7th Grade Math Quiz: Represent Proportional Relationships By Equations
20 questions · exam conditions
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Represent Proportional Relationships By EquationsQuestion 1 of 20

A bus travels 180 miles in 3 hours at a constant speed. Let dd be the distance (in miles) and let tt be the time (in hours). Which equation models this proportional relationship?

d=t+60d=t+60
d=60td=60t
d=180td=180t
t=60dt=60d
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7th Grade Math Quiz

7th Grade Math Quiz: Represent Proportional Relationships By Equations

Practice Represent Proportional Relationships By Equations in 7th Grade Math with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Represent Proportional Relationships By Equations, giving you a quick way to practice the rules, question types, and explanations that matter most for 7th Grade Math.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

A bus travels 180 miles in 3 hours at a constant speed. Let dd be the distance (in miles) and let tt be the time (in hours). Which equation models this proportional relationship?

  1. d=t+60d=t+60
  2. d=60td=60t (correct answer)
  3. d=180td=180t
  4. t=60dt=60d
Explanation: This question tests writing equations y=kx for proportional relationships from tables, graphs, contexts, or verbal descriptions, identifying k and defining variables contextually. Proportional equation y=kx: k is constant of proportionality (unit rate, ratio y/x). From table: calculate k from any pair (14/2=7, k=7 gives y=7x), from graph: k=slope (or read y when x=1: if graph through (1,7), k=7), from context: stated rate is k ("$3 per pound" → k=3, equation c=3p where c=cost, p=pounds). Variables: choose meaningful (c for cost, n for number, d for distance) and define in context. For example: context "apples $3/lb" write c=3p (c=cost dollars, p=pounds), k=3 from $/lb rate; or table x:2,4,6 y:10,20,30 find k=10/2=5, write y=5x; or graph through origin with slope 8 write y=8x. The correct equation is d=60t, where d is distance in miles and t is time in hours, with k=60 from 180 miles / 3 hours. A common error is using total like d=180t without dividing, reversing like t=60d, or additive d=t+60. To write the equation: (1) identify proportional relationship (context says "constant speed"), (2) find k (ratio 180/3=60), (3) choose variables (d for distance, t for time), (4) write d=60t, (5) define variables (d=distance in miles, t=time in hours), (6) verify (for t=3, d=60×3=180, matches✓). Multiple representations: equation d=60t matches a table with ratio 60, a graph through origin with slope 60, and verbal "60 miles per hour"—all show same k=60.

Question 2

A teacher buys markers in bulk. The total cost cc (in dollars) is proportional to the number of marker packs pp. If 7 packs cost $28, which equation represents the relationship?

  1. c=p+28c=p+28
  2. c=4pc=4p (correct answer)
  3. c=28pc=28p
  4. p=4cp=4c
Explanation: This question tests writing equations y=kx for proportional relationships from tables, graphs, contexts, or verbal descriptions, identifying k and defining variables contextually. Proportional equation y=kx: k is constant of proportionality (unit rate, ratio y/x). From table: calculate k from any pair (14/2=7, k=7 gives y=7x), from graph: k=slope (or read y when x=1: if graph through (1,7), k=7), from context: stated rate is k ("$3 per pound" → k=3, equation c=3p where c=cost, p=pounds). Variables: choose meaningful (c for cost, n for number, d for distance) and define in context. For example: context "apples $3/lb" write c=3p (c=cost dollars, p=pounds), k=3 from $/lb rate; or table x:2,4,6 y:10,20,30 find k=10/2=5, write y=5x; or graph through origin with slope 8 write y=8x. The correct equation is c=4p, where c is total cost in dollars and p is number of packs, with k=4 from 28/7=4. A common error is using total like c=28p without dividing, additive c=p+28, or reversing p=4c. To write the equation: (1) identify proportional relationship (context says "proportional to the number"), (2) find k (ratio 28/7=4), (3) choose variables (c for cost, p for packs), (4) write c=4p, (5) define variables (c=total cost in dollars, p=number of marker packs), (6) verify (for p=7, c=4×7=28, matches✓). Multiple representations: equation c=4p matches a table with ratio 4, a graph through origin with slope 4, and verbal "$4 per pack"—all show same k=4.

Question 3

A printer produces pages at a constant rate. The equation p=18tp = 18t represents the number of pages pp printed after tt minutes. How many pages will be printed in the first 2.52.5 minutes, and what does this demonstrate about proportional relationships?

  1. 3636 pages; it shows that doubling the time doubles the output in proportional relationships
  2. 4545 pages; it shows that the constant rate applies to any time interval in proportional relationships (correct answer)
  3. 20.520.5 pages; it shows that fractional inputs produce fractional outputs in proportional relationships
  4. 7272 pages; it shows that proportional relationships always involve whole number coefficients and results
Explanation: The correct answer is B. Using p=18tp = 18t with t=2.5t = 2.5: p=18(2.5)=45p = 18(2.5) = 45 pages. This demonstrates that the constant rate of 1818 pages per minute applies to any time interval, including fractional times. Choice A gives the wrong calculation (18×2=3618 × 2 = 36). Choice C gives an incorrect sum (18+2.518 + 2.5). Choice D uses incorrect multiplication (18×418 × 4) and makes a false claim about whole numbers.

Question 4

A car travels at a constant speed of 5555 miles per hour. Let dd be the distance (in miles) and let hh be the time (in hours). Which equation represents this proportional relationship?

  1. d=55+hd=55+h
  2. h=55dh=55d
  3. d=h+55d=h+55
  4. d=55hd=55h (correct answer)
Explanation: This question tests writing equations y=kx for proportional relationships from tables, graphs, contexts, or verbal descriptions, identifying k and defining variables contextually. Proportional equation y=kx: k is constant of proportionality (unit rate, ratio y/x). From table: calculate k from any pair (14/2=7, k=7 gives y=7x), from graph: k=slope (or read y when x=1: if graph through (1,7), k=7), from context: stated rate is k ("$3 per pound" → k=3, equation c=3p where c=cost, p=pounds). Variables: choose meaningful (c for cost, n for number, d for distance) and define in context. For example, context "apples $3/lb" write c=3p (c=cost dollars, p=pounds), k=3 from $/lb rate; or table x:2,4,6 y:10,20,30 find k=10/2=5, write y=5x; or graph through origin with slope 8 write y=8x. The correct equation is d=55h with proper k=55 and variables d for distance and h for hours. A common error is reversing variables like h=55d instead of d=55h, using wrong form like d=h+55 not proportional, or d=55+h which is additive. To write the equation: (1) identify proportional relationship (context says "55 miles per hour"), (2) find k (stated rate of 55), (3) choose variables (d for distance, h for hours), (4) write d=55h, (5) define variables (d=distance in miles, h=time in hours), (6) verify (substitute h=1, d=55×1=55, reasonable? yes✓). Multiple representations: equation d=55h matches table of multiples of 55, graph through origin with slope 55, verbal "55 mph"—all show same k=55. Mistakes: wrong form (additive d=h+55 not multiplicative), variables reversed, k wrong, forgetting to define variables.

Question 5

A bus travels 45 miles in 1.5 hours at a constant rate. Let dd be distance (miles) and tt be time (hours). Which equation models the proportional relationship?

  1. d=30td=30t (correct answer)
  2. d=t+30d=t+30
  3. d=45td=45t
  4. t=30dt=30d
Explanation: This question tests writing equations y=kx for proportional relationships from tables, graphs, contexts, or verbal descriptions, identifying k and defining variables contextually. Proportional equation y=kx: k is constant of proportionality (unit rate, ratio y/x). From table: calculate k from any pair (14/2=7, k=7 gives y=7x), from graph: k=slope (or read y when x=1: if graph through (1,7), k=7), from context: stated rate is k ("$3 per pound" → k=3, equation c=3p where c=cost, p=pounds). Variables: choose meaningful (c for cost, n for number, d for distance) and define in context. For example, bus 45 miles in 1.5 hours, k=45/1.5=30, write d=30t (d=miles, t=hours); or table x:2,4,6 y:10,20,30 find k=10/2=5, write y=5x; or graph through origin with slope 8 write y=8x. The correct equation is d=30t, with k=30 from calculated rate. A common error is wrong k like d=45t using total without dividing, reversing t=30d, or additive d=t+30. To write: (1) identify proportional from constant rate, (2) find k=30, (3) choose d and t, (4) write d=30t, (5) define d as miles and t as hours, (6) verify t=1.5, d=30×1.5=45. Multiple representations: d=30t matches given point, graph slope 30, verbal "30 mph"—all k=30. Mistakes: wrong k calculation, reversed, added terms.

Question 6

A runner runs at a constant speed of 6 miles per hour. Let dd be the distance (in miles) and let hh be the time (in hours). Which equation models this proportional relationship?

  1. h=6dh=6d
  2. d=h+6d=h+6
  3. d=6h+2d=6h+2
  4. d=6hd=6h (correct answer)
Explanation: This question tests writing equations y=kx for proportional relationships from tables, graphs, contexts, or verbal descriptions, identifying k and defining variables contextually. Proportional equation y=kx: k is constant of proportionality (unit rate, ratio y/x). From table: calculate k from any pair (14/2=7, k=7 gives y=7x), from graph: k=slope (or read y when x=1: if graph through (1,7), k=7), from context: stated rate is k ("$3 per pound" → k=3, equation c=3p where c=cost, p=pounds). Variables: choose meaningful (c for cost, n for number, d for distance) and define in context. For example: context "apples $3/lb" write c=3p (c=cost dollars, p=pounds), k=3 from $/lb rate; or table x:2,4,6 y:10,20,30 find k=10/2=5, write y=5x; or graph through origin with slope 8 write y=8x. The correct equation is d=6h, where d is the distance in miles and h is the time in hours, with k=6 from the 6 miles per hour speed. A common error is reversing variables like h=6d instead of d=6h, using a non-proportional form like d=h+6, or adding constants like d=6h+2 when the relationship passes through the origin. To write the equation: (1) identify proportional relationship (context says "constant speed of 6 miles per hour"), (2) find k (stated rate of 6), (3) choose variables (d for distance, h for hours), (4) write d=6h, (5) define variables (d=distance in miles, h=time in hours), (6) verify (for h=2, d=6×2=12, reasonable? yes✓). Multiple representations: equation d=6h matches a table where distances are multiples of 6, a graph through origin with slope 6, and verbal "6 miles per hour"—all show same k=6.

Question 7

A recipe uses 2.52.5 cups of flour for each batch of cookies. Let ff be the number of cups of flour and let bb be the number of batches. Which equation shows the proportional relationship?

  1. f=2.5b+1f=2.5b+1
  2. f=b+2.5f=b+2.5
  3. f=2.5bf=2.5b (correct answer)
  4. b=2.5fb=2.5f
Explanation: This question tests writing equations y=kx for proportional relationships from tables, graphs, contexts, or verbal descriptions, identifying k and defining variables contextually. Proportional equation y=kx: k is constant of proportionality (unit rate, ratio y/x). From table: calculate k from any pair (14/2=7, k=7 gives y=7x), from graph: k=slope (or read y when x=1: if graph through (1,7), k=7), from context: stated rate is k ("$3 per pound" → k=3, equation c=3p where c=cost, p=pounds). Variables: choose meaningful (c for cost, n for number, d for distance) and define in context. For example, context "apples $3/lb" write c=3p (c=cost dollars, p=pounds), k=3 from $/lb rate; or table x:2,4,6 y:10,20,30 find k=10/2=5, write y=5x; or graph through origin with slope 8 write y=8x. The correct equation is f=2.5b with proper k=2.5 and variables f for flour and b for batches. A common error is reversing variables like b=2.5f instead of f=2.5b, wrong form like f=b+2.5 not proportional, or including intercept like f=2.5b+1. To write the equation: (1) identify proportional relationship (context says "2.5 cups per batch"), (2) find k (stated rate of 2.5), (3) choose variables (f for flour, b for batches), (4) write f=2.5b, (5) define variables (f=cups of flour, b=number of batches), (6) verify (b=1, f=2.5×1=2.5, yes✓). Multiple representations: equation f=2.5b matches table of multiples of 2.5, graph with slope 2.5, verbal "2.5 per batch"—all show k=2.5. Mistakes: wrong form (additive), variables reversed, k wrong, undefined variables.

Question 8

A runner travels at a constant speed of 6 miles per hour. Let dd be the distance (in miles) and hh be the time (in hours). Which equation models the relationship?

  1. d=6hd=6h (correct answer)
  2. d=h+6d=h+6
  3. h=6dh=6d
  4. d=6h+6d=6h+6
Explanation: This question tests writing equations y=kx for proportional relationships from tables, graphs, contexts, or verbal descriptions, identifying k and defining variables contextually. Proportional equation y=kx: k is constant of proportionality (unit rate, ratio y/x). From table: calculate k from any pair (14/2=7, k=7 gives y=7x), from graph: k=slope (or read y when x=1: if graph through (1,7), k=7), from context: stated rate is k ("$3 per pound" → k=3, equation c=3p where c=cost, p=pounds). Variables: choose meaningful (c for cost, n for number, d for distance) and define in context. For example, in context of speed at 6 mph, write d=6h (d=distance in miles, h=time in hours), k=6 from miles per hour; or table x:2,4,6 y:10,20,30 find k=10/2=5, write y=5x; or graph through origin with slope 8 write y=8x. The correct equation is d=6h, with k=6 and variables d for distance and h for hours. A common error is wrong form like d=h+6 not proportional, reversing variables like h=6d, or including intercept like d=6h+6. To write the equation: (1) identify proportional from "constant speed," (2) find k=6 as rate, (3) choose d and h, (4) write d=6h, (5) define d as miles and h as hours, (6) verify with h=1, d=6. Multiple representations: d=6h matches table of multiples of 6, graph with slope 6 through origin, verbal "6 miles per hour"—all k=6. Mistakes: additive form, reversed variables, added constants.

Question 9

A recipe uses 2 cups of flour for each batch of muffins. Let ff be the number of cups of flour and bb be the number of batches. Which equation represents this proportional relationship?

  1. f=2bf=2b (correct answer)
  2. f=b+2f=b+2
  3. f=2b+2f=2b+2
  4. b=2fb=2f
Explanation: This question tests writing equations y=kx for proportional relationships from tables, graphs, contexts, or verbal descriptions, identifying k and defining variables contextually. Proportional equation y=kx: k is constant of proportionality (unit rate, ratio y/x). From table: calculate k from any pair (14/2=7, k=7 gives y=7x), from graph: k=slope (or read y when x=1: if graph through (1,7), k=7), from context: stated rate is k ("$3 per pound" → k=3, equation c=3p where c=cost, p=pounds). Variables: choose meaningful (c for cost, n for number, d for distance) and define in context. For example, recipe 2 cups flour per batch, write f=2b (f=cups of flour, b=batches), k=2 from cups per batch; or table x:2,4,6 y:10,20,30 find k=10/2=5, write y=5x; or graph through origin with slope 8 write y=8x. The correct equation is f=2b, with k=2 and variables f for flour and b for batches. A common error is reversing like b=2f, using additive f=b+2, or including intercept f=2b+2. To write: (1) identify proportional from "2 cups for each batch," (2) find k=2 as rate, (3) choose f and b, (4) write f=2b, (5) define f as cups and b as batches, (6) verify b=1, f=2. Multiple representations: f=2b matches table multiples of 2, graph slope 2, verbal "2 per batch"—all k=2. Mistakes: reversed variables, wrong form, added constants.

Question 10

A recipe uses 3 cups of flour for every 2 batches of cookies. Let ff be the number of cups of flour and let bb be the number of batches. Which equation represents this proportional relationship?

  1. f=b+32f=b+\frac{3}{2}
  2. b=32fb=\frac{3}{2}f
  3. f=32bf=\frac{3}{2}b (correct answer)
  4. f=23bf=\frac{2}{3}b
Explanation: This question tests writing equations y=kx for proportional relationships from tables, graphs, contexts, or verbal descriptions, identifying k and defining variables contextually. Proportional equation y=kx: k is constant of proportionality (unit rate, ratio y/x). From table: calculate k from any pair (14/2=7, k=7 gives y=7x), from graph: k=slope (or read y when x=1: if graph through (1,7), k=7), from context: stated rate is k ("$3 per pound" → k=3, equation c=3p where c=cost, p=pounds). Variables: choose meaningful (c for cost, n for number, d for distance) and define in context. For example: context "apples $3/lb" write c=3p (c=cost dollars, p=pounds), k=3 from $/lb rate; or table x:2,4,6 y:10,20,30 find k=10/2=5, write y=5x; or graph through origin with slope 8 write y=8x. The correct equation is f=(3/2)b, where f is the cups of flour and b is the number of batches, with k=3/2 from 3 cups per 2 batches. A common error is reversing the ratio like f=(2/3)b, reversing variables like b=(3/2)f, or using additive form like f=b+(3/2) instead of multiplicative. To write the equation: (1) identify proportional relationship (context says "3 cups for every 2 batches"), (2) find k (ratio 3/2), (3) choose variables (f for flour, b for batches), (4) write f=(3/2)b, (5) define variables (f=cups of flour, b=number of batches), (6) verify (for b=2, f=(3/2)×2=3, matches✓). Multiple representations: equation f=(3/2)b matches a table with ratios of 3/2, a graph through origin with slope 3/2, and verbal "3 cups per 2 batches"—all show same k=3/2.

Question 11

A proportional relationship is graphed on the coordinate plane. The line passes through the points (0,0)(0,0) and (1,7)(1,7). Which equation represents the relationship between yy and xx?

  1. y=7xy=7x (correct answer)
  2. y=7x+2y=7x+2
  3. x=7yx=7y
  4. y=x+7y=x+7
Explanation: This question tests writing equations y=kx for proportional relationships from tables, graphs, contexts, or verbal descriptions, identifying k and defining variables contextually. Proportional equation y=kx: k is constant of proportionality (unit rate, ratio y/x). From table: calculate k from any pair (14/2=7, k=7 gives y=7x), from graph: k=slope (or read y when x=1: if graph through (1,7), k=7), from context: stated rate is k ("$3 per pound" → k=3, equation c=3p where c=cost, p=pounds). Variables: choose meaningful (c for cost, n for number, d for distance) and define in context. For example, context "apples $3/lb" write c=3p (c=cost dollars, p=pounds), k=3 from $/lb rate; or table x:2,4,6 y:10,20,30 find k=10/2=5, write y=5x; or graph through origin with slope 8 write y=8x. The correct equation is y=7x with proper k=7 from the slope through (0,0) and (1,7). A common error is including intercept like y=x+7 or y=7x+2 when proportional must pass through origin, reversing variables like x=7y, or wrong form. To write the equation: (1) identify proportional relationship (graph through origin), (2) find k (slope = 7/1=7), (3) choose variables (y and x), (4) write y=7x, (5) define if needed, (6) verify (x=1, y=7×1=7, matches point✓). Multiple representations: y=7x matches table of multiples of 7, graph with slope 7, verbal rate 7—all show k=7. Mistakes: additive form, meaningless variables, wrong k, undefined in context.

Question 12

A proportional relationship is shown on a graph by a line that passes through the origin and the point (3,15)(3, 15). Which equation represents the relationship between yy and xx?

  1. y=15xy=15x
  2. y=5xy=5x (correct answer)
  3. y=x+5y=x+5
  4. x=5yx=5y
Explanation: This question tests writing equations y=kx for proportional relationships from tables, graphs, contexts, or verbal descriptions, identifying k and defining variables contextually. Proportional equation y=kx: k is constant of proportionality (unit rate, ratio y/x). From table: calculate k from any pair (14/2=7, k=7 gives y=7x), from graph: k=slope (or read y when x=1: if graph through (1,7), k=7), from context: stated rate is k ("$3 per pound" → k=3, equation c=3p where c=cost, p=pounds). Variables: choose meaningful (c for cost, n for number, d for distance) and define in context. For example, graph through origin and (3,15), slope=15/3=5, write y=5x; or table x:2,4,6 y:10,20,30 find k=10/2=5, write y=5x; or context apples $3/lb write c=3p. The correct equation is y=5x, with k=5 from slope. A common error is wrong k like y=15x using point without dividing, reversing like x=5y, or additive y=x+5. To write: (1) identify proportional from line through origin, (2) find k=slope=5, (3) use x,y, (4) write y=5x, (5) no further definition needed, (6) verify with (3,15): 5×3=15. Multiple representations: y=5x matches graph slope 5, table like x=1,y=5, verbal "y is 5 times x"—all k=5. Mistakes: wrong k, reversed variables, non-proportional form.

Question 13

A school store sells pencils for $0.50 each. Let $mbethetotalcost(indollars)andbe the total cost (in dollars) andp$ be the number of pencils. Which equation represents the relationship?

  1. m=0.5p+0.5m=0.5p+0.5
  2. m=0.5pm=0.5p (correct answer)
  3. m=p+0.5m=p+0.5
  4. p=0.5mp=0.5m
Explanation: This question tests writing equations y=kx for proportional relationships from tables, graphs, contexts, or verbal descriptions, identifying k and defining variables contextually. Proportional equation y=kx: k is constant of proportionality (unit rate, ratio y/x). From table: calculate k from any pair (14/2=7, k=7 gives y=7x), from graph: k=slope (or read y when x=1: if graph through (1,7), k=7), from context: stated rate is k ("$3 per pound" → k=3, equation c=3p where c=cost, p=pounds). Variables: choose meaningful (c for cost, n for number, d for distance) and define in context. For example, pencils 0.50each,writem=0.5p(m=costindollars,p=pencils),k=0.5fromdollarsperpencil;ortablex:2,4,6y:10,20,30findk=10/2=5,writey=5x;orgraphthroughoriginwithslope8writey=8x.Thecorrectequationism=0.5p,withk=0.5andvariablesmformoneyandpforpencils.Acommonerroriswrongformlikem=p+0.5,reversingp=0.5m,orinterceptm=0.5p+0.5.Towrite:(1)identifyfrom"0.50 each, write m=0.5p (m=cost in dollars, p=pencils), k=0.5 from dollars per pencil; or table x:2,4,6 y:10,20,30 find k=10/2=5, write y=5x; or graph through origin with slope 8 write y=8x. The correct equation is m=0.5p, with k=0.5 and variables m for money and p for pencils. A common error is wrong form like m=p+0.5, reversing p=0.5m, or intercept m=0.5p+0.5. To write: (1) identify from "0.50 each," (2) find k=0.5, (3) choose m and p, (4) write m=0.5p, (5) define m as dollars and p as pencils, (6) verify p=2, m=1. Multiple representations: m=0.5p matches table like p=1,m=0.5, graph slope 0.5, verbal "half dollar per pencil"—all k=0.5. Mistakes: additive, reversed, extra terms.

Question 14

A movie theater charges $9 per ticket. Let $cbethetotalcost(indollars)andbe the total cost (in dollars) andt$ be the number of tickets. Which equation represents this proportional relationship?

  1. c=t+9c=t+9
  2. c=9tc=9t (correct answer)
  3. c=9t+9c=9t+9
  4. t=9ct=9c
Explanation: This question tests writing equations y=kx for proportional relationships from tables, graphs, contexts, or verbal descriptions, identifying k and defining variables contextually. Proportional equation y=kx: k is constant of proportionality (unit rate, ratio y/x). From table: calculate k from any pair (14/2=7, k=7 gives y=7x), from graph: k=slope (or read y when x=1: if graph through (1,7), k=7), from context: stated rate is k ("$3 per pound" → k=3, equation c=3p where c=cost, p=pounds). Variables: choose meaningful (c for cost, n for number, d for distance) and define in context. For example, in the context of movie tickets at $9 each, write c=9t (c=total cost in dollars, t=number of tickets), where k=9 from the dollars per ticket rate; or table x:2,4,6 y:10,20,30 find k=10/2=5, write y=5x; or graph through origin with slope 8 write y=8x. The correct equation is c=9t, with k=9 and variables c for total cost and t for tickets. A common error is using the wrong form like c=t+9 which is not proportional, or reversing variables like t=9c, or including an intercept like c=9t+9 when it should pass through the origin. To write the equation: (1) identify the proportional relationship from the context "charges 9perticket,"(2)findk=9asthestatedrate,(3)choosevariablescforcostandtfortickets,(4)writec=9t,(5)definecastotalcostindollarsandtasnumberoftickets,(6)verifybysubstitutingt=1,c=9×1=9,whichisreasonable.Multiplerepresentations:equationc=9tmatchesatablewherecostsaremultiplesof9,agraphthroughoriginwithslope9,andtheverbal"9 per ticket," (2) find k=9 as the stated rate, (3) choose variables c for cost and t for tickets, (4) write c=9t, (5) define c as total cost in dollars and t as number of tickets, (6) verify by substituting t=1, c=9×1=9, which is reasonable. Multiple representations: equation c=9t matches a table where costs are multiples of 9, a graph through origin with slope 9, and the verbal "9 per ticket"—all show k=9. Mistakes include using additive forms like c=t+9 instead of multiplicative, reversing variables, or adding unnecessary constants.

Question 15

Two students are modeling the same proportional relationship between gallons of gas gg and total driving distance dd in miles. Student A writes d=28gd = 28g while Student B writes g=d28g = \frac{d}{28}. Which statement best describes these equations?

  1. Only Student A is correct; Student B should have written g=28dg = 28d for the relationship
  2. Only Student B is correct; Student A confused the independent and dependent variables completely
  3. Both students are correct; they represent the same proportional relationship expressed in different equivalent forms (correct answer)
  4. Neither student is correct; proportional relationships cannot be written with division or fractions in the equations
Explanation: The correct answer is C. Both equations represent the same proportional relationship. Student A's equation d=28gd = 28g shows distance as a function of gallons (2828 miles per gallon). Student B's equation g=d28g = \frac{d}{28} is the inverse, showing gallons as a function of distance. These are equivalent: solving d=28gd = 28g for gg gives g=d28g = \frac{d}{28}. Choice A and B incorrectly claim only one is right. Choice D makes a false statement about proportional relationships.

Question 16

At a school fundraiser, a student earns $2.50 for each box of candy sold. Let $mbethemoneyearned(indollars)andletbe the money earned (in dollars) and letb$ be the number of boxes sold. Which equation represents this proportional relationship?

  1. m=2.50b+5m=2.50b+5
  2. b=2.50mb=2.50m
  3. m=2.50bm=2.50b (correct answer)
  4. m=b+2.50m=b+2.50
Explanation: This question tests writing equations y=kxy=kx for proportional relationships from tables, graphs, contexts, or verbal descriptions, identifying kk and defining variables contextually. Proportional equation y=kxy=kx: kk is constant of proportionality (unit rate, ratio y/xy/x). From table: calculate kk from any pair (14/2=714/2=7, k=7k=7 gives y=7xy=7x), from graph: k=k=slope (or read yy when x=1x=1: if graph through (1,7)(1,7), k=7k=7), from context: stated rate is kk ("$3 per pound" → $k=3,equation, equation c=3pwherewherec=cost,cost, p=pounds).Variables:choosemeaningful(pounds). Variables: choose meaningful (cforcost,for cost,nfornumber,for number,dfordistance)anddefineincontext.Forexample:context"apples$3/lb"write$c=3p for distance) and define in context. For example: context "apples $3/lb" write $c=3p (c=c=cost dollars, p=p=pounds), k=3k=3 from $/lb rate; or table xx:2,4,6 yy:10,20,30 find k=10/2=5k=10/2=5, write y=5xy=5x; or graph through origin with slope 8 write y=8xy=8x. The correct equation is m=2.50bm=2.50b, where mm is money earned in dollars and bb is boxes sold, with k=2.50k=2.50 from $2.50 per box. A common error is using additive form like $m=b+2.50,reversingvariableslike, reversing variables like b=2.50m,oraddingextraconstantslike, or adding extra constants like m=2.50b+5.Towritetheequation:(1)identifyproportionalrelationship(contextsays"$2.50foreachbox"),(2)find$k. To write the equation: (1) identify proportional relationship (context says "$2.50 for each box"), (2) find $k (stated rate of 2.50), (3) choose variables (mm for money, bb for boxes), (4) write m=2.50bm=2.50b, (5) define variables (m=m=money earned in dollars, b=b=number of boxes), (6) verify (for b=2b=2, m=2.50×2=5m=2.50×2=5, reasonable? yes✓). Multiple representations: equation m=2.50bm=2.50b matches a table with multiples of 2.50, a graph through origin with slope 2.50, and verbal "$2.50 per box"—all show same $k=2.50$.

Question 17

A gym charges a proportional fee based on the number of classes taken. The fee is $9 per class. Let ff be the total fee (in dollars) and let cc be the number of classes. Using the proportional equation, what is the fee for 77 classes?

  1. $16
  2. $63 (correct answer)
  3. $9
  4. $72
Explanation: This question tests writing equations y=kxy=kx for proportional relationships from tables, graphs, contexts, or verbal descriptions, identifying kk and defining variables contextually, and applying to find values. Proportional equation y=kxy=kx: kk is constant of proportionality (unit rate, ratio y/xy/x). From table: calculate kk from any pair (14/2=714/2=7, k=7k=7 gives y=7xy=7x), from graph: k=k=slope (or read yy when x=1x=1: if graph through (1,7)(1,7), k=7k=7), from context: stated rate is kk ("$3 per pound" → $k=3,equation, equation c=3pwherewherec=cost,cost, p=pounds).Variables:choosemeaningful(pounds). Variables: choose meaningful (cforcost,for cost,nfornumber,for number,dfordistance)anddefineincontext.Forexample,context"apples$3/lb"write$c=3p for distance) and define in context. For example, context "apples $3/lb" write $c=3p (c=c=cost dollars, p=p=pounds), k=3k=3 from $/lb rate; or table xx:2,4,6 yy:10,20,30 find k=10/2=5k=10/2=5, write y=5xy=5x; or graph through origin with slope 8 write y=8xy=8x. The correct equation is f=9cf=9c with k=9k=9, and for 7 classes, f=9×7=63f=9\times7=63. A common error is wrong calculation like 72ifusing8instead,ornon−proportionalforms.Tosolve:(1)identifyproportional("72 if using 8 instead, or non-proportional forms. To solve: (1) identify proportional ("9 per class"), (2) find k=9k=9, (3) variables ff fee, cc classes, (4) write f=9cf=9c, (5) define (f=f=fee in dollars, c=c=classes), (6) substitute c=7c=7, f=63✓f=63\checkmark. Multiple representations: f=9cf=9c matches table multiples of 9, graph slope 9, verbal "$9 per"—all $k=9.Mistakes:wrongform,miscalculation,wrong. Mistakes: wrong form, miscalculation, wrong k$, no verification.

Question 18

A movie theater charges $8 per ticket. Let tt be the total cost (in dollars) and let nn be the number of tickets. Which equation represents this proportional relationship?

  1. t=8nt=8n (correct answer)
  2. t=n+8t=n+8
  3. t=8n+5t=8n+5
  4. n=8tn=8t
Explanation: This question tests writing equations y=kx for proportional relationships from tables, graphs, contexts, or verbal descriptions, identifying k and defining variables contextually. Proportional equation y=kx: k is constant of proportionality (unit rate, ratio y/x). From table: calculate k from any pair (14/2=7, k=7 gives y=7x), from graph: k=slope (or read y when x=1: if graph through (1,7), k=7), from context: stated rate is k ("$3 per pound" → k=3, equation c=3p where c=cost, p=pounds). Variables: choose meaningful (c for cost, n for number, d for distance) and define in context. For example, in the context of a movie theater charging 8perticket,writet=8n(t=totalcostindollars,n=numberoftickets),wherek=8fromthedollarsperticketrate;ortablex:2,4,6y:10,20,30findk=10/2=5,writey=5x;orgraphthroughoriginwithslope8writey=8x.Thecorrectequationist=8nwithproperk=8andvariablestfortotalcostandnfornumberoftickets.Acommonerrorisreversingvariablesliken=8tinsteadoft=8n,usingawrongformliket=n+8whichisnotproportional,orincludinganinterceptliket=8n+5whenitshouldpassthroughtheorigin.Towritetheequation:(1)identifyproportionalrelationship(contextsays"8 per ticket, write t=8n (t=total cost in dollars, n=number of tickets), where k=8 from the dollars per ticket rate; or table x:2,4,6 y:10,20,30 find k=10/2=5, write y=5x; or graph through origin with slope 8 write y=8x. The correct equation is t=8n with proper k=8 and variables t for total cost and n for number of tickets. A common error is reversing variables like n=8t instead of t=8n, using a wrong form like t=n+8 which is not proportional, or including an intercept like t=8n+5 when it should pass through the origin. To write the equation: (1) identify proportional relationship (context says "8 per ticket"), (2) find k (stated rate of 8), (3) choose variables (t for cost, n for tickets), (4) write t=8n, (5) define variables (t=total cost in dollars, n=number of tickets), (6) verify (substitute n=1, t=8×1=8, reasonable? yes✓). Multiple representations: equation t=8n matches a table where costs are multiples of 8, a graph through origin with slope 8, and verbal "$8 per ticket"—all show same k=8. Mistakes include wrong form (additive t=n+8 not multiplicative), variables reversed (n=8t), k wrong, or forgetting to define variables in context.

Question 19

A gym charges $12 per month for a membership with no starting fee. Let $Tbethetotalcost(dollars)andbe the total cost (dollars) andm$ be the number of months. Which equation represents the relationship, and what is the cost for 7 months?

  1. m=12Tm=12T; T=7T=7
  2. T=12m+12T=12m+12; T=96T=96
  3. T=m+12T=m+12; T=19T=19
  4. T=12mT=12m; T=84T=84 (correct answer)
Explanation: This question tests writing equations y=kxy=kx for proportional relationships from tables, graphs, contexts, or verbal descriptions, identifying kk and defining variables contextually. Proportional equation y=kxy=kx: kk is constant of proportionality (unit rate, ratio y/xy/x). From table: calculate kk from any pair (14/2=714/2=7, k=7k=7 gives y=7xy=7x), from graph: k=k=slope (or read yy when x=1x=1: if graph through (1,7)(1,7), k=7k=7), from context: stated rate is kk ("$3 per pound" → $k=3,equation, equation c=3pwherewherec=cost,cost, p=pounds).Variables:choosemeaningful(pounds). Variables: choose meaningful (cforcost,for cost,nfornumber,for number,dfordistance)anddefineincontext.Forexample,gym$12permonthnofee,write$T=12m for distance) and define in context. For example, gym $12 per month no fee, write $T=12m (T=T=total cost in dollars, m=m=months), k=12k=12 from dollars per month; or table xx:2,4,6 yy:10,20,30 find k=10/2=5k=10/2=5, write y=5xy=5x; or graph through origin with slope 8 write y=8xy=8x. The correct equation is T=12mT=12m with T=84T=84 for 7 months, k=12k=12 and variables TT for total and mm for months. A common error is adding fee like T=12m+12T=12m+12, reversing m=12Tm=12T, or additive T=m+12T=m+12. To write: (1) identify from "$12 per month no fee," (2) find $k=12,(3)choose, (3) choose Tandandm,(4)write, (4) write T=12m,(5)define, (5) define Tasdollarsandas dollars andmasmonths,(6)verifyas months, (6) verifym=7,, T=84.Multiplerepresentations:. Multiple representations: T=12mmatchestablemultiplesof12,graphslope12,verbal"12permonth"—allmatches table multiples of 12, graph slope 12, verbal "12 per month"—allk=12$. Mistakes: adding intercepts, reversed, wrong form.

Question 20

A landscaping company's profit PP (in dollars) is proportional to the number of lawns nn they service each week. When they service 2424 lawns, their profit is $960. If their goal is to earn $1400 profit next week, which equation should they use to find how many lawns to service?

  1. P=n+40P = n + 40 where 4040 represents the additional profit per lawn above fixed costs
  2. 1400=24n1400 = 24n where the constant 2424 represents the previous number of lawns
  3. 1400=960n1400 = 960n where the constant 960960 represents the base profit amount
  4. 1400=40n1400 = 40n where the constant 4040 represents dollars per lawn serviced (correct answer)
Explanation: When you see that one quantity is "proportional" to another, this means they have a direct relationship where one equals a constant times the other. Here, profit equals some constant times the number of lawns: P=k⋅nP = k \cdot n, where kk is the constant rate. To find this constant rate, use the given information: when n=24n = 24 lawns, P=$960P = \$960. Substituting: 960=k⋅24960 = k \cdot 24, so k=960÷24=40k = 960 ÷ 24 = 40 dollars per lawn. This means the company earns $40 profit for each lawn they service. Now you can set up the equation for their goal: if they want $1400 profit, then $1400=40n1400 = 40n ,where, where nn $ is the unknown number of lawns needed. Choice A is wrong because proportional relationships are multiplicative ( P = k \cdot n ), not additive ( P = n + \text{constant} ). The "+40" suggests adding a fixed amount rather than multiplying by a rate. Choice B incorrectly uses 24 as the multiplier, but 24 was the number of lawns in the given example, not the profit rate. This confuses the input with the constant. Choice C uses 960 as the multiplier, but 960 was the profit amount from the example, not the rate per lawn. This treats the output as the constant rate. Choice D correctly identifies 40 as the dollars earned per lawn serviced, making 1400 = 40n the right equation. Study tip: In proportional relationships, always find the constant rate by dividing the given output by the given input, then use that rate in your equation.