7th Grade Math Quiz: Represent Proportional Relationships By Equations

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.

Question 1

1. $36$ pages; it shows that doubling the time doubles the output in proportional relationships
2. $45$ pages; it shows that the constant rate applies to any time interval in proportional relationships (correct answer)
3. $20.5$ pages; it shows that fractional inputs produce fractional outputs in proportional relationships
4. $72$ pages; it shows that proportional relationships always involve whole number coefficients and results

Explanation: The correct answer is B. Using $p = 18t$ with $t = 2.5$: $p = 18(2.5) = 45$ pages. This demonstrates that the constant rate of $18$ pages per minute applies to any time interval, including fractional times. Choice A gives the wrong calculation ($18 × 2 = 36$). Choice C gives an incorrect sum ($18 + 2.5$). Choice D uses incorrect multiplication ($18 × 4$) and makes a false claim about whole numbers.

Question 2

1. $f = 1.5s$; flour needed is $1$ cup exactly (correct answer)
2. $s = 1.5f$; flour needed is $\frac{4}{9}$ cup exactly
3. $f = 1.5s$; flour needed is $\frac{4}{9}$ cup exactly
4. $f = s + 1.5$; flour needed is $\frac{13}{6}$ cups exactly

Explanation: The correct answer is A. Since flour is $1.5$ times the sugar, $f = 1.5s$. With $s = \frac{2}{3}$: $f = 1.5 × \frac{2}{3} = \frac{3}{2} × \frac{2}{3} = 1$ cup. Choice B reverses the relationship. Choice C has the right equation but wrong calculation ($\frac{4}{9}$ instead of $1$). Choice D uses addition instead of multiplication and gets $\frac{2}{3} + 1.5 = \frac{13}{6}$.

Question 3

1. Only Student A is correct; Student B should have written $g = 28d$ 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 $d = 28g$ shows distance as a function of gallons ($28$ miles per gallon). Student B's equation $g = \frac{d}{28}$ is the inverse, showing gallons as a function of distance. These are equivalent: solving $d = 28g$ for $g$ gives $g = \frac{d}{28}$. Choice A and B incorrectly claim only one is right. Choice D makes a false statement about proportional relationships.

Question 4

1. $f=2.5b+1$
2. $f=b+2.5$
3. $f=2.5b$ (correct answer)
4. $b=2.5f$

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 5

1. $f=2b$ (correct answer)
2. $f=b+2$
3. $f=2b+2$
4. $b=2f$

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 6

1. $f=b+\frac{3}{2}$
2. $b=\frac{3}{2}f$
3. $f=\frac{3}{2}b$ (correct answer)
4. $f=\frac{2}{3}b$

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 7

1. $c=t+9$
2. $c=9t$ (correct answer)
3. $c=9t+9$
4. $t=9c$

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 $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 8

1. $h=6d$
2. $d=h+6$
3. $d=6h+2$
4. $d=6h$ (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 9

1. $d=55+h$
2. $h=55d$
3. $d=h+55$
4. $d=55h$ (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 10

1. $d=30t$ (correct answer)
2. $d=t+30$
3. $d=45t$
4. $t=30d$

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 11

1. $c=p+28$
2. $c=4p$ (correct answer)
3. $c=28p$
4. $p=4c$

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 12

1. $y=x+\frac{14}{4}$
2. $y=14x$
3. $x=\frac{14}{4}y$
4. $y=\frac{14}{4}x$ (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 y=(14/4)x, with k=14/4 from the slope through (0,0) and (4,14). A common error is using wrong k like y=14x without dividing, additive form like y=x+(14/4), or reversing like x=(14/4)y. To write the equation: (1) identify proportional relationship (graph through origin), (2) find k (slope=14/4), (3) choose variables (y and x as given), (4) write y=(14/4)x, (5) define variables if needed, (6) verify (for x=4, y=(14/4)×4=14, matches point✓). Multiple representations: equation y=(14/4)x matches graph with slope 14/4, a table with ratio 14/4, and verbal description—all show same k=14/4.

Question 13

1. $m=0.5p+0.5$
2. $m=0.5p$ (correct answer)
3. $m=p+0.5$
4. $p=0.5m$

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$0.50 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

1. $m=2.50b+5$
2. $b=2.50m$
3. $m=2.50b$ (correct answer)
4. $m=b+2.50$

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$. The correct equation is$m=2.50b$, where$m$is money earned in dollars and$b$is boxes sold, with$k=2.50$from$2.50 per box. A common error is using additive form like $m=b+2.50$, reversing variables like $b=2.50m$, or adding extra constants like $m=2.50b+5$. To write the equation: (1) identify proportional relationship (context says "$2.50 for each box"), (2) find$k$(stated rate of 2.50), (3) choose variables ($m$for money,$b$for boxes), (4) write$m=2.50b$, (5) define variables ($m=$money earned in dollars,$b=$number of boxes), (6) verify (for$b=2$,$m=2.50×2=5$, reasonable? yes✓). Multiple representations: equation$m=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 15

1. $y=7x$ (correct answer)
2. $y=7x+2$
3. $x=7y$
4. $y=x+7$

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 16

1. \16$
2. \63$ (correct answer)
3. \9$
4. \72$

Explanation: This question tests writing equations $y=kx$ for proportional relationships from tables, graphs, contexts, or verbal descriptions, identifying $k$ and defining variables contextually, and applying to find values. 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=9c$with$k=9$, and for 7 classes,$f=9\times7=63$. A common error is wrong calculation like$72 if using 8 instead, or non-proportional forms. To solve: (1) identify proportional ("$9 per class"), (2) find$k=9$, (3) variables$f$fee,$c$classes, (4) write$f=9c$, (5) define ($f=$fee in dollars,$c=$classes), (6) substitute$c=7$,$f=63\checkmark$. Multiple representations:$f=9c$matches table multiples of 9, graph slope 9, verbal "$9 per"—all $k=9$. Mistakes: wrong form, miscalculation, wrong $k$, no verification.

Question 17

1. $t=8n$ (correct answer)
2. $t=n+8$
3. $t=8n+5$
4. $n=8t$

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$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 18

1. The distance increases by $4.5$ miles for each additional hour. (correct answer)
2. The distance starts at $4.5$ miles when $t=0$.
3. The distance increases by $t$ miles for each additional $4.5$ hours.
4. The relationship is not proportional because $4.5$ is a decimal.

Explanation: This question tests writing equations y=kx for proportional relationships from tables, graphs, contexts, or verbal descriptions, identifying k and defining variables contextually, and interpreting meaning. 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 interpretation of d=4.5t is distance increases by 4.5 miles per hour, with k=4.5 as rate. A common error is reversing like increases by t per 4.5 hours, assuming intercept like starts at 4.5 when t=0 (but it's 0), or thinking not proportional due to decimal. To interpret: (1) identify proportional (form y=kx), (2) find k=4.5 (miles per hour), (3) variables d distance, t time, (4) equation d=4.5t, (5) define (d=miles, t=hours), (6) verify (t=1, d=4.5, rate matches✓). Multiple representations: d=4.5t matches table multiples of 4.5, graph slope 4.5, verbal "4.5 mph"—all k=4.5. Mistakes: reversed meaning, assuming intercept, wrong form, misinterpreting decimal.

Question 19

1. $d=t+60$
2. $d=60t$ (correct answer)
3. $d=180t$
4. $t=60d$

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 20

1. $d=6h$ (correct answer)
2. $d=h+6$
3. $h=6d$
4. $d=6h+6$

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.