Math in Focus

Practice the problems of Math in Focus Grade 7 Workbook Answer Key Chapter 6 Lesson 6.3 Alternate Interior, Alternate Exterior, and Corresponding Angles to score better marks in the exam.

Math in Focus Grade 7 Course 2 B Chapter 6 Lesson 6.3 Answer Key Alternate Interior, Alternate Exterior, and Corresponding Angles

Math in Focus Grade 7 Chapter 6 Lesson 6.3 Guided Practice Answer Key

Hands-On Activity

Materials:

• protractor

Explore The Angles Formed By Parallel Lines And A Transversal Using A Protractor

Work in pairs.

Step 1.
On a piece of paper, draw a pair of parallel lines, \(\overleftrightarrow{A B}\) and \(\overleftrightarrow{C D}\), as shown. Draw a transversal \(\overleftrightarrow{P Q}\) that intersects the pair of parallel lines. Use a protractor to measure the angles in the diagram.

Step 2.
Record your results in a table like the one shown.

Math Journal
What do you notice about the measures of these angle pairs: Alternate interior angles, alternate exterior angles, and corresponding angles? Make a conjecture about the angle measures for each pair.

Math Journal
Compare the conjecture that you have made for corresponding angles, alternate interior angles, and alternate exterior angles with the conjectures made by other students. What do you observe?

When two parallel lines are cut by a transversal,

• the alternate interior angles are always congruent (alt. int. ∠s, || lines).
• the alternate exterior angles are always congruent (alt. ext. ∠s, || lines).
• the corresponding angles are always congruent (corr. ∠s, || lines).

Use the diagram at the right to complete the following questions.

Question 1.
\(\overleftrightarrow{\mathrm{AD}}\), \(\overleftrightarrow{\mathrm{EH}}\), \(\overleftrightarrow{\mathrm{XY}}\), and \(\overleftrightarrow{\mathrm{WZ}}\) are straight lines and \(\overleftrightarrow{\mathrm{XY}}\) is parallel to \(\overleftrightarrow{\mathrm{WZ}}\). Identify all the pairs of angles formed by the intersection of \(\overleftrightarrow{\mathrm{AD}}\) with \(\overleftrightarrow{\mathrm{XY}}\) and \(\overleftrightarrow{\mathrm{WZ}}\).

Answer:
\(\overline{X Y}\) || \(\overline{W Z}\) a) We are given:
∠ABX and ∠BCW Pairs of corresponding angles:
∠XBC and ∠WCD
∠EFB and ∠FGC
∠BFG and ∠CGH
∠XBC and ∠BCG b) Alternate interior angles:
∠FBC and ∠BCW

a) Corresponding angles: ∠ABX and ∠BCW;
Answer:
∠XBC and ∠WCD
∠EFB and ∠FGC
∠BFG and ∠CGH

b) Alternate interior angles: ∠XBC and ∠BCG;
Answer:
∠FBC and ∠BCW

Question 2.
Name another transversal of the parallel lines in the diagram.
Answer:
\(\overline{X Y}\) || \(\overline{W Z}\) we are given
A transversal of the parallel lines \(\overline{X Y}\) and \(\overline{W Z}\) other than \(\overline{A D}\) is \(\overline{E H}\).
\(\overline{E H}\)

Question 3.
Identify one pair of each of the following angles formed by the intersection of \(\overleftrightarrow{\mathrm{EH}}\) with \(\overleftrightarrow{\mathrm{XY}}\) and \(\overleftrightarrow{\mathrm{WZ}}\).
a) Corresponding angles
Answer:
\(\overline{X Y}\) || \(\overline{W Z}\) We are given:
∠EFY and ∠FOZ a) Identify a pair of corresponding angles formed by the intersection of \(\overline{E H}\) with \(\overline{X Y}\) and \(\overline{W Z}\):

b) Alternate interior angles
Answer:
∠BFG and ∠FGZ b) Identify a pair of alterrate interior angles formed by the intersection of \(\overline{E H}\) with \(\overline{X Y}\) and \(\overline{W Z}\):

c) Alternate exterior angles
Answer:
∠EFY and ∠CGR c) Identity a pair of aiterrate exterior angles formed by the intersection of \(\overline{E H}\) with \(\overline{X Y}\) and \(\overline{W Z}\):
a) ∠EFY and ∠FGZ
b) ∠BFG and ∠FGZ
c) ∠EFY and ∠CGH

Complete.

Question 4.
In the diagram, \(\overleftrightarrow{\mathrm{MN}}\) is parallel to \(\overleftrightarrow{\mathrm{PQ}}\). Find the measures of ∠1, ∠2, and ∠3.

Answer:
\(\overline{M N}\) || \(\overline{P Q}\) We are given:
m∠1 = 81° Alternate interior ∠s:
m∠2 = 81° : Corresponding ∠s:
m∠3 + m∠2 = 180° Supplementary ∠s:
m∠3 + 81° = Substitute m∠2 = 81°:
m∠3 + 81° — 81° = 180° — 81° Subtract 81° from bothsides:
m∠3 = 99° simplify
81° ; 81° ; m∠2 ; 81° ; 81° ; 81° ; 81° ; 81° ; 99°

Math in Focus Course 2B Practice 6.3 Answer Key

\(\overleftrightarrow{\mathrm{MN}}\) is parallel to \(\overleftrightarrow{\mathrm{PQ}}\). Identify each pair of angles as corresponding, alternate interior, alternate exterior angles, or none of the above.

Question 1.
∠3, ∠6
Answer:
\(\overline{M N}\) || \(\overline{P Q}\) We are given:
∠3 and ∠6 are alternate interior angles because they lie inside the pair of parallel lines and on the opposite sides of the transversal.
Alternate interior angles

Question 2.
∠5, ∠7
Answer:
\(\overline{M N}\) || \(\overline{P Q}\) We are given:
∠5 and ∠7 are corresponding angles because they lie in matching corners on same side of the transversal \(\overline{S T}\).
corresponding angles

Question 3.
∠1, ∠2
Answer:
\(\overline{M N}\) || \(\overline{P Q}\) We are given:
∠1 and ∠2 are neither alternate interior, nor alternate exterior or corresponding, therefore none of the above.
None of the above

Question 4.
∠1, ∠8
Answer:
\(\overline{M N}\) || \(\overline{P Q}\) We are given:
∠1 and ∠8 are alternate exterior angles because they lie outside the pair of parallel lines and on the opposite sides of the transversal \(\overline{S T}\).
Alternate exterior angles

Question 5.
∠8, ∠6
Answer:
\(\overline{M N}\) || \(\overline{P Q}\) We are given:
∠8 and ∠6 are corresponding angles because they lie in matching corners on same side of the transversal \(\overline{S T}\).
corresponding angles

Question 6.
∠4, ∠7
Answer:
\(\overline{M N}\) || \(\overline{P Q}\) We are given:
∠4 and ∠7 are neither alternate interior, nor alternate exterior or corresponding, therefore none of the above.
None of the above

Question 7.
∠2, ∠7
Answer:
\(\overline{M N}\) || \(\overline{P Q}\) We are given:
∠2 and ∠7 are alternate interior angles because they lie inside the pair of parallel lines and on the opposite sides of the transversal \(\overline{S T}\).
Alternate interior angles

Question 8.
∠6, ∠7
Answer:
\(\overline{M N}\) || \(\overline{P Q}\) We are given:
∠6 and ∠7 are neither alternate interior, nor alternate exterior or corresponding, therefore none of the above.
None of the above

\(\overleftrightarrow{\mathrm{AB}}\) is parallel to \(\overleftrightarrow{\mathrm{CD}}\). Use the diagram to answer the following.

Question 9.
Name two angles that have the same measure as ∠2.
Answer:
\(\overline{A B}\) || \(\overline{C D}\) We are given:
m∠7 = m∠2 ∠2 and ∠7 are alternate exterior angles because they lie outside the pair of parallel lines and on the opposite sides of the transversal \(\overline{E F}\).
m∠6 = m∠2 ∠2 and ∠6 are corresponding angles because they lie in matching corners on the same side of the transversal \(\overline{E F}\).
∠6, ∠7

Question 10.
Name an angle that is supplementary to ∠6.
Answer:
\(\overline{A B}\) || \(\overline{C D}\) We are given:
∠5, ∠8, ∠3, ∠1 We identify supplementary angles to ∠6:
∠5, ∠8, ∠3, ∠1

Question 11.
If m∠4 = 46°, find m∠5.
Answer:
\(\overline{A B}\) || \(\overline{C D}\) We are given:
m∠6 = m∠4 Alternate interior ∠s:
m∠6 = 46° substitute:
m∠5 + m∠6 = 180° supplementary ∠s:
m∠5 + 46° = 180° substitute
m∠5 + 46° – 46° = 180° – 46° subtract 46° from both sides
m∠5 = 134° simplify
134°

Question 12.
If m∠1 = 131°, find m∠7.
Answer:
\(\overline{A B}\) || \(\overline{C D}\) We are given:
Corresponding ∠s:
m∠5 = m∠1 corresponding ∠s:
m∠5 = 131° Substitute:
m∠5 + m∠7 = 180° Supplementary ∠s:
131° + m∠7 = 180° Substitute:
131° + m∠7 – 131° = 180° – 131° Subtract 131° from both sides:
m∠7 = 49° Simplify:
49°

Find the measure of each numbered angle.

Question 13.
\(\overleftrightarrow{M N}\) is parallel to \(\overleftrightarrow{P Q}\).

Answer:
\(\overline{M N}\) || \(\overline{P Q}\) We are given:
m∠1 = 78° vertical ∠s:
m∠1 = m∠2 Alternate interior ∠s:
m∠2 = 78° Substitute:
m∠3 = m∠2 vertical ∠s:
m∠3 = 78° Substitute:
m∠1 = 78° ;m∠2 = 78° ; m∠3 = 78°

Question 14.
\(\overleftrightarrow{M N}\) is parallel to \(\overleftrightarrow{P Q}\).

Answer:
\(\overline{M N}\) || \(\overline{P Q}\) We are given:
m∠1 = 107° vertical ∠s:
m∠1 = m∠2 corresponding ∠s:
m∠1 = m∠2 vertical ∠s:
m∠2 = 107° Substitute:
m∠1 = 107° ;m∠2 = 107°

Question 15.
\(\overleftrightarrow{P Q}\) is parallel to \(\overleftrightarrow{R S}\).

Answer:
\(\overline{P Q}\) || \(\overline{R S}\) We are given:
m∠1 = 87° Alternate interior ∠s:
m∠2 = 52° corresponding ∠s:
m∠1 = 87° ; m∠2 = 52°

Question 16.
\(\overleftrightarrow{P Q}\) is parallel to \(\overleftrightarrow{R S}\).

Answer:
\(\overline{P Q}\) || \(\overline{R S}\) We are given:
m∠1 = 48° Alternate exterior ∠s:
m∠1 + m∠2 = 180° supplementary ∠s:
48° + m∠2 = 180° substitute
48° + m∠2 – 48° = 180° – 48° subtract 48° from both sides
m∠2 = 132° simplify
m∠1 = 48° ; m∠2 = 132°

Question 17.
\(\overleftrightarrow{A B}\) is parallel to \(\overleftrightarrow{C D}\) and \(\overleftrightarrow{M N}\) is parallel to \(\overleftrightarrow{P Q}\).

Answer:
\(\overline{M N}\) || \(\overline{P Q}\) We are given:
m∠1 = 126° Alternate interior ∠s:(transversal \(\overline{A B}\))
\(\overline{A B}\) || \(\overline{C D}\) we are given
m∠2 = m∠1 Alternate exterior ∠s:(transversal \(\overline{P Q}\))
m∠2 = 126° substitute
m∠1 = 126° ; m∠2 = 126°

Question 18.
\(\overleftrightarrow{A B}\) is parallel to \(\overleftrightarrow{C D}\) and \(\overleftrightarrow{M N}\) is parallel to \(\overleftrightarrow{P Q}\).

Answer:
\(\overline{M N}\) || \(\overline{P Q}\) We are given:
m∠1 = 32° Alternate interior ∠s:(transversal \(\overline{G F}\))
\(\overline{A B}\) || \(\overline{C D}\) we are given
m∠2 = 105° corresponding ∠s:(transversal \(\overline{M N}\))
m∠1 = 32° ; m∠2 = 105°

Find the value of each variable.

Question 19.
\(\overleftrightarrow{P Q}\) is parallel to \(\overleftrightarrow{R S}\).

Answer:
\(\overline{P Q}\) || \(\overline{R S}\) We are given:
a = 142 Alternate interior ∠s (transversal \(\overline{T U}\)):
a = 142
2b° + a° = 180° Supplementary ∠s
2b° + 142° = 180° Substitute:
2b° + 142° — 142° = 180° — 142° Subtract 142° from both sides:
2b° = 38°
\(\frac{2 b^{\circ}}{2}\) = \(\frac{38^{\circ}}{2}\) Divide by 2:
b = 19 simplify
a = 142
b = 19

Question 20.
\(\overleftrightarrow{P Q}\) is parallel to \(\overleftrightarrow{R S}\).

Answer:
\(\overline{P Q}\) || \(\overline{R S}\) We are given:
Let’s note by X the intersection between the lines \(\overline{T U}\) and \(\overline{R S}\).
m∠TXS = Corresponding ∠s (transversal \(\overline{T U}\)):
m∠TXS + m∠TXR = 180° Supplementary ∠s
5w° + 3w° = 180° Substitute:
8w° = 180° Simplify:
\(\frac{8 w^{\circ}}{8}\) = \(\frac{180^{\circ}}{8}\) Divide by 8
w = 22.5° simplify
w = 22.5°

MathJournal
Determine whether \(\overleftrightarrow{A B}\) is parallel to \(\overleftrightarrow{C D}\). Use the fact that two lines are parallel if a pair of corresponding angles formed by a transversal are congruent. Explain your answer.

Question 21.

Answer:
m∠BMN = 66° ∠BMN and ∠MNC are alternate interior ∠s We have:
m∠BMN ≠ m∠MNC As 66° ≠ 65°, we have

This means the lines \(\overline{A B}\) and \(\overline{C D}\) are not parallel:

Question 22.

Answer:
m∠MND = 90° ∠RMB and ∠MND are corresponding ∠s. We have:
m∠MND = 90°
m∠RMB = m∠MND We get:
\(\overline{A B}\) = \(\overline{C D}\) This means the lines \(\overline{A B}\) and \(\overline{C D}\) are parallel:
\(\overline{A B}\) || \(\overline{C D}\)

Question 23.

Answer:
m∠AMN + m∠NMB = 180° Supptementary ∠s
108° + m∠NMB = 180° Substitute:
108° + m∠NMB — 108° = 180° — 108° Subtract 108° from both sides:
m∠NMB = 72° Simplify:
m∠NMB = 72° ∠SND and ∠NMB are corresponding ∠s. We have:
m∠SND = m∠NMB we got
\(\overline{A B}\) = \(\overline{C D}\) This means the lines \(\overline{A B}\) and \(\overline{C D}\) are parallel
\(\overline{A B}\) || \(\overline{C D}\)

\(\overline{M N}\) is parallel to \(\overline{P Q}\). Find each unknown angle measure.

Question 24.

Answer:
m∠QMN = m∠PQM AS \(\overline{M N}\) || \(\overline{M N}\), ∠PQM and ∠QMN are alternate interior ∠s. We hava
m∠QMN = 34° Substitute
∠RMQ + m∠QMN + m∠1 = 180° ∠RMS straight angle
22° + 34° + m∠1 = 180°
56° + m∠1 = 180° Simplify
56° + m∠1 – 56° = 180° – 56° subtract 56° from both sides
m∠1 = 124° simplify
m∠1 = 124°

Question 25.

Answer:
m∠1 = m∠NPQ As \(\overline{M N}\) and \(\overline{P Q}\), ∠1 and ∠NPQ are alternate interior ∠s. We have:
m∠1 = 25° Substitute:
m∠MPN + m∠NPQ + m∠2 = 180° ∠MPS straight:
90° + 25° + m∠2 = 180° Substitute:
115° + m∠2 = 180° Simplify:
115° + m∠2 – 115° = 180° – 115° subtract 115° from both sides
m∠2 = 65° simplify
m∠1 = 25°
m∠2 = 65°

Question 26.

Answer:
m∠TRN = 114° As \(\overline{M N}\) || \(\overline{P Q}\), a pair of corresponding angles is:
62° + m∠3 = 114° Substitute:
62° + m∠3 – 62° = 114° — 62° Subtract 62° from both sides:
m∠3 = 52° simplify
m∠2 = m∠3 alternate interior ∠s
m∠2 = 52°
m∠1 = 114° Alternate exterior ∠s
m∠1 = 114°
m∠2 = 52°
m∠3 = 52°

Question 27.

Answer:
m∠NPQ = m∠MNP As \(\overline{M N}\) || \(\overline{P Q}\), ∠MNP and ∠NPQ are alternate interior angles
m∠NPQ = 101° Substitute:
m∠1 + m∠NPQ + m∠QPS = 180° ∠MPS is a straight line
m∠1 + 101° + 21° = 180° substitute
m∠1 + 122° = 180° simplify
m∠1 + 122° – 122° = 180° – 122° subtract 122° from both sides
m∠1 = 58° simplify
m∠1 = 58°

Question 28.

Answer:
m∠DCQ = m∠DBN As \(\overline{M N}\) || \(\overline{P Q}\), a pair of a corresponding angles is :
m∠DCQ = 76° Substitute
m∠1 + m∠DCQ = 180° ∠ACD is straight
m∠1 + 76° = 180° substitute
m∠1 + 76° – 76° = 180° – 76° subtract 76° from both sides
m∠1 = 104° simplify
m∠MFH + m∠HFN = 180° m∠MFH straight line
m∠MFH + 100° = 180° substitute
m∠MFH + 100° – 100° = 180° Subtract 100° from both sides:
m∠MFH = 80° Simplify:
m∠PGH = m∠MFH ∠MFH and ∠PGH corresponding angles:
m∠2 = 80° Substitute:
m∠1 = 104°
m∠2 = 80°

Question 29.

Answer:
m∠1 + 103° = 180° Straight ∠
m∠1 + 103° — 103° = 180° — 103° Subtract 103° from both sides:
m∠1 = 77° Simplify:
m∠1 = m∠2 + 16° As \(\overline{M N}\) || \(\overline{P Q}\), a pair of alternate interior angles is:
77° = m∠2 + 16° Substitute:
77° – 16° = m∠2 + 16 — 16° Subtract 16° from both sides:
m∠2 = 61° Simplify:
m∠1 = 77°
m∠2 = 61°

\(\overline{\mathbf{A B}}\) is parallel to \(\overline{\mathbf{C D}}\). Find the value of x.

Solve. Show your work.

Question 30.

Answer:
27° + x° = 42 As \(\overline{A B}\) || \(\overline{C D}\), a pair of alternate interior angles is:
27° + x° — 27° = 42° — 27° Subtract 27° from both sides:
x° = 15° Simplify:
x = 15

Question 31.

Answer:

we are given
Let X be the intersection of the Lines \(\overline{A B}\) and \(\overline{P D}\), Y be the intersection of the lines \(\overline{C D}\) and \(\overline{B Q}\) and Z be the intersection of the lines \(\overline{P D}\) || \(\overline{B Q}\).
110° + m∠PDY = 180° ∠CDY is straight
110° + m∠PDY = 180° – 110° subtract 110° from both sides
m∠PDY = 70° simplify
m∠DZY + 154° = 180° ∠PZD is straight
m∠DZY + 154° – 154 = 180° — 134° Subtract 154° from both sides:
m∠DZY = 26° Simplify:
m∠DZY + m∠PDY + m∠BYD = 180° Sum of the angles in ∆ZDY:
26° + 70° + m∠BYD = 180° Substitute:
96° + m∠BYD = 180° Simplify:
96° + m∠BYD – 96° = 180° — 96° Subtract 96° from both sides:
m∠BYD = 84° Simplify:
m∠BYD + m∠DYQ = 180° ∠BYQ is straight:
84° + m∠DYQ = 180° Substitute:
84° + m∠DYQ – 84° = 180° – 84° Subtract 84° from both sides:
m∠DYQ = 96° Simplify:
m∠ABQ = m∠DYQ As \(\overline{A B}\) || \(\overline{C D}\), a pair of corresponding angles is:
x° = 96° Substitute:
x = 96

Question 32.

Answer:
m∠BCD = m∠ABC As \(\overline{A B}\) || \(\overline{C D}\), a pair of alternate interior angles is:
m∠BCD = 108° Substitute:
m∠BCD + m∠DCQ + m∠QCP = 180° ∠BCP is straight:
108° + x° + 35° = 180° Substitute:
143° + x° = 180° Simplify:
143° + x° — 143° = 180° – 143° Substitute:
x = 37 Simplify:
x = 37

Question 33.

Answer:

we are given
Let X be the intersection of the lines \(\overline{A B}\) and \(\overline{P C}\), Y be the intersection of the lines \(\overline{C D}\) and \(\overline{A P}\).
150° + m∠XCY = 180° ∠YCD is straight:
150° + m∠XCY — 150° = 180° — 150° Subtract 150° from both sides:
m∠XCY = 30° Simplify:
m∠XAP + 132° = 180° ∠XAB is straight:
m∠XAP = 48° simplify
m∠XAP = m∠PYC As \(\overline{A B}\) || \(\overline{C D}\), a pair of alternate interior angles is:
m∠PYC = 48° Substitute:
m∠PYC + m∠XCY + m∠YPC = Sum of the ∠s in a triangle:
48° + 30° + m∠YPC = 180° Substitute:
78° + m∠YPC = Simplify:
78° + m∠YPC — 78° = 180° — 78° Subtract 78° from both sides:
m∠YPC = 102° Simplify:
m∠YPC + x° = 180° ∠APY straight:
102 + x° = 180° Substitute:
102° + x° – 102° = 180° — 102 Subtract 102° from both sides:
x° = 78° simplify
x = 78

Solve. Show your work.

Question 34.
In the diagram below, \(\overleftrightarrow{M N}\) is parallel to \(\overleftrightarrow{P Q}\). m∠1 = (x + 28)° and m∠2 = (3x + 14)°. Write and solve an equation to find the measures of ∠1 and ∠2.

Answer:
m∠1 = m∠2 As \(\overline{M N}\) || \(\overline{P Q}\), a pair of alternate interior angles is:
x + 28 = 3x + 11 substitute
x + 28 – x = 3x + 14 – x Subtract x from both sides:
28 = 2x + 14 simplify
28 — 14 = 2x + 14 — 14 Subtract 14 from both sides:
2x = 14 Simplify:
\(\frac{2 x}{2}\) = \(\frac{14}{2}\) Divide by 2:
x = 7 Simplify:
m∠1 = m∠2 = (7 + 28)° = 35° Determine m∠1, m∠2:
x + 28 = 3x + 14
m∠1 = 35° ;m∠2 = 35°

Question 35.
Math Journal
In a plane, if a line is perpendicular to one of two parallel lines, is it also perpendicular to the other? Explain your reasoning.
Answer:
\(\overline{A B}\) || \(\overline{C D}\) We are given:
\(\overline{E F}\) ⊥ \(\overline{A B}\)
m∠EMB = 90°: \(\overline{E F}\) ⊥ \(\overline{A B}\)
m∠MND = m∠EMB As \(\overline{A B}\) || \(\overline{C D}\), a pair of corresponding angles is:
m∠MND = 90° Substitute:
\(\overline{E F}\) ⊥ \(\overline{C D}\) m∠MND = 90°
yes

Question 36.
The diagram below contains examples of parallel lines cut by transversals. Line MN is parallel to line PQ and line AB is parallel to line CO.

a) Name two pairs of corresponding angles.
Answer:
∠1 and ∠3 a) As \(\overline{A B}\) || \(\overline{P Q}\), pairs of corresponding angles are (transversal \(\overline{A B}\)):
∠2 and ∠4
∠5 and ∠7

b) Name all the angles that have the same measure as ∠1.
Answer:
m∠5 = m∠1 b) vertical ∠s:
m∠3 = m∠1 corresponding ∠s:
m∠8 = m∠3 = m∠1
m∠7 = m∠5 = m∠1
a) ∠1 and ∠3
∠2 and ∠4
∠5 and ∠7
b) ∠1, ∠3, ∠5, ∠7 = ∠8

Question 37.
The two mirrors used in a periscope are parallel to each other as shown. m∠1 = 3x°, m∠2 = (60 – x)°, and m∠3 = 90°. Write and solve an equation to find the value of x. Then find the measure of ∠4.

Answer:
m∠1 = 3x° We are given:
m∠2 = (60 – x)°
m∠3 = 90°
m∠1 = m∠2 Alternate interior ∠s
3x° = (60 – x)°
3x = 60 — x Substitute:
3x + x = 60 – x + x Add x to both sides:
4x = 60 Simplify:
\(\frac{4 x}{4}\) = \(\frac{60}{4}\) Divide by 4:
x = 15 Simplify:
m∠2 + m∠3 + m∠4 = 180° Straight ∠:
Substitute:
(60 – x)° + 90° + m∠4 = 180°
(60 – 15)° + 90° + m∠4 = 180°
45° + 90° + m∠4 = 180°
135° + m∠4 = 180° Simplify:
135° + m∠4 – 135° = 180° — 135° Subtract 135° from both sides:
m∠4 = 45° Simplify:
x = 15
m∠4 = 45°

Question 38.
MathJournal
Use a diagram to illustrate each of the following: transversal, corresponding angles, alternate exterior angles, and alternate interior angles. Label your diagram and explain which angles are congruent.
Answer:

We draw the diagram of 2 parallel lines cut by a transversal:
m∠1 = m∠5 Pairs of corresponding angles:
m∠2 = m∠6
m∠3 = m∠7
m∠4 = m∠8
m∠1 = m∠7 Pairs of alternate exterior aigles
m∠2 = m∠8
m∠3 = m∠5 Pairs of alternate Interior angles
m∠4 = m∠6
See diagram

Practice the problems of Math in Focus Grade 7 Workbook Answer Key Chapter 6 Lesson 6.2 Angles that Share a Vertex to score better marks in the exam.

Math in Focus Grade 7 Course 2 B Chapter 6 Lesson 6.2 Answer Key Angles that Share a Vertex

Math in Focus Grade 7 Chapter 6 Lesson 6.2 Guided Practice Answer Key

Complete.

Question 1.
Find the value of p in the diagram.

Answer:
p° = 20°

Explanation:

7 p°+ 75° + 145° = 360°
7 p° + 220° = 360°
7p° + 220° – 360° = 360°-360°
7p° = 140°
p°  = 140 ÷ 7
p° = 20°

Solve.

Question 2.
\(\overleftrightarrow{B E}\) and \(\overleftrightarrow{C A}\) are straight lines. Find the value of each variable.


Answer:
r°=28°, q°=28°, 3q°=84°

Explanation:


∠BOA + ∠AOE = 180°
152°+r°=180°
r = 180°-152°
r= 28°
∠BOC + ∠COD + ∠DOE = 180°
q°+ 68°+3q° = 180°
4q°+68° = 180°
4q° = 180°-68°
4q° = 112°
q°= 112÷4
q°= 28°
3q°= 84°

Question 3.
\(\overleftrightarrow{P Q}\) is a straight line. Find the value of each variable.

Answer:
m°=31°, n°=17°, 3m°=93°

Explanation:

m°+149° = 180°
m° = 180°-149°
m° = 31°
3m°= 93°
n°+3m°+70°=180°
n°+93°+70°=180°
n°=180°-163°
n°= 17°

Complete.

Question 4.
In the diagram at the right, the ratio a: b: c = 1: 3 : 5. Find the values of a, b, and c.
The ratio a : b : c = 1 : 3 : 5. So, b = 3 • a and c = 5 • a.


Answer:
a°= 40°, b°=120°, c°=200°

Explanation:
The ratio given is a : b : c = 1 : 3 : 5. Let a=1a, b=3a, c=5a

Technology Activity

Materials:

• geometry software

Explore The Relationship Among Vertical Angles Using Geometry Software

Work in pairs.

Step 1.
Construct intersecting line segments, \(\overline{A B}\) and \(\overline{C D}\), as shown.

Step 2.
Select ∠AOD and find its measure.

Step 3.
Select ∠COB and find its measure.

Step 4.
Then select ∠AOC and ∠BOD and find their measures.

Step 5.
Select point D and drag it so that you change the measures of ∠AOD and ∠AOC. As the measure of ∠AOD changes, what do you notice about the measure of ∠AOC?

Math Journal
Describe what you notice about the measures of the vertical angles.
Answer:
Vertical angles are always equal to one another and are always congruent. The four angles all together always sum to a full angle 360°.

Complete.

Question 5.
\(\overleftrightarrow{A B}\) and \(\overleftrightarrow{C D}\) are straight lines. Find the value of y.

Answer:
3y°=120°

Explanation:

∠COB = ∠AOD
3y°=120°
y°= 120°÷3
y° = 40°

Complete.

Question 6.
In the diagram, two straight lines intersect to form angles 1, 2, 3, and 4. Find the value of each variable if m∠1 = 114°.

m∠1 + m∠2 = 180° Supp. ∠S

Answer:
r°=66°, p°=66°, q°=114°

Explanation:


114°+r°=180°
r°=180°-114
r°=66°
∠3=∠1
q°=114°
∠4=∠2
p°=r°
as r°=66°
p°=66°

Math in Focus Course 2B Practice 6.2 Answer Key

Find the value of each variable.

Question 1.

Answer:
x°=307°

Explanation:

x°+53°=360°
x°=360°-53°
x°=307°

Question 2.

Answer:
x°=257°

Explanation:

x°+33°=290°
x°=290°-33°
x°=257°

Question 3.

Answer:
x°=100°

Explanation:

x°+110°+150°=360°
x°+260°=360°
x°=100°

Question 4.

Answer:
x°=95°

Explanation:

∠AOD+∠BOC+∠AOD = 360°
x°+110°+95°+60°=360°
x°+265°=360°
x°=360°-265°
x°=95°

Question 5.
\(\overleftrightarrow{A B}\) and \(\overleftrightarrow{C D}\) are straight lines.

Answer:
p°= 34°

Explanation:

∠COA = ∠BOD
34° = p°
Hence p° = 34°

Question 6.
\(\overleftrightarrow{A B}\) and \(\overleftrightarrow{C D}\) are straight lines.

Answer:
r°=16°

Explanation:

In the image given below, we can observe that CD and AB are two straight lines. Here, ∠AOB and ∠COD are vertical angles.
a°= 164°
a°+r°=180°
164°+r°=180°
r°=180°-164°
r°=16°

Question 7.
\(\overleftrightarrow{A B}\), \(\overleftrightarrow{C D}\) and \(\overleftrightarrow{E F}\) are straight lines.

Answer:
a° = 35°, b° = 106°

Explanation:

∠FOD = ∠COE
a° = 35° (vertically opposite angles)
∠EOB = ∠AOF
b° = 106° (Vertically opposite angles)

Question 8.
\(\overleftrightarrow{A B}\), \(\overleftrightarrow{C D}\) and \(\overleftrightarrow{E F}\) are straight lines.

Answer:
b° = 126°, a° = 31°

Explanation:
∠AOF = ∠EOB
b° = 126° (vertically opposite angles)
∠AOC = ∠DOB
31° = a° (vertically opposite angles)

Question 9.
\(\overleftrightarrow{A B}\), \(\overleftrightarrow{C D}\) and \(\overleftrightarrow{E F}\) are straight lines.

Answer:
x° = 119°, y°= 122°

Explanation:

∠COE + ∠AOC = 180°
y°+ 58° = 180°
y° = 180°-58°
y° = 122°
∠DOB + ∠FOD = 180°
x° + 61° = 180°
x° = 180°-61°
x° = 119°

Question 10.
\(\overleftrightarrow{A B}\) and \(\overleftrightarrow{C D}\) are straight lines.

Answer:
w + 90° = 168° Vertical ∠s:
w° + 90° – 90° = 168° – 90° Subtract 90° from both sides:
w° = 78° Simplify:
w = 78

Find the value of k.

Question 11.
The ratio ∠1 : ∠2 : ∠3 : ∠4 = 3 : 2 : 1 : 3.

Answer:
k° = 40°

Explanation:

Let the assume given ratio be x
Now, the ratio is 3x: 2x: x: 3x
The sum of all angles at one point is 360°
Now add 3x+ 2x + x + 3x = 360°
9x = 360°
x = 360 ÷ 9
x° = 40°
3x° = 120°
2x° = 80°
3x° = 120°
Now we need to calculate the numbered angle 3 which is named as k° that is 40°

Question 12.
\(\overleftrightarrow{A B}\) and \(\overleftrightarrow{C D}\) are straight lines.

Answer:
4k°= 84°, 3k° = 63

Explanation:

∠COE +∠EOF = 174°
4k°+ 90° = 174°
4k° = 174-90
4k°= 84°
k° = 84÷4
k° = 21°
3k°= 21×3
3k° = 63°

Question 13.

Answer:
2k°+5k°+3k°+130°=180°
10k°+130°=360°
10k° = 360°-130°
10k° = 230°
k° = 23°
IF k°=23° then 2k° = 2×23 = 46°, 3k° = 3×23=69°
5k° = 5×23=115°

Name the pairs of vertical angles.

Question 14.

Answer:
AE and DB are a pair of non-adjacent angles that are formed when two lines intersect.

Explanation:
Vertical lines are a pair of non-adjacent angles formed when two lines intersect.

Question 15.

Answer:
Vertical angles are a pair of opposite angles formed by intersecting lines.
∠MKF and ∠NKG Lines \(\overline{M N}\) and \(\overline{F G}\) intersect in K.
∠FKN and ∠MKC We identify the pairs of vertical angles:
∠MKF and ∠NKG
∠FKN and ∠MKC

Question 16.
\(\overleftrightarrow{P S}\) and \(\overleftrightarrow{R N}\) are straight lines.

Answer:
Vertical angles are a pair of opposite angles formed by ¡ntersecting tines.
∠SOR ard ∠NOP Lines \(\overline{P S}\) and \(\overline{R N}\) intersect in O.
∠SON and ∠ROP We identify the pairs of vertical angles:
∠SOR and ∠NOP
∠SON ard ∠ROP

Find the value of each variable.

Question 17.

Answer:
64°+23°+3e°=360°
87°+3e°=360°
3e°=360°-87°
3e°=273°
e°=273°÷ 3°
e° = 91°

Question 18.
\(\overleftrightarrow{A B}\) and \(\overleftrightarrow{C D}\) are straight lines.

Answer:
2p°=88, q°= 92°, r°=2°

Explanation:

∠DOA = ∠BOC
2p°=88
p°= 88÷2
p°= 44
∠DOE+∠EOB+∠BOC= 180°
r°+2p°+88 = 180°
r°+88+88 = 180°
r°+178°=180°
r°=180°-178°
r°=2°
2p°+q°=180°
88+q°=180°
q°=180-88
q°= 92°

Question 19.
\(\overleftrightarrow{A B}\), \(\overleftrightarrow{C D}\) and \(\overleftrightarrow{E F}\) are straight lines. The ratio a : b : c = 1 : 2 : 2.

Answer:
a° = 36°, b = 72°, c = 72°

Explanation:

Let the ratio be x
The sum of the ratio be x + 2x + 2x = 5x
∠AOC = ∠DOB (vertically opposite angles are equal)
∠FOD + ∠DOB + ∠EOB = 180°
c°+a°+b° = 180°
2x + x° + 2x° = 180°
5x° = 180°
x° = 180÷5
x° = 36°
2x°= 72°

Question 20.
\(\overleftrightarrow{A B}\) and \(\overleftrightarrow{C D}\) are straight lines.

Answer:
f° = 40° Vertical ∠s:
f = 40
e° + f° = 180° Straight ∠:
e° + 40° = 1800 Substitute:
e° + 40° – 40° = 180° – 40° Subtract 40° from both sides:
e = 140 Simplify:
d° + 90° + 15° = e° Vertical ∠s:
d° + 105° = 140° Simplify:
d° + 105° — 105° = 140° – 105° Subtract 105° from both sides:
d = 35 Simplify:
f = 40
e = 140
d = 35

Answer each of the following.

Question 21.
In the diagram, the ratio p: q: r = 1: 2 : 3. Find the values of p, q, and r.

Answer:
p = 60°, q = 120°, r = 180°

Explanation:

p: q: r = 1: 2 : 3
x+2x+3x = 360°
6x = 360°
x = 360° ÷ 6
x = 60°
2x= 120°
3x = 180°
p = 60°, q = 120°, r = 180°

Question 22.
If ∠P and ∠N are angles at a point and m∠P = 149°, what is the m∠N?
Answer:
m∠N = 31°

Explanation:
The sum of angles at a point is 180°
Given angles are ∠P = 149°
Here we need to calculate ∠N
∠P + ∠N = 180°
149° + ∠N = 180°
∠N = 180°-149°
∠N = 31°

Question 23.
If 67°, 102°, 1 5°, and x° are angles at a point, what is the value of x?
Answer:
x° = 176°

Explanation:
67°+102°+15°+x°=360°
184°+x° = 360°
x°=360° -184°
x° = 176°

In the diagram below, \(\overleftrightarrow{M P}\) and \(\overleftrightarrow{Q R}\) are straight lines. Answer each of the following.

Question 24.
Name the angle that is vertical to ∠MNR.
Answer:
∠QNP

Explanation:
Vertical angles mean the angles that are opposite to each other. By observing from the given figure the angle ∠QNP is vertical to ∠MNR.

Question 25.
What kind of angles are ∠RNP and ∠PNS?
Answer:
∠RNP and ∠PNS are adjacent angles because they have a common vertex (N) and a common side (\(\overline{N P}\)) and they do not overlap.
Adjacent angles

Question 26.
Find the measure of ∠QNS.
Answer:
m∠QNS + m∠SNP = m∠MNR Vertical ∠s:
m∠QNS + 61° = Substitute:
m∠QNS + 61° – 61° = 138° – 61° Subtract 61° from both sides:
m∠QNS = 77° Simplify:
m∠QNS = 77°

Question 27.
Find the measure of ∠PNR.
Answer:
m∠MNP = 180° Straight ∠
m∠PNR + 138° = 180° Substitute:
m∠PNR + 138° — 138° = 180° — 138° Subtract 138° from both sides:
m∠PNR = 42° Simplify:
m∠PNR = 42°

Use an equation to find the value of each variable.

Question 28.
\(\overleftrightarrow{\mathrm{AB}}\) is a straight line.

Answer:
x°=30°, y°=150°, w°=120°

Explanation:

∠COE + ∠BOE =90°
2x°+x°= 90°
3x°=90
x°=30°
2x°=60°
∠AOC + ∠COE = 180°
w°+2x°=180°
w°+60°=180°
w°=180°-60°
w°=120°
∠AOE + ∠BOE = 180°
y°+x°=180°
y°+30°=180°
y°=180°-30°
y°=150°

Question 29.
\(\overleftrightarrow{\mathrm{AB}}\) and \(\overleftrightarrow{\mathrm{CD}}\) is a straight line.

Answer:
3b° = 90°, b° = 30°, c° = 60°

Explanation:

3b°=90°
b°=90÷3
b°=30°
∠DOB = ∠AOC (vertically opposite angles)
∠AOC = c
∠DOB = c°
∠AOD = 120°
∠DOB + ∠AOD = 180°
c°+120° = 180°
c° = 180°- 120°
c° = 60°

\(\overleftrightarrow{\mathrm{AB}}\) and \(\overleftrightarrow{\mathrm{CD}}\) are straight lines.

Question 30.
The ratio ∠1 : ∠2 = 3 : 1

Answer:
p°= 45°, q°= 135°

Explanation:
Let the given ratio be x
The sum of angles in a traingle is 180°
∠AOD +∠AOC = 180°
The ratio ∠1 : ∠2 = 3x : 1x
3x + x = 180°
4x = 180°
x°= 180÷4
x°= 45°
3x = 135°
∠1 = ∠3 vertically opposite angles
∠2 = ∠4 vertically opposite angles
∠3 = q°= 135°
∠4= p°= 45°

Question 31.

Answer:
p°=50°, q°= 80°, 2p° = 100°

Explanation:

∠AOC = ∠DOB
2p° = p+50
2p°-p° = 50
p°=50°
∠DOB = 100°
∠DOB+∠COB = 180°
P+50 = 50+50 = 100°
100°+q°=180°
q°=180°-100°
q°= 80°

Solve.

Question 32.
The diagram below shows the flag of the Philippines, m∠ADB = 60° and m∠ADC = m∠BDC. Find the measures of ∠ADC and ∠BDC.

Answer:
∠ADB = 60°, ∠BDC = 150°

Explanation:
∠ADB = 60°
∠ADC = ∠BDC
∠ADB + ∠ADC +∠BDC = 360°
60° + ∠ADC + ∠ADC = 360°
60 + 2 ∠ADC = 360°
2 ∠ADC = 360°-60°
2∠ADC = 300°
∠ADC = 300÷2
∠ADC = 150°
∠BDC = 150°

Question 33.
Math Journal
The diagram shows a pattern on a carpet.
a) Are ∠4 and ∠6 vertical angles? Explain why or why not.
Answer:
No.
When two lines cross then vertical angles are opposite to each other.

Explanation:
The angles ∠4 and ∠6 are not vertical angles. Because the vertical angles are opposite to each other but by observing the angles from the given figure the angles given are not opposite to each other.

b) Suppose m∠4 = m∠6. Are ∠4 and ∠5 supplementary angles? Explain your answer.

Answer:
Yes, angles ∠4 and ∠5 are supplementary angles.

Explanation:
Two angles are supplementary when if they add up to 180°. If we add up the given angles ∠4 and ∠5 then the angles add up to 180°

Practice the problems of Math in Focus Grade 7 Workbook Answer Key Chapter 6 Lesson 6.1 Complementary, Supplementary, and Adjacent Angles to score better marks in the exam.

Math in Focus Grade 7 Course 2 B Chapter 6 Lesson 6.1 Answer Key Complementary, Supplementary, and Adjacent Angles

Math in Focus Grade 7 Chapter 6 Lesson 6.1 Guided Practice Answer Key

Technology Activity

Materials:

• geometry software

Explore The Relationship Of Complementary Angles Using Geometry Software

Work in pairs.

Step 1.
Construct \(\overline{\mathrm{AB}}\). Then construct a second line segment, \(\overline{\mathrm{BC}}\), that is perpendicular to \(\overline{\mathrm{AB}}\). Finally, construct line segment \(\overline{\mathrm{BD}}\).

Step 2.
Select ∠ABC and find
Step 3.
Select ∠ABD and find its measure. Then select ∠DBC and find its measure.
Step 4.
Use the calculate function of the program to find the sum of the measures of ∠ABD and ∠DBC. What do you notice about the sum of their measures?
Step 5.
Select the point D and drag it so that you change the measures of ∠ABD and ∠DBC. Record your results in a table as shown below.

Step 6.
As the angle measures change, how does the sum of the angle measures change?

Math Journal
Describe what you notice about the sum of the measures of complementary angles.

Solve.

Question 1.
Name three pairs of complementary angles.

Answer:
m∠ABC = 61° We have:
m∠PQR = 29°
m∠ABC + m∠PQR = 61° + 29° = 90° we add the measures of the two angles:
Since the sum of their measures is 90° ∠ABC and ∠PQR are complementary angles
m∠DEF = 38° We have:
m∠MNO = 52°
m∠DEF + m∠MNO = 38° + 52° = 90° we add the measures of the two angles:
Since the sum of their measures is 90°, ∠DEF and ∠MNO are complementary angles
m∠GIH = 21° We have:
m∠JKL = 69°
m∠GIH + m∠JKL = 21° + 69° = 90° We add the measures of the two angles:
Since the sum of their measures is 90°, ∠GIH and ∠JKL are complementary angles
∠ABC and ∠PQR
∠DEF and ∠MNO
∠GIH and ∠JKL
Copy and complete the table.

Question 2.
Angles A and 8 are complementary. Find m∠B for each measure of ∠A.

Answer:
m∠A + m∠B = 90° We are given:

We determine each measure of ∠A:
62° ;17° ;54° ; 75°

Technology Activity

Materials:

• geometry software

Explore The Relationship Of Supplementary Angles Using Geometry Software

Work in pairs.

Step 1.
Construct \(\overline{P R}\). Then construct a second line segment, \(\overline{S Q}\), as shown below.

Step 2.
Select ∠PQR and find its measure.
Step 3.
Select ∠SQP and find its measure. Then select ∠SQR and find its measure.
Step 4.
Use the calculate function of the program to find the sum of the measures of ∠SQP and ∠SQR. What do you notice about the sum of their measures?
Step 5.
Select the point S and drag it so that you change the measures of ∠SQP and ∠SQR. Record your results in a table as shown below.

Step 6.
As the angle measures change, how does the sum of the angle measures change?

MathJournal
Describe what you notice about the sum of the measures of supplementary angles.

Solve.

Question 3.
Tell whether each pair of angles is supplementary.
a) m∠X = 32° and m∠Y = 108°
Answer:
m∠X = 32° a) We are given:
m∠Y = 108°
m∠X + m∠Y = 32° + 108° = 140° Add the measures of the two angles:
As m∠X + m∠Y ≠ 180°. ∠X and ∠Y are not supplementary

b) m∠A = 45° and m∠B = 45°
Answer:
m∠A = 45° b) We are given:
m∠B = 45°
m∠A + m∠B = 45° + 45° = 90° Add the measures of the two angles:
As m∠A + m∠B ≠ 180°, ∠A and ∠B are not supplementary.

c) m∠D = 12° and m∠E = 168°
Answer:
m∠D = 12° c) We are given:
m∠E = 168°
m∠D + m∠E = 12° + 168° = 180° Add the measures of the two angles:
As m∠D + m∠E = 180°, ∠D and ∠E are supplementary.

d) m∠V = 85° and m∠W = 95°
Answer:
m∠V = 85° d) We are given:
m∠W = 95°
m∠V + m∠W = 85° + 95° = 180° Add the measures of the two angles:
As m∠V + m∠W = 180°. ∠V and ∠W are supplementary.

Copy and complete the table.

Question 4.
Angles A and B are supplementary. Find m∠B for each measure of ∠A.

Answer:
m∠A + m∠B = 180° We are given:

We determine each measure of ∠A:
98°; 154°; 44°; 75°

Complete.

Question 5.
In the diagram, m∠ABC = 90°. Find the value of x.

Answer:

We are given
m∠ABD + m∠DBC = 90° Complementary angles:
x° + 23° = 90° Substitute:
x + 23° – 23° = 90° — 23° Subtract 23° from bothsides:
x = 67 Simplify:
x = 67

Given that \(\overleftrightarrow{P Q}\) is a straight line, find the value of y.

Question 6.

Answer:

we are given
m∠PQR + m∠ROQ = 180° Adjacent angles on a straight line:
37° + y° = 180° Substitute:
37° + y° – 37° = 180° – 37° Subtract 37° from both sides:
y = 143 Simplify:
y = 143

Question 7.

Answer:

we are given
m∠POS + m∠SQR + m∠ROQ = 180° Adjacent angles on a straight line:
2y° + 70° + 3y° = 180° substitute
5y° + 70° = 180° Simplify:
5y° + 70° – 70° = 180° — 70° Subtract 70° from both sides:
5g = 110 Simplify:
\(\frac{5 y}{5}\) = \(\frac{110}{5}\) Divide both sides by 5:
y = 22 Simplify:
y = 22

Complete.

Question 8.
In the diagram, m∠PQR = 90° and the ratio x : y = 1: 4. Find the values of x and y.
Method I

Use bar models.

Answer:

m∠PQR = 90°
x : y = 1 : 4
we are given

x° + y° = 90° Method 1; use bar models
Complementary angles:

We use bar models:
5 units → 90 We have:
1 unit → \(\frac{90}{5}\) = 18
x = 18
y = 4 ∙ 18 = 72

Method 2; use a variable to represent the measure of the angle
y = 4 • x = 4x The ratio x : y = 1 : 4 So we have:
x° + y° = 90° Complementary angles
x + 4x = 90 Substitute:
5x = 90 Simplify:
\(\frac{5 x}{5}\) = \(\frac{90}{5}\) Divide both sides by 5:
x = 18 Simplify:
y = 4 • x Substitute:
= 4 • 18
= 72 Simptify:
x = 18; y = 72

Math in Focus Course 2B Practice 6.1 Answer Key

Tell whether each pair of angles is complementary.

Question 1.
m∠A = 25° and m∠B = 65°
Answer:
m∠A = 25° We are given:
m∠B = 65°
m∠A + m∠B = 25° + 65° = 90° We compute the sum:
Because the sum of the measures of the two angles is exactly 90°, the given angles are complementary.
Yes

Question 2.
m∠C = 105° and m∠D = 7°
Answer:
m∠C = 105° We are given:
m∠D = 7°
m∠C + m∠D = 105° + 7° = 112° We compute the sum:
Because the sum of the measures of the two angles is not exactly 90°, the given angles are not complementary.
No

Question 3.
m∠E = 112° and m∠F = 68°
Answer:
m∠E = 112° We are given:
m∠F = 68°
m∠E + m∠F = 112° + 68° = 180° We compute the sum:
Because the sum of the measures of the two angles is not exactly 90° the given angles are not complementary.
No

Question 4.
m∠G = 45° and m∠H = 45°
Answer:
m∠G = 45° We are given:
m∠H = 45°
m∠G + m∠H = 45° + 45° = 90° We compute the sum:
Because the sum of the measures of the two angles is exactly 90°, the given angles are complementary.
Yes

Tell whether each pair of angles is supplementary.

Question 5.
m∠A = 130° and m∠8 = 50°
Answer:
m∠A = 130° We are given:
m∠B = 50°
m∠A + m∠B = 130° + 50° = 180° We compute the sum:
Because the sum of the measures of the two angles is exactly 180°, the given angles are supplementary.
Yes

Question 6.
m∠C = 90° and m∠D = 80°
Answer:
m∠C = 90° We are given:
m∠D = 80°
m∠C + m∠D = 90° + 80° = 170° We compute the sum:
Because the sum of the measures of the two angles is exactly 180°, the given angles are not supplementary.
No

Question 7.
m∠E = 120° and m∠F = 60°
Answer:
m∠E = 120° We are given:
m∠F = 60°
m∠E + m∠F = 120° + 60° = 180° We compute the sum:
Because the sum of the measures of the two angles is exactly 180°, the given angles are supplementary.
yes

Question 8.
m∠G = 60° and m∠H = 30°
Answer:
m∠G = 60° We are given:
m∠H = 30°
m∠G + m∠H = 60° + 30° = 90° We compute the sum:
Because the sum of the measures of the two angles is exactly 180°, the given angles are not supplementary.
yes

Find the measure of the complement of the angle with the given measure.

Question 9.
19°
Answer:
m∠θ = 19° We are given:
90° – m∠θ = 90° – 19 = 71° We determine the measure of the complement of the angle with the given
measure:
71°

Question 10.
64°
Answer:
m∠θ = 64° We are given:
90° – m∠θ = 90° – 64° = 26° We determine the measure of the complement of the angle with the given
measure:
26°

Question 11.
7°
Answer:
m∠θ = 7° We are given:
90° – m∠θ = 90° – 7° = 83° We determine the measure of the complement of the angle with the given
measure:
83°

Question 12.
35°
Answer:
m∠θ = 35° We are given the angle:
90° – m∠θ = 90° – 35° = 55° We determine the measure of the complement of the angle with the given
measure:
55°

Find the measure of the supplement of the angle with the given measure.

Question 13.
78°
Answer:
m∠θ = 78° We are given the angle:
180° – m∠θ = 180° – 78° = 102° We determine the measure of the complement of the angle with the given measure:
102°

Question 14.
4°
Answer:
m∠θ = 4° We are given the angle:
180° – m∠θ = 180° – 78° = 102° We determine the measure of the complement of the angle with the given measure:
176°

Question 15.
153°
Answer:
m∠θ = 153° We are given the angle:
180° – m∠θ = 180° – 153° = 27° We determine the measure of the complement of the angle with the given measure:
27°

Question 16.
101°
Answer:
m∠θ = 101° We are given the angle:
180° – m∠θ = 180° – 101° = 79° We determine the measure of the complement of the angle with the given measure:
79°

∠ABD and ∠DBC are complementary angles. Find the value of x.

Question 17.

Answer:
m∠ABD + m∠DBC = 90° ∠ABD and ∠DBC complementary:
65 + x = 90 Substitute:
65 + x – 65 = 90 – 65 Subtract 65 from both sides:
x = 25 Simplify:
x = 25

Question 18.

Answer:
m∠ABD + m∠DBC = 90° ∠ABD and ∠DBC complementary:
8 + x = 90 Substitute:
8 + x – 8 = 90 – 8 Subtract 8 from both sides:
x = 82 Simplify:
x = 82

∠PQS and ∠SQR are complementary angles. Find the value of m.

Question 19.

Answer:
m∠PQS + m∠SQR = 90° ∠PQS and ∠SQR supplementary:
m + 29 = 180 Substitute:
m + 29 – 29 = 180 – 29 Subtract 29 from both sides:
x = 151 Simplify:
x = 151

Question 20.

Answer:
m∠PQS + m∠SQR = 180° ∠PQS and ∠SQR supplementary:
152 + m = 180 Substitute:
152 + m – 152 = 180 – 152 Subtract 152 from both sides:
m = 28 Simplify:
m = 28

Answer each of the following.

Question 21.
The measure of an angle is 7°. Find the measure of its complement.
Answer:
m∠θ = 7° We are given the angle:
90° – m∠θ = 90° – 7° = 83° We determine the measure of the complement of the angle with the given measure:
83°

Question 22.
The measure of an angle is 84°. Find the measure of its supplement.
Answer:
m∠θ = 84° We are given the angle:
180° – m∠θ = 180° – 84° = 96° We determine the measure of the supplement of the angle with the given measure:
83°

Question 23.
Math Journal
a) Find the measures of the complement and the supplement of each of the following angles, where possible.
m∠W = 2° m∠X = 40° m∠Y = 32° m∠Z = 115°
Answer:
m∠W = 2° a) we are given the angle:
90° – m∠W = 90° – 2° = 88° We determine the complement of ∠W:
180° – m∠W = 180° – 2° = 178° We determine the supplement of ∠W:
m∠X = 40° We are given the angle:
90° — m∠X = 90° – 40° = 50° We determine the complement of ∠X:
180° — m∠X = 180° – 40° = 140° We determine the supplement of ∠X:
m∠Y = 32° We are given the angle:
90° — m∠Y = 90° — 32° = 58° We determine the complement of ∠Y:
180° — m∠Y = 180° — 32° = 148° We determine the supplement of ∠Y:
m∠Z = 115° We are given the angle:
As m∠Z > 90°: ∠Z has no complement
180° — m∠Z = 180 — 115° = 65° We determinethe supplement of ∠Z:

b) Which angle in a) does not have both a complement and a supplement?
Answer:
∠Z does not have both complement and supplement

c) In general, what must be true about the measure of an angle that has both a complement and a supplement?
Answer:
In order that an angle has both complement and supplement, its measure must be greater or equal than 0° and less or equal than 90°.

Question 24.
Math Journal Identify all the angles in each diagram. Tell which angles are adjacent. Explain your reasoning.

The measure of ∠ABC = 90°. Find the value of x.
Answer:
∠GOH We identify all the angles in the first diagram:
∠HOK
∠GOK
∠GOH and ∠HOK The adjacent angles in the diagram is the pair of angles which have a common side and a common vertex and don’t overlap:

∠WOX We identify all the angles in the second diagram:
∠XOY
∠YOZ
∠WOY
∠XOZ
∠WOZ
∠WOX and ∠XOY (common side OX, common vertex O) We identify the adjacent angles in the second diagram:
∠XOY and ∠YOZ (common side OY, common vertex O)
∠WOY and ∠YOZ (common side OY, common vertex O)
∠WOX and ∠XOZ (common side OX, common vertex O)

Question 25.

Answer:
m∠ABC = 90° We are given:
x° + 42° + 30° = 90° we have
x + 72° = 90°
x + 72 – 72 = 90 – 72 Subtract 72 from both sides
x = 18 simplify
x = 18

Question 26.

Answer:
m∠ABC = 90° We are given:
2x° + 22° = 90° we have
2x + 22 – 22 = 90 – 22 Subtract 22 from both sides
2x = 68 simplify
\(\frac{2 x}{2}\) = \(\frac{68}{2}\) Divide by 2:
x = 34 simplify:
x = 34

Question 27.

Answer:
m∠ABC = 90° We are given:
2x° + 45° + 21° = 90° we have
2x° + 66 = 90°
2x + 66 — 66 = 90 — 66 Subtract 66 from both sides:
2x = 24 Simplify:
\(\frac{2 x}{2}\) = \(\frac{24}{2}\) Divide by 2:
x = 12 Simplify:
x = 12

Question 28.

Answer:
m∠ABC = 90° We are given:
18° + 3x° + 21° = 90° we have
39° + 3x° = 90°
39 + 3x — 39 = 90 — 39 Subtract 39 from both sides:
3x = 51 Simplify:
\(\frac{3 x}{3}\) = \(\frac{51}{3}\) Divide by 3:
x = 17 Simplify:
x = 17

\(\overleftrightarrow{P R}\) is a straight line. Find the value of m.

Question 29.

Answer:
m∠PQR = 90° Because \(\overline{P R}\) is a straight line, we have
m∠PQT + m∠TQS + m∠SQR = m∠PQR we have
49° + m° + 57° = 180° substitute
106° + m° = 180°
106 + m – 106 = 180 – 106 subtract 106 from both sides
m = 74

Question 30.

Answer:
m∠PQR = 180° Because \(\overline{P R}\) is a straight line, we have
m∠PQT + m∠TQS + m∠SQR = m∠PQR we have
20° + m° + 87° = 180° substitute
150° + m° = 180°
150 + m – 150 = 180 – 150 subtract 150 from both sides
m = 30 Simplify:
m = 30

In the diagram, the ratio a: b = 2: 3. Find the values of a and b.

Question 31.

Answer:
a : b = 2 : 3 we are given
\(\frac{a}{b}\) = \(\frac{2}{3}\) we rewrite the ratio
3a = 2b cross-multiply
\(\frac{3 a}{3}\) = \(\frac{2 b}{3}\) Divide both sides by 3
a = \(\frac{2 b}{3}\)
m∠RQS + m∠SQP = 90° ∠RQS and ∠SQP are complementary
b + a = 90
b + \(\frac{2 b}{3}\) = 90 substitute
\(\frac{3 b+2 b}{3}\) = 90 Determine b
\(\frac{5 b}{3}\) = 90
5b = 3 ∙ 90
5b = 270
\(\frac{5 b}{5}\) = \(\frac{270}{5}\)
b = 54
a = \(\frac{2 \cdot 54}{3}\) = 36 Determine a
a = 36
b = 54

Question 32.
\(\overleftrightarrow{P R}\) is a straight line.

Answer:
a : b = 2 : 3 we are given
\(\frac{a}{b}\) = \(\frac{2}{3}\) we rewrite the ratio
3a = 2b cross-multiply
\(\frac{3 a}{3}\) = \(\frac{2 b}{3}\) Divide bothsides by 3
a = \(\frac{2 b}{3}\)
m∠RQS + m∠SQP = 180° ∠RQS and ∠SQP are supplementary
b + a = 180 substitute
b + \(\frac{2 b}{3}\) = 180
\(\frac{3 b+2 b}{3}\) = 180
\(\frac{5 b}{3}\) = 180
5b = 3 ∙ 180
5b = 540
\(\frac{5 b}{5}\) = \(\frac{540}{5}\)
b = 108
a = \(\frac{2 \cdot 108}{3}\) = 72
a = 72
b = 108

Solve.

Question 33.
The diagram shows the pattern on a stained glass window. \(\overleftrightarrow{A C}\) is a straight line. ∠EBD and ∠DBA are complementary angles and m∠DBA = 30°. Find the measures of ∠EBD and ∠CBD.

Answer:

we are given
m∠EBD + m∠DBA = 90° ∠EBD and ∠DBA complementary:

m∠EBD + 30° = 90° Substitute:
m∠EBD + 30° — 30° = 90° – 30° Subtract 30° from both sides:
m∠EBD = 60° Simplify:
m∠CBD + m∠DBA = 180° \(\overline{A C}\) straight line:
m∠CBD + 30° = 180° Substitute
m∠CBD + 30° – 30° = 180° – 30° Subtract 30° from both sides
m∠CBD = 150° Simplify
m∠EBD = 60°
m∠CBD = 150°

Question 34.
The diagram shows a kite. The two diagonals \(\overline{M P}\) and \(\overline{Q T}\) are perpendicular to each other. Identify all pairs of complementary angles and all pairs of supplementary angles that are not pairs of right angles.

Answer:
\(\overline{M P}\) ⊥ \(\overline{Q T}\) We are given:
∠MNR and ∠RNQ We identify the pairs of complementary angles:
∠QNS and ∠SNP
∠NMQ and ∠MQN
∠NQP and ∠QPN
∠TMN and ∠MTN
∠NTP and ∠NPT
∠MNR and ∠RNQ We identify the pairs of supplementary angLes that are not pairs of right angles:
∠MNS and ∠SNP
Complementary angles: ∠MNR and ∠RNQ
∠QNS and ∠SNP
∠NMQ and ∠MQN
∠NQP and ∠QPN
∠TMN and ∠MTN
∠NTP and ∠NPT
Supplementary angles: ∠MNR and ∠RNP
∠MNS and ∠SNP

Practice the problems of Math in Focus Grade 7 Workbook Answer Key Chapter 6 Angle Properties and Straight Lines to score better marks in the exam.

Math in Focus Grade 7 Course 2 B Chapter 6 Answer Key Angle Properties and Straight Lines

Math in Focus Grade 7 Chapter 6 Quick Check Answer Key

Tell whether each angle is an acute, right, obtuse, or straight angle.

Question 1.

Answer:
Right angle triangle.

Explanation:
A right-angled triangle is a triangle with one of the angles as 90 degrees.

Question 2.

Answer:
Acute angle triangle.

Explanation:
An acute-angled triangle is a type of triangle in which all the three internal angles of the triangle measures less than 90°.

Question 3.

Answer:
90° < 154° < 180° We have
The given angle is obtuse because Its measure is greater than 90°, but Less than 180°.
obtuse

Question 4.

Answer:
straight angle.

Explanation:
The 180-degree angle is known as a straight angle. The sides of the angle are opposite to each other and they make a straight angle.

Question 5.
m∠w = 86°
Answer:
Acute angle

Explanation:

Question 6.
m∠y = 90°
Answer:
Right angle triangle.

Explanation:

Identify each pair of parallel line segments.

Question 7.

Answer:
RS∥UT

Explanation:
The lines will not meet each other at any point in the given figure. Hence parallel line will never intersect.

Question 8.

Answer:
WX∥ZY

Explanation:
The lines that do not intersect or meet each other at any point in a plane. Parallel lines will never intersect.

Identify each pair of perpendicular line segments.

Question 9.
ABCD is a rectangle.

Answer:
There are four right angles, and 2 pairs of perpendicular lines.

Explanation:
A rectangle has lines that are perpendicular to two other lines and it also has four right angles.

Question 10.

Answer:
The given figure is a right-angle triangle and has two perpendicular lines.

Explanation:
A triangle has three sides and three angles. Only one type of triangle is a right-angle triangle that has two perpendicular line segments.

Question 11.

Answer:
The given figure has two perpendicular line segments.

Explanation:
Perpendicular lines are lines, segments that intersect to form right angles.

Practice the problems of Math in Focus Grade 7 Workbook Answer Key Chapter 5 Review Test to score better marks in the exam.

Math in Focus Grade 7 Course 2 A Chapter 5 Review Test Answer Key

Concepts and Skills

Tell whether each table, graph, or equation represents a direct proportion, an inverse proportion, or neither.

Question 1.

Answer:

Direct Proportion :
For each pair of values x and y →
\(\frac{y}{x}\) = \(\frac{4.5}{3}\) = 1.5
\(\frac{y}{x}\) = \(\frac{7.5}{5}\) = 1.5
\(\frac{y}{x}\) = \(\frac{10.5}{7}\) = 1.5
\(\frac{y}{x}\) is a constant so x and y is direct proportional

Inverse Proportion :
For each pair of values x and y →
xy = 3 ∙ 4.5 = 13.5
xy = 5 . 7.5 = 37.5
xy = 7 ∙ 10.5 = 73.5
xy is not a constant so y is not inversely proportional to x

NOTE :
Direct Proportion : y is directly proportional to x then they have a constant of proportionality k, and it can be rewritten as y = kx or \(\frac{y}{x}\) = k
Inverse Proportion : x and y are in inverse proportion then their product is a constant of proportionality k, and it can be rewritten a. xy = k or y = \(\frac{k}{x}\)
Direct proportion

Question 2.

Answer:

Direct Proportion :
For each pair of valties x and y →
\(\frac{y}{x}\) = \(\frac{50}{2}\) = 25
\(\frac{y}{x}\) = \(\frac{25}{4}\) = 6.25
\(\frac{y}{x}\) = \(\frac{12.5}{8}\) = 1.56
\(\frac{y}{x}\) is not a constant so x and y does not represent direct proportion

Inverse Proportion :
For each pair of values x and y →
xy = 2 ∙ 50 = 100
xy = 4 ∙ 25 = 100
xy = 8 ∙ 12.5 = 100
The product of x and y is a constant so y is inversely proportional to x

NOTE :
Direct Proportion : y is directly proportional to x then they have a constant of proportionality k, and it can be rewritten as y = kx or \(\frac{y}{x}\) = k
Inverse Proportion : x and y are in inverse proportion then their product is a constant of proportionality k, and it can be rewritten as xy = k or y = \(\frac{k}{x}\)
Inverse Proportion

Question 3.

Answer:

Direct Proportion :
For each pair of values x and y →
\(\frac{y}{x}\) = \(\frac{12}{6}\) = 2
\(\frac{y}{x}\) = \(\frac{9}{8}\) = 1.125
\(\frac{y}{x}\) = \(\frac{3.5}{2.4}\) = 0.145
\(\frac{y}{x}\) is not a constant so x and y is not direct proportion.

Inverse Proportion :
For each pair of values x and y →
xy = 6 ∙ 12 = 72
xy = 8 ∙ 9 = 72
xy = 24 ∙ 3.5 = 84
The product of z and y is not a constant so y is not inversely proportional to z

NOTE :
Direct Proportion : y is directly proportional to x then they have a constant of proportionality k, and it can be rewritten as y = kx or \(\frac{y}{x}\) = k
Inverse Proportion : x and y are in inverse proportion then their product is a constant of proportionality k, and it can be rewritten as xy = k or y = \(\frac{k}{x}\)
Neither Direct Proportion nor Inverse Proportion

Question 4.

Answer:

Direct Proportion :
For each pair of values x and y →
\(\frac{y}{x}\) = \(\frac{5}{2.5}\) = 2
\(\frac{y}{x}\) = \(\frac{10}{5}\) = 2
\(\frac{y}{x}\) = \(\frac{15}{7.5}\) = 2
\(\frac{y}{x}\) is a constant so x and y is direct proportional

Inverse Proportion :
For each pair of values x and y →
xy = 2.5 ∙ 5 = 12.5
xy = 5 ∙ 10 = 50
xy = 7.5 ∙ 15 = 112.5
The product of x and y is not a constant so x is not inversely proportional to x

NOTE :
Direct Proportion : y is directly proportional to x then they have a constant of proportionality k, and it can be rewritten as y = kx or \(\frac{y}{x}\) = k
Inverse Proportion : x and y are in inverse proportion then their product is a constant of proportionality k, and it can be rewritten as xy = k or y = \(\frac{k}{x}\)
Direct Proportion

Question 5.

Answer:
Direct Proportion :
The given graph is a straight line that does not lie along
the x — axis or y — axis but does not pass through the origin.
So, the given graph does not represents a direct proportion.

Inverse Proportion :
$Use\graph\: $
x ∙ y = 3 ∙ 1 = 2
x ∙ y = 2 ∙ 2 = 4
x ∙ y = 1 ∙ 3 = 3
The product of x and y is not a constant so x and y does not represent a inverse proportion.
Neither direct proportion nor inverse proportion

Question 6.

Answer:
Direct Proportion :
The given graph is a straight line that does not lie along the x — axis or y — axis and it pass through the origin.
So. the given graph represents a direct proportion.

Inverse Proportion :
$Use\graph\: $
x ∙ y = 1 ∙ 1 = 1
x ∙ y = 2 ∙ 2 = 4
x ∙ y = 3 ∙ 3 = 9
The product of x and y is not a constant so x and y does not ‘represent a inverse proportion.
Direct Proportion

Question 7.

Answer:
Direct Proportion :
The given graph is not a straight line and it lie along the x — axis or y — axis and does not pass through the origin.
So. the given graph does not ‘represents a direct proportion.

Inverse Proportion:
$Use\ graph\: $
x ∙ y = 4 ∙ 1 = 4
x ∙ y = 2 ∙ 2 = 4
x ∙ y = 1 ∙ 4 = 4
The product of x and y is a constant so x and y represents a inverse proportion.
Inverse Proportion

Question 8.

Answer:
Direct Proportion :
The given graph is not a straight line and it lie along the x — axis but it pass through the origin.
So, the given graph does not represents a direct proportion.

Inverse Proportion :
$Use\graph\: $
x ∙ y = 1 ∙ 2 = 2
x ∙ y = 2 ∙ 4 = 8
x ∙ y = 3 ∙ 2 = 6
The product of x and y is not a constant so x and y does not represent a inverse proportion.
Neither direct proportion nor inverse proportion

Question 9.
y = \(\frac{1}{2}\)x + 5
Answer:
y = \(\frac{1}{2} x\) + 5
Direct Proportion:
y = \(\frac{1}{2} x\) + 5
Because the original equation y = y = \(\frac{1}{2} x\) + 5 cannot be rewritten as an equivalent
equation in the form y = kx, it does not represent a direct proportion.

Inverse Proportion :
y = \(\frac{1}{2} x\) + 5

Because the original equation y = \(\frac{1}{2} x\) + 5 cannot be rewritten as an equivalent
equation in the form xy = k, it does not represent an inverse proportion.

NOTE :
Direct Proportion : y is directly proportional to x then they have a constant of proportionality k, and it can be rewritten as y = kx or \(\frac{y}{x}\) = k
Inverse Proportion : x and y are in inverse proportion then their product is a constant of proportionality k, and it can be rewritten xy = k or y = \(\frac{k}{x}\)
Neither Direct Proportion nor Inverse Proportion

Question 10.
\(\frac{y}{4}\) = 7x
Answer:
\(\frac{y}{4}\) = 7x
4 ∙ \(\frac{y}{4}\) = 4 ∙ 7x
y = 28x
Because the original equation \(\frac{y}{4}\) = x can be rewritten as an equivalent equation in the form y = kx, it represent a direct proportion.

Inverse Proportion :
\(\frac{y}{4}\) = 7x
4 ∙ \(\frac{y}{4}\) = 4 ∙ 7x
y = 28x
Because the original equation \(\frac{y}{4}\) = x cannot be re’wrítten as an equivalent
equation in the form xy = k, it does not represent an inverse proportion.

NOTE :
Direct Proportion : y is directly proportional to x then they have a constant of proportionality k. and it can be rewritten as y = kx or \(\frac{y}{x}\) = k
Inverse Proportion : x and y are in inverse proportion then their product is a constant of proportionality k. and it can be rewritten as xy = k or y = \(\frac{k}{x}\)
Direct Proportion

Question 11.
-3 + x = y
Answer:
-3 + x = y
Direct Proportion :
-3 + x = y
-3 + x – x = y – x Subtract x from both sides
y – x = -3
Because the original equation -3 + x = y cannot be rewritten as an equivalent equation in the form y = kx, it does not represent a direct proportion.
Inverse Proportion :
-3 + x = y
-3 + x — x = y — x Subtract x from both sides
y – x = -3
Because the original equation — 3 + x = y cannot be rewritten as an equivalent equation in the form xy = k, it does not represent an inverse proportion.

NOTE :
Direct Proportion : y is directly proportional to z then they have a constant
of proportionality k, and it crin be rewritten as y = kz or \(\frac{y}{x}\) = k
Inverse Proportion : z and y are in inverse proportion thìen their product is a constant of proportionality k, and it can be rewritten as zy = k or y = \(\frac{k}{x}\)
Neither Direct Proportion nor inverse Proportion

Question 12.
\(\frac{y}{2}\) = \(\frac{3}{x}\)
Answer:
\(\frac{y}{2}\) = \(\frac{3}{x}\)
Direct Proportion :
\(\frac{y}{2}\) = \(\frac{3}{x}\)
x ∙ y = 3 ∙ 2 Cross Multiplication
xy = 6
Because the original equation \(\frac{y}{2}\) = \(\frac{3}{x}\) cannot be rewritten as an equivalent equation in the form y = kx, it does not represent a direct proportion.

Inverse Proportion :
\(\frac{y}{2}\) = \(\frac{3}{x}\)
x ∙ y = 3 ∙ 2 Cross Multiplication
xy = 6
Because the original equation \(\frac{y}{2}\) = \(\frac{3}{x}\) can be rewritten as an equivalent equation in the form xy = k, it does represent an inverse proportion.

NOTE :
Direct Proportion : y is directly proportional to x then they have a constant of proportionality k, and it crin be rewritten as y = kx or \(\frac{y}{x}\) = k
Inverse Proportion : x and y are in inverse proportion then their product is a constant
of proportionality k. and it can be rewritten as xy = k or y = \(\frac{k}{x}\)
Inverse proportion

In each table, y is directly proportional to x. Find the constant of proportionality. Then copy and complete the table.

Question 13.

Answer:

y is directly proportional to z :
\(\frac{y}{x}\) = \(\frac{16}{4}\)
\(\frac{y}{x}\) = 4 Simplify
x ∙ \(\frac{y}{x}\) = 4 ∙ x Multiply both sides by x
y = 4x
Find missing value in the table.
When y = 25 find x.
y = 4x
25 = 4x Evaluate y = 25
\(\frac{25}{4}\) = \(\frac{4x}{4}\) Divide both sides by 4
x = 6.25 Simplify
When x = 2 find y,
y = 4x
y = 4 ∙ 2 Evaluate x = 2
y = 8 Simplify

The missing number are 8 and 6.25

Question 14.

Answer:

y is directly proportional to x :
\(\frac{y}{x}\) = \(\frac{2.5}{5}\)
\(\frac{y}{x}\) = \(\frac{1}{2}\) Simplify
x ∙ \(\frac{y}{x}\) = \(\frac{1}{2}\) ∙ x Multiply both sides by x
y = \(\frac{1}{2}\)x

Find missing value in the table.
When y = 3 find x,
y = \(\frac{1}{2}\)x
3 = \(\frac{1}{2}\)x Evaluate y = 3
2 ∙ 3 = 2 ∙ \(\frac{1}{2}\)x Multiply both sides by 2
x = 6 Simplify
When x = 3 find y,
y = \(\frac{1}{2}\)x
y = \(\frac{1}{2}\) ∙ 3 Evaluate x = 3
y = 1.5 simplify

The missing number are y = 1.5 and x = 6

In each table, y is inversely proportional to x. Find the constant of proportionality. Then copy and complete the table.

Question 15.

Answer:

y is inversely proportional to x :
xy = 2 ∙ 30
xy = 60 Simplify
Find missing value in the table.
When x = 4 find y.
xy = 60
4 ∙ y = 60 Evaluate x = 4
\(\frac{4 y}{4}\) = \(\frac{60}{4}\) Divide both sides by 4
y = 15 Simplify
When y = 10 find x.
x ∙ 10 = 60 Evaluate y = 10
\(\frac{10 x}{10}\) = \(\frac{60}{10}\) Divide both sides by 10
x = 6 Simplify

The missing number are y = 15 and x = 6

Question 16.

Answer:

y is inversely proportional to x :
xy = 5 ∙ 1.6
xy = 8 Simplify
Find missing value in the table.
When x = 2.5 find y.
xy = 8
2.5 ∙ y = 8 Evaluate x = 2.5
\(\frac{2.5 y}{2.5}\) = \(\frac{8}{2.5}\) Divide both sides by 2.5
y = 3.2 simplify
when y = 2 find x,
x ∙ 2 = 8 Evaluate y = 2
\(\frac{2 x}{2}\) = \(\frac{8}{2}\) Divide both sides by 2.
x = 4 simplify

The missing number are y = 3.2 and x = 4

Solve using proportionality reasoning.

Question 17.
y is directly proportional to x, and y = 56 when x = 7. Find the value of y when x = 4.
Answer:
x = 7 and y = 56
y is directly proportional to x:
\(\frac{y}{x}\) = \(\frac{56}{7}\)
\(\frac{y}{x}\) = 8
x ∙ \(\frac{y}{x}\) = 8 ∙ x Multiply both sides by x
y = 8x
Find the value of y when x = 4
We know y = 8x,
y = 8 ∙ 4 Evaluate y = 8x when x = 4
y = 32 Simplify
y = 32

Question 18.
y is inversely proportional to x, and y = 12 when x = 4. Find the value of x when y = 8.
Answer:
x = 4 and y = 12
y is inversely proportional to x :
xy = 4 ∙ 12
xy = 48
\(\frac{x y}{y}\) = \(\frac{48}{y}\) Divide both sides by y
x = \(\frac{48}{y}\)
Find the value of x when y = 8
We know x = \(\frac{48}{y}\),
x = \(\frac{48}{8}\) Evaluate x = \(\frac{48}{y}\) when y = 8
x = 6 simplify
x = 6

Problem Solving

Use a proportion to solve each question. Show your work.

Question 19.
The graph shows that the cost of gasoline, y dollars, is directly proportional to x gallons of gasoline.

a) Find the constant of proportionality. What does this value represent in the context of the problem?
Answer:
y is directly proportional to x :
\(\frac{y}{x}\) = \(\frac{4}{1}\)
\(\frac{y}{x}\) = 4 simplify
Constant of Proportionality = 4
It represents the unit cost per gallons of gasoline.

b) Write a direct proportion equation.
Answer:
y is directly proportional to x :
\(\frac{y}{x}\) = \(\frac{4}{1}\) = 4 Simplify
\(\frac{y}{x}\) = \(\frac{8}{2}\) = 4 Simplify
\(\frac{y}{x}\) = \(\frac{12}{3}\) = 4 Simplify
x ∙ \(\frac{y}{x}\) = 4 ∙ x Multiply both sides by x
y = 4x

c) If Umberto spent $24 for gasoline, how many gallons of gasoline did he buy?
Answer:
y = 4x
24 = 4x Evaluate y = 24
\(\frac{24}{4}\) = \(\frac{4 x}{4}\) Divide both sides by 4
x = 6
Umberto will get 6 gallons of gasoline if he spent $24.

Question 20.
Out of every $100 that Judy earns at her part-time job, she saves $25 for college. The amount she saves is directly proportional to the amount she earns. If she earns $3,880 in one year, how much will she save for college?
Answer:
Amount of money she saves = y
Amount of money she earns = x
Find Direct Proportion Equation :
Amount of money she saves is directly proportional to Amount of money she earns :
\(\frac{y}{x}\) = \(\frac{25}{100}\)
\(\frac{y}{x}\) = \(\frac{1}{4}\) simplify
x ∙ \(\frac{y}{x}\) = x ∙ \(\frac{1}{4}\) Multiply both sides x
y = \(\frac{1}{4}\)x

If she earns $3880 in one year, how much she will save :
y = \(\frac{1}{4}\)x
y = \(\frac{1}{4}\) ∙ 3880 Evaluate x = 3880
y = 970
She will save $970 in one year if she earns $3880.

Question 21.
Harry made fruit punch using 2 parts orange juice to 3 parts soda water. The amount of soda water that Harry used is directly proportional to the amount of orange juice he used. How many cups of orange juice should Harry use with 18 cups of soda water?
Answer:
Amount of soda water = y
Amount of orange juice = x
Find Direct Proportion Equation :
Amount of soda water is directly proportional to Amount of orange juice :
\(\frac{y}{x}\) = \(\frac{3}{2}\)
\(\frac{y}{x}\) = 1.5 simplify
x ∙ \(\frac{y}{x}\) = x ∙ 1.5 Multiply both sides by x
y = 1.5x
How many cups of orange juice should Harry use with 18 cups of soda water :
y = 1.5z
18 = 1.5x Evaluate y = 18
\(\frac{18}{1.5}\) = \(\frac{1.5 x}{1.5}\) Divide both sides by 1.5
x = 12
Harry should ‘use 12 cups of orange juice with 18 cups of soda water.
12 cups

Use a graph paper. Solve.

Question 22.
An initial amount of money deposited in a bank account that earns interest is called the principal. In the table below, P stands for the principal, and l stands for the interest earned by that principal for a period of one year at a particular bank. P is directly proportional to l. Graph the direct proportion relationship between P and l. Use 1 unit on the horizontal axis to represent 1 dollar and 1 unit on the vertical axis to represent $50.

Answer:

a) Using the graph, find the interest earned when the principal is $350.
Answer:
$From\ the\ graph,\ for\ $350\ Principal,\ Interest\ earned\ is\ $7 $

b) Write an equation relating P and I. Then find the principal when the interest earned is $15.
Answer:
P is directly proportional to I :
\(\frac{P}{I}\) = \(\frac{100}{2}\)
\(\frac{P}{I}\) = 50 Simplifying
I ∙ \(\frac{P}{I}\) = I ∙ 50 Multiplying both sides by I
P = 50I
Find the principal when the Interest earned is $15 :
P = 50I
P = 50 ∙ 15 Substituting I = 15
P = 750

Use a proportion to solve each question. Show your work.

Question 23.
The time taken by some students to deliver 500 flyers, t hours, is inversely proportional to the number of students, n. The graph shows the relationship between n and t.

a) Find the constant of proportionality graphically.
Answer:
$Since,\ the\ line\ is\ passing\ through\ the\ point\ (6,\ 2) $
So, the constant of proportionality is 6 × 2 = 12
Constant of Proportionality = 12

b) Write an equation relating n and t.
Answer:
Number of hours (t) to delivers the flyers is inversely proportional to the ‘number of students
n ∙ t = 6 ∙ 2
nt = 12

c) Describe the relationship between the number of students and the time needed to deliver the flyers.
Answer:
Since t is inversely proportional to n
The number of students increases as the amount of time to deliver the flyers decreases.

d) Explain what the point (6, 2) represents in this situation.
Answer:
6 students take 2 hours to deliver 500 flyers.

Question 24.
Jerry has set aside a certain amount of money to download applications for his new smart phone. The number of applications he can afford to download is inversely proportional to the cost of each download. With his money, he can afford to download 12 applications that cost $2 each. How many applications can he afford to download if he finds less expensive applications that cost only $1.50 each?
Answer:
Number of application he can afford to download = x
Cost of each download = y
Find Inverse Proportion equation :
$He\ has\ the\ amount\ of\ money\ to\ download\ 12\ application\ which\ cost\ $2\ each\
Number of applications he can afford to download is inversely proportional to Cost of each download \ :
xy=12\cdot2
xy = 24 $
How many applications he can afford to download if cost of each application is $1.50 :
xy = 24
1.50 = 24 Substitute y = 1.50
\(\frac{1.50 x}{1.50}\) = \(\frac{24}{1.50}\) Divide both sides by 1.50
x = 16
$He\ can\ afford\ to\ download\ 16\ applications\ if\ each\ applications\ cost\ $1.50 $
16 application

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

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

Hands-On Activity

Materials

• algebra tiles

RECOGNIZE INVERSE PROPORTION

Work in pairs.
There are 6 ways of forming a rectangle with 12 algebra tiles. The diagram shows two possible ways.

Step 1: Form a different rectangle by rearranging the 12 algebra tiles. Record your results in a table like the one shown.

Step 2: Complete the table by repeating Step 1:

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

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

Copy and complete.

Question 1.
Some friends want to share the cost of buying a present. The table shows the amount of money that each person has to contribute, y dollars, and the number of people sharing the cost, x. Tell whether y is inversely proportional to x. If so, find the constant of proportionality.

For each pair of values, x and y:
1 • 180 = • 90 = . • =
The value of x increases as the value of y decreases, and the product of x and y is a value. So, y is inversely proportional to x.
The constant of proportionality is .
Answer:

For each pair of values, x and y ;
1. 180 = 180
2. 90 = 180
3. 60 = 180
The va1te of increases as the value of y decreases, and the product of x and y is a constant value.
So y is inversely proportional to x.
constant of proportionality = 180
Yes, y is proportional to x
Constant of proportionality = 180

Question 2.
Henry drove from Town A to Town B. The table shows the time he took, y hours, if he traveled at various speeds, x miles per hour. Tell whether x and y are in inverse proportion. If so, find the constant of proportionality.

For each pair of values, x and y:
40 • 9 = 50 • = • =
The value of x increases as the value of y decreases, but the product of x and y is value. So, y is inversely proportional to x.
Answer:

For each pair of values, x and y :
40 ∙ 9 = 360
50 ∙ 7\(\frac{1}{3}\) = 366.66
60 ∙ 6 = 360
The value of x increases as the value of y decreases, but the product of x and y is not a constant value.
So y is not inversely proportional to z.
No, y is not proportional to x

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

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

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

Solve an inverse proportion problem graphically.

Question 5.
The amount of time needed for volunteers to pick up trash on a beach is inversely proportional to the number of volunteers. The graph shows the amount of time, y hours, needed by x volunteers.

a) Find the constant of proportionality graphically. Then write an inverse proportion equation.
Use the point (, ) from the graph to find the constant of proportionality:
x • y = • Choose the point (, ).
xy = Multiply.
The constant of proportionality is .
The inverse proportion equation is .
Answer:
Use the point (6, 10) from the graph to find the constant of proportionality.
x ∙ y = 6 10 Choose the point (6, 10)
xy = 60 Multiply
The constant of proportionality is 60
The inverse proportion equation is xy = 60

b) Explain what the point (6, 10) represents in this situation.
It means that volunteers can clean the beach in hours.
Answer:
It means that 6 volunteers can beach in 10 hours.

Copy and complete.

Question 6.
y is inversely proportional to x, and y = 3 when x = 5.
Answer:
y = 3 when x = 5
a) Find the value of the constant
Constant of proportionality:
x ∙ y = ∙
The constant of proportionality is .
Answer:
Constant of proportionality :
x ∙ y = 5 ∙ 3 = 15

b) Write an inverse proportion equation that relates x and y.
Inverse proportion equation:
= or =
The inverse proportion equation is = or = .
Answer:
Inverse proportion equation :
xy = 15 or y = \(\frac{15}{x}\)

c) Find the value of x when y = 10.
When x = 10 and y = \(\frac{?}{x}\), y = \(\frac{?}{?}\) Evaluate y = \(\frac{?}{x}\) when x = 10.
y = . Simplify.
The value of y is .
Answer:
When x = 10 and y = \(\frac{15}{x}\),
y = \(\frac{15}{10}\)
y = 1.5

Solve.

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

Math in Focus Course 2A Practice 5.4 Answer Key

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

Question 1.

Answer:

For each pair of values, x and y :
25 . 2 = 50
10 . 5 = 50
5 . 10 = 50
The value of x decreases as the value of y increases and the product of x and y is a constant value.
So y is inversely proportional to x.
Constant of Proportionality = 50
Yes, y is proportional to x
Constant of proportionality = 50

Question 2.

Answer:

For each pair of values, x and y :
7 . 30 = 210
5 . 60 = 300
3 . 70 = 210
The value of x decreases as the value of y increases, but the product of x and y is not a constant value.
So y is not inversely proportional to x.
No, y is not proportional to x

Question 3.

Answer:

For each pair of values, x and y :
4 ∙ 16 = 64
6 ∙ 24 = 144
8 ∙ 32 = 236
The value of x increases as well as the value of y increases, and the product of x and y is also not a constant value.
So y is not inversely proportional to x.
No, y is not proportional to x

Question 4.

Answer:

For each pair of values, x and y :
6 ∙ 2 = 12
3 ∙ 4 = 12
1 ∙ 12 = 12
The value of decreases as the value of y increases, and the product of x and y is a constant value.
So y is inversely proportional to x.
Constant of Proportionality = 12
Yes, y is proportional to x
Constant of proportionality = 12

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

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

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

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

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

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

Question 9.

Answer:
Find the constant of proportionality graphically. Then write an inverse proportion equation.
Use the point (4, 0.5) f rom the graph to find the constant of proportionality.
x ∙ y = 4 ∙ 0.5 Choose the point (4, 0.5)
xy = 2 Multiply
The constant of proportionality is 2
The inverse proportion eqnation is xy = 2
Constant proportionality = 2; xy = 2

Question 10.

Answer:
Find the constant of proportionality graphically. Then write an inverse proportion equation.
Use the paint (2, 5) from the graph to find the constant of proportionality.
x ∙ y = 2 ∙ 5 Choose the point (2, 5)
xy = 10 Multiply
The constant of proportionality is 10
The inverse proportion equation is xy = 10
Constant of proportionality = 10; xy = 10

Question 11.

Answer:
Find the constant of proportionality graphically. Then write an inverse proportion equation.
Use the point (0.5, 8) from the graph to find the constant of proportionality.
x ∙ y = 0.5 ∙ 8 Choose the point (0.5, 8)
xy = 4 Multiply
The constant of proportionality is 4
The inverse proportion equation is xy = 4
Constant of proportionality = 4 ; xy = 4

Question 12.

Answer:
Find the constant of proportionality graphically. Then write an inverse proportion equal ion.
Use the point (\(\frac{1}{2}\), 30) from the graph to find the constant of proportionality.
x ∙ y = \(\frac{1}{2}\) ∙ 30 Choosethe point (\(\frac{1}{2}\), 3o)
xy = 15 Multiply
The constant of proportionality is 15
The inverse proportion equation is xy = 15
Constant of proportionality = 15; xy = 15

Solve. Show your work.

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

Question 14.
The workers at a bakery must mix 12 batches of bagel dough every hour to meet the needs of customers. The number of batches of bagel dough, b, that each worker needs to mix in one hour is inversely proportional to the number of workers, n. The graph shows the relationship between b and n.

a) Find the constant of proportionality from the graph. Then write an inverse proportion equation.
Answer:
Use the point (6, 2) from the graph to find the on stunt of proportionality.
n ∙ b = 6 ∙ 2 Choose the point (6, 2)
bn = 12 Multiply
The constant of proportionality is 12
The inverse proportion equation is bn = 12

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

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

Question 15.
A rectangle has a fixed area that does not change. The length, l, of the rectangle is inversely proportional to its width, w. The graph shows the relationship between l and w.

a) Find the constant of proportionality from the graph. Then write an inverse proportion equation.
Answer:
Use the point (3, 8) from the graph to find the constant of proportionality.
w ∙ l = 3 ∙ 8 Choose the point (3, 8)
lw = 24 Multiply
The constant of proportionality is 24
The inverse proportion equation is lw = 24

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

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

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

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

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

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

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

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

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

Question 19.
The number of hours it takes to mow nine lawns is inversely proportional to the number of gardeners. It takes three gardeners four hours to mow nine lawns. How many hours would it take one gardener to mow the same nine lawns?

Answer:
Let x be the number of hours.
Let y be the number of gardeners.
The numbers of hours to mow nine lawns is inversely proportional to the number of gardeners.
Constant of proportionality :
x ∙ y = 4 ∙ 3
xy = 12
Inverse proportion equation :
xy = 12 write an inverse equation
When y = 1 and xy = 12 :
1 ∙ x = 12 Evaluate xy = 12 ‘when y = 1
x = 12 Simplify
It takes one gardeners 12 hours to mow nine lawns..
xy = 12 Write an inverse equation
12 hours

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

Question 21.
Math journal Each table shows the price, y dollars, that x people have to pay to rent a guest house for one day. Describe how the two tables are alike, and how they are different. Be sure to discuss inverse proportion.

Answer:

For each pair of values, x and y ;
1 ∙ 240 = 240
2 ∙ 120 = 240
3 ∙ 80 = 240
4 ∙ 60 = 240
The value of x increases as the value of y decreases, and the product of x and y is a constant value.
So y is inversely proportional to x.
Constant of Proportionality = 240

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

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

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

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

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

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

Brain @ Work

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

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

Cost of book in United States:

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

Question 2.
Johnny leaves Town P to drive to Town Q, a distance of 350 miles. He hopes to use only 12 gallons of gasoline. After traveling 150 miles, he checks his gauge and estimates that he has used 5 gallons of gasoline. At this rate, will he be able to reach Town Q before stopping for gasoline? Justify your answer.

Answer:
Distance between Town P and Town Q = 350 miles
Estimated gasoline he wili requin =12 galions
After traveling 150 miles he used 5 gallons of gasoline
Gasoline requires per miles = \(\frac{5}{150}\) = \(\frac{1}{30}\) gallons
Distance left to travel = 350 – 150 = 200 miles
Gasoline required for remaining distance
= 200 • \(\frac{1}{30}\)
= 6\(\frac{2}{3}\) gallons

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

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

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

Solve.
Question 1.
At a factory, the number of cars produced is directly proportional to the number of hours factory workers are making the cars. It takes 45 hours to make 60 cars. Use a proportion to find how long it will take to make 250 cars.
Method 1
Use a proportion.

Method 2
Use a direct proportion equation.
Let x be the number of hours.
Let y be the number of cars. Define the variables.
Constant of proportionality:
\(\frac{y}{x}=\frac{?}{?}\)
Substitute y = and x =
Simplify.

Direct proportion equation:
y = x Write an equation.
When y = 250 and y = x, 250 = • x Evaluate y = x when y = 250.
• x = 250 Write an equivalent equation.
\(\frac{? \cdot x}{?}=\frac{250}{?}\) Divide both sides by .
x = Simplify.
It takes hours to produce 250 cars.
Answer:
It takes 187.5 hours to produce 250 cars.

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

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

Solve.
Question 2.
The number of pears for sale at an orchard, P, is directly proportional to the number of crates used to pack the pears, C. The table shows the relationship between the total number of pears for sale and the number of crates.

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

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

b) Find the missing values in the table.
Answer:

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

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

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

Math in Focus Course 2A Practice 5.3 Answer Key

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

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

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

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

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

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

 

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

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

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

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

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

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

 

In each table, b is directly proportional to a. Copy and complete the table.
Question 3.

Answer:

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

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

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

Question 4.

Answer:

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

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

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

 

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

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

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

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

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

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

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

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

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

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

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

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

 

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

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

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

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

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

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

 

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

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

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

Explanation:
Direct variation equation:
C = kn.

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

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

 

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

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

 

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

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

 

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

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

 

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

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

 

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

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

 

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

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

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

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

 

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

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

 

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

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

 

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

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

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

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

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

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

 

Question 20.
An initial amount of money deposited in a bank account that earns interest is called the principal. In the table below, P stands for principal, and P stands for the interest earned by that principal for a period of one year at a particular bank. At this bank, the interest earned for a period of one year is directly proportional to the principal amount deposited.

a) Write a direct proportion equation that relates l and P.
Answer:
Direct proportion equation: P = kI

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

b) Copy and complete the table.
Answer:

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

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

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

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

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

Question 22.
Math Journal Jenny wants to buy some blackberries. Three stores sell blackberries at different prices:

Which store has the best deal? Give your reasons.
Answer:
Store A sells the best deal because cost of 16 strawberries is $0.15 ounces less than other two stores.

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

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

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

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

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

Question 1.

Answer:
The given graph is a straight line through the origin, and
It does not lie along the x — axis or y — axis.
So, It represents a direct proportion.

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

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

Question 2.

Answer:
The given graph is a straight line that does not lie along
the x — axis or y — axis but does not pass through the origin.
So, the given graph does not represents a direct proportion.
The graph does not represent a Direct Proportion

Question 3.

Answer:
3.
The given graph passes through the origin that does not lie along
the x — axis or y — axis but It is not a straight line.
So, the given graph does not represents a direct proportion.
The graph does not represent a Direct Proportion

Complete.

Question 4.
Ms. Gray is driving on a long distance trip. The distance she travels is directly proportional to time she travels. The graph shows the distance she travels, y miles, after t hours.

a) Find the constant of proportionality. What is Ms. Gray’s driving speed in miles per hour?
Constant of proportionality: \(\frac{?}{?}\) =
The constant of proportionality is . So, Ms. Gray’s driving speed is miles per hour.
Answer:

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

b) Write a direct proportion equation.
The direct proportion equation is y = t.
Answer:
\(\frac{y}{t}\) = \(\frac{50}{1}\)
t ∙ \(\frac{y}{t}\) = t ∙ \(\frac{50}{1}\)

c) Explain what the point (7, 350) represents in this situation.
It means that Ms. Gray travels miles in hours.
Answer:
It means that Ms. Gray travels 350 miles in 7 hours.

d) Find the distance traveled in 3 hours.
From the graph, the distance traveled is miles.
Answer:
From the graph, the distance traveled is 150 miles.

e) How long does it take Ms. Gray to travel 400 miles?
From the graph, it takes her hours to travel 400 miles.
Answer:
From the graph, it takes her 8 hours to travel 400 miles.

Math in Focus Course 2A Practice 5.2 Answer Key

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

Question 1.

Answer:
The given graph is a straight line that does not lie along
the x — axis or y — axis but does not pass through the origin.
So, the given graph does not represents a direct proportion.
The graph does not represent a Direct Proportion

Question 2.

Answer:
The given graph is a straight line through the origin, and
It does not lie along the x — axis or y — axis.
So, it represents a direct proportion.
Because the graph passes through (1, 500),
The constant of proportionality is 500
The direct proportion equation is :
\(\frac{y}{x}\) = \(\frac{500}{1}\)
x ∙ \(\frac{y}{x}\) = x ∙ \(\frac{500}{1}\)
y = 500x
The graph represents a Direct Proportion

Question 3.

Answer:
In the given graph there is no straight line which show that there is no relation between x — axis and y — axis.
So, the given graph does not represents a direct proportion.
The graph does not represent a Direct Proportion

Question 4.

Answer:
The given graph is a straight line through the origin, and
It does not lie along the x — axis or y – axis.
So, It represents a direct proportion.
Because the graph passes through (1, 100),
The constant of proportionality is 100
The direct proportion equation is :
x ∙ \(\frac{y}{x}\) = x ∙ \(\frac{10}{1}\)
y = 100x
The graph represent a Direct Proportion

Solve. Show your work.

Question 5.
The cost of staying at a motel is directly proportional to the number of nights you stay. The graph shows the cost of staying at a motel, y dollars, for x nights.

a) Find the constant of proportionality. What does this value represent in this situation?
Answer:
\(\frac{y}{x}\) = \(\frac{50}{1}\)
\(\frac{y}{x}\) = 50
The Constant of Proportionality is 50

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

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

The line should be Straight line passes through the origin.

Question 7.
When you travel to another country, you can exchange U.S. dollars for the currency of that country. The amount of the new currency you get for your dollars depends on the exchange rate. The graph shows the amount of Mexican pesos, y, you could get if you were to exchange x U.S. dollars for pesos.

Answer:

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

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

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

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

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

Use graph paper. Solve.

Question 8.
Beth works at a pottery studio. She is making ceramic pots to sell at a craft fair. Graph the relationship between the number of ceramic pots she makes, y, and the number of days she works at the studio, x. Use 1 unit on the horizontal axis to represent 1 day and 1 unit on the vertical axis to represent 5 ceramic pots.

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

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

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

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

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

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

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

Hands-On Activity

Identify Direct Proportion In An Experiment

Work in pairs

Step 1: Make a table like the one shown.

Step 2: Tape a yardstick to the
Step 3: Stand 1 foot away from the yardstick. Look at the yardstick through the cardboard tube. How many inches of the yardstick can you see? Record the number of inches in the table.
Step 4: Repeat Step 3 for the other values of L shown in the table. Then complete the table.

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

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

Copy and complete to determine whether y is directly proportional to x.
Question 1.
The table shows the distance traveled by a school bus, y miles, after x hours.

For each pair of values, x and y:

So, the distance traveled by the school bus is to the number of hours it has traveled.
The constant of proportionality is . and represents the speed of the bus.
The direct proportion equation is .
Answer:
The constant of proportionality is 50 miles per hour and represents the speed of the bus.
The direct proportion equation: Distance = speed × time.

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

 

Question 2.
The table shows the number of pitches made, y, in x innings of a baseball game.

For each pair of values, x and y:

So, the number of pitches made is to the number of innings of a baseball game.
Answer:
The number of pitches made is 95  to the number of innings of a baseball game.

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

 

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

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

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

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

 

Copy and complete.
Question 5.
The table shows the number of baseballs, y, made in x days. The number of baseballs made is directly proportional to the number of days of production. Find the constant of proportionality and tell what it represents in this situation. Then write a direct proportion equation.

Constant of proportionality:
The constant of proportionality is and represents .
The direct proportion equation is .
Answer:
Constant of proportionality: 56.
The constant of proportionality is 56 and represents baseballs produced.
The direct proportion equation: y = kx.

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

Question 6.
A cafeteria sells sandwiches for $4 each. The amount Jason pays for some sandwiches is directly proportional to the number he buys. Write an equation that represents the direct proportion.
Let be the number of sandwiches.
Let be the amount Jason pays.
Cost per sandwich: $ per sandwich
The direct proportion equation is = .
Answer:
Cost per sandwich: $4 per sandwich
The direct proportion equation: Cost of sandwiches a cafeteria sells × Number of sandwiches Jason buys = Total amount of sandwiches Jason buys.

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

Question 7.
q is directly proportional to p, and p = 12 when q = 24. Find the constant of proportionality. Then write a direct proportion equation.
Constant of proportionality: \(\frac{q}{p}=\frac{?}{?}\)
= Write in simplest form.
The constant of proportionality is .
The direct proportion equation is .
Answer:
The constant of proportionality: 2.
The direct proportion equation: q = kp.

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

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

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

Math in Focus Course 2A Practice 5.1 Answer Key

Tell whether y is directly proportional to x. If so, find the constant of proportionality. Then write a direct proportion equation.
Question 1.

Answer:
Constant of proportionality: 5.

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

Question 2.

Answer:
Constant of proportionality: 65, 25, 11.67.

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

 

Question 3.

Answer:
Constant of proportionality: 6.67, 6.67, 5.56.

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

 

Question 4.

Answer:
Constant of proportionality: 25.

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

 

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

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

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

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

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

Explanation:
p = 0.25q
=> y = kx

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

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

Solve. Show your work.
Question 9.
The table shows the distance traveled, d miles, and the amount of gasoline used, n gallons. Tell whether d is directly proportional to n. If so, give the constant of proportionality and tell what it represents in this situation. Then write a direct proportion equation.

Answer:
Constant of proportionality = 20.
Direct proportion equation: d = kn.

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

Question 10.
The table shows the number of points scored, y, in x basketball games. Tell whether y is directly proportional to x. If so, give the constant of proportionality and tell what it represents in this situation. Then write a direct proportion equation.

Answer:
Constant of proportionality: 24, 24, 26.67.
Direct proportion equation: y = kx.

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

Question 11.
The table shows the number of tennis balls produced, y, by x machines. Tell whether y is directly proportional to x. If so, give the constant of proportionality and tell what it represents in this situation. Then write a direct proportion equation.

Answer:
Constant of proportionality: 20.
Direct proportion equation: y = kx.

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

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

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

 

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

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

 

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

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

 

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

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

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

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

 

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

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

 

Question 18.
Karl hikes 3 miles in 45 minutes. Given that his distance is directly proportional to the time he walks, find the constant of proportionality and write an equation to represent the direct proportion.

Answer:
Constant of proportionality: speed = 0.67 miles per hour.
Direct proportion equation: Distance = speed × time.

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

 

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

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

 

Question 20.
Math journal Each table shows the cost of placing an advertisement in a newspaper, C dollars, for t days. Describe how the two tables are alike, and how they are different. Be sure to discuss direct proportion in your answer.

Answer:
Constant of proportionality: Cost of placing an advertisement in a newspaper per day = 20 dollars.
Direct proportion equation: Total cost of advertisement ( c dollars) = Cost of placing an advertisement in a newspaper per day × Number of days.

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

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

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

Math in Focus Grade 7 Chapter 5 Quick Check Answer Key

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

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

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

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

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

 

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

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

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

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

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

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

 

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

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

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

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

 

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

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

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

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

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

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

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

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

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

Use the coordinate plane below.

Question 12.
Give the coordinates of points P, Q, R. S, and T.
Answer:
coordinates of point of  P = (1,2)
coordinates of point of Q = (2,4)
coordinates of point of R = (5,6)
coordinates of point of  S = (5,2)
coordinates of point of T = (6,3)

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

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

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

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

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