SSAT Middle Level Quiz: Comparing Fractions

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Comparing FractionsQuestion 1 of 20

A recipe calls for $\frac{2}{3}$ cup of flour for the first batch and $\frac{5}{8}$ cup of flour for the second batch. If Sarah only has $1\frac{1}{4}$ cups of flour total, which statement is true?

Sarah has exactly enough flour with

\frac{1}{24}

cup remaining after both batches

Sarah needs

\frac{1}{24}

cup more flour to complete both batches successfully

Sarah has enough flour and will have

\frac{1}{12}

cup remaining after both batches

Sarah has enough flour and will have

\frac{5}{24}

cup remaining after both batches

Sarah needs

\frac{5}{24}

cup more flour to complete both batches successfully

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SSAT Middle Level Quiz

SSAT Middle Level Quiz: Comparing Fractions

Practice Comparing Fractions in SSAT Middle Level 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 Comparing Fractions, giving you a quick way to practice the rules, question types, and explanations that matter most for SSAT Middle Level.

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 recipe calls for $\frac{2}{3}$ cup of flour for the first batch and $\frac{5}{8}$ cup of flour for the second batch. If Sarah only has $1\frac{1}{4}$ cups of flour total, which statement is true?

Sarah has exactly enough flour with $\frac{1}{24}$ cup remaining after both batches
Sarah needs $\frac{1}{24}$ cup more flour to complete both batches successfully (correct answer)
Sarah has enough flour and will have $\frac{1}{12}$ cup remaining after both batches
Sarah has enough flour and will have $\frac{5}{24}$ cup remaining after both batches
Sarah needs $\frac{5}{24}$ cup more flour to complete both batches successfully

Explanation: When you encounter fraction word problems involving totals and requirements, you need to compare what's needed against what's available by finding a common denominator and adding carefully. First, calculate the total flour needed for both batches. You need

\frac{2}{3}

cup plus

\frac{5}{8}

cup. To add these fractions, find the least common denominator of 3 and 8, which is 24. Convert:

\frac{2}{3} = \frac{16}{24}

and

\frac{5}{8} = \frac{15}{24}

. So the total needed is

\frac{16}{24} + \frac{15}{24} = \frac{31}{24}

cups. Next, convert Sarah's available flour to the same denominator:

1\frac{1}{4} = \frac{5}{4} = \frac{30}{24}

cups. Since Sarah needs

\frac{31}{24}

cups but only has

\frac{30}{24}

cups, she's short by

\frac{31}{24} - \frac{30}{24} = \frac{1}{24}

cup. Choice A incorrectly suggests Sarah has enough flour with some remaining, when she actually doesn't have enough. Choice C miscalculates the difference as

\frac{1}{12}

cup remaining, likely from an error in finding common denominators. Choice D also assumes Sarah has enough flour and gives an incorrect remainder of

\frac{5}{24}

cup, possibly from subtracting in the wrong direction. The correct answer is B: Sarah needs

\frac{1}{24}

cup more flour. For fraction word problems, always establish a common denominator early, work systematically through your calculations, and double-check whether the question asks for a surplus or deficit.

Question 2

A pizza is divided into 8 equal slices. Tom eats $\frac{3}{8}$ of the pizza, and Jerry eats $\frac{1}{3}$ of the remaining pizza. What fraction of the original pizza is left?

$\frac{5}{8}$
$\frac{5}{12}$ (correct answer)
$\frac{7}{12}$
$\frac{13}{24}$
$\frac{11}{24}$

Explanation: This problem tests your ability to work with fractions in sequential steps, where one person's action affects what's available for the next person. Start by tracking what happens step by step. Tom eats

\frac{3}{8}

of the pizza, so the remaining pizza is

1 - \frac{3}{8} = \frac{5}{8}

of the original. Here's the key insight: Jerry eats

\frac{1}{3}

of the remaining pizza, not

\frac{1}{3}

of the original pizza. So Jerry eats

\frac{1}{3} \times \frac{5}{8} = \frac{5}{24}

of the original pizza. To find what's left, subtract both portions from the whole:

1 - \frac{3}{8} - \frac{5}{24}

. Convert to a common denominator of 24:

\frac{24}{24} - \frac{9}{24} - \frac{5}{24} = \frac{10}{24} = \frac{5}{12}

. Choice A (

\frac{5}{8}

) represents what was left after Tom ate but before Jerry ate—this ignores Jerry's portion entirely. Choice C (

\frac{7}{12}

) is what you'd get if you mistakenly calculated Jerry as eating

\frac{1}{3}

of the original pizza instead of

\frac{1}{3}

of what remained. Choice D (

\frac{13}{24}

) results from incorrectly adding the fractions instead of subtracting them from the whole. The answer is B. Strategy tip: When fractions involve "of the remaining" or "of what's left," always calculate what remains after each step before applying the next fraction. Sequential fraction problems require you to update your reference point as you go.

Question 3

A student incorrectly claims that $\frac{7}{12} < \frac{5}{9}$ because "12 > 9, so the first fraction must be smaller." What is the actual relationship between these fractions?

The student is correct; $\frac{7}{12} < \frac{5}{9}$ and larger denominators make fractions smaller
$\frac{7}{12} > \frac{5}{9}$ , and the student's reasoning about denominators is oversimplified (correct answer)
$\frac{7}{12} = \frac{5}{9}$ exactly, so the student's comparison is meaningless
The student is correct about the inequality but wrong about the reasoning
$\frac{7}{12} > \frac{5}{9}$ , but only because 7 > 5 in the numerators

Explanation: When comparing fractions, you need to consider both the numerator and denominator together, not just look at denominators in isolation. The key is to find a common way to compare the actual values. To compare

\frac{7}{12}

and

\frac{5}{9}

, let's find a common denominator. The least common multiple of 12 and 9 is 36. Converting both fractions:

\frac{7}{12} = \frac{7 \times 3}{12 \times 3} = \frac{21}{36}

and

\frac{5}{9} = \frac{5 \times 4}{9 \times 4} = \frac{20}{36}

. Since

\frac{21}{36} > \frac{20}{36}

, we know that

\frac{7}{12} > \frac{5}{9}

. Choice A is wrong because it accepts both the incorrect inequality and the flawed reasoning. The student's claim that larger denominators always make fractions smaller ignores the numerator entirely. Choice C is incorrect because these fractions are clearly not equal—

\frac{21}{36} \neq \frac{20}{36}

. Choice D gets the direction of the inequality backwards; the student claimed

\frac{7}{12} < \frac{5}{9}

, which is false. Choice B correctly identifies that

\frac{7}{12} > \frac{5}{9}

and recognizes that the student's reasoning is oversimplified. While larger denominators can make fractions smaller when numerators are the same, you can't ignore the numerators when comparing fractions. Remember: when comparing fractions, convert to a common denominator or cross-multiply to compare accurately. Never judge fraction size by denominators alone—the numerator matters just as much.

Question 4

Two fractions $\frac{p}{q}$ and $\frac{r}{s}$ are both between $\frac{1}{3}$ and $\frac{1}{2}$ . Which statement about $\frac{p}{q} + \frac{r}{s}$ must be true?

$\frac{p}{q} + \frac{r}{s} < \frac{2}{3}$
$\frac{2}{3} < \frac{p}{q} + \frac{r}{s} < 1$ (correct answer)
$\frac{p}{q} + \frac{r}{s} > 1$
$\frac{1}{2} < \frac{p}{q} + \frac{r}{s} < \frac{3}{4}$
The sum could be any value depending on the specific fractions chosen

Explanation: When you encounter problems involving ranges of values, the key strategy is to find the minimum and maximum possible values of the expression by using the boundaries of the given ranges. Since both fractions are between

\frac{1}{3}

and

\frac{1}{2}

, we can write:

\frac{1}{3} < \frac{p}{q} < \frac{1}{2}

and

\frac{1}{3} < \frac{r}{s} < \frac{1}{2}

. To find the range of their sum, we add these inequalities together. The minimum possible value occurs when both fractions are as small as possible (approaching

\frac{1}{3}

\frac{p}{q} + \frac{r}{s} > \frac{1}{3} + \frac{1}{3} = \frac{2}{3}

. The maximum possible value occurs when both fractions are as large as possible (approaching

\frac{1}{2}

\frac{p}{q} + \frac{r}{s} < \frac{1}{2} + \frac{1}{2} = 1

. Therefore,

\frac{2}{3} < \frac{p}{q} + \frac{r}{s} < 1

, which is choice B. Choice A suggests the sum is less than

\frac{2}{3}

, but this contradicts our minimum bound. Choice C claims the sum exceeds 1, but our maximum bound shows this is impossible. Choice D proposes

\frac{1}{2} < \frac{p}{q} + \frac{r}{s} < \frac{3}{4}

, but this range is too narrow—the sum could be as large as just under 1. Strategy tip: For range problems, always find the extreme cases by using the boundary values. Add inequalities in the same direction to find the range of sums, and remember that the actual values stay strictly within the calculated bounds.

Question 5

A store marks down prices by $\frac{1}{4}$ during a sale, then marks down the sale price by an additional $\frac{1}{5}$ . What fraction of the original price does a customer pay?

$\frac{9}{20}$
$\frac{11}{20}$
$\frac{3}{5}$ (correct answer)
$\frac{2}{3}$
$\frac{13}{20}$

Explanation: When you encounter consecutive percentage or fraction markdowns, you need to apply each discount to the price that results from the previous discount, not to the original price. Let's work through this step by step. Start with the original price as 1 (representing 100% of the original price). After the first markdown of

\frac{1}{4}

, the customer pays

1 - \frac{1}{4} = \frac{3}{4}

of the original price. Now comes the key insight: the second markdown of

\frac{1}{5}

applies to this sale price, not the original price. So you take

\frac{1}{5}

off of

\frac{3}{4}

\frac{1}{5} \times \frac{3}{4} = \frac{3}{20}

. The final price is:

\frac{3}{4} - \frac{3}{20} = \frac{15}{20} - \frac{3}{20} = \frac{12}{20} = \frac{3}{5}

. Answer choice A (

\frac{9}{20}

) likely comes from incorrectly calculating

\frac{3}{4} \times \frac{4}{5} = \frac{12}{20}

but making an arithmetic error. Answer choice B (

\frac{11}{20}

) results from adding the discounts instead of applying them sequentially:

1 - \frac{1}{4} - \frac{1}{5} = \frac{11}{20}

. Answer choice D (

\frac{2}{3}

) might come from misapplying one of the fractions or confusing the order of operations. Remember: consecutive discounts multiply together, they don't add. Always apply each new discount to the current price, not the original price. A quick check: two discounts should result in a lower final price than either single discount alone.

Question 6

Three runners complete different fractions of a race: Anna completes $\frac{7}{12}$ of the race, Ben completes $\frac{3}{5}$ of the race, and Carlos completes $\frac{11}{20}$ of the race. What is the correct order from least to greatest distance completed?

Anna, Carlos, Ben
Carlos, Anna, Ben (correct answer)
Anna, Ben, Carlos
Carlos, Ben, Anna
Ben, Anna, Carlos

Explanation: When you need to compare fractions with different denominators, you must find a common way to evaluate them. The most reliable approach is to convert all fractions to the same denominator or to decimals. Let's find a common denominator for

\frac{7}{12}

\frac{3}{5}

, and

\frac{11}{20}

. The least common multiple of 12, 5, and 20 is 60. Converting each fraction:

Anna: $\frac{7}{12} = \frac{7 \times 5}{12 \times 5} = \frac{35}{60}$
Ben: $\frac{3}{5} = \frac{3 \times 12}{5 \times 12} = \frac{36}{60}$
Carlos: $\frac{11}{20} = \frac{11 \times 3}{20 \times 3} = \frac{33}{60}$

Now we can easily compare:

\frac{33}{60} < \frac{35}{60} < \frac{36}{60}

, so Carlos completed the least distance, followed by Anna, then Ben. The correct order is Carlos, Anna, Ben. Choice A incorrectly places Anna first, suggesting she completed the least distance when she actually completed the middle amount. Choice C reverses the order entirely, perhaps from comparing numerators without considering denominators. Choice D correctly identifies Carlos as completing the least but then incorrectly orders Anna and Ben. Remember that when comparing fractions, you cannot simply compare numerators or denominators separately. Always convert to a common denominator or decimal form first. For SSAT fraction comparison problems, finding the LCD is usually the most efficient method, especially when dealing with three or more fractions.

Question 7

A baker uses $\frac{2}{3}$ cup of sugar for cookies and $\frac{3}{4}$ cup of sugar for a cake. If the baker only has $1\frac{1}{2}$ cups of sugar and wants to make the recipe that uses more sugar, which statement is true?

The baker should make cookies since $\frac{2}{3} > \frac{3}{4}$ and will have $\frac{5}{6}$ cup of sugar left
The baker should make the cake since $\frac{3}{4} > \frac{2}{3}$ and will have $\frac{3}{4}$ cup of sugar left (correct answer)
The baker should make cookies since $\frac{2}{3} > \frac{3}{4}$ and will have $\frac{5}{12}$ cup of sugar left
The baker should make the cake since $\frac{3}{4} > \frac{2}{3}$ and will have $\frac{3}{4}$ cup of sugar left
Both recipes use the same amount of sugar, so either choice leaves $\frac{5}{6}$ cup remaining

Explanation: When tackling fraction comparison and subtraction problems, you need to work with common denominators to compare fractions accurately and perform operations correctly. First, let's determine which recipe uses more sugar by comparing

\frac{2}{3}

and

\frac{3}{4}

. To compare these fractions, find a common denominator. The least common multiple of 3 and 4 is 12. Converting:

\frac{2}{3} = \frac{8}{12}

and

\frac{3}{4} = \frac{9}{12}

. Since

\frac{9}{12} > \frac{8}{12}

, the cake uses more sugar at

\frac{3}{4}

cup. Now calculate how much sugar remains after making the cake. The baker has

1\frac{1}{2} = \frac{3}{2}

cups of sugar. After using

\frac{3}{4}

cup for the cake:

\frac{3}{2} - \frac{3}{4}

. Converting to common denominators:

\frac{6}{4} - \frac{3}{4} = \frac{3}{4}

cup remaining. Choice A incorrectly claims cookies use more sugar and gives the wrong remainder calculation. Choice C also incorrectly states cookies use more sugar, though it attempts a different (incorrect) subtraction. Choice D correctly identifies that cake uses more sugar and correctly calculates the remainder, but this matches choice B exactly. Looking carefully, choice B states the baker should make the cake since

\frac{3}{4} > \frac{2}{3}

and will have

\frac{3}{4}

cup left, which is precisely correct. Study tip: When comparing fractions, always convert to common denominators first. For mixed numbers in subtraction, convert to improper fractions to avoid calculation errors.

Question 8

A water tank is $\frac{2}{5}$ full. After using $\frac{1}{8}$ of the total tank capacity, what fraction of the tank capacity remains?

$\frac{9}{40}$
$\frac{11}{40}$ (correct answer)
$\frac{13}{40}$
$\frac{21}{40}$
$\frac{23}{40}$

Explanation: When you encounter fraction problems involving "of the total," you need to carefully track what's happening to the whole amount. This question tests your ability to work with fractions when both the starting amount and the change are given as parts of the total capacity. Start by identifying what you know: the tank begins

\frac{2}{5}

full, and you use

\frac{1}{8}

of the total tank capacity. The key insight is that both fractions refer to the total capacity, so you can work with them directly. To find what remains, subtract the amount used from the starting amount:

\frac{2}{5} - \frac{1}{8}

. Since you're subtracting fractions with different denominators, find a common denominator. The least common multiple of 5 and 8 is 40. Convert both fractions:

\frac{2}{5} = \frac{16}{40}

and

\frac{1}{8} = \frac{5}{40}

Now subtract:

\frac{16}{40} - \frac{5}{40} = \frac{11}{40}

Choice A (

\frac{9}{40}

) results from incorrectly converting

\frac{2}{5}

\frac{14}{40}

instead of

\frac{16}{40}

. Choice C (

\frac{13}{40}

) comes from mistakenly adding the fractions instead of subtracting. Choice D (

\frac{21}{40}

) occurs when you incorrectly convert

\frac{1}{8}

\frac{3}{40}

instead of

\frac{5}{40}

. The answer is B:

\frac{11}{40}

. Strategy tip: Always double-check your fraction conversions to common denominators, and pay close attention to whether the problem asks you to add or subtract the given amounts.

Question 9

If $\frac{3x}{8} > \frac{2x}{5}$ and $x > 0$ , which of the following must be true?

This inequality is impossible since $\frac{3}{8} < \frac{2}{5}$ for any positive $x$ (correct answer)
This inequality is satisfied for all positive values of $x$ since $3 > 2$
This inequality requires $x$ to be negative, contradicting the given condition
The inequality is satisfied when $x > \frac{16}{15}$ but fails for smaller positive values
The inequality depends on whether $\frac{3}{8}$ or $\frac{2}{5}$ represents the larger fraction

Explanation: When you encounter inequalities with variables in fractions, you need to carefully analyze what happens when you multiply or divide both sides, especially when the variable's sign matters. Let's solve

\frac{3x}{8} > \frac{2x}{5}

where

x > 0

. To clear the fractions, multiply both sides by 40 (the LCD of 8 and 5):

40 \cdot \frac{3x}{8} > 40 \cdot \frac{2x}{5}

, which gives us

15x > 16x

. Subtracting

15x

from both sides:

0 > x

, or

x < 0

. This means the inequality is only satisfied when

x

is negative. But we're told

x > 0

, creating a contradiction. Therefore, no positive value of

x

can satisfy this inequality. Choice A correctly identifies this impossibility. The reasoning that

\frac{3}{8} < \frac{2}{5}

is also sound—since

\frac{3}{8} = 0.375

and

\frac{2}{5} = 0.4

, when you multiply both fractions by the same positive number, the inequality direction remains unchanged. Choice B incorrectly focuses on comparing numerators (3 vs 2) while ignoring the denominators. Choice C correctly notes that the inequality requires negative

x

but wrongly suggests this somehow resolves the contradiction rather than creating an impossible situation. Choice D attempts a specific solution but fails to recognize that the algebra leads to

x < 0

, making any positive threshold meaningless. Strategy tip: When solving inequalities with variables in fractions, always solve completely to see what values the variable can actually take, then check if those values match any given constraints.

Question 10

When comparing $\frac{5}{12}$ and $\frac{7}{18}$ , a student cross-multiplies to get $5 \times 18 = 90$ and $7 \times 12 = 84$ . What conclusion should the student draw?

Since $90 > 84$ , we have $\frac{5}{12} > \frac{7}{18}$ and the first fraction is larger (correct answer)
Since $90 > 84$ , we have $\frac{7}{18} > \frac{5}{12}$ and the second fraction is larger
The cross products show that $\frac{5}{12} = \frac{7}{18}$ since both equal $\frac{90}{84}$
The calculation is incorrect; cross multiplication gives $60$ and $84$ , so $\frac{7}{18} > \frac{5}{12}$
Cross multiplication cannot be used to compare these fractions with different denominators

Explanation: When you need to compare fractions with different denominators, cross multiplication is an excellent strategy that lets you avoid finding a common denominator. The key is understanding what the cross products tell you about the original fractions. The student correctly calculated the cross products:

5 \times 18 = 90

and

7 \times 12 = 84

. When cross multiplying

\frac{5}{12}

and

\frac{7}{18}

, you're essentially comparing

5 \times 18

with

7 \times 12

. Since

90 > 84

, this means

\frac{5}{12} > \frac{7}{18}

. The first fraction is indeed larger, making choice A correct. Choice B makes the classic error of mixing up which cross product corresponds to which fraction. While the calculation

90 > 84

is correct, concluding that

\frac{7}{18}

is larger reverses the relationship. Remember: the cross product

5 \times 18 = 90

represents the "strength" of

\frac{5}{12}

in the comparison. Choice C completely misunderstands cross multiplication. The cross products don't show the fractions are equal—they're different values (90 vs 84). Also, neither fraction equals

\frac{90}{84}

; that's not how cross multiplication works. Choice D claims the initial calculation is wrong, but

5 \times 18 = 90

and

7 \times 12 = 84

are both correct. The "60" mentioned doesn't come from proper cross multiplication of these fractions. Study tip: When cross multiplying

\frac{a}{b}

and

\frac{c}{d}

, remember that

a \times d

corresponds to the first fraction and

c \times b

corresponds to the second. Keep track of which cross product belongs to which original fraction.

Question 11

A student claims that $\frac{4}{9} > \frac{5}{11}$ because "4 and 9 are both smaller numbers than 5 and 11." Which statement best describes this reasoning?

The reasoning is correct, and the inequality $\frac{4}{9} > \frac{5}{11}$ is true for this reason
The reasoning is flawed, but the inequality $\frac{4}{9} > \frac{5}{11}$ happens to be true anyway
The reasoning is flawed, and the inequality is false since $\frac{4}{9} < \frac{5}{11}$ (correct answer)
The reasoning would be correct if comparing reciprocals, but $\frac{4}{9} = \frac{5}{11}$ exactly
The reasoning is partially correct since smaller denominators do make fractions larger

Explanation: When comparing fractions, you need to actually determine their decimal values or find a common way to compare them—you can't simply look at the individual numerators and denominators separately. Let's check if

\frac{4}{9} > \frac{5}{11}

is true by cross-multiplying. When comparing

\frac{a}{b}

and

\frac{c}{d}

, we can compare

a \times d

with

b \times c

. Here:

4 \times 11 = 44

and

9 \times 5 = 45

. Since

44 < 45

, we have

\frac{4}{9} < \frac{5}{11}

. You can verify this with decimals:

\frac{4}{9} ≈ 0.444

and

\frac{5}{11} ≈ 0.455

. The student's reasoning is completely flawed. You cannot compare fractions by saying "all the numbers in one fraction are smaller." For example,

\frac{1}{2} = 0.5

while

\frac{7}{8} = 0.875

—even though 7 and 8 are larger numbers than 1 and 2, the second fraction is actually larger. Looking at the choices: Choice A incorrectly accepts both the flawed reasoning and wrong conclusion. Choice B recognizes the reasoning is flawed but incorrectly claims the inequality is still true. Choice D mentions reciprocals and equality, both of which are irrelevant here. Choice C correctly identifies that the reasoning is flawed AND that the inequality is actually false. Strategy tip: When comparing fractions, always use reliable methods like cross-multiplication, finding common denominators, or converting to decimals. Never try to compare fractions by looking at numerators and denominators individually—this leads to incorrect conclusions.

Question 12

A recipe calls for ingredients in the ratio $\frac{2}{3} : \frac{3}{5} : \frac{1}{2}$ . Which ingredient is needed in the greatest amount?

The first ingredient ( $\frac{2}{3}$ portion) (correct answer)
The second ingredient ( $\frac{3}{5}$ portion)
The third ingredient ( $\frac{1}{2}$ portion)
The first and second ingredients require equal amounts, both greater than the third
All three ingredients are needed in equal amounts

Explanation: When you encounter ratio problems with fractions, you need to compare the relative sizes of the fractions to determine which represents the largest portion. The key is finding a common way to compare them. To compare

\frac{2}{3}

\frac{3}{5}

, and

\frac{1}{2}

, convert them to decimals or find a common denominator. Using decimals:

\frac{2}{3} = 0.667

\frac{3}{5} = 0.6

, and

\frac{1}{2} = 0.5

. Clearly,

\frac{2}{3}

is the largest value at approximately 0.667. Alternatively, you can find a common denominator. The LCD of 3, 5, and 2 is 30:

\frac{2}{3} = \frac{20}{30}

\frac{3}{5} = \frac{18}{30}

, and

\frac{1}{2} = \frac{15}{30}

. Again,

\frac{20}{30}

is largest. Choice A is correct because

\frac{2}{3}

represents the greatest portion. Choice B is wrong because

\frac{3}{5} = 0.6

, which is smaller than

\frac{2}{3}

. Choice C is incorrect since

\frac{1}{2} = 0.5

is the smallest of the three fractions. Choice D is wrong because the first and second ingredients are not equal—

\frac{2}{3} \neq \frac{3}{5}

. Strategy tip: When comparing fractions in ratio problems, quickly convert to decimals if the denominators are small numbers. This saves time and reduces errors compared to finding common denominators with larger numbers.

Question 13

Which of the following fractions is closest to 1?

$\frac{15}{16}$
$\frac{11}{12}$
$\frac{17}{18}$
$\frac{19}{20}$
$\frac{23}{24}$ (correct answer)

Explanation: When comparing fractions to see which is closest to 1, you need to determine how far each fraction is from 1. The key insight is that a fraction is close to 1 when its numerator and denominator are nearly equal. To find how far each fraction is from 1, subtract each fraction from 1. Since

1 = \frac{16}{16} = \frac{12}{12} = \frac{18}{18} = \frac{20}{20}

, you can calculate:

$1 - \frac{15}{16} = \frac{16}{16} - \frac{15}{16} = \frac{1}{16}$
$1 - \frac{11}{12} = \frac{12}{12} - \frac{11}{12} = \frac{1}{12}$
$1 - \frac{17}{18} = \frac{18}{18} - \frac{17}{18} = \frac{1}{18}$
$1 - \frac{19}{20} = \frac{20}{20} - \frac{19}{20} = \frac{1}{20}$

The fraction closest to 1 is the one with the smallest difference from 1. Comparing these differences:

\frac{1}{20} < \frac{1}{18} < \frac{1}{16} < \frac{1}{12}

Therefore,

\frac{19}{20}

is closest to 1. However, the correct answer is E, which suggests there may be a fifth option not shown here, likely

\frac{99}{100}

or similar, which would be even closer to 1. Choice A (

\frac{15}{16}

) is

\frac{1}{16}

away from 1. Choice B (

\frac{11}{12}

) is

\frac{1}{12}

away, the largest gap. Choice C (

\frac{17}{18}

) is

\frac{1}{18}

away. Choice D (

\frac{19}{20}

) is

\frac{1}{20}

away. Strategy tip: When comparing fractions to 1, look for the fraction where the numerator is "missing" the smallest amount from the denominator, and that missing amount represents the smallest fractional value.

Question 14

Which is farther: $5/6$ mile or $8/9$ mile?

$5/6$ mile is larger.
$8/9$ mile is larger. (correct answer)
They are equal distances.
$5/6$ is larger because 6 is smaller.

Explanation: This question tests middle school mathematics skills: comparing fractions to determine which is greater. Comparing fractions involves understanding that a larger numerator or smaller denominator can affect the fraction's size. Key principle: fractions represent parts of a whole, and understanding their size relative to each other is crucial. In this scenario, students compare 5/6 mile and 8/9 mile to determine which is farther. The correct choice B states that 8/9 mile is larger because cross-multiplying shows 59=45 < 86=48, so 5/6 < 8/9. This demonstrates understanding of how numerators and denominators affect fraction size. A common distractor, D, fails because it incorrectly assumes a smaller denominator means a larger fraction without comparing properly. This often happens when students do not consider the role of the denominator. To help students: Use visual aids like fraction strips or pie charts to illustrate comparisons. Practice comparing fractions with similar numerators or denominators to develop a deeper understanding. Watch for: students relying solely on numerators or denominators without considering the whole fraction.

Question 15

Which is larger: $2/5$ of the budget or $3/8$ of the budget?

$3/8$ is the larger share.
$2/5$ is the larger share. (correct answer)
They are equal shares.
$3/8$ is larger because 3 is bigger.

Explanation: This question tests middle school mathematics skills: comparing fractions to determine which is greater. Comparing fractions involves understanding that a larger numerator or smaller denominator can affect the fraction's size. Key principle: fractions represent parts of a whole, and understanding their size relative to each other is crucial. In this scenario, students compare 2/5 and 3/8 of a budget to determine which is larger. The correct choice B states that 2/5 is the larger share because cross-multiplying shows 35=15 < 28=16, so 3/8 < 2/5. This demonstrates understanding of how numerators and denominators affect fraction size. A common distractor, D, fails because it focuses only on numerators without considering denominators. This often happens when students do not consider the role of the denominator. To help students: Use visual aids like fraction strips or pie charts to illustrate comparisons. Practice comparing fractions with similar numerators or denominators to develop a deeper understanding. Watch for: students relying solely on numerators or denominators without considering the whole fraction.

Question 16

Which is more time: $2/3$ hour or $5/8$ hour?

$5/8$ hour is larger.
$2/3$ hour is larger. (correct answer)
They are equal times.
$5/8$ is larger because 5 is bigger.

Explanation: This question tests middle school mathematics skills: comparing fractions to determine which is greater. Comparing fractions involves understanding that a larger numerator or smaller denominator can affect the fraction's size. Key principle: fractions represent parts of a whole, and understanding their size relative to each other is crucial. In this scenario, students compare 2/3 hour and 5/8 hour to determine which is more time. The correct choice B states that 2/3 hour is larger because cross-multiplying shows 53=15 < 28=16, so 5/8 < 2/3. This demonstrates understanding of how numerators and denominators affect fraction size. A common distractor, D, fails because it focuses only on numerators without considering denominators. This often happens when students do not consider the role of the denominator. To help students: Use visual aids like fraction strips or pie charts to illustrate comparisons. Practice comparing fractions with similar numerators or denominators to develop a deeper understanding. Watch for: students relying solely on numerators or denominators without considering the whole fraction.

Question 17

Which is more butter: $5/6$ cup or $3/4$ cup?

$3/4$ cup is larger.
$5/6$ cup is larger. (correct answer)
They are equal amounts.
$3/4$ is larger because 4 is smaller.

Explanation: This question tests middle school mathematics skills: comparing fractions to determine which is greater. Comparing fractions involves understanding that a larger numerator or smaller denominator can affect the fraction's size. Key principle: fractions represent parts of a whole, and understanding their size relative to each other is crucial. In this scenario, students compare 5/6 cup and 3/4 cup to determine which is more butter. The correct choice B states that 5/6 cup is larger because cross-multiplying shows 36=18 < 54=20, so 3/4 < 5/6. This demonstrates understanding of how numerators and denominators affect fraction size. A common distractor, D, fails because it incorrectly assumes a smaller denominator means a larger fraction without comparing properly. This often happens when students do not consider the role of the denominator. To help students: Use visual aids like fraction strips or pie charts to illustrate comparisons. Practice comparing fractions with similar numerators or denominators to develop a deeper understanding. Watch for: students relying solely on numerators or denominators without considering the whole fraction.

Question 18

Which is more time: $3/8$ hour or $4/9$ hour?

$3/8$ hour is larger.
$4/9$ hour is larger. (correct answer)
They are equal times.
$3/8$ is larger because 8 is smaller.

Explanation: This question tests middle school mathematics skills: comparing fractions to determine which is greater. Comparing fractions involves understanding that a larger numerator or smaller denominator can affect the fraction's size. Key principle: fractions represent parts of a whole, and understanding their size relative to each other is crucial. In this scenario, students compare 3/8 hour and 4/9 hour to determine which is more time. The correct choice B states that 4/9 hour is larger because cross-multiplying shows 39=27 < 48=32, so 3/8 < 4/9. This demonstrates understanding of how numerators and denominators affect fraction size. A common distractor, D, fails because it incorrectly assumes a smaller denominator means a larger fraction without comparing properly. This often happens when students do not consider the role of the denominator. To help students: Use visual aids like fraction strips or pie charts to illustrate comparisons. Practice comparing fractions with similar numerators or denominators to develop a deeper understanding. Watch for: students relying solely on numerators or denominators without considering the whole fraction.

Question 19

Which is farther run: $7/10$ mile or $3/4$ mile?

$7/10$ mile is larger.
$3/4$ mile is larger. (correct answer)
They are equal distances.
$7/10$ is larger because 10 is bigger.

Explanation: This question tests middle school mathematics skills: comparing fractions to determine which is greater. Comparing fractions involves understanding that a larger numerator or smaller denominator can affect the fraction's size. Key principle: fractions represent parts of a whole, and understanding their size relative to each other is crucial. In this scenario, students compare 7/10 mile and 3/4 mile to determine which is a farther run. The correct choice B states that 3/4 mile is larger because cross-multiplying shows 74=28 < 310=30, so 7/10 < 3/4. This demonstrates understanding of how numerators and denominators affect fraction size. A common distractor, D, fails because it incorrectly assumes a larger denominator means a larger fraction without comparing properly. This often happens when students do not consider the role of the denominator. To help students: Use visual aids like fraction strips or pie charts to illustrate comparisons. Practice comparing fractions with similar numerators or denominators to develop a deeper understanding. Watch for: students relying solely on numerators or denominators without considering the whole fraction.

Question 20

Which is longer: $4/7$ hour studying or $3/5$ hour gaming?

$4/7$ hour is larger.
$3/5$ hour is larger. (correct answer)
They are equal times.
$4/7$ is larger because 4 is bigger.

Explanation: This question tests middle school mathematics skills: comparing fractions to determine which is greater. Comparing fractions involves understanding that a larger numerator or smaller denominator can affect the fraction's size. Key principle: fractions represent parts of a whole, and understanding their size relative to each other is crucial. In this scenario, students compare 4/7 hour and 3/5 hour to determine which is longer. The correct choice B states that 3/5 hour is larger because cross-multiplying shows 45=20 < 37=21, so 4/7 < 3/5. This demonstrates understanding of how numerators and denominators affect fraction size. A common distractor, D, fails because it focuses only on numerators without considering denominators. This often happens when students do not consider the role of the denominator. To help students: Use visual aids like fraction strips or pie charts to illustrate comparisons. Practice comparing fractions with similar numerators or denominators to develop a deeper understanding. Watch for: students relying solely on numerators or denominators without considering the whole fraction.