Question 4.

Answer:

Explanation:
Let the side EF be the base of triangle and a perpendicular segment is drawn between the base and the opposite vertex which will be the height of the triangle,which is DC.
Therefore, the side EF can be marked as b and the side DC as h.

Question 5.

Answer:

Explanation:
Let the side GH be the base of triangle and the perpendicular segment between the base and the opposite vertex will be the height of the triangle,which is HI.
Therefore, the side GH can be marked as b and the side HI as h.

Question 6.

Answer:

Explanation:
Let the side JK be the base of triangle and the perpendicular segment between the base and the opposite vertex will be the height of the triangle,which is JL.
Therefore, the side JK can be marked as b and the side JL as h.

Find the area of each triangle.

Question 7.

Answer:
Let us assume the side measuring 25 cm be the base of the triangle and the perpendicular segment between the base and the opposite vertex will be the height of the triangle which measures 4 cm.
Thus, the base of triangle = 25 cm
and the height of the triangle = 4 cm
Area of triangle = \(\frac{1}{2}\)×b×h
= \(\frac{1}{2}\)×25×4
= \(\frac{1}{2}\)×100
= 100÷2
= 50
Therefore, the area of the triangle will be 50 sq.cm

Question 8.

Answer:
Let us assume the horizontal side measuring 15.5 cm be the base of the triangle and the perpendicular segment between the base and the opposite vertex will be the height of the triangle which measures 6 cm.
Thus, the base of triangle = 15.5 cm
and the height of the triangle = 6 cm
Area of triangle = \(\frac{1}{2}\)×b×h
= \(\frac{1}{2}\)×15.5×6
= \(\frac{1}{2}\)×93
= 93÷2
= 46.5
Therefore, the area of the triangle will be 46.5 sq.cm

The area of each triangle is 76 square inches. Find the height and round your answer to the nearest tenth of an inch.

Question 9.

Answer:
Given that the area of triangle is 76 sq.in and base is 15.7 in
We now that, Area of triangle = \(\frac{1}{2}\)×b×h
76 = \(\frac{1}{2}\)×15.7×h
76 = (15.7÷2)×h
76 = 7.85 × h
76 ÷ 7.85 = (7.85 × h)÷7.85
9.6 = h
The value nearest to tenth will be 10 in.
Therefore, the height of the given triangle will be 10 in.

Question 10.

Answer:
Given that the area of triangle is 76 sq.in and base is 9.3 in
We now that, Area of triangle = \(\frac{1}{2}\)×b×h
76 = \(\frac{1}{2}\)×9.3×h
76 = (9.3÷2)×h
76 = 4.65 × h
76 ÷ 4.65 = (4.65×h)÷4.65
16.3 = h
The value nearest to tenth will be 16 in.
Therefore, the height of the given triangle will be 16 in.

The area of each triangle is 45 square centimeters. Find the base and round your answer to the nearest tenth of a centimeter if necessary.

Question 11.

Answer:
Given that the area of triangle is 45 sq.cm and height is 7.2 cm
We now that, Area of triangle = \(\frac{1}{2}\)×b×h
45 = \(\frac{1}{2}\)×b×7.2
45 = (7.2 ÷ 2)×b
45 = 3.6 × b
45 ÷ 3.6 = (3.6 × b)÷3.6
12.5 = b
The value nearest to tenth will be 12 in.
Therefore, the base of the given triangle will be 12 in.

Question 12.

Answer:
Given that the area of triangle is 45 sq.cm and height is 16.6 cm
We now that, Area of triangle = \(\frac{1}{2}\)×b×h
45 = \(\frac{1}{2}\)×b×16.6
45 = 8.3 × b
45 ÷ 8.3 = (8.3 × b) ÷ 8.3
5.4 = b
The value nearest to base will be 5 cm.
Therefore, the base of the given triangle will be 5 cm.

Question 13.

Answer:
Given that the area of triangle is 45 sq.cm and height is 7.5 cm
We now that, Area of triangle = \(\frac{1}{2}\)×b×h
45 = \(\frac{1}{2}\)×b×7.5
45 = 3.75 × b
45 ÷ 3.75 = (3.75 × b)÷ 3.75
12 = b
Therefore, the base of the given triangle will be 12 cm.

Question 14.

Answer:
Given that the area of triangle is 45 sq.cm and height is 15 cm
We now that, Area of triangle = \(\frac{1}{2}\)×b×h
45 = \(\frac{1}{2}\)×b×15
45 = 7.5 × b
45 ÷ 7.5 = (7.5×b) ÷ 7.5
6 = b
Therefore, the base of the given triangle will be 6 cm.

Find the area of the shaded region.

Question 15.

Answer:
Total Area:
Given rectangle measures 20m wide and 40m long.
Area of rectangle = length×width
= 20×40
= 800 sq.m
Larger triangle:
Let us assume the 40 m line segment as base and the perpendicular segment between the base and the opposite vertex will be the height of the triangle which measures 20 m.
Area of larger triangle = \(\frac{1}{2}\)×b×h
= \(\frac{1}{2}\)×40×20
= \(\frac{1}{2}\)×800
= 800÷2
= 400 sq.m
Smaller triangle:
Let us assume the 20 m line segment as base and the perpendicular segment between the base and the opposite vertex will be the height of the triangle which measures 7 m.
Area of larger triangle = \(\frac{1}{2}\)×b×h
= \(\frac{1}{2}\)×20×7
= \(\frac{1}{2}\)×140
= 140÷2
= 70 sq.m
Area of shaded region = Total Area – (Area of larger triangle + Area of smaller triangle)
= 800 – (400+70)
= 800 – 470
= 330 sq.m

Question 16.

Answer:
Larger triangle:
Let us assume the 24 in line segment as base and the perpendicular segment between the base and the opposite vertex will be the height of the triangle which measures (12-5)=7 in.
Area of larger triangle = \(\frac{1}{2}\)×b×h
= \(\frac{1}{2}\)×24×7
= \(\frac{1}{2}\)×168
= 168÷2
= 84 sq.in
Smaller triangle:
Let us assume the 12 in line segment as base and the perpendicular segment between the base and the opposite vertex will be the height of the triangle which measures 10 in.
Area of larger triangle = \(\frac{1}{2}\)×b×h
= \(\frac{1}{2}\)×12×10
= \(\frac{1}{2}\)×120
= 120÷2
= 60 sq.m
Area of shaded region = Area of larger triangle + Area of smaller triangle
= 84 + 60
= 144 sq.in

Question 17.

Answer:

First triangle:
Let us assume the base of first triangle be 12 cm and the perpendicular segment between the base and the opposite vertex will be the height of the triangle which measures 34 cm.
Area of first triangle = \(\frac{1}{2}\)×b×h
= \(\frac{1}{2}\)×12×34
= \(\frac{1}{2}\)×408
= 408÷2
= 204 sq.cm
Second triangle:
Let us assume the base of second triangle be 12 cm and the perpendicular segment between the base and the opposite vertex will be the height of the triangle which measures 10 cm.
Area of second triangle = \(\frac{1}{2}\)×b×h
= \(\frac{1}{2}\)×12×10
= \(\frac{1}{2}\)×120
= 120÷2
= 60 sq.cm
Third triangle:
Let us assume the base of first triangle be 24 cm and the perpendicular segment between the base and the opposite vertex will be the height of the triangle which measures 24 cm.
Area of first triangle = \(\frac{1}{2}\)×b×h
= \(\frac{1}{2}\)×24×24
= \(\frac{1}{2}\)×576
= 576÷2
= 288 sq.cm
Total Area:
The rectangle measures 24 cm wide and 34 cm high
Area of rectangle = length×width
= 24×34
= 816 sq.cm
Area of shaded region will be =  Total Area – (Area of first triangle + Area of second triangle + Area of third trinagle)
= 816 – (204+60+288)
= 816 – 552
= 264 sq.cm

Question 18.

Answer:

Let us assume the base of first triangle be 6 ft and the perpendicular segment between the base and the opposite vertex will be the height of the triangle which measures 24 ft.
Area of first shaded triangle = \(\frac{1}{2}\)×b×h
= \(\frac{1}{2}\)×6×24
= 144÷2
= 72 sq.ft
Area of second shaded triangle = \(\frac{1}{2}\)×b×h
= \(\frac{1}{2}\)×24×12
= 288÷2
= 144 sq.ft
Area of shaded region = Area of first triangle + Area of second triangle
= 72+144
= 216 sq.ft

Question 19.

Answer:

Area of larger triangle:
Let us assume the base of first triangle be 6 cm and the perpendicular segment between the base and the opposite vertex will be the height of the triangle which measures 12 cm.
Area of larger triangle = \(\frac{1}{2}\)×b×h
= \(\frac{1}{2}\)×6×12
= 72÷2
= 36 sq.cm
Area of smaller triangle:
Let us assume the base of first triangle be 6 cm and the perpendicular segment between the base and the opposite vertex will be the height of the triangle which measures 6 cm.
= \(\frac{1}{2}\)×6×6
= 36÷2
= 18 sq.cm
Area of shaded region = Area of larger triangle + Area of smaller triangle
= 36 + 18
= 54 sq.cm

Use graph paper. Solve.

Question 20.
The coordinates of the vertices of a triangle are A (4, 7), B (4, 1), and C (8, 1). Find the area of triangle ABC.
Answer:

Given that the vertices of a triangle are A (4, 7), B (4, 1), and C (8, 1).
Let BC be the base of the triangle and the perpendicular segment between the base and the opposite vertex will be the height of the triangle which is AB.
BC measures 4 units and AB measures 6 units.
Area of triangle ABC = \(\frac{1}{2}\)×b×h
= \(\frac{1}{2}\)×4×6
= 24 ÷ 2
= 12 sq.units

Question 21.
The coordinates of the vertices of a triangle are D (1, 7), E (-3, 2), and F (6, 2). Find the area of triangle DEF.
Answer:

Given that the vertices of a triangle are D (1, 7), E (-3, 2), and F (6, 2).
Let EF measuring 9 units be the base of the triangle and the perpendicular segment between the base and the opposite vertex will be the height of the triangle which measures 5 units.
Area of triangle DEF = \(\frac{1}{2}\)×b×h
= \(\frac{1}{2}\)×9×5
= 45 ÷ 2
= 22.5 sq.units

Question 22.
The coordinates of the vertices of a triangle are J (-5, 2), K (1, -2), and L (5, -2). Find the area of triangle JKL.
Answer:

Given that the vertices of a triangle are J (-5, 2), K (1, -2), and L (5, -2).
Let LK measuring 4 units be the base of the triangle and the perpendicular segment between the base and the opposite vertex will be the height of the triangle which measures 4 units.
Area of triangle JKL= \(\frac{1}{2}\)×b×h
= \(\frac{1}{2}\)×4×4
= 16 ÷ 2
= 8 sq.units

Question 23.
The area of triangle MNP is 1 7.5 square units. The coordinates of M are (-9, 5), and the coordinates of N are (-2, 0). The height of triangle MNP is 5 units and is perpendicular to the x-axis. Point P lies to the right of point N. Given that \(\overline{N P}\) is the base of the triangle, find the coordinates of point P.
Answer:
Given that the area of triangle MNP is 17.5 square units and height of the triangle is 5 units.
Area of triangle = \(\frac{1}{2}\)×b×h
17.5 = \(\frac{1}{2}\)×b×5
17.5 = 2.5×b
(17.5÷2.5) = (2.5xb)÷2.5
7 = b
Thus, the base of the triangle will be 7 units.

Explanation:
From the above calculations, the base of the triangle measures 7 units and lies on the right of the point N.
Move 7 units to the right from the point N and mark the point. The marked point is the required P point.
Thus, the coordinates of P will be (5,0)

Question 24.
The coordinates of the vertices of a triangle are X (1, 2), Y (-6, -2), and Z (1, -4). Find the area of triangle XYZ. (Hint: Use the vertical side as the base.)
Answer:

Given that the vertices of a triangle are X (1, 2), Y (-6, -2), and Z (1, -4).
As given the base of the triangle is the vertical side, XZ which measures 6 units.
The perpendicular segment between the base and the opposite vertex will be the height of the triangle which measures 5 units.
Area of triangle = \(\frac{1}{2}\)×b×h
= \(\frac{1}{2}\)×6×5
= \(\frac{1}{2}\)×30
= 15 sq.units

Question 25.
The coordinates of the vertices of a triangle are P (-2, 6), Q (-4, 2), and R (5, 1). Find the area of triangle PQR. (Hint: Draw a rectangle around triangle PQR.)
Answer:

Part 1:
From the given figure, we can observe that the rectangle around triangle PQR measures 5 units long and 9 units wide.
Area of rectangle = length× width
= 5×9
= 45 sq.units
The total area of the figure is 45 sq.units
Part 2:
Area of first triangle:
Let us assume the base of first triangle be 2 units and the perpendicular segment between the base and the opposite vertex will be the height of the triangle which measures 4 units
Area of first triangle = \(\frac{1}{2}\)×b×h
= \(\frac{1}{2}\)×2×4
= \(\frac{1}{2}\)×8
= 8÷2
= 4 sq.units
Part 3:
Area of second triangle:
Let us assume the base of second triangle be 7 units and the perpendicular segment between the base and the opposite vertex will be the height of the triangle which measures 5 units
Area of first triangle = \(\frac{1}{2}\)×b×h
= \(\frac{1}{2}\)×7×5
= \(\frac{1}{2}\)×35
= 35÷2
= 17.5 sq.units
Part 4:
Area of third triangle:
Let us assume the base of third triangle be 9 units and the perpendicular segment between the base and the opposite vertex will be the height of the triangle which measures 1 units
Area of first triangle = \(\frac{1}{2}\)×b×h
= \(\frac{1}{2}\)×9×1
= 9÷2
= 4.5 sq.units
Part 5:
Area of the triangle = Total Area-(Area of first triangle+Area of second triangle+Area of third triangle)
= 45-(4+17.5+4.5)
= 45-26
= 19 sq.units

Find the area of the shaded region for questions 26 to 29.

Question 26.
Figure DGHJ is a trapezoid.

Answer:
Given that figure DGHJ is a trapezoid measuring 28 ft high and two parallel sides with 24ft and 56 ft long.
Area of trapezoid = \(\frac{1}{2}\)×(a+b)×h
= \(\frac{1}{2}\)×(24+56)×28
= \(\frac{1}{2}\)×80×28
= 2240÷2
= 1120 sq.ft
The unshaded figure seems to be square as each angle is 90 degrees. Thus, all sides of it will be equal.
Area of square = 24×24
= 576 sq.ft
Area of the shaded region = Total Area – Area of the unshaded region
= 1120-576
= 544 sq.ft

Question 27.
Figure ABCD is a rectangle. The length of \(\overline{Z B}\) is \(\frac{3}{7}\) the length of \(\overline{A B}\).

Answer:
Part 1:
From the given figure, we can observe that the area of shaded region will be the sum of areas of square and triangle.
Given that the figure ABCD is a rectangle and the length of \(\overline{Z B}\) is \(\frac{3}{7}\) the length of \(\overline{A B}\).
Length of AB = 35cm
Length of ZB will be \(\frac{3}{7}\) × AB
= \(\frac{3}{7}\) × 35
= 3×5
= 15 cm.
Area of square = 15×15
= 225 sq.cm
Area of triangle = \(\frac{1}{2}\)×b×h
= \(\frac{1}{2}\)×15×35
= 525÷2
= 262.5
Part 2:
Total Area of shaded region = Area of sqaure + Area of triangle
= 225 + 262.5
= 487.5

Question 28.
The area of triangle POS is \(\frac{7}{12}\) of the area of trapezoid PRST.

Answer:
Part 1:
Area of first triangle SQR:
Let us assume the base of first triangle be 14 units and the perpendicular segment between the base and the opposite vertex will be the height of the triangle which measures 22 units
Area of first triangle = \(\frac{1}{2}\)×b×h
= \(\frac{1}{2}\)×14×22
= \(\frac{1}{2}\)×308
= 154 sq.in
Part 2:
Area of second triangle PTS:
Let us assume the base of first triangle be 10 units and the perpendicular segment between the base and the opposite vertex will be the height of the triangle which measures 22 units
Area of first triangle = \(\frac{1}{2}\)×b×h
= \(\frac{1}{2}\)×10×22
= \(\frac{1}{2}\)×220
= 110 sq.in
Part 3:
Given that the area of triangle PQS is \(\frac{7}{12}\) of the area of trapezoid PRST.Let us assue the area of trapezoid as ‘x’ inches.
Then the area of triangle PQS will be \(\frac{7}{12}\) of the area of trapezoid PRST.
Area of triangle PQS = \(\frac{7}{12}\) × x
Total Area will be the area of trapezoid.
Total Area = Area of triangle PQS + Area of triangle SQR + Area of triangle PTS
x = (\(\frac{7}{12}\) × x)+ 154 + 110
x = (\(\frac{7}{12}\) × x)+ 264
x = \(\frac{7x}{12}\)+ 264
x-\(\frac{7x}{12}\) = \(\frac{7x}{12}\)–\(\frac{7x}{12}\)+264
(\(\frac{12}{12}\)×x)-\(\frac{7x}{12}\)  = 264
(\(\frac{12x}{12}\))-\(\frac{7x}{12}\) = 264
\(\frac{5x}{12}\) = 264
\(\frac{5x}{12}\) × 12 = 264 × 12
5x = 3168
x = 633.6
Thus, the area of trapezoid = 633.6 sq.in
Area of triangle = \(\frac{7}{12}\) × 633.6
= 4435.2÷12
= 369.6 sq.in

Question 29.
Figure EFHL is a parallelogram. The length of \(\overline{F G}\) is \(\frac{5}{8}\) the length of \(\overline{F H}\).

Answer:

Given that:
Part 1:
The length of \(\overline{F G}\) is \(\frac{5}{8}\) the length of \(\overline{F H}\).
Length of FH be ‘x’ ft.Length of FG will be \(\frac{5x}{8}\)
FH length = FG length + GH length
x = \(\frac{5x}{8}\) + 3
x – \(\frac{5x}{8}\) = \(\frac{5x}{8}\)–\(\frac{5x}{8}\)+3
(\(\frac{8}{8}\)×x)- \(\frac{5x}{8}\) = \(\frac{5x}{8}\)–\(\frac{5x}{8}\)+3
\(\frac{3x}{8}\) = 3
\(\frac{3x}{8}\)÷\(\frac{3}{8}\) = 3÷\(\frac{3}{8}\)
x = 8 ft
Length of FH be 8ft and length of FG will be \(\frac{5}{8}\) × 8 = 5 units.
In the figure, we can observe two squares, larger square with 8 ft length and smaller square with length of 3 ft.
Area of larger square = 8×8 = 64 sq.ft
Area of smaller square = 3×3 = 9 sq.ft
Figure EFHL is a parallelogram.
Part 2:
Area of first shaded region = Area of larger sqaure – Area of smaller square
= 64-9
= 55 sq.ft
Area of triangle LEH
Let us assume EL as base and EH.
Area of triangle mesauring 3 ft wide and 3 ft high = \(\frac{1}{2}\)×b×h
= \(\frac{1}{2}\)×3×3
= 4.5 sq.ft
There are 2 such triangles = 4.5+4.5 = 9 sq.ft
Part 3:
Area of shaded region will be 55+9 = 64 sq.ft

Practice the problems of Math in Focus Grade 6 Workbook Answer Key Chapter 10 Lesson 10.2 Area of Parallelograms and Trapezoids to score better marks in the exam.

Math in Focus Grade 6 Course 1 B Chapter 10 Lesson 10.2 Answer Key Area of Parallelograms and Trapezoids

Math in Focus Grade 6 Chapter 10 Lesson 10.2 Guided Practice Answer Key

Complete to find the sum of the bases, height, and area of each trapezoid. Each square measures 1 unit by 1 unit.

Question 1.

Base = units
Height = units
Area = bh
= ∙
= units²
Answer:
The base in the above figure is occupying 4 units, so it will measure 4 units and the height is occupying 7 units, so it will measure 7 units.
Base = 4 units
Height = 7 units
Area of rectangle = base×height
= 4×7
= 28 sq.units

Question 2.

Base = units
Height = units
Area = bh
= ∙
= units²
Answer:
The base in the above figure is occupying 3 units, so it will measure 4 units and the height is occupying 7 units, so it will measure 7 units.
Base = 3 units
Height = 7 units
Area of rectangle = base×height
= 3×7
= 21 sq.units

Question 3.

Base = units
Height = units
Area = bh
= ∙
= units²
Answer:
The base in the above figure is occupying 4 units, so it will measure 4 units and the height is occupying 8 units, so it will measure 7 units.
Base = 4 units
Height = 8 units
Area of rectangle = base×height
= 4×8
= 32 sq.units

Question 4.

Base = units
Height = units
Area = bh
= ∙
= units²
Answer:
The base in the above figure is occupying 2 units, so it will measure 4 units and the height is occupying 8 units, so it will measure 7 units.
Base = 4 units
Height = 8 units
Area of rectangle = base×height
= 4×8
= 32 sq.units

Complete to find the area of each trapezoid.

Question 5.

Base = units
Height = units
Area = bh
= ∙
= cm²
Answer:
In the given figure, base measures 21cm and the height measures 12cm.
Base = 21cm
Height = 12cm
Area = base × height
= 21 × 12
= 252 sq.cm

Question 6.

Base = in.
Height = in.
Area = bh
= ∙
= in²
Answer:
In the given figure, base measures 24 in and the height measures 14.5 in.
Base = 24 in
Height = 14.5 in
Area = base × height
= 24 × 14.5
= 348 sq.in

Complete to find the sum of the bases, height, and area of each trapezoid. Each square measures 1 unit by 1 unit.

Question 7.

Height = units
Sum of bases = +
= units
Area = \(\frac{1}{2}\)h(sum of bases)
= ∙ ∙
= units²
Answer:
The lengths will be equal to the number of squares each occupies because each square has length of 1 unit.
Given figure has two bases measuring 7 units and 3 units each. The height is measuring 3 units.
Height = 3 units
Sum of bases = (7+3)
= 10 units
Area = \(\frac{1}{2}\)h(sum of bases)
= \(\frac{1}{2}\)×3×10
= \(\frac{1}{2}\)×30
= 30÷2
= 15 sq.units

Question 8.

Height = units
Sum of bases = +
= units
Area = \(\frac{1}{2}\)h(sum of bases)
= ∙ ∙
= units²
Answer:
The lengths will be equal to the number of squares each occupies because each square has length of 1 unit.
Given figure has two vertical bases measuring 1 unit and 3 units each. The horizontal height is measuring 4 units.
Height = 4 units
Sum of bases = 1+3
= 4 units
Area = \(\frac{1}{2}\)h(sum of bases)
= \(\frac{1}{2}\)×4×4
= \(\frac{1}{2}\)×16
= 16÷2
= 8 sq.units

Question 9.

Height = units
Sum of bases = +
= units
Area = \(\frac{1}{2}\)h(sum of bases)
= ∙ ∙
= units²
Answer:
The lengths will be equal to the number of squares each occupies because each square has length of 1 unit.
Given figure has two vertical bases measuring 4 units and 7 units each. The horizontal height is measuring 7 units.
Height = 7 units
Sum of bases = 4+7
= 11 units
Area = \(\frac{1}{2}\)h(sum of bases)
= \(\frac{1}{2}\)×7×11
= \(\frac{1}{2}\)×77
= 77÷2
= 38.5 sq.units

Question 10.

Height = units
Sum of bases = +
= units
Area = \(\frac{1}{2}\)h(sum of bases)
= ∙ ∙
= units²
Answer:
The lengths will be equal to the number of squares each occupies because each square has length of 1 unit.
Given figure has two vertical bases measuring 5 units and 3 units each. The horizontal height is measuring 7 units.
Height = 7 units
Sum of bases = 5+3
= 8 units
Area = \(\frac{1}{2}\)h(sum of bases)
= \(\frac{1}{2}\)×7×8
= \(\frac{1}{2}\)×56
= 56÷2
= 28 sq.units

Complete to find the area of each trapezoid.

Question 11.

Height = ft
Sum of bases = +
= ft
Area = \(\frac{1}{2}\)h(sum of bases)
= ∙ ∙
= ft²
Answer:
Given figure has two vertical parallel sides measuring 25ft and 13ft each. The height of the trapezoid is 39ft.
Height = 39ft
Sum of bases = 25+13
= 38 ft
Area = \(\frac{1}{2}\)h(sum of bases)
= \(\frac{1}{2}\)×39×38
= 1482÷2
= 741 sq.ft

Question 12.

Height = cm
Sum of bases = +
= cm
Area = \(\frac{1}{2}\)h(sum of bases)
= ∙ ∙
= cm²
Answer:
Given figure has two horizontal parallel sides measuring 10.6cm and 21cm each. The height of the trapezoid is 13cm.
Height = 13cm
Sum of bases = 10.6+21
= 31.6 cm
Area = \(\frac{1}{2}\)h(sum of bases)
= \(\frac{1}{2}\)×31.6×13
= 410.8÷2
= 205.4 sq.cm

Complete to find the area of triangle ABD.

Question 13.
The area of trapezoid ABCD is 1,248 square centimeters.

Area of trapezoid ABCD = \(\frac{1}{2}\)h(b₁ + b₂)
= \(\frac{1}{2}\) ∙ h ∙ ( + )
= \(\frac{1}{2}\) ∙ h ∙
= \(\frac{1}{2}\) ∙ ∙ h
= ∙ h
Since area of trapezoid = cm²
• h = area of trapezoid
• h =
h =
= cm
Height of trapezoid =
= cm
Area of triangle ABD = \(\frac{1}{2}\)bh
= \(\frac{1}{2}\) • •
= cm²
The area of triangle ABD is square centimeters.

Answer:
Given that the area of trapezoid ABCD is 1,248 sq.cm
From the figure, we can observe two horizontal parallel sides measuring 22 cm and 38 cm.
Sum of the sides = 22+38 = 60cm
Area of trapezoid ABCD = \(\frac{1}{2}\)×h×(b₁ + b₂)
= \(\frac{1}{2}\) × h × 60
= \(\frac{60}{2}\) × h
= 30×h
Since area of trapezoid = 1,248 cm²
30×h = 1248
(30×h)÷30 = 1248÷30
h = 41.6 cm
Height of trapezoid = 41.6 cm
Area of triangle ABD = \(\frac{1}{2}\)×b×h
Let us assume the base as AD measuring 22cm and the height is 41.6cm
= \(\frac{1}{2}\)×b×h
= \(\frac{1}{2}\)×22×41.6
= 915.2÷2
= 457.6 sq.cm
The area of triangle ABD is 457.6 square centimeters.

Math in Focus Course 1B Practice 10.2 Answer Key

Copy each parallelogram. Label a base and a height for each. Use b and h.

Question 1.

Answer:

Explanation:
In parallelogram, any side can be the base of the paralleogram
But here, Let us assume the side BC as base of the paralleogram and label it as ‘b’.
The height of a parallelogram is the perpendicular distance between the base side and the parallel side opposite to it, the height is labelled as ‘h’.

Question 2.

Answer:

Explanation:
In parallelogram, any side can be the base of the paralleogram
But here, Let us assume the side XY as base of the paralleogram and label it as ‘b’.
The height of a parallelogram is the perpendicular distance between the base side and the parallel side opposite to it, the height is labelled as ‘h’.

Question 3.

Answer:

Explanation:
In parallelogram, any side can be the base of the paralleogram
But here, Let us assume the side QR as the base of the paralleogram and label it as ‘b’.
The height of a parallelogram is the perpendicular distance between the base side and the parallel side opposite to it, the height is labelled as ‘h’.

Question 4.

Answer:

Explanation:
In parallelogram, any side can be the base of the paralleogram
But here, Let us assume the side MN as the base of the paralleogram and label it as ‘b’.
The height of a parallelogram is the perpendicular distance between the base side and the parallel side opposite to it, the height is labelled as ‘h’.

Find the area of each parallelogram.

Question 5.

Answer:
From the given figure the base length is 28 in and the height is 15 in.
Base = 28 in
Height = 15 in
Area of the parallelogram = base × height
= 28×15
= 420 sq.in

Question 6.

Answer:
From the given figure the base length is 18 in and the height is 10 in.
Base = 18 in
Height = 10 in
Area of the parallelogram = base × height
= 18×10
= 180 sq.in

Copy each trapezoid. Label the height and bases. Use h₁, b₁ and b₂.

Question 7.

Answer:

Explanation:
The parallel sides are called the bases of the trapezium.
Let b1​ and b2​ be the lengths of these bases.
The distance between the bases is called the height of the trapezium. Let $h$ be this height.

Question 8.

Answer:

Explanation:
The parallel sides are called the bases of the trapezium.
Let b1​ and b2​ be the lengths of these bases.
The distance between the bases is called the height of the trapezium. Let $h$ be this height.

Question 9.

Answer:

Explanation:
The parallel sides are called the bases of the trapezium.
Let b1​ and b2​ be the lengths of these bases.
The distance between the bases is called the height of the trapezium. Let $h$ be this height.

Question 10.

Answer:

Explanation:
The parallel sides are called the bases of the trapezium.
Let b1​ and b2​ be the lengths of these bases.
The distance between the bases is called the height of the trapezium. Let $h$ be this height.

Find the area of each trapezoid.

Question 11.

Answer:
Given figure has two horizontal bases measuring 15 cm and 7 cm each. The height is measuring 10 cm.
Height = 10 cm
Sum of bases = 15+ 7
= 22 cm
Area = \(\frac{1}{2}\)h(sum of bases)
= \(\frac{1}{2}\) × 10 × 22
= \(\frac{1}{2}\) × 220
= 220÷2
= 110 sq.cm

Question 12.

Answer:
Given figure has two horizontal bases measuring 12 ft and 24 ft each. The height is measuring 17 ft.
Height = 17 ft
Sum of bases = 12+ 24
= 36 ft
Area = \(\frac{1}{2}\)h(sum of bases)
= \(\frac{1}{2}\) × 17 × 36
= \(\frac{1}{2}\) × 612
= 612÷2
= 306 sq.ft

The area of each parallelogram is 64 square inches. Find the height. Round your answer to the nearest tenth of an inch.

Question 13.

Answer:
Given that the area of parallelogram is 64 sq.in and the base is of 7 in length.
Area of parallelogram = base × height
64 = 7×height
64÷7 = (7×height)÷7
9.1 = height
Rounding off 9.1 to the nearest tenth of an inch
Height of the given parallelogram is 9 in.

Question 14.

Answer:
Given that the area of parallelogram is 64 sq.in and the base is of 6 in length.
Area of parallelogram = base × height
64 = 6×height
64÷6 = (6×height)÷6
10.6 = height
Rounding off 10.6 to the nearest tenth of an inch
Height of the given parallelogram is 11 in.

The area of each trapezoid is 42 square centimeters. Find the height. Round your answer to the nearest tenth of a centimeter.

Question 15.

Answer:
Given that the area of trapezoid is 42 sq.cm. The two parallel sides are measuring 5 cm and 7.5 cm.
Sum of the sides = 5+7.5 = 12.5 cm
Area of trapezoid = \(\frac{1}{2}\)×h×(b₁ + b₂)
42 = \(\frac{1}{2}\) × h × 12.5
42 = 6.25 × h
42 ÷6.25 = (6.25 × h)÷6.25
6.7 = h
Rounding off 6.7 to the nearest tenth of an cm.
Height of the given trapezoid is 7 cm.

Question 16.

Answer:
Given that the area of trapezoid is 42 sq.cm. The two parallel sides are measuring 5 cm and 7.5 cm.
Sum of the sides = 5.7+10 = 15.7 cm
Area of trapezoid = \(\frac{1}{2}\)×h×(b₁ + b₂)
42 = \(\frac{1}{2}\) × h × 15.7
42 = 7.8 × h
42 ÷7.8 = (7.8 × h)÷7.8
5.3 = h
Rounding off 5.3 to the nearest tenth of an cm.
Height of the given trapezoid is 5 cm.

Solve.

Question 17.
The area of trapezoid ABCO is 503.25 square centimeters. Find the length of \(\overline{B C}\).

Answer:
Given that the area of trapezoid ABCD is 503.25 sq.cm
From the given figure, length of the height measures 18.3 cm and length of a side is 34 cm
Area of trapezoid = \(\frac{1}{2}\)×h×(b₁ + b₂)
503.25 = \(\frac{1}{2}\) × 18.3 × (34 + b₂)
503.25 = 9.1× (34 + b₂)
503.25÷9.1 = 9.1× (34 + b₂)÷9.1
55.3 = 34 + b₂
55.3 – 34 = 34 – 34 + b₂
21.3 = b₂
The length of second side BC will be 21.3 cm

Question 18.
The area of trapezoid EFGH is 273 square centimeters. Find the area of triangle EGH.

Answer:
Given that the area of trapezoid EFGH is 273 sq.cm. The length of two parallel bases will be 14 cm and 25 cm.
Area of trapezoid = \(\frac{1}{2}\)×h×(b₁ + b₂)
Sum of bases = 14+25 = 39cm
273 = \(\frac{1}{2}\)×h×39
273 = 19.5×h
273 ÷ 19.5 = (19.5×h)÷ 19.5
14 = h
The height of the trapezoid is 14 cm.
Area of triangle EGH = \(\frac{1}{2}\)×b×h
The length of base is 14cm
= \(\frac{1}{2}\)×14×14
= 196÷2
= 98 sq.cm

Solve. Use graph paper.

Question 19.
The coordinates of the vertices of a parallelogram are P (0, 5), Q (-3, 0), R (2, 0), and S (5, 5). Find the area of parallelogram PQRS.
Answer:

From the figure, we can get the height of the parallelogram which is 5 units.
Let us consider QR as base which measures 5 units.
Area of parallelogram = base×height
= 5×5
= 25 sq.units

Question 20.
Three out of the four coordinates of the vertices of parallelogram WXYZ are W(0, 1), X(-4, -4), and Y (-1, -4). Find the coordinates of Z. Then find the area of the parallelogram.
Answer:

Given that three vertices of parallelogram WXYZ are W(0, 1), X(-4, -4), and Y (-1, -4).
Let XY be the base of parallelogram,since XY is measuring 3 units. The distance between W and Z will also be 3 units.
Thus, the point Z will be (3,1)
The height of the parallelogram is 5 units.
Area of parallelogram = base × height
= 3×5
= 15 sq.units

Question 21.
The coordinates of the vertices of trapezoid EFGH are E (-3, 3), F (-3, 0), G (1, -4), and H (1, 4). Find the area of the trapezoid.
Answer:

The two parallel sides EF and HG are measuring 3 units and 8 units respectively. The height is of 4 units length.
Sum of the sides = 3+8 = 11 units
Area of trapezoid = \(\frac{1}{2}\)×h×(b₁ + b₂)
= \(\frac{1}{2}\)×4×11
= 44 ÷2
= 22 sq.units

Question 22.
Three out of the four coordinates of the vertices of trapezoid ABCD are A (0, 1), B (-4, -4), and C(-1, -4). \(\overline{\mathrm{AD}}\) is parallel to \(\overline{\mathrm{BC}}\). AD is 6 units. The point D lies to the right of point A. Find the coordinates of point D. Then find the area of the trapezoid.
Answer:

Given that the three vertices of trapezoid ABCD are A (0, 1), B (-4, -4), and C(-1, -4). AD is parallel to BC and length of AD is 6 units.
Therefore, the point D will be 6 units away from A which can be pointed at (-6,1)
The two parallel sides DA and BC are measuring 6 units and 3 units respectively. The height is of 5 units length.
Sum of the sides = 6+3 = 9 units
Area of trapezoid = \(\frac{1}{2}\)×h×(b₁ + b₂)
= \(\frac{1}{2}\)×5×9
= 45÷2
= 22.5 sq.units

Solve.

Question 23.
Parallelogram PQRT is made up of isosceles triangle PST and trapezoid PQRS. Find the area of parallelogram PQRT.

Answer:
Given that, PST is an isosceles triangle. Therefore the length of PT and TS will be 15cm.
Area of triangle = \(\frac{1}{2}\)×b×h
= \(\frac{1}{2}\)×15×15
= \(\frac{1}{2}\)×225
= 225÷2
= 112.5 sq.cm
Length of TR = Length of RS + Length of ST
Length of TR = 5+15 = 20 cm
Since PQRT is parallelogram, TR will be parallel to PQ. Thus, PQ = 20cm.
Area of trapezoid = \(\frac{1}{2}\)×h×(b₁ + b₂)
Sum of bases = SR + PQ
= 5 +15
= 20 cm
The height is of length 14.5 cm.
Area of trapezoid = \(\frac{1}{2}\)×14.5×20
= \(\frac{1}{2}\)×290
= 290÷2
= 145 sq.cm
Area of parallelogram PQRT = Area of isosceles triangle + Area of trapezoid
= 112.5 + 145
= 257.5 sq.cm

Go through the Math in Focus Grade K Workbook Answer Key Chapter 20 Money to finish your assignments.

Math in Focus Kindergarten Chapter 20 Answer Key Money

Lesson 1 Coin Values

Match.

Answer:

Explanation:
1 quarter dollar is equal to 25 cents
1 dime is equal to 10 cents
1 nickel is equal to 5 cents
1 penny is equal to 1 cent

Lesson 2 Counting Coins

How many pennies do you need? Color.

Question 1.

Answer:

Explanation:
The cost of the toy airplane is 6 cents
so, colored 6 cents

Question 2.

Answer:

Explanation:
The cost of the toys are 6 + 2 = 8
so, colored the 8 cents

Question 3.

Answer:

Explanation:
The cost of the toys are 3 + 5 = 8 cents
so, colored 8 cents

Question 4.

Answer:

Explanation:
The cost of the toys are 2 + 4 + 3 = 9 cents
so, colored 9 cents

How much is needed? Circle the purse.

Question 1.

Answer:

Explanation:
The cost of the toy house is 9 cents
5 + 4 = 9 cents
so, colored the first purse

Question 2.

Answer:

Explanation:
The cost of the toys are 4 cents and 2 cents
4 + 2 = 6 and also
5 + 1 = 6
so circled 6 cents purse

Go through the Math in Focus Grade 6 Workbook Answer Key Chapter 5 Lesson 5.1 Rates and Unit Rates to finish your assignments.

Math in Focus Grade 6 Course 1 A Chapter 5 Lesson 5.1 Answer Key Rates and Unit Rates

Math in Focus Grade 6 Chapter 5 Lesson 5.1 Guided Practice Answer Key

State whether each statement is expressed as a unit rate.

Question 1.
A monkey plucked 4 coconuts per minute.
Answer:
yes, it can be expressed in unit ratio.
Explanation:
Statement is expressed as a unit rate.
per minute is unit

Question 2.
Mary paid $2 for a bottle of orange juice.
Answer:
Yes, it can be expressed in unit ratio.
Explanation:
$2 for a bottle of orange juice.
$ is unit per bottle juice.

Question 3.
A basketball team scored 294 points in 6 games.
Answer:
yes, it can be expressed in unit ratio.
Explanation:
points are the units for games

Question 4.
Douglas reads 3 books in a week.
Answer:
Yes, it can be expressed in unit ratio.
Explanation:
Week is unit for reading a book.

Solve.

Question 5.
A photocopy machine can print 405 copies in 9 minutes. What is the rate at which the machine prints the copies?

Answer:
45 copies per min
Explanation:
Question 6.
A sports center buys 1.25 acres of land for a new swimming complex. What is the cost per acre if the sports center pays $100,000 for the land?
Cost per acre = Total cost ÷ Total number of acres
= $100,000 ÷
= $\(\frac{100,000}{?}\)
= $\(\frac{?}{?}\)
= $
The unit cost of the piece of land is $ per acre.
Answer:
$ 80,000 per acre.
Explanation:
Cost per acre = Total cost ÷ Total number of acres
= $100,000 ÷ 1.25
= $\(\frac{100,000}{1.25}\)
= $\(\frac{1,000,000}{125}\)
= $ 80,000
The unit cost of the piece of land is $ 80000 per acre.

Question 7.
A few years ago, when the fuel tank in Sally’s car was completely empty, she paid $63 to fill the tank with 22.5 gallons of gasoline. What was the cost per gallon?
Cost per gallon = Cost of gasoline ÷ Volume of gasoline filled
=$ ÷
= $
The unit cost of the gasoline was $ per gallon.
Answer:
$2.8
Explanation:
Cost per gallon = Cost of gasoline ÷ Volume of gasoline filled
=$ 63 ÷ 22.5
= $ 2.8
The unit cost of the gasoline was $ 2.8 per gallon.

Question 8.
The table shows the costs of some food items Billy bought from a supermarket.

Answer:

Which type of food costs the most per pound?
Cost of potatoes per pound = Cost of potatoes ÷ Weight of potatoes
= $ ÷
= $
The unit cost of the potatoes is $ per pound.
Cost of carrots per pound = Cost of carrots ÷ Weight of carrots
= $ ÷
= $
The unit cost of the carrots is $ per pound.
Cost of onions per pound = Cost of onions ÷ Weight of onions
= $ ÷
= $
The unit cost of the onions is $ per pound.
Comparing the unit costs of the food items, the unit cost of the is the greatest.
Unit cost of < Unit cost of < Unit cost of
$ < $ < $
So, the cost the most per pound.
Answer:
$1.25 cost the most per pound.
Explanation:
Cost of potatoes per pound = Cost of potatoes ÷ Weight of potatoes
= $ 4 ÷ 5
= $ 0.8
The unit cost of the potatoes is $ 0.8 per pound.
Cost of carrots per pound = Cost of carrots ÷ Weight of carrots
= $3 ÷ 5
= $0.6
The unit cost of the carrots is $ 0.6 per pound.
Cost of onions per pound = Cost of onions ÷ Weight of onions
= $2.5 ÷ 2
= $ 1.25
The unit cost of the onions is $ 1.25 per pound.
Comparing the unit costs of the food items, the unit cost of the 1.25 is the greatest.
Unit cost of  Carrot < Unit cost of Potatoes < Unit cost of Onions
$0.6 < $0.8 < $1.25
So, the $1.25 cost the most per pound.

Question 9.
A truck traveled a distance of 280 kilometers in 4 hours. Find the speed of the truck.
Method 1

The speed of the truck is 70kilometers per hour.

Method 2
Speed of truck = \(\frac{\text { Distance }}{\text { Time }}\)
= \(\frac{?}{?}\)
= km/h
The speed of the truck is kilometers per hour.
Answer:
70 kilometers per hour.
Explanation:

Speed of truck = \(\frac{\text { Distance }}{\text { Time }}\)
= \(\frac{280}{4}\)
= 70 km/h
The speed of the truck is 70 kilometers per hour.

Question 10.
A car travels \(\frac{1}{4}\) kilometer in \(\frac{1}{2}\) minute. Find the speed of the car in kilometers per minute.
Answer:
\(\frac{1}{2}\) kilometers per minute.
Explanation:
Speed of car = \(\frac{\text { Distance }}{\text { Time }}\)
Speed of car = \(\frac{1}{4}\) ÷ \(\frac{1}{2}\)
= \(\frac{1}{4}\) X \(\frac{2}{1}\)
= \(\frac{1}{2}\) kilometers per minute.

Math in Focus Course 1A Practice 5.1 Answer Key

Solve. Show your work.

Question 1.
A machine can print 300 T-shirts in 10 minutes. How many T-shirts can the machine print in 1 minute?
Answer:
30 T-shirts
Explanation:
A machine can print 300 T-shirts in 10 minutes.
Number of T-shirts can the machine print in 1 minute
= \(\frac{300}{10}\)
= 30 T-shirts can the machine print in 1 minute

Question 2.
Alisa types 900 words in 20 minutes. What is her typing speed in words per minute?
Answer:
45 words per minute
Explanation:
Alisa types 900 words in 20 minutes.
Her typing speed in words per minute is,
= \(\frac{900}{20}\)
= 45 words per minute

Question 3.
A 2-liter bottle is filled completely with water from a faucet in 10 seconds. How much water is filled into the bottle each second?
Answer:
200 ml bottle each second
Explanation:
A 2-liter bottle is filled completely with water from a faucet in 10 seconds.
water is filled into the bottle each second
= \(\frac{2000}{10}\)
= 200 ml bottle each second

Question 4.
Bill is paid $200 for 5 days of work. How much is he paid per day?
Answer:
$40 per day
Explanation;
Bill is paid $200 for 5 days of work.
Total amount he paid per day
= \(\frac{200}{5}\)
= $40 per day

Question 5.
Janice swims 450 meters in 5 minutes. Find her swimming speed in meters per minute.
Answer:
90 meters per minute.
Explanation:
Janice swims 450 meters in 5 minutes.
Her swimming speed in meters per minute.
= \(\frac{450}{5}\)
= 90 meters per minute.

Question 6.
A garden snail moves \(\frac{1}{6}\) foot in \(\frac{1}{3}\) hour. Find the speed of the snail in feet per hour.
Answer:
\(\frac{1}{2}\) feet per hour
Explanation:
A garden snail moves \(\frac{1}{6}\) foot in \(\frac{1}{3}\) hour.
The speed of the snail in feet per hour.
\(\frac{1}{6}\) ÷ \(\frac{1}{3}\)
= \(\frac{1}{6}\) X \(\frac{3}{1}\)
= \(\frac{1}{2}\) feet per hour

Question 7.
Math Journal A plumber pays $3.60 for 60 centimeters of pipe. Explain how the plumber can use the unit cost of the pipe to find the cost of buying 100 meters of the same kind of pipe. Show the calculations the plumber needs to make.
Answer:
The cost of 100 meters pipe = $600
Explanation:
A plumber pays $3.60 for 60 centimeters of pipe.
1 cm pipe = 3.60 ÷ 60
= 0.6 cm
1m = 100 cm
S0, 0.6 cm = 600

Question 8.
Rovan can make 48 tarts per hour. Copy and complete the table.

Answer:

a) At this rate, how many tarts can Rovan make in 6 hours?
Answer:
288 Tarts
Explanation:
From the above given information,
Number of tarts Rovan can make in 6 hours
6 x 48 = 288 tarts

b) At this rate, how long will she take to make 120 tarts?
Answer:
2.5 hours
Explanation:
From the above given information,
Number of hours she take to make tarts = 120 ÷ 48
= 2.5 hours

Question 9.
A sprinkler system is designed to water \(\frac{5}{8}\) acres of land in \(\frac{1}{4}\) hour. How many acres of land can it water in 1 hour?
Answer:
\(\frac{5}{2}\) per hour
Explanation:
A sprinkler system is designed to water \(\frac{5}{8}\) acres of land in \(\frac{1}{4}\) hour. Total acres of land can it water in 1 hour?
= \(\frac{5}{8}\) ÷ \(\frac{1}{4}\)
= \(\frac{5}{8}\) X \(\frac{4}{1}\)
= \(\frac{20}{8}\)
= \(\frac{5}{2}\)

Question 10.
The table shows the costs of three types of meat John bought at a supermarket. Copy and complete the table.

Which type of meat costs the least per pound?
Answer:
Chicken.
Explanation:

From the above table Chicken costs the least amount per pound.

Question 11.
Math Journal The table shows the data about distances and times for three sprinters.

Who is the fastest sprinter? Justify your answer.
Answer:
Smith is the fastest sprinter
Explanation:

Question 12.
A supermarket sells the three brands of rice shown in the table below.

Raimondo wants to buy 30 kilograms of rice.
a) Which brand of rice should he buy to get the best deal, assuming that all three brands are of the same quality?
Answer:
Brand A
Explanation:

b) How much will he save if he buys the cheapest brand of rice as compared to the most expensive one?
Answer:
$306
Explanation:
Cheapest brand price = $72
Most expensive brand price = $378
378 – 72 = 306

Go through the Math in Focus Grade 6 Workbook Answer Key Chapter 4 Lesson 4.2 Equivalent Ratios to finish your assignments.

Math in Focus Grade 6 Course 1 A Chapter 4 Lesson 4.2 Answer Key Equivalent Ratios

Math in Focus Grade 6 Chapter 4 Lesson 4.2 Guided Practice Answer Key

Complete.

Question 1.

The ratio of the number of pencils to the number of erasers is : .
Answer:
1 : 2
Explanation:
The ratio of the number of pencils to the number of erasers is
10 : 20 = 1 : 2

Question 2.

The ratio of the number of groups of pencils to the number of groups of erasers is : .
Answer:
1 : 2
Explanation:

The ratio of the number of groups of pencils to the number of groups of erasers is
5 : 10 = 1 : 2

Question 3.

The ratio of the number of groups of pencils to the number of groups of erasers is : .
Answer:
1 : 2
Explanation:

The ratio of the number of groups of pencils to the number of groups of erasers is = 1 : 2

Question 4.

The ratio : , : , and : are equivalent ratios.
Answer:
1 : 2
Explanation:
The ratio 10 : 20, 5 : 10, and  1:2 are equivalent ratios.

Question 5.
Express the ratio 12 : 64 in simplest form.

Answer:
3 : 16
Explanation:
the ratio 12 : 64 in simplest form.

Question 6.
Express the ratio 7 kg : 21 g in simplest form.
7 kg = g
7 kg : 21 g = g : g
= ÷ : ÷
= :
Answer:
1000 : 3
Explanation:
7 kg = 7000 g
7 kg : 21 g = 7000 g : 21 g
= 7000 ÷ 7 : 21 ÷ 7
= 1000 : 3

State whether each pair of ratios are equivalent.

Question 7.
7 : 8 and 8 : 7
Answer:
NO, pair of ratios are not equivalent.
Explanation:
7 : 8 and 8 : 7 are not equivalent

Question 8.
5 : 9 and 15 : 27
Answer:
YES, pair of ratios are equivalent
Explanation:
5 : 9 and 15 : 27
5 : 9 and 15 ÷ 3 : 27 ÷  3
5 : 9 and 5 : 9
5 : 9 and 15 : 27 are equivalent

Question 9.
12 : 13 and 24 : 39
Answer:
NO, pair of ratios are not equivalent.
Explanation:
12 : 13 and 24 : 39
For and 12, 24 common factor is 2,
where as 13 and 39 has common factor 13 .
12 : 13 and 24 : 39 are not equivalent ratios.

Question 10.
4 : 24 and 8 : 48
Answer:
YES, pair of ratios are equivalent.
Explanation:
4 : 24 and 8 : 48
4 : 24 and 8 ÷ 2 : 48 ÷ 2
4 : 24 and 8 : 48 are equivalent ratios.

Complete.

Question 11.
Use multiplication to find three ratios equivalent to 7 : 8.

Answer:

Explanation:

Question 12.
Use division to find all the whole number ratios that are equivalent to 24 : 96.

Answer:
24 : 96
Explanation:
24 ÷ 2 : 96 ÷ 2  => 12 : 48
24 ÷ 3 : 96 ÷ 3  => 8  : 32
24 ÷ 4 : 96 ÷ 4  => 6 : 24
24 ÷ 6 : 96 ÷ 6  => 4 : 16
24 ÷ 8 : 96 ÷ 8  => 3 : 12

Find the missing term in each pair of equivalent ratios.

Question 13.

Answer:
30 : 35
Explanation:
6 : 7
multiply both the numbers with 5

Question 14.

Answer:
4 : 6
Explanation:
4 : 6
multiply both the numbers with 7

Question 15.
48 : 64 = : 8
Answer: 6
Explanation:
48 : 64
divided both the numbers with 8

Question 16.
4 : 9 = 36 :
Answer: 81
Explanation:
4 : 9
multiply both the numbers with 9

Solve.

Question 17.
Selena and Drew each has a summer job. The table shows the amount of money they earn, based on the number of hours they work.

a) Express the ratio of Selena’s earnings to Drew’s earnings in simplest form.
Answer:
31 : 33
Explanation:
Selena and Drew each has a summer job.
for 1 day Selena earns $31
for 1 day Drew earns $33
ratio of Selena’s earnings to Drew’s earnings = 31 : 33

b) If Selena works 4 days, she will earn $.
If Selena works 30 days, she will earn $.
Answer:
$930
Explanation:
If Selena works 4 days, she will earn $124.
for 1 day Selena earns $31
for 4 days = 31 x 4 = 124
If Selena works 30 days, she will earn $930.
31 x 30 = 930

c) If Drew works 4 days, he will earn $.
If Drew works 30 days, he will earn $.
Answer:
If Drew works 4 days, he will earn $132.
If Drew works 30 days, he will earn $990.
Explanation:
for 1 day Drew earns $33
for 4 days = 33 x4 = 132
for 30 days = 30 x 33 = 990

Question 18.
A school organized a paper recycling competition. The table shows the amount of oil and the amount of water saved by recycling paper.

Answer:

a) How many gallons of water will be saved if 1 ton of paper is recycled?
Answer:
7,000gal
Explanation:
for 2 tons the amount of water saved is 14,000
So, for 1 ton 14000 ÷ 2 = 7000

b) Express the ratio of the amount of oil saved to the amount of water saved in simplest form.
Answer:
19 : 350
Explanation:
Weight of paper recycled for 1 ton =
Amount of oil saved : Amount of water saved
380 : 7000
= 19 : 350

c) How many gallons of oil will be saved if 2 tons of paper are recycled?
Answer:
760gal
Explanation:
for 1 ton of paper recycle need 380gal of oil saved.
for 2 tons oil saved = 380 x 2 = 760

d) How many gallons of water will be saved if 3 tons of paper are recycled?
Answer:
21,000gal
Explanation:
for 1 ton of paper recycle need 7,000gal of water saved.
for 2 tons water saved = 7000 x 3 = 21,000

e) How many gallons of oil and water will be saved if 4 tons of paper are recycled?

Answer:

Explanation:
for 1 ton of paper recycle need 7,000gal of water saved.
for 4 tons water saved = 7000 x 4 = 28,000gal
for 1 ton of paper recycle need 380gal of oil saved.
for 4 tons oil saved = 380 x 4 = 1520gal

Math in Focus Course 1A Practice 4.2 Answer Key

Express each ratio in simplest form.

Question 1.
13 : 39
Answer:
1 : 3
Explanation:
13 : 39
common factor for both the numbers is 13
so, 1 : 3

Question 2.
16 : 40
Answer:
2 : 5
Explanation:
16 : 40
common factor for both the numbers is 8
so, 2 : 5

Question 3.
25 : 15
Answer:
5 : 3
Explanation:
25 : 15
common factor for both the numbers is 5
so, 5 : 3

Question 4.
56 : 21
Answer:
8 : 3
Explanation:
56 : 21
common factor for both the numbers is 7
so, 8 : 3

Question 5.
30 : 54
Answer:
5 : 9
Explanation:
30 : 54
common factor for both the numbers is 6
so, 5 : 9

Question 6.
72 : 48
Answer:
8 : 6
Explanation:
72 : 38
common factor for both the numbers is 8
so, 8 : 6

Question 7.
26 cm : 4 m
Answer:
13cm : 200cm
Explanation:
26cm : 4m
convert m in cm
1m = 100 cm
4m = 400cm
26cm : 400cm
common factor for both the numbers is 2
so, 13 : 200

Question 8.
9 kg : 36 g
Answer:
250g : 1g
Explanation:
9kg : 36g
convert kg to g
1kg = 100o g
9kg = 9000g
250g : 1g
common factor for both the numbers is 9
so, 250 : 1

Question 9.
35 min : 2 h
Answer:
7min : 24min
Explanation:
35m : 2h
convert hr in min
1hr = 60min
2hr = 120min
35min : 120min
common factor for both the numbers is 5
so, 7 : 24

State whether each pair of ratios are equivalent.

Question 10.
11 : 17 and 17 : 11
Answer:
No, pair of ratios are not equal
Explanation:
11 : 17 and 17 : 11
the given numbers are not equivalent to their ratios.

Question 11.
7 : 11 and 21 : 33
Answer:
YES, pair of ratios are equal.
Explanation:
7 : 11 and 21 : 33
The common factor for both the numbers is 3
So, both are equivalent.

Question 12.
15 : 35 and 25 : 45
Answer:
NO, pair of ratios are not equal.
Explanation:
15 : 35 and 25 : 45
3 : 7  and 5 : 9
The common factor for both the numbers is 5
So, both are not equivalent.

Question 13.
15 : 20 and 20 : 25
Answer:
NO, pair of ratios are not equal.
Explanation:
15 : 20 and 20 : 25
3 : 4 and 4 : 5
So, both are not equivalent.

Question 14.
38 : 19 and 2 : 1
Answer:
NO, pair of ratios are not equal.
Explanation:
38 : 19 and 2 : 1
2:1 and 1:2 [19 x 2 =38 and 19 x 1 = 19]
So, both are not equivalent.

Question 15.
12 : 8 and 18 : 12
Answer:
YES, pair of ratios are equal.
Explanation:
12 : 8 and 18 : 12
3 : 2 and 3 : 2
So, both are equivalent.

Find the missing term in each pair of equivalent ratio.

Question 16.
7 : 9 = 49 :
Answer: 63
Explanation:
product of extremes = product of means
a:b = c:d
\(\frac{a}{b}\) = \(\frac{c}{d}\)
ad = bc
a = 7, b = 9, c = 49, d = x
7 x x = 9 x 49
7x = 441
x = \(\frac{441}{7}\)
x = 63

Question 17.
12 : 5 = 144 :
Answer: 60
Explanation:
product of extremes = product of means
a:b = c:d
\(\frac{a}{b}\) = \(\frac{c}{d}\)
ad = bc
a =12, b = 5, c = 144, d = x
12 x x = 5 x 144
12x = 720
x = \(\frac{720}{12}\)
x = 60

Question 18.
4 : 15 = 48 :
Answer: 180
Explanation:
product of extremes = product of means
a:b = c:d
\(\frac{a}{b}\) = \(\frac{c}{d}\)
ad = bc
a = 4, b = 15, c = 48, d = x
4 x x = 15 x 48
4x = 720
x = \(\frac{720}{4}\)
x = 180

Question 19.
7 : 13 = 77 :
Answer: 143
Explanation:
product of extremes = product of means
a:b = c:d
\(\frac{a}{b}\) = \(\frac{c}{d}\)
ad = bc
a = 7, b = 13, c = 77, d = x
7 x x = 13 x 77
7x = 1,001
x = \(\frac{1001}{7}\)
x = 143

Question 20.
45 : 36 = : 12
Answer: 15
Explanation:
product of extremes = product of means
a:b = c:d
\(\frac{a}{b}\) = \(\frac{c}{d}\)
ad = bc
a = 45, b = 36, c = x, d = 12
45 x 12 = 36 x x
36x = 540
x = \(\frac{540}{36}\)
x = 15

Question 21.
30 : 48 = : 8
Answer: 5
Explanation:
product of extremes = product of means
a:b = c:d
\(\frac{a}{b}\) = \(\frac{c}{d}\)
ad = bc
a = 30, b = 48, c = x, d = 8
30 x 8 = 48 x x
48x = 240
x = \(\frac{240}{48}\)
x = 5

Question 22.
72 : 84 = : 7
Answer: 6
Explanation:
product of extremes = product of means
a:b = c:d
\(\frac{a}{b}\) = \(\frac{c}{d}\)
ad = bc
a = 72, b = 84, c = x, d = 7
72 x 7 = 84 x x
504 = 84x
x = \(\frac{504}{84}\)
x = 6

Question 23.
121 : 88 = : 8
Answer: 11
Explanation
product of extremes = product of means
a:b = c:d
\(\frac{a}{b}\) = \(\frac{c}{d}\)
ad = bc
a = 121, b = 88, c = x, d = 8
121 x 8 = 88 x x
88x = 968
x = \(\frac{968}{88}\)
x = 11

Find the equivalent ratios.

Question 24.
Use multiplication to find three ratios equivalent to 8 : 12.
Answer:
The three equivalent ratios of 8 : 12 are 16 : 24; 24: 36 and 32 : 48.
Explanation:
first we need to write the given ratio as fraction = 8/12
= (8 × 2)/(12 × 2)
= 16/24
= 16 : 24 (one equivalent ratio),
So, 16 : 24 is an equivalent ratio of 8 : 12.
Similarly again, we need to write the given ratio 8 : 12 as fraction to get another equivalent ratio = 8/12
= (8 × 3)/(12 × 3)
= 24/36 is another equivalent ratio.
Similarly again, we need to write the given ratio 8 : 12 as fraction to get another equivalent ratio = 8/12
= (8 × 4)/(12 × 4)
= 32/48 is another equivalent ratio.
Therefore, the three equivalent ratios of 8 : 12 are 16 : 24; 24: 36 and 32 : 48.

Question 25.
Use division to find all the whole number ratios equivalent to 168 : 56.
Answer:
84 : 28; 56: 18 and 42 : 14.
Explanation:
first we need to write the given ratio as fraction = 168/56
= (168 ÷ 2)/(56 ÷ 2)
= 84/28
So, 168 : 56 is an equivalent ratio of 84 : 28.
Similarly again, we need to write the given ratio 168 : 56 as fraction to get another equivalent ratio = 168/56
= (168 ÷ 3)/(56 ÷ 3)
= 56/18
So, 168 : 56 is an equivalent ratio of 56 : 18.
Similarly again, we need to write the given ratio 168 : 56 as fraction to get another equivalent ratio = 168/56
= (168 ÷ 4)/(56 ÷ 4)
= 42/14
So, 168 : 56 is an equivalent ratio of 42 : 14.
Therefore, the two equivalent ratios of 168 : 56 are 84 : 28; 56: 18 and 42 : 14.

Copy and complete.

Question 26.
A manufacturer’s instruction states that 3 cups of cleaning agent should be diluted with 5 cups of water before use. Copy and complete the table.

Answer:

Explanation:
A manufacturer’s instruction states that 3 cups of cleaning agent should be diluted with 5 cups of water before use.
ratio of cleaning agent and water = 3 : 5

Find the missing term of each pair of equivalent ratio.

Question 27.
63 : 27 = 49 :
Answer: 21
Explanation:
the product of extremes = the product of means
a:b :: c:d  = > a x d = b x c
(27 x 49) /63 = 21
63 : 27 = 49 : 21

Question 28.
81 : 18 = 36 :
Answer: 8
Explanation:
the product of extremes = the product of means
a:b :: c:d  = > a x d = b x c
(18 x 36) /81 = 648/81 = 8
81 : 18 = 36 : 8

Question 29.
24 : 96 = 5 :
Answer: 20
Explanation:
the product of extremes = the product of means
a:b :: c:d  = > a x d = b x c
(96 x 5) /24 = 480/24 = 20
24 : 96 = 5 : 20

Question 30.
72 : 24 = 15 :
Answer: 5
Explanation:
the product of means = the product of extremes
a:b :: c:d  = > a x d = b x c
(24 x 15) /72 = 360/72 = 5
72 : 24 = 15 : 5

Question 31.
60 : 144 = : 60
Answer: 25
Explanation:
the product of means = the product of extremes
a:b :: c:d  = > a x d = b x c
(60 x 60) /144= 3600/144 = 25
60 : 144 = 25 : 60

Question 32.
125 : 80 = : 48
Answer: 75
Explanation:
the product of extremes = the product of means
a:b :: c:d  = > a x d = b x c
(125 x 48) /80 = 6000/80 = 75
125 : 80 = 75 : 48

Question 33.
90 : 15 = : 7
Answer: 42
Explanation:
the product of extremes = the product of means
a:b :: c:d  = > a x d = b x c
(90 x 7) /15 = 630/15 = 42
90 : 15 = 42 : 7

Question 34.
98 : 112 = 63 :
Answer: 72
Explanation:
the product of extrems = the product of means
a:b :: c:d  = > a x d = b x c
(112 x 63) /98 = 7056/98 = 72
98 : 112 = 63 : 72

Solve.

Question 35.
Judy uses 5 ounces of lemonade concentrate for every 9 ounces of orange juice concentrate to make a fruit punch.

a) Find the ratio of the number of ounces of orange juice concentrate to the number of ounces of lemonade concentrate she uses.
Answer:
9 : 5
Explanation:
Judy uses 5 ounces of lemonade concentrate for every 9 ounces of orange juice concentrate to make a fruit punch.
The ratio of the orange juice and lemonade = 9 : 5

b) If Judy uses 36 ounces of orange juice concentrate to make the fruit punch, how many ounces of lemonade concentrate does she use?
Answer:
20 oz
Explanation:
If Judy uses 36 ounces of orange juice concentrate to make the fruit punch,
how many ounces of lemonade concentrate does she use
9 : 5 = 36 : ?
the product of means = the product of extremes
a:b :: c:d  = > a x d = b x c
(36 x 5) /9 = 180/9 = 20
9 : 5 = 36 : 20

c) If Judy uses 45 ounces of lemonade concentrate to make the fruit punch, how many ounces of orange juice concentrate does she use?
Answer:
81 oz
Explanation:
If Judy uses 45 ounces of lemonade concentrate to make the fruit punch,
how many ounces of orange juice concentrate does she use
9 : 5 = ? : 45
the product of means = the product of extremes
a:b :: c:d  = > a x d = b x c
(9 x 45) /5 = 405/5 = 81
9 : 5 = 81 : 45

Question 36.
In a science experiment, Farah mixed a salt solution and vinegar in the ratio 3 : 7.
a) If she used 262.8 milliliters of salt solution1 how much vinegar did she use?
Answer:
183.9ml
Explanation:
The ratio of vinegar to total mixture is 7 : 10
The 10 is 3 + 7.  Out of 10 ml, 3 will be Salt and 7 will be Vinegar.
7/10 = V/262.8
1839.6 = 10V
V = 183.9 ml

b) If 0.56 liter of vinegar was used, how much salt solution did she use?
Answer:
78.84 ml
Explanation:
The ratio of salt to total mixture is 3 : 10
The 10 is 3 + 7.  Out of 10 ml, 3 will be Salt and 7 will be Vinegar.
3/10 = Salt/262.8
262.8 x 3 = 10S
788.4 = 10S
S = 788.4/10
S = 78.84 ml

Question 37.
A fruit seller packs different fruits into baskets of the same size. The ratio of the weight of bananas to the weight of apples to the weight of pears s the same for all the baskets. The table shows the different weights of fruits in the baskets. Copy and complete the table.

Answer:

Explanation:
A fruit seller packs different fruits into baskets of the same size.
The ratio of the weight of bananas to the weight of apples to the weight of pears are the same for all the baskets.
The table shows the different weights of fruits in the baskets.
Find the largest common factor, of all the fruits.
46 apples = 8 x 6
30 pears = 5 x 6
24 bananas = 4 x 6
6 baskets each with 8 apples, 5 pears and 4 bananas.

Go through the Math in Focus Grade 6 Workbook Answer Key Chapter 3 Review Test to finish your assignments.

Math in Focus Grade 6 Course 1 A Chapter 3 Review Test Answer Key

Concepts and Skills

Divide.

Question 1.
15 ÷ \(\frac{1}{3}\)

Answer: 45
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.

\(\frac{15}{1}\) ÷ \(\frac{1}{3}\)

\(\frac{15}{1}\) x \(\frac{3}{1}\)

15 x 3 = 45

Question 2.
24 ÷ \(\frac{1}{6}\)

Answer: 144
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.

24 ÷ \(\frac{1}{6}\)

24 x 6 = 144

Question 3.
\(\frac{3}{8}\) ÷ \(\frac{3}{4}\)

Answer:
\(\frac{1}{2}\)

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.

\(\frac{3}{8}\) ÷ \(\frac{3}{4}\)

\(\frac{3}{8}\) x \(\frac{4}{3}\)

\(\frac{12}{24}\) = \(\frac{1}{2}\)

Question 4.
\(\frac{7}{12}\) ÷ \(\frac{1}{3}\)

Answer:

\(\frac{7}{3}\)

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.

\(\frac{7}{12}\) ÷ \(\frac{1}{3}\)

\(\frac{7}{12}\) x \(\frac{3}{1}\)

\(\frac{21}{12}\) = \(\frac{7}{3}\)

Multiply.

Question 5.
0.3 × 8
Answer: 2.4
Explanation:
Multiplication of decimals is done by ignoring the decimal point and multiply the numbers,
then the number of decimal places in the product is equal to the total number of decimal places in both the given numbers.
0.3 x 8 = 2.4

Question 6.
6 × 0.7
Answer: 4.2
Explanation:
Multiplication of decimals is done by ignoring the decimal point and multiply the numbers,
then the number of decimal places in the product is equal to the total number of decimal places in both the given numbers.
6 x 0.7 = 4.2

Question 7.
0.28 × 6
Answer: 1.68
Explanation:
Multiplication of decimals is done by ignoring the decimal point and multiply the numbers,
then the number of decimal places in the product is equal to the total number of decimal places in both the given numbers.
0.28 x 6 = 1.68

Question 8.
7 × 0.068
Answer: 0.476
Explanation:
Multiplication of decimals is done by ignoring the decimal point and multiply the numbers,
then the number of decimal places in the product is equal to the total number of decimal places in both the given numbers.
7 x 0.068 = 0.476

Question 9.
0.3 × 0.6
Answer: 0.18
Explanation:
Multiplication of decimals is done by ignoring the decimal point and multiply the numbers,
then the number of decimal places in the product is equal to the total number of decimal places in both the given numbers.
0.3 x 0.6 = 0.18

Question 10.
0.5 × 0.8
Answer: 0.4
Explanation:
Multiplication of decimals is done by ignoring the decimal point and multiply the numbers,
then the number of decimal places in the product is equal to the total number of decimal places in both the given numbers.
0.5 x 0.8 = 0.4

Question 11.
5.7 × 0.4
Answer: 2.28
Explanation:
Multiplication of decimals is done by ignoring the decimal point and multiply the numbers,
then the number of decimal places in the product is equal to the total number of decimal places in both the given numbers.
5.7 x 0.4 = 2.28

Question 12.
9.3 × 0.89
Answer: 8.277
Explanation:
Multiplication of decimals is done by ignoring the decimal point and multiply the numbers,
then the number of decimal places in the product is equal to the total number of decimal places in both the given numbers.
9.3 x 0.89 = 8.277

Divide.

Question 13.
6 ÷ 0.6
Answer: 10
Explanation:
To divide a number by a decimal number,
multiply the divisor by as many tens as necessary until we get a whole number,
then remember to multiply the dividend by the same number of tens.
6 ÷ 0.6 = 10

Question 14.
8 ÷ 0.4
Answer: 20
Explanation:
To divide a number by a decimal number,
multiply the divisor by as many tens as necessary until we get a whole number,
then remember to multiply the dividend by the same number of tens.

Question 15.
35 ÷ 0.7
Answer: 50
Explanation:
To divide a number by a decimal number,
multiply the divisor by as many tens as necessary until we get a whole number,
then remember to multiply the dividend by the same number of tens.

Question 16.
88 ÷ 0.2
Answer: 440
Explanation:
To divide a number by a decimal number,
multiply the divisor by as many tens as necessary until we get a whole number,
then remember to multiply the dividend by the same number of tens.

Question 17.
5 ÷ 0.25
Answer: 20
Explanation:
To divide a number by a decimal number,
multiply the divisor by as many tens as necessary until we get a whole number,
then remember to multiply the dividend by the same number of tens.

Question 18.
8 ÷ 0.16
Answer: 50
Explanation:
To divide a number by a decimal number,
multiply the divisor by as many tens as necessary until we get a whole number,
then remember to multiply the dividend by the same number of tens.

Question 19.
96 ÷ 0.16
Answer: 600
Explanation:
To divide a number by a decimal number,
multiply the divisor by as many tens as necessary until we get a whole number,
then remember to multiply the dividend by the same number of tens.

Question 20.
396 ÷ 0.36
Answer: 1,100
Explanation:
To divide a number by a decimal number,
multiply the divisor by as many tens as necessary until we get a whole number,
then remember to multiply the dividend by the same number of tens.

Question 21.
0.87 ÷ 0.03
Answer: 29
Explanation:
To divide a decimal number by a decimal number,
multiply the divisor by as many tens as necessary until we get a whole number,
then remember to multiply the dividend by the same number of tens.

Question 22.
0.98 ÷ 0.7
Answer: 1.4
Explanation:
To divide a decimal number by a decimal number,
multiply the divisor by as many tens as necessary until we get a whole number,
then remember to multiply the dividend by the same number of tens.

Problem Solving

Solve. Show your work.

Question 23.
In January, Jane volunteered at a hospital for a total of 12 hours. She spent \(\frac{4}{5}\) hour at the hospital every time she volunteered. How many times did Jane volunteer in January?
Answer:
15 times
Explanation:
In January, Jane volunteered at a hospital for a total of 12 hours.
She spent \(\frac{4}{5}\) hour at the hospital every time she volunteered.
Number times Jane volunteer in January
\(\frac{12}{1}\) ÷ \(\frac{4}{5}\)

\(\frac{12}{1}\) x \(\frac{5}{4}\)
3 x 5 = 15

Question 24.
Paul is making loaves of raisin bread to sell at a fundraising event. The recipe calls for \(\frac{1}{3}\) cup of raisins for each loaf, and Paul has 3\(\frac{1}{4}\) cups of raisins.
a) How many loaves can Paul make?
Answer:
Paul can make 9 loaves.
Explanation:
Paul has 3\(\frac{1}{4}\) cups of raisins.

The recipe calls for \(\frac{1}{3}\) cup of raisins for each loaf,

Number of loaves can Paul make are

\(\frac{13}{4}\) ÷ \(\frac{1}{3}\)

\(\frac{13}{4}\) x \(\frac{3}{1}\)

39 ÷ 4 = 9\(\frac{3}{4}\)

b) How many cups of raisins will he have left over?
Answer:
\(\frac{3}{4}\)

Explanation:
Paul has 3\(\frac{1}{4}\) cups of raisins.

The recipe calls for \(\frac{1}{3}\) cup of raisins for each loaf,
Number of loaves can Paul make are

\(\frac{13}{4}\) ÷ \(\frac{1}{3}\)

\(\frac{13}{4}\) x \(\frac{3}{1}\)

39 ÷ 4 = 9\(\frac{3}{4}\)

Number of loaves left over are \(\frac{3}{4}\)

Question 25.
Jane has a dog that eats 0.8 pound of dog food each day. She buys a 40-pound bag of dog food. How many days will this bag of dog food last?
Answer:
50 days
Explanation:
Jane has a dog that eats 0.8 pound of dog food each day.
She buys a 40-pound bag of dog food.
Number of days will this bag of dog food last

Question 26.
Mervin had some cartons of milk. He sold \(\frac{2}{5}\) of the cartons of milk in the morning. He then sold \(\frac{3}{4}\) of the remainder in the afternoon. 24 more cartons of milk were sold in the afternoon than in the morning. How many cartons of milk did Mervin have at first?
Answer:
480 cartons of milk.
Explanation:
total 5 parts
each part can be sub divided in to 4 parts each
in each  box 24 cartons of milk
if total boxes are 20 and each box contains 24 cartons of milk
then the total number of milk is 24 x 20 = 480

Question 27.
Alice baked a certain number of pies. She gave \(\frac{1}{8}\) of the pies to her friends and \(\frac{1}{4}\) of the remainder to her neighbor. She was left with 63 pies. How many pies did Alice bake at first?
Answer:
96 pies
Explanation:
Let the original count of pies be represented by $x$She gave \(\frac{1}{8}\) of the pies to her friend

Question 28.
At a concert, \(\frac{2}{5}\) of the people were men. There were 3 times as many women as children. If there were 45 more men than children, how many people were there at the concert?
Answer:
180 people
Explanation:
Let people= p
men = \(\frac{2}{5}\)p
Let children = c
women = 3c
men = \(\frac{2}{5}\)p = c+45
people = men + women + children
= \(\frac{2}{5}\)p + 3c + c

= \(\frac{2}{5}\)p + 4c

= \(\frac{3}{5}\)p = 4c
\(\frac{2}{5}\)p = c + 45
Children = 27
people = 180
So, 180 people at the show.
women = 3c = 3 x 27 = 81
men = \(\frac{2}{5}\)p = c+45
= 72

Question 29.
\(\frac{3}{4}\) of the students in a school were girls and the rest were boys. \(\frac{2}{3}\) of the girls and \(\frac{1}{2}\) of the boys attended the school carnival. Find the total number of students in the school if 330 students did not attend the carnival.
Answer:
Total number of students 880
Explanation:
girls = 3boys,
since \(\frac{3}{4}\) = 3 x \(\frac{1}{4}\)

\(\frac{2}{3}\) girls + \(\frac{1}{2}\) boys attended

\(\frac{2}{3}\) x 3boys + \(\frac{1}{2}\) boys

= \(\frac{5}{2}\) boys attended

subtract that from the total (boys + girls) students:
boys + girls – \(\frac{5}{2}\) boys = 330

4b – \(\frac{5}{2}\) boys = 330

\(\frac{3}{2}\) boys = 330
boys = 220
so, girls = 3b
= 220 x 3 = 660
Boys + Girls = (220 + 660) = 880
There are 880 students
660 girls and 220 boys
440 girls and 110 boys attended = 550
the remaining 330 did not attend.

Question 30.
At a baseball game, there were three times as many males as females. \(\frac{5}{6}\) of the males were boys and the rest were men. \(\frac{2}{3}\) of the females were girls and the rest were women. Given that there were 121 more boys than girls, how many adults were there at the baseball game?
Answer:
55 adults
Explanation:
Number of females x
Number of girls = \(\frac{2}{3}\) x – girls

Number of women = \(\frac{1}{3}\) x

Number of males = 3x

Number of boys = \(\frac{5}{6}\) x 3x

= \(\frac{5}{2}\) x – boys

Number of men = \(\frac{1}{6}\) x 3x

= \(\frac{1}{2}\) x – men
So, \(\frac{5}{2}\) – 121 = \(\frac{2}{3}\) x

\(\frac{15}{6}\)x – \(\frac{4}{6}\)x = 121

\(\frac{11}{6}\)x = 121
x = 66
Men = \(\frac{1}{2}\) x 66 = 33

Women = \(\frac{1}{3}\) x 66 = 22
Adults = Men + Women
Adults = 33 + 22 = 55
Question 31.
Mr. Thomas spent $1,600 of his savings on a television set and \(\frac{2}{5}\) of the remainder on a refrigerator. He had \(\frac{1}{3}\) of his original amount of savings left.
a) What was Mr. Thomas’s original savings?
Answer:
$3600
b) What was the cost of the refrigerator?
Answer:
$2400
Explanation:
Let $\mathrm{x\; is\; the\; savings,}$x−(1600+(2x− \(\frac{1600}{5}\)) = x3
Because x is total amount of savings and he spent $1600
So, now he has $2÷5$ of the total amount – $1600
($x÷3$ Because he has $1÷3$ of his original amount of savings)
$x+(-1600-\frac{2x-\; \backslash (\backslash frac\{3200\}\{5\}\backslash )}{})$ = $\frac{x}{3}$3x+3(−1600+−2x+ \(\frac{3200}{5}\))=x
3x+(−4800+−6x+ \(\frac{9600}{5}\))=x
3x+(−4800−1.2x+1920)=x
2x=4800+1.2x−1920.
8x=2880
x=3600

Question 32.
Sue buys 8.5 pounds of chicken to make tacos. She uses 0.3 pound of chicken for each taco.
a) How many tacos can Sue make?
Answer:
Sue can make 28 tacos
b) How many pounds of chicken are left over?
Answer:
\(\frac{1}{3}\) left over
Explanation:
Sue buys 8.5 pounds of chicken to make tacos.
She uses 0.3 pound of chicken for each taco.
\(\frac{8.5}{0.3}\)
= 28 \(\frac{1}{3}\)

Go through the Math in Focus Grade 6 Workbook Answer Key Chapter 3 Lesson 3.2 Multiplying Decimals to finish your assignments.

Math in Focus Grade 6 Course 1 A Chapter 3 Lesson 3.2 Answer Key Multiplying Decimals

Math in Focus Grade 6 Chapter 3 Lesson 3.2 Guided Practice Answer Key

Complete.

Question 1.
Multiply 0.9 by 4.

Answer:
The multiplication of  0.9 by 4 is 3.6.

Explanation:
The multiplication of  0.9 by 4 is 3.6.

Question 2.
Multiply 0.025 by 3.

Answer:

Explanation:
The multiplication of 0.025 by 3 is 0.075.

Multiply.

Question 3.

Answer:
The multiplication of 0.07 × 9 is 0.63.

Explanation:
The multiplication of 0.07 × 9 is 0.63.

Question 4.

Answer:
The multiplication of 0.14 × 3 is 0.42.

Explanation:
The multiplication of 0.14 × 3 is 0.42.

Question 5.

Answer:
The multiplication of 0.045 × 7 is 0.315.

Explanation:
The multiplication of 0.045 × 7 is 0.315.

Write in vertical form. Then multiply and decide where to place the decimal point.

Question 6.
0.32 × 8
Answer:
0.32 × 8 = 2.56.

Explanation:
The multiplication of 0.32 × 8 is 2.56.

Question 7.
9 × 0.24
Answer:
9 × 0.24 = 2.16.

Explanation:
The multiplication of 9 × 0.24 is 2.16.

Question 8.
0.057 × 6
Answer:
0.057 × 6 = 0.342.

Explanation:
The multiplication of 0.057 × 6 is 0.342.

Complete.

Question 9.
Find 0.3 × 0.6.

Answer:
The multiplication of 0.3 × 0.6 is 0.18.

Explanation:
The multiplication of 0.3 × 0.6 is
0.3 × 0.6 = \(\frac{3}{10}\) × \(\frac{6}{10}\)
= \(\frac{18}{100}\)
= 0.18.

Question 10.
Find 0.9 × 0.8.

Answer:
The multiplication of 0.9 × 0.8 is 0.72.

Explanation:
The multiplication of 0.9 × 0.8 is

Complete.

Question 11.
Find 3.2 × 0.6

Answer:
The multiplication of 3.2 × 0.6 is 1.92.

Explanation:
The multiplication of 3.2 × 0.6 is

Write in vertical form. Then multiply and decide where to place the decimal point.

Question 12.
4.3 × 5.7
Answer:
The multiplication of 4.3 × 5.7 is 24.51.

Explanation:
The multiplication of 4.3 × 5.7 is

Complete.

Question 13.
Find 0.89 × 0.4

Answer:
The multiplication of 0.89 × 0.4 is 0.356.

Explanation:
The multiplication of 0.89 × 0.4 is

Write in vertical form. Then multiply and decide where to place the decimal point.

Question 14.
4.3 × 5.7
Answer:
The multiplication of 4.3 × 5.7 is 24.51.

Explanation:
The multiplication of 4.3 × 5.7 is

Hands-On Activity

Materials

• graph paper
• ruler

Finding the factors of a decimal.

Step 1.
Draw four 10 × 10 squares on graph paper.

Step 2.
Mark each side from 0 to 1 as shown.

 

Step 3.
a) Find two decimals that give a product of 0.12.
× = 0.12
Show and shade 0.12 on the grids in two different ways.
Example
0.2 × 0.6 = 0.12

Answer:

b) Find two decimals that give a product of 0.36. Show and shade 0.36 on the grids in two different ways.
Answer:
0.6 × 0.6 = 0.36.

Explanation:
The two decimals that give a product of 0.36 is 0.6 × 0.6.

Math in Focus Course 1A Practice 3.2 Answer Key

Complete.

Question 1.
0.3 × 4 is the same as groups of .
Answer:
0.3 groups of 4.

Explanation:
Here 0.3 × 4 is the same as 0.3 groups of 4.

Question 2.
7 × 0.8 is the same as groups of .
Answer:
7 groups of 0.8.

Explanation:
Here 7 × 0.8 is the same as 7 groups of 0.8.

Write a multiplication statement that represents each number line.

Question 3.

Answer:
3×0.5 = 1.5.

Explanation:
The multiplication statement that represents the number line is 3×0.5 = 1.5.

Question 4.

Answer:
6×0.9 = 5.4.

Explanation:
The multiplication statement that represents the number line is 6×0.9 = 5.4.

Write in vertical form. Then multiply.

Question 5.
0.9 × 12
Answer:
The multiplication of 0.9 × 12 is 10.8.

Explanation:
The multiplication of 0.9 × 12 is
9                          0.9
×  12                       × 12
_____               ———
108                     10.8

Question 6.
0.47 × 5
Answer:
The multiplication of 0.47 × 5 is 2.35.

Explanation:
The multiplication of 0.47 × 5 is
47 × 5 = 235
0.47 × 5 = 2.35

Question 7.
0.063 × 9
Answer:
0.063 × 9 = 0.67.

Explanation:
The multiplication of 0.063 × 9 is
63 × 9 = 567
0.063 × 9 = 0.567.

Question 8.
00.85 × 11
Answer:
The multiplication of 00.85 × 11 is  9.35.

Explanation:
The multiplication of 00.85 × 11 is
85 × 11 = 935,
00.85 × 11 = 9.35.

Question 9.
00.1 × 0.2
Answer:
The multiplication of 00.1 × 0.2 is 0.02.

Explanation:
The multiplication of 00.1 × 0.2 is
1 × 2 = 2,
00.1 × 0.2 = 0.02.

Question 10.
0.2 × 0.3
Answer:
The multiplication of 0.2 × 0.3 is

Explanation:
The multiplication of 0.2 × 0.3 is
2 × 3 = 6
0.2 × 0.3 = 0.06.

Question 11.
0.4 × 0.4
Answer:
The multiplication of 0.4 × 0.4 is 0.16.

Explanation:
The multiplication of 0.4 × 0.4 is
4 × 4 = 16,
0.4 × 0.4 = 0.16.

Question 12.
0.6 × 0.7
Answer:
The multiplication of 0.6 × 0.7 is 0.42.

Explanation:
The multiplication of 0.6 × 0.7 is
6 × 7 = 42,
0.6 × 0.7 = 0.42.

Question 13.
0.7 × 0.9
Answer:
The multiplication of 0.7 × 0.9 is 0.63.

Explanation:
The multiplication of 0.7 × 0.9 is
7 × 9 = 63,
0.7 × 0.9 = 0.63.

Multiply mentally.

Question 14.
0.7 × 8
Answer:
The mental multiplication of 0.7 × 8 is 5.6.

Explanation:
The mental multiplication of 0.7 × 8 is
0.7 × 8 = \(\frac{7}{10}\) × 8
= \(\frac{7}{5}\) × 4
= \(\frac{28}{5}\)
= 5.6.

Question 15.
0.9 × 9
Answer:
The mental multiplication of 0.9 × 9 is 8.1.

Explanation:
The mental multiplication of 0.9 × 9 is
0.9 × 9 = \(\frac{9}{10}\) × 9
= \(\frac{81}{10}\)
= 8.1.

Question 16.
0.9 × 11
Answer:
The mental multiplication of 0.9 × 11 is 9.9.

Explanation:
The mental multiplication of 0.9 × 11 is
0.9 × 11 = \(\frac{9}{10}\) × 11
= \(\frac{99}{10}\)
= 9.9.

Question 17.
0.7 × 0.4
Answer:
The mental multiplication of 0.7 × 0.4 is \(\frac{14}{50}\).

Explanation:
The mental multiplication of 0.7 × 0.4 is
0.7 × 0.4 = \(\frac{7}{10}\) × \(\frac{4}{10}\)
= \(\frac{7}{10}\) × \(\frac{2}{5}\)
= \(\frac{14}{50}\).

Question 18.
0.8 × 0.6
Answer:
The mental multiplication of 0.8 × 0.6 is 0.48.

Explanation:
The mental multiplication of 0.8 × 0.6 is
0.8 × 0.6 = \(\frac{8}{100}\) × \(\frac{6}{100}\)
= \(\frac{48}{100}\)
= 0.48.

Question 19.
0.3 × 0.9
Answer:
The mental multiplication of 0.3 × 0.9 is 0.27.

Explanation:
The mental multiplication of 0.3 × 0.9 is
0.3 × 0.9 = \(\frac{3}{100}\) × \(\frac{9}{100}\)
= \(\frac{27}{100}\)
= 0.27.

Question 20.
0.7 × 0.7
Answer:
The mental multiplication of 0.7 × 0.7 is 0.49.

Explanation:
The mental multiplication of 0.7 × 0.7 is
0.7 × 0.7 = \(\frac{7}{100}\) × \(\frac{7}{100}\)
= \(\frac{49}{100}\)
= 0.49.

Question 21.
0.5 × 0.9
Answer:
The mental multiplication of 0.5 × 0.9 is 0.45.

Explanation:
The mental multiplication of 0.5 × 0.9 is
0.5 × 0.9 = \(\frac{5}{10}\) × \(\frac{9}{10}\)
= \(\frac{45}{100}\)
= 0.45.

Question 22.
0.8 × 0.9
Answer:
The mental multiplication of 0.8 × 0.9 is 0.72.

Explanation:
The mental multiplication of 0.5 × 0.9 is
0.8 × 0.9 = \(\frac{8}{10}\) × \(\frac{9}{10}\)
= \(\frac{72}{100}\)
= 0.72.

Question 23.
0.15 × 6
Answer:
The mental multiplication of 0.15 × 6 is 0.9.

Explanation:
The mental multiplication of 0.15 × 6 is
0.15 × 6 = \(\frac{15}{100}\) × \(\frac{6}{10}\)
= \(\frac{90}{100}\)
= 0.9

Question 24.
0.22 × 4
Answer:
The mental multiplication of 0.22 × 4 is 0.88.

Explanation:
The mental multiplication of 0.22 × 4 is
0.22 × 4 = \(\frac{22}{100}\) ×4
= \(\frac{88}{100}\)
= 0.88.

Question 25.
0.25 × 3
Answer:
The mental multiplication of 0.25 × 3 is 0.75.

Explanation:
The mental multiplication of 0.25 × 3 is
0.25 × 3 = \(\frac{25}{100}\) × 3
= \(\frac{75}{100}\)
= 0.75.

Question 26.
0.032 × 5
Answer:
The mental multiplication of 0.032 × 5 is 0.16.

Explanation:
The mental multiplication of 0.032 × 5 is
0.032 × 5 = \(\frac{32}{1000}\) × 5
= \(\frac{160}{1000}\)
= 0.16.

Question 27.
0.04 1 × 8
Answer:
The mental multiplication of 0.04 1 × 8 is 0.328.

Explanation:
The mental multiplication of 0.04 1 × 8 is
0.04 1 × 8 = \(\frac{41}{1000}\) × 8
= \(\frac{41}{1000}\) × 8
= \(\frac{328}{1000}\)
= 0.328.

Question 28.
0.055 × 9
Answer:
The mental multiplication of 0.055 × 9 is 0.495.

Explanation:
The mental multiplication of 0.055 × 9 is
0.055 × 9 = \(\frac{55}{1000}\) × 9
= \(\frac{495}{1000}\)
= 0.495.

Write in vertical form. Then multiply.

Question 29.
1.2 × 0.6
Answer:
The multiplication of 1.2 × 0.6 is 0.72.

Explanation:
The multiplication of 1.2 × 0.6 is
12 × 06 = 72,
1.2 × 0.6 = 0.72.

Question 30.
0.89 × 1.2
Answer:
The multiplication of 0.89 × 1.2 is 1.068.

Explanation:
The multiplication of 0.89 × 1.2 is
89 × 12 = 1,068,
0.89 × 1.2 = 1.068.

Question 31.
2.3 × 1.5
Answer:
The multiplication of 2.3 × 1.5 is 3.45.

Explanation:
The multiplication of 2.3 × 1.5 is
23 × 15 = 345,
2.3 × 1.5 = 3.45.

Question 32.
3.4 × 6.7
Answer:
The multiplication of 3.4 × 6.7 is 22.78.

Explanation:
The multiplication of 3.4 × 6.7 is
34 × 67 = 2,278.
3.4 × 6.7 = 22.78.

Question 33.
4.9 × 6.3
Answer:
The multiplication of 4.9 × 6.3 is 30.87.

Explanation:
The multiplication of 4.9 × 6.3 is
49 × 63 = 3,087,
4.9 × 6.3 = 30.87.

Question 34.
5.8 × 7.8
Answer:
The multiplication of 5.8 × 7.8 is 45.24.

Explanation:
The multiplication of 5.8 × 7.8 is
58 × 78 = 4,524,
5.8 × 7.8 = 45.24.

Question 35.
0.46 × 1.3
Answer:
The multiplication of 0.46 × 1.3 is 0.598.

Explanation:
The multiplication of 0.46 × 1.3 is
46 × 13 = 598.
0.46 × 1.3 = 0.598.

Question 36.
0.705 × 0.5
Answer:
The multiplication of 0.705 × 0.5 is 0.3525.

Explanation:
The multiplication of 0.705 × 0.5 is
705 × 5 = 3,525,
0.705 × 0.5 = 0.3525.

Question 37.
0.597 × 0.21
Answer:
The multiplication of 0.597 × 0.21 is 0.12537.

Explanation:
The multiplication of 0.597 × 0.21 is
597 × 21 = 12,537,
0.597 × 0.21 = 0.12537.

Question 38.
Math Journal Your friend knows how to find the product \(\frac{57}{100}\) × \(\frac{3}{10}\). However, your friend does not know how to find the product 0.57 × 0.3. Write an explanation that will help your friend understand how to multiply the two decimals.
Answer:
Here the product of \(\frac{57}{100}\) × \(\frac{3}{10}\) is \(\frac{171}{100}\) which is 1.71 and to find the product of 0.57 × 0.3 first we will multiply 57 × 3 which is 171 and then we will place decimal, so the product will be 1.71.

Go through the Math in Focus Grade 6 Workbook Answer Key Chapter 3 Lesson 3.1 Dividing Fractions to finish your assignments.

Math in Focus Grade 6 Course 1 A Chapter 3 Lesson 3.1 Answer Key Dividing Fractions

Math in Focus Grade 6 Chapter 3 Lesson 3.1 Guided Practice Answer Key

Solve.

Question 1.
Rina cut 3 paper squares into a number of equal pieces. Each piece was \(\frac{1}{6}\) of a square. Into how many pieces did Rina cut the 3 paper squares?


The model shows that:
Number of one-sixths in 1 paper square =
Number of one-sixths in 3 paper squares = ×

Answer:
Rina cut the 3 paper squares into 18 pieces.

Explanation:
Given that Rina cut 3 paper squares into a number of equal pieces and each piece was \(\frac{1}{6}\) of a square. So the number of many pieces did Rina cut the 3 paper squares is 3 ÷ \(\frac{1}{6}\)
= 3 × 6
= 18 pieces.

Divide.

Question 2.
3 ÷ \(\frac{1}{5}\) =
Answer:
3 ÷ \(\frac{1}{5}\) = 15.

Explanation:
The division of 3 ÷ \(\frac{1}{5}\)  is 3 × 5 = 15.

Question 3.
7 ÷ \(\frac{1}{4}\) =
Answer:
7 ÷ \(\frac{1}{4}\) = 28.

Explanation:
The division of 7 ÷ \(\frac{1}{4}\) is 7 × 4 = 28.

Question 4.
4 ÷ \(\frac{1}{2}\) =
Answer:
4 ÷ \(\frac{1}{2}\) = 8.

Explanation:
The division of 4 ÷ \(\frac{1}{2}\) is 4 × 2 = 8.

Question 5.
5 ÷ \(\frac{1}{3}\) =
Answer:
5 ÷ \(\frac{1}{3}\) = 15.

Explanation:
The division of 5 ÷ \(\frac{1}{3}\) is 5 × 3 = 15.

Question 6.
6 ÷ \(\frac{1}{5}\) =
Answer:
6 ÷ \(\frac{1}{5}\) = 15.

Explanation:
The division of 6 ÷ \(\frac{1}{5}\) is 6 × 5 = 15.

Question 7.
8 ÷ \(\frac{1}{8}\) =
Answer:
8 ÷ \(\frac{1}{8}\) = 64.

Explanation:
The division of 8 ÷ \(\frac{1}{8}\) is 8 × 8 = 64.

Hands-On

Activity Materials:

• 5 paper strips

Dividing Whole Numbers By A Fraction

Use 5 paper strips of the same size and length. Each strip represents 1 whole.

Step 1.
Take 2 paper strips. Divide each of them into thirds using vertical lines and place them as shown. Then find 2 ÷ \(\frac{2}{3}\)

Refer to your model.
Complete:
There are two-thirds in the 2 paper strips.
So, 2 ÷ \(\frac{2}{3}\) = .

Step 2.
Divide each of the other 3 paper strips into fourths using vertical lines and place them as shown. Then find 3 ÷ \(\frac{3}{4}\).

Refer to your model.
Complete:
There are three-fourths in the 3 paper strips.
So, 3 ÷ \(\frac{3}{4}\) = .

Complete:

Question 8.
Find 7 ÷ \(\frac{3}{4}\)

Question 9.
Mrs. Johnson bought 6 pizzas. She cut them into many equal pieces for the students in her class. Each piece was \(\frac{3}{10}\) of a whole pizza. How many students were there in the class if each child received only one piece of pizza?

Answer:
There were 20 students in the class.

Explanation:
Given that Mrs. Johnson bought 6 pizzas and she cut them into many equal pieces for the students in her class. Each piece was \(\frac{3}{10}\) of a whole pizza. So the number of students were there in the class is 6 ÷ \(\frac{3}{10}\) which is 6 × \(\frac{10}{3}\)
= 2 × 10
= 20 students.

Divide. Express the quotient in simplest form.

Question 10.
4 ÷ \(\frac{4}{7}\)
Answer:
4 ÷ \(\frac{4}{7}\) = 7.

Explanation:
The division of 4 ÷ \(\frac{4}{7}\) is 4 × \(\frac{7}{4}\) = 7.

Question 11.
6 ÷ \(\frac{2}{7}\)
Answer:
6 ÷ \(\frac{2}{7}\) = 21.

Explanation:
The division of 6 ÷ \(\frac{2}{7}\) is 6 × \(\frac{7}{2}\) which is 3 × 7 = 21.

Question 12.
9 ÷ \(\frac{3}{8}\)
Answer:
9 ÷ \(\frac{3}{8}\) = 24.

Explanation:
The division of 9 ÷ \(\frac{3}{8}\) is 9 × \(\frac{8}{3}\) which is 3 × 8 = 24.

Question 13.
5 ÷ \(\frac{10}{13}\)
Answer:
6\(\frac{1}{2}\).

Explanation:
The division of 5 ÷ \(\frac{10}{13}\) is 5 × \(\frac{13}{10}\) which is \(\frac{13}{2}\) = 6\(\frac{1}{2}\).

Question 14.
10 ÷ \(\frac{5}{14}\)
Answer:
10 ÷ \(\frac{5}{14}\) = 28.

Explanation:
The division of 10 ÷ \(\frac{5}{14}\) is 10 × \(\frac{14}{5}\) which is 2 × 14 = 28.

Question 15.
12 ÷ \(\frac{9}{10}\)
Answer:
12 ÷ \(\frac{9}{10}\) = 13 \(\frac{1}{3}\).

Explanation:
The division of 12 ÷ \(\frac{9}{10}\) is 12 × \(\frac{10}{9}\) which is \(\frac{40}{3}\) = 13 \(\frac{1}{3}\).

Hands-On Activity

Materials:

• 2 paper strips

Use 2 paper strips of the same size and length. Each strip represents 1 whole.

Step 1.
Take 1 paper strip and divide it into halves using vertical lines.
Then find \(\frac{1}{2}\) ÷ \(\frac{1}{4}\)
Refer to your model.
Complete:
There are fourths in a half.

So, \(\frac{1}{2}\) ÷ \(\frac{1}{4}\) =

Step 2.
Find \(\frac{1}{2}\) ÷ \(\frac{1}{4}\) by multiplication.
\(\frac{1}{2}\) ÷ \(\frac{1}{4}\) = Rewrite using the reciprocal of the divisor.
= Multiply.

Step 3.
Divide the other paper strip into one-thirds using vertical lines.
Then find \(\frac{2}{3}\) ÷ \(\frac{1}{6}\).
Refer to your model.
Complete.

Step 4.
Find \(\frac{2}{3}\) ÷ \(\frac{1}{6}\) by multiplication.
\(\frac{2}{3}\) ÷ \(\frac{1}{6}\) = Rewrite using the reciprocal of the divisor
= Multiply.

Hands-On Activity

Dividing Fractions with Remainder

A pitcher contains \(\frac{4}{5}\) quart of orange juice.

Step 1.
Copy the model and divide
Complete: \(\frac{4}{5}\) qt = qt

Step 2.
Use the model to answer this question.
Into how many glasses, each containing \(\frac{3}{10}\) quart, can the orange juice be poured .
How many quarts of orange juice will be left in the pitcher? qt

Step 3.
Now find the number of glasses by division. Express your answer as a mixed number.

The answer 2\(\frac{2}{3}\) means there are 2 glasses of orange juice, each containing \(\frac{3}{10}\) quart, and a remaining glass of orange juice that contains of \(\frac{2}{3}\) of \(\frac{3}{10}\) quart.
How many quarts of orange juice will be left in the pitcher.

Complete.

Question 16.
Adam had \(\frac{5}{7}\) liter of water. He used the water to fill a few glasses completely. The capacity of each glass was \(\frac{2}{7}\) liter. How many glasses of water did Adam fill?

Answer:

Question 17.
Lina had \(\frac{2}{3}\) of a pizza. She cut it into pieces that were each \(\frac{1}{9}\) of the whole pizza. Into how many pieces did she cut it?

Answer:

Divide. Express the quotient in simplest form.

Question 18.
\(\frac{2}{3}\) ÷ \(\frac{1}{6}\)
Answer:
\(\frac{2}{3}\) ÷ \(\frac{1}{6}\)

Explanation:
The simplest form of \(\frac{2}{3}\) ÷ \(\frac{1}{6}\) is \(\frac{2}{3}\) × \(\frac{6}{1}\) = 4.

Question 19.
\(\frac{3}{5}\) ÷ \(\frac{1}{10}\)
Answer:
\(\frac{3}{5}\) ÷ \(\frac{1}{10}\) = 6.

Explanation:
The simplest form of \(\frac{3}{5}\) ÷ \(\frac{1}{10}\) is \(\frac{3}{5}\) × \(\frac{10}{1}\) which is 3 × 2 = 6.

Question 20.
\(\frac{3}{4}\) ÷ \(\frac{1}{2}\)
Answer:
\(\frac{3}{4}\) ÷ \(\frac{1}{2}\) = 1 \(\frac{1}{2}\).

Explanation:
The simplest form of
\(\frac{3}{4}\) ÷ \(\frac{1}{2}\) = \(\frac{3}{4}\) × \(\frac{2}{1}\) which is \(\frac{3}{2}\) = 1 \(\frac{1}{2}\).

Question 21.
\(\frac{1}{6}\) ÷ \(\frac{2}{3}\)
Answer:
\(\frac{1}{6}\) ÷ \(\frac{2}{3}\) = \(\frac{1}{4}\).

Explanation:
The simplest form of
\(\frac{1}{6}\) ÷ \(\frac{2}{3}\) = \(\frac{1}{6}\) × \(\frac{3}{2}\)
= \(\frac{1}{4}\).

Question 22.
\(\frac{5}{8}\) ÷ \(\frac{15}{16}\)
Answer:
\(\frac{5}{8}\) ÷ \(\frac{15}{16}\) = \(\frac{2}{3}\).

Explanation:
The simplest form of
\(\frac{5}{8}\) ÷ \(\frac{15}{16}\) = \(\frac{5}{8}\) × \(\frac{16}{15}\)
= \(\frac{2}{3}\).

Question 23.
\(\frac{7}{16}\) ÷ \(\frac{5}{12}\)
Answer:
\(\frac{7}{16}\) ÷ \(\frac{5}{12}\) = 1\(\frac{1}{20}\).

Explanation:
The simplest form of
\(\frac{7}{16}\) ÷ \(\frac{5}{12}\) = \(\frac{7}{16}\) × \(\frac{12}{5}\)
= \(\frac{21}{20}\)
= 1\(\frac{1}{20}\).

Complete.

Question 24.
Divide \(\frac{5}{9}\) by 4\(\frac{1}{3}\).

Answer:

Caution
Before finding the reciprocal of a whole number or a mixed number, you need to first write ¡t as an improper fraction.

Question 25.
Divide 2\(\frac{3}{5}\) by 1\(\frac{8}{9}\).

2\(\frac{3}{5}\) and 1\(\frac{8}{9}\) are both mixed numbers. Express the mixed numbers as improper fractions.

Answer:

Hands-On Activity

Division Involving Whole Numbers and Fractions

Work in pairs.

Step 1.
Find each quotient.

a) 4 ÷ \(\frac{2}{5}\) and \(\frac{2}{5}\) ÷ 4
Answer:
4 ÷ \(\frac{2}{5}\) = 10.
\(\frac{2}{5}\) ÷ 4 = \(\frac{1}{10}\).

Explanation:
The quotient of 4 ÷ \(\frac{2}{5}\) is
4 ÷ \(\frac{2}{5}\) = 4 × \(\frac{5}{2}\)
= 2 × 5
= 10.
And the quotient of \(\frac{2}{5}\) ÷ 4 is
\(\frac{2}{5}\) ÷ 4 = \(\frac{2}{5}\) × \(\frac{1}{4}\)
= \(\frac{1}{10}\).

b) \(\frac{1}{4}\) ÷ \(\frac{2}{3}\) and \(\frac{2}{3}\) ÷ \(\frac{1}{4}\)
Answer:
\(\frac{1}{4}\) ÷ \(\frac{2}{3}\) = \(\frac{3}{8}\).
\(\frac{2}{3}\) ÷ \(\frac{1}{4}\) = 2 \(\frac{2}{3}\).

Explanation:
The quotient of \(\frac{1}{4}\) ÷ \(\frac{2}{3}\) is
\(\frac{1}{4}\) ÷ \(\frac{2}{3}\) = \(\frac{1}{4}\) × \(\frac{3}{2}\)
= \(\frac{3}{8}\).
The quotient of \(\frac{2}{3}\) ÷ \(\frac{1}{4}\) is
\(\frac{2}{3}\) ÷ \(\frac{1}{4}\) = \(\frac{2}{3}\) × \(\frac{4}{1}\)
= \(\frac{8}{3}\)
= 2 \(\frac{2}{3}\).

c) \(\frac{4}{5}\) ÷ \(\frac{3}{10}\) and \(\frac{3}{10}\) ÷ \(\frac{4}{5}\)
Answer:
\(\frac{4}{5}\) ÷ \(\frac{3}{10}\) = 2\(\frac{2}{3}\).
\(\frac{3}{10}\) ÷ \(\frac{4}{5}\) = \(\frac{3}{8}\).

Explanation:
The quotient of \(\frac{4}{5}\) ÷ \(\frac{3}{10}\) is
\(\frac{4}{5}\) ÷ \(\frac{3}{10}\) = \(\frac{4}{5}\) × \(\frac{10}{3}\)
= \(\frac{8}{3}\)
= 2\(\frac{2}{3}\).
And the quotient of \(\frac{3}{10}\) ÷ \(\frac{4}{5}\) is
\(\frac{3}{10}\) ÷ \(\frac{4}{5}\) = \(\frac{3}{10}\) × \(\frac{5}{4}\)
= \(\frac{3}{8}\).

d) \(\frac{5}{8}\) ÷ \(\frac{3}{4}\) and \(\frac{3}{4}\) ÷ \(\frac{5}{8}\)
Answer:
\(\frac{5}{8}\) ÷ \(\frac{3}{4}\) = \(\frac{5}{6}\).
\(\frac{3}{4}\) ÷ \(\frac{5}{8}\) = 1\(\frac{1}{5}\).

Explanation:
The quotient of \(\frac{5}{8}\) ÷ \(\frac{3}{4}\) is
\(\frac{5}{8}\) ÷ \(\frac{3}{4}\) = \(\frac{5}{8}\) × \(\frac{4}{3}\)
= \(\frac{5}{6}\).
The quotient of \(\frac{3}{4}\) ÷ \(\frac{5}{8}\) is
\(\frac{3}{4}\) ÷ \(\frac{5}{8}\) = \(\frac{3}{4}\) × \(\frac{8}{5}\)
= \(\frac{6}{5}\)
= 1\(\frac{1}{5}\).

Step 2.
What do you observe about the products of each pair of quotients?

Step 3.
Given that \(\frac{6}{7}\) ÷ 9 = \(\frac{2}{21}\) and \(\frac{10}{11}\) ÷ \(\frac{5}{6}\) = \(\frac{12}{11}\), find the following quotients mentally.

a) 9 ÷ \(\frac{6}{7}\)
Answer:
9 ÷ \(\frac{6}{7}\) = 10\(\frac{1}{2}\).

Explanation:
The quotient of 9 ÷ \(\frac{6}{7}\) is
9 ÷ \(\frac{6}{7}\) = 9 × \(\frac{7}{6}\)
= \(\frac{21}{2}\)
= 10\(\frac{1}{2}\).

b) \(\frac{5}{6}\) ÷ \(\frac{10}{11}\)
Answer:
\(\frac{5}{6}\) ÷ \(\frac{10}{11}\) = \(\frac{11}{12}\).

Explanation:
The quotient of \(\frac{5}{6}\) ÷ \(\frac{10}{11}\) is
\(\frac{5}{6}\) ÷ \(\frac{10}{11}\) = \(\frac{5}{6}\) × \(\frac{11}{10}\)
= \(\frac{11}{12}\).

Math Journal Explain in words the meaning of each division statement.

a) 4 ÷ \(\frac{2}{5}\) and \(\frac{2}{5}\) ÷ 4
Answer:
4 ÷ \(\frac{2}{5}\) = 10.
\(\frac{2}{5}\) ÷ 4 = \(\frac{1}{10}\).

Explanation:
The quotient of 4 ÷ \(\frac{2}{5}\) is
4 ÷ \(\frac{2}{5}\) = 4 × \(\frac{5}{2}\)
= 2 × 5
= 10.
And the quotient of \(\frac{2}{5}\) ÷ 4 is
\(\frac{2}{5}\) ÷ 4 = \(\frac{2}{5}\) × \(\frac{1}{4}\)
= \(\frac{1}{10}\).

b) \(\frac{1}{4}\) ÷ \(\frac{2}{3}\) and \(\frac{2}{3}\) ÷ \(\frac{1}{4}\)
Answer:
\(\frac{1}{4}\) ÷ \(\frac{2}{3}\) = \(\frac{3}{8}\).
\(\frac{2}{3}\) ÷ \(\frac{1}{4}\) = 2\(\frac{2}{3}\).

Explanation:
The quotient of \(\frac{1}{4}\) ÷ \(\frac{2}{3}\) is
\(\frac{1}{4}\) ÷ \(\frac{2}{3}\) = \(\frac{1}{4}\) × \(\frac{3}{2}\)
= \(\frac{3}{8}\).
And the quotient of \(\frac{2}{3}\) ÷ \(\frac{1}{4}\) is
\(\frac{2}{3}\) ÷ \(\frac{1}{4}\) = \(\frac{2}{3}\) × \(\frac{4}{1}\)
= \(\frac{8}{3}\)
= 2\(\frac{2}{3}\).

Math in Focus Course 1A Practice 3.1 Answer Key

Divide. Express the quotient ¡n simplest form. Use models to help you.

Question 1.
1 ÷ \(\frac{1}{4}\)

Answer:
1 ÷ \(\frac{1}{4}\) = 4.

Explanation:
The simplest form of 1 ÷ \(\frac{1}{4}\) is
1 ÷ \(\frac{1}{4}\) = 1 × \(\frac{4}{1}\)
= 4.

Question 2.
3 ÷ \(\frac{3}{5}\)

Answer:
3 ÷ \(\frac{3}{5}\) = 5.

Explanation:
The simplest form of 3 ÷ \(\frac{3}{5}\) is
3 ÷ \(\frac{3}{5}\) = 3 × \(\frac{5}{3}\)
= 5.

Question 3.
\(\frac{3}{4}\) ÷ \(\frac{1}{8}\)

Answer:
\(\frac{3}{4}\) ÷ \(\frac{1}{8}\) = 6.

Explanation:
The simplest form of \(\frac{3}{4}\) ÷ \(\frac{1}{8}\) is
\(\frac{3}{4}\) ÷ \(\frac{1}{8}\) = \(\frac{3}{4}\) × \(\frac{8}{1}\)
= 6.

Question 4.
\(\frac{2}{3}\) ÷ \(\frac{2}{9}\)

Answer:
\(\frac{2}{3}\) ÷ \(\frac{2}{9}\) = 3.

Explanation:
The simplest form of \(\frac{2}{3}\) ÷ \(\frac{2}{9}\) is
\(\frac{2}{3}\) ÷ \(\frac{2}{9}\) = \(\frac{2}{3}\) × \(\frac{9}{2}\)
= 3.

Draw a model to find each quotient.

Question 5.
1 ÷ \(\frac{1}{5}\)
Answer:
1 ÷ \(\frac{1}{5}\) = 5.

Explanation:
The quotient of 1 ÷ \(\frac{1}{5}\) is
1 ÷ \(\frac{1}{5}\) = 1 × \(\frac{5}{1}\)
= 5.

Question 6.
4 ÷ \(\frac{8}{9}\)
Answer:
4 ÷ \(\frac{8}{9}\) = 4\(\frac{1}{2}\).

Explanation:
The quotient of 4 ÷ \(\frac{8}{9}\) is
4 ÷ \(\frac{8}{9}\) = 4 × \(\frac{9}{8}\)
= \(\frac{9}{2}\)
= 4\(\frac{1}{2}\).

Question 7.
\(\frac{2}{5}\) ÷ \(\frac{3}{10}\)
Answer:
\(\frac{2}{5}\) ÷ \(\frac{3}{10}\) = 1\(\frac{1}{3}\).

Explanation:
The quotient of \(\frac{2}{5}\) ÷ \(\frac{3}{10}\) is
\(\frac{2}{5}\) ÷ \(\frac{3}{10}\) = \(\frac{2}{5}\) × \(\frac{10}{3}\)
= \(\frac{4}{3}\)
= 1\(\frac{1}{3}\).

Question 8.
\(\frac{3}{4}\) ÷ \(\frac{3}{16}\)
Answer:
\(\frac{3}{4}\) ÷ \(\frac{3}{16}\) = 4.

Explanation:
The quotient of \(\frac{3}{4}\) ÷ \(\frac{3}{16}\) is
\(\frac{3}{4}\) ÷ \(\frac{3}{16}\) = \(\frac{3}{4}\) × \(\frac{16}{3}\)
= 4.

Find each quotient. Express your answer in its simplest form.

Question 9.
4 ÷ \(\frac{1}{7}\)
Answer:
4 ÷ \(\frac{1}{7}\) = 28.

Explanation:
The quotient of 4 ÷ \(\frac{1}{7}\) is
4 ÷ \(\frac{1}{7}\) = 4 × \(\frac{7}{1}\)
= 28.

Question 10.
12 ÷ \(\frac{1}{3}\)
Answer:
12 ÷ \(\frac{1}{3}\) = 36.

Explanation:
The quotient of 12 ÷ \(\frac{1}{3}\) is
12 ÷ \(\frac{1}{3}\) = 12 × \(\frac{3}{1}\)
= 36.

Question 11.
9 ÷ \(\frac{3}{4}\)
Answer:
9 ÷ \(\frac{3}{4}\) = 12.

Explanation:
The quotient of 9 ÷ \(\frac{3}{4}\) is
9 ÷ \(\frac{3}{4}\) = 9 × \(\frac{4}{3}\)
= 3 × 4
= 12.

Question 12.
10 ÷ \(\frac{4}{5}\)
Answer:
10 ÷ \(\frac{4}{5}\) = 12\(\frac{1}{2}\).

Explanation:
The quotient of 10 ÷ \(\frac{4}{5}\) is
10 ÷ \(\frac{4}{5}\) = 10 × \(\frac{5}{4}\)
= \(\frac{25}{2}\)
= 12\(\frac{1}{2}\).

Question 13.
\(\frac{1}{2}\) ÷ \(\frac{1}{8}\)
Answer:
\(\frac{1}{2}\) ÷ \(\frac{1}{8}\) = 4.

Explanation:
The quotient of \(\frac{1}{2}\) ÷ \(\frac{1}{8}\) is
\(\frac{1}{2}\) ÷ \(\frac{1}{8}\) = \(\frac{1}{2}\) × \(\frac{8}{1}\)
= 4.

Question 14.
\(\frac{1}{4}\) ÷ \(\frac{1}{2}\)
Answer:
\(\frac{1}{4}\) ÷ \(\frac{1}{2}\) = \(\frac{1}{2}\).

Explanation:
The quotient of \(\frac{1}{4}\) ÷ \(\frac{1}{2}\) is
\(\frac{1}{4}\) ÷ \(\frac{1}{2}\) = \(\frac{1}{4}\) × \(\frac{2}{1}\)
= \(\frac{1}{2}\).

Question 15.
\(\frac{3}{5}\) ÷ \(\frac{11}{15}\)
Answer:
\(\frac{3}{5}\) ÷ \(\frac{11}{15}\) = 1\(\frac{4}{11}\).

Explanation:
The quotient of \(\frac{3}{5}\) ÷ \(\frac{11}{15}\) is
\(\frac{3}{5}\) ÷ \(\frac{11}{15}\) = \(\frac{3}{5}\) × \(\frac{15}{11}\)
= \(\frac{15}{11}\)
= 1\(\frac{4}{11}\).

Question 16.
\(\frac{2}{3}\) ÷ \(\frac{10}{13}\)
Answer:
\(\frac{2}{3}\) ÷ \(\frac{10}{13}\) = \(\frac{13}{15}\).

Explanation:
The quotient of \(\frac{2}{3}\) ÷ \(\frac{10}{13}\) is
\(\frac{2}{3}\) ÷ \(\frac{10}{13}\) = \(\frac{2}{3}\) × \(\frac{13}{10}\)
= \(\frac{13}{15}\).

Question 17.
\(\frac{5}{6}\) ÷ \(\frac{7}{12}\)
Answer:
\(\frac{5}{6}\) ÷ \(\frac{7}{12}\) = 1\(\frac{3}{7}\).

Explanation:
The quotient of \(\frac{5}{6}\) ÷ \(\frac{7}{12}\) is
\(\frac{5}{6}\) ÷ \(\frac{7}{12}\) = \(\frac{5}{6}\) × \(\frac{12}{7}\)
= \(\frac{10}{7}\)
= 1\(\frac{3}{7}\).

Question 18.
\(\frac{3}{4}\) ÷ \(\frac{9}{16}\)
Answer:
\(\frac{3}{4}\) ÷ \(\frac{9}{16}\) = 1\(\frac{1}{3}\).

Explanation:
The quotient of \(\frac{3}{4}\) ÷ \(\frac{9}{16}\) is
\(\frac{3}{4}\) ÷ \(\frac{9}{16}\) = \(\frac{3}{4}\) × \(\frac{16}{9}\)
= \(\frac{4}{3}\)
= 1\(\frac{1}{3}\).

Find each quotient. Express your answer in its simplest form.

Question 19.
\(\frac{1}{3}\) ÷ \(\frac{7}{4}\)
Answer:
\(\frac{1}{3}\) ÷ \(\frac{7}{4}\) = \(\frac{4}{21}\).

Explanation:
The quotient of \(\frac{1}{3}\) ÷ \(\frac{7}{4}\) is
\(\frac{1}{3}\) ÷ \(\frac{7}{4}\) = \(\frac{1}{3}\) × \(\frac{4}{7}\)
= \(\frac{4}{21}\).

Question 20.
\(\frac{1}{2}\) ÷ \(\frac{8}{4}\)
Answer:
\(\frac{1}{2}\) ÷ \(\frac{8}{4}\) = \(\frac{1}{4}\).

Explanation:
The quotient of \(\frac{1}{2}\) ÷ \(\frac{8}{4}\) is
\(\frac{1}{2}\) ÷ \(\frac{8}{4}\) = \(\frac{1}{2}\) × \(\frac{4}{8}\)
= \(\frac{1}{4}\).

Question 21.
\(\frac{1}{9}\) ÷ \(\frac{14}{3}\)
Answer:
\(\frac{1}{9}\) ÷ \(\frac{14}{3}\) = \(\frac{1}{42}\).

Explanation:
The quotient of \(\frac{1}{9}\) ÷ \(\frac{14}{3}\) is
\(\frac{1}{9}\) ÷ \(\frac{14}{3}\) = \(\frac{1}{9}\) × \(\frac{3}{14}\)
= \(\frac{1}{42}\).

Question 22.
\(\frac{5}{8}\) ÷ \(\frac{21}{4}\)
Answer:
\(\frac{5}{8}\) ÷ \(\frac{21}{4}\) = \(\frac{5}{42}\).

Explanation:
The quotient of \(\frac{5}{8}\) ÷ \(\frac{21}{4}\) is
\(\frac{5}{8}\) ÷ \(\frac{21}{4}\) = \(\frac{5}{8}\) × \(\frac{4}{21}\)
= \(\frac{5}{42}\).

Question 23.
3\(\frac{1}{2}\) ÷ 2\(\frac{1}{8}\)
Answer:
3\(\frac{1}{2}\) ÷ 2\(\frac{1}{8}\) = 1 \(\frac{11}{17}\).

Explanation:
The quotient of 3\(\frac{1}{2}\) ÷ 2\(\frac{1}{8}\) is
3\(\frac{1}{2}\) ÷ 2\(\frac{1}{8}\) = \(\frac{7}{2}\) ÷ \(\frac{17}{8}\)
= \(\frac{7}{2}\) × \(\frac{8}{17}\)
= \(\frac{28}{17}\)
= 1 \(\frac{11}{17}\).

Question 24.
5\(\frac{1}{4}\) ÷ 3\(\frac{1}{2}\)
Answer:
5\(\frac{1}{4}\) ÷ 3\(\frac{1}{2}\) = \(\frac{3}{2}\).

Explanation:
The quotient of 5\(\frac{1}{4}\) ÷ 3\(\frac{1}{2}\) is
5\(\frac{1}{4}\) ÷ 3\(\frac{1}{2}\) = \(\frac{21}{4}\) ÷ \(\frac{7}{2}\)
= \(\frac{21}{4}\) × \(\frac{2}{7}\)
= \(\frac{3}{2}\).

Question 25.
7\(\frac{3}{5}\) ÷ 8\(\frac{11}{15}\)
Answer:
7\(\frac{3}{5}\) ÷ 8\(\frac{11}{15}\) = \(\frac{114}{131}\).

Explanation:
The quotient of 7\(\frac{3}{5}\) ÷ 8\(\frac{11}{15}\) is
7\(\frac{3}{5}\) ÷ 8\(\frac{11}{15}\) =\(\frac{38}{5}\) ÷ \(\frac{131}{15}\)
= \(\frac{38}{5}\) × \(\frac{15}{131}\)
= \(\frac{114}{131}\).

Question 26.
12\(\frac{2}{3}\) ÷ 5\(\frac{11}{13}\)
Answer:
12\(\frac{2}{3}\) ÷ 5\(\frac{11}{13}\) = 2\(\frac{1}{6}\).

Explanation:
The quotient of 12\(\frac{2}{3}\) ÷ 5\(\frac{11}{13}\) is
12\(\frac{2}{3}\) ÷ 5\(\frac{11}{13}\) = \(\frac{26}{3}\) ÷ \(\frac{76}{13}\)
= \(\frac{38}{3}\) × \(\frac{13}{76}\)
= \(\frac{13}{6}\)
= 2\(\frac{1}{6}\).

Solve. Show your work.

Question 27.
6 pizzas were shared equally among a group of children. Each child got \(\frac{1}{9}\) of a pizza. How many children were in the group?
Answer:
The number of children were in the group is 54 children.

Explanation:
Given that 6 pizzas were shared equally among a group of children and each child got \(\frac{1}{9}\) of a pizza. So the number of children were in the group is 6 ÷ \(\frac{1}{9}\)
= 6 × 9
= 54 children.

Question 28.
A rectangle has an area of 15 square meters. It ¡s divided into parts, each with an area of \(\frac{3}{8}\) square meter. Into how many parts has the rectangle been divided?
Answer:
The number of parts has the rectangle been divided is 40 square meters.

Explanation:
Given that a rectangle has an area of 15 square meters and it ¡s divided into parts, each with an area of \(\frac{3}{8}\) square meter. So the number of parts has the rectangle been divided is 15 ÷ \(\frac{3}{8}\)
= 15 × \(\frac{8}{3}\)
= 5 × 8
= 40 square meters.

Question 29.
How many \(\frac{3}{8}\)-cup servings are in a pitcher containing 6\(\frac{3}{4}\) cups of orange juice?
Answer:
The number of serving pitchers be 18 cups.

Explanation:
Given that 1 cup contains \(\frac{3}{8}\) servings are in a pitcher containing and the total is 6\(\frac{3}{4}\) which is \(\frac{27}{4}\). So the number of serving pitchers be \(\frac{27}{4}\) ÷ \(\frac{3}{8}\)
= \(\frac{27}{4}\) × \(\frac{8}{3}\)
= 9 × 2
= 18 cups.

Question 30.
Maria buys 8\(\frac{1}{3}\) pounds of beef to make tacos for a party. She uses \(\frac{5}{9}\) pound of beef for each taco. How many tacos can Maria make?
Answer:
The number of tacos can Maria make is 15 tacos.

Explanation:
Given that Maria buys 8\(\frac{1}{3}\) pounds of beef to make tacos for a party which is \(\frac{25}{3}\) and she uses \(\frac{5}{9}\) pound of beef for each taco. So the number of tacos can Maria make is \(\frac{25}{3}\) ÷ \(\frac{5}{9}\)
= \(\frac{25}{3}\) × \(\frac{9}{5}\)
= 5 × 3
= 15.

The capacity of a large milk carton is 1\(\frac{1}{2}\) liters. A dozen large cartons are poured into a container and then poured into small cartons that each hold \(\frac{3}{10}\) liter. How many small cartons of milk can be filled?
Answer:
The number of small cartons of milk that can be filled is 60 cartons.

Explanation:
Given that the capacity of a large milk carton is 1\(\frac{1}{2}\) liters which is \(\frac{3}{2}\), so in pounds, it will be 12 × \(\frac{3}{2}\)
= 6 × 3
= 18.
And a dozen large cartons are poured into a container and then poured into small cartons that each hold \(\frac{3}{10}\) liter. Here, the total number of small cartons of milk can be filled = Total number of pounds of a dozen large cartons ÷ small carton capacity which is
= 18 ÷ \(\frac{3}{10}\)
= 18 × \(\frac{10}{3}\)
= 6 × 10
= 60.
The number of small cartons of milk that can be filled is 60 cartons.