ISEE Lower Level Quantitative Reasoning Quiz: Comparing Fractions

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 ISEE Lower Level Quantitative Reasoning.

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. The $\frac{1}{2}$ meter rope
2. The $\frac{5}{9}$ meter rope (correct answer)
3. The $\frac{4}{7}$ meter rope
4. All the ropes must be the same length.

Explanation: To find the rope with the middle length, we must order the fractions $\frac{5}{9}$, $\frac{1}{2}$, and $\frac{4}{7}$. One way is to convert them to decimals: $\frac{1}{2} = 0.5$, $\frac{5}{9} = 0.555...$, and $\frac{4}{7} \approx 0.571$. Ordering these from least to greatest gives 0.5, 0.555..., 0.571. This corresponds to the order $\frac{1}{2}$, $\frac{5}{9}$, $\frac{4}{7}$. The fraction in the middle is $\frac{5}{9}$.

Question 2

1. The dinosaur exhibit
2. The space exhibit (correct answer)
3. An equal number of students chose each exhibit.
4. There is not enough information to tell.

Explanation: To solve this, we must compare the fractions $\frac{1}{3}$ and $\frac{3}{8}$. The total number of students, 24, is extra information. To compare the fractions directly, we find a common denominator, which is 24. Dinosaur exhibit: $\frac{1}{3} = \frac{8}{24}$. Space exhibit: $\frac{3}{8} = \frac{9}{24}$. Since $9 > 8$, $\frac{9}{24} > \frac{8}{24}$, which means a larger fraction of students chose the space exhibit. Therefore, the space exhibit was chosen by more students.

Question 3

1. Candidate B received a greater fraction of the votes.
2. Candidate A received a greater fraction of the votes. (correct answer)
3. Both candidates received the same fraction of the votes.
4. Candidate A received $\frac{4}{5}$ fewer votes than Candidate B.

Explanation: To compare $\frac{9}{10}$ and $\frac{13}{15}$, we can find a common denominator, which is 30. Convert the fractions: Candidate A: $\frac{9}{10} = \frac{27}{30}$. Candidate B: $\frac{13}{15} = \frac{26}{30}$. Since $27 > 26$, we know that $\frac{27}{30} > \frac{26}{30}$. Therefore, Candidate A received a greater fraction of the votes.

Question 4

1. Liam
2. Noah
3. Olivia (correct answer)
4. Emma

Explanation: To determine who is in the lead, we must find the largest fraction among $\frac{3}{4}$, $\frac{2}{3}$, $\frac{5}{6}$, and $\frac{7}{12}$. A common denominator for these fractions is 12. Convert each fraction: Liam: $\frac{3}{4} = \frac{9}{12}$. Noah: $\frac{2}{3} = \frac{8}{12}$. Olivia: $\frac{5}{6} = \frac{10}{12}$. Emma: $\frac{7}{12}$. Comparing the numerators, 10 is the largest. Therefore, Olivia has painted the most and is in the lead.

Question 5

1. $\frac{1}{2} - \frac{1}{10}$
2. $\frac{1}{5} + \frac{1}{20}$
3. $\frac{3}{4} - \frac{1}{2}$
4. $\frac{1}{3} + \frac{1}{12}$ (correct answer)

Explanation: When comparing fractions through addition and subtraction, you need to find common denominators and calculate the actual values to determine which expression yields the largest result. Let's work through each expression systematically. For choice A: $\frac{1}{2} - \frac{1}{10}$, convert to the common denominator 10: $\frac{5}{10} - \frac{1}{10} = \frac{4}{10} = 0.4$ For choice B: $\frac{1}{5} + \frac{1}{20}$, convert to the common denominator 20: $\frac{4}{20} + \frac{1}{20} = \frac{5}{20} = 0.25$ For choice C: $\frac{3}{4} - \frac{1}{2}$, convert to the common denominator 4: $\frac{3}{4} - \frac{2}{4} = \frac{1}{4} = 0.25$ For choice D: $\frac{1}{3} + \frac{1}{12}$, convert to the common denominator 12: $\frac{4}{12} + \frac{1}{12} = \frac{5}{12} ≈ 0.417$ Comparing the results: A gives 0.4, B gives 0.25, C gives 0.25, and D gives approximately 0.417. Choice D produces the largest value. Choice A is close but falls short of D's value. Choices B and C both equal 0.25, making them tied for the smallest values. The key trap here is that addition doesn't automatically create larger results than subtraction—the actual fractional values matter more than the operations. When comparing fraction expressions, always calculate the final decimal values rather than making assumptions based on whether you're adding or subtracting. Convert everything to a common form for easy comparison.

Question 6

1. The first tank (correct answer)
2. The second tank
3. Both tanks had the same fraction of water used.
4. It is not possible to determine from the information given.

Explanation: First, calculate the fraction of water used from each tank. For the first tank, the amount used is $\frac{4}{5} - \frac{1}{3}$. The common denominator is 15: $\frac{12}{15} - \frac{5}{15} = \frac{7}{15}$. For the second tank, the amount used is $\frac{5}{6} - \frac{1}{2}$. The common denominator is 6: $\frac{5}{6} - \frac{3}{6} = \frac{2}{6} = \frac{1}{3}$. Now, compare the fractions of water used: $\frac{7}{15}$ and $\frac{1}{3}$. The common denominator is 15: $\frac{1}{3} = \frac{5}{15}$. Since $\frac{7}{15} > \frac{5}{15}$, the first tank had a greater fraction of its water used.

Question 7

1. There is more red paint than blue paint.
2. The amount of red paint is exactly half the amount of blue paint.
3. There is more blue paint than red paint. (correct answer)
4. The amounts of red paint and blue paint are equal.

Explanation: To compare the fractions $\frac{5}{12}$ and $\frac{4}{9}$, find a common denominator. The least common multiple of 12 and 9 is 36. Convert each fraction to have a denominator of 36. For the red paint: $\frac{5}{12} = \frac{5 \times 3}{12 \times 3} = \frac{15}{36}$. For the blue paint: $\frac{4}{9} = \frac{4 \times 4}{9 \times 4} = \frac{16}{36}$. Since $16 > 15$, $\frac{16}{36} > \frac{15}{36}$. Thus, there is more blue paint than red paint.

Question 8

1. The fifth-grade class
2. It depends on the total number of events.
3. They were equally close to being finished.
4. The sixth-grade class (correct answer)

Explanation: When comparing fractions to see which represents being "closer to finished," you need to determine which fraction is larger. Since both classes had the same number of events, you can directly compare $\frac{5}{8}$ and $\frac{2}{3}$. To compare fractions with different denominators, find a common denominator. The least common multiple of 8 and 3 is 24. Converting both fractions: $\frac{5}{8} = \frac{5 \times 3}{8 \times 3} = \frac{15}{24}$ and $\frac{2}{3} = \frac{2 \times 8}{3 \times 8} = \frac{16}{24}$. Since $\frac{16}{24} > \frac{15}{24}$, the sixth-grade class completed more of their events and was closer to being finished. Choice A is incorrect because $\frac{5}{8}$ is actually smaller than $\frac{2}{3}$, so the fifth-grade class was further from completion. Choice B is wrong because the problem states both grades had the same number of events, making the total number irrelevant to the comparison. Choice C is incorrect because the fractions are not equal—when converted to the same denominator, $\frac{15}{24} \neq \frac{16}{24}$. Choice D is correct because $\frac{2}{3} > \frac{5}{8}$. When comparing fractions on the ISEE, always convert to a common denominator or use cross-multiplication to avoid errors. Don't let different denominators fool you into thinking you can't make a direct comparison—there's always a way to determine which fraction is larger.

Question 9

1. The person who took slices from the pie cut into 12 slices got more pie.
2. The person who took slices from the pie cut into 10 slices got more pie. (correct answer)
3. Both people got the same amount of pie.
4. It is impossible to tell who got more pie without knowing the size of the pies.

Explanation: The first person takes $\frac{3}{10}$ of a pie. The second person takes $\frac{3}{12}$ of a pie. We need to compare $\frac{3}{10}$ and $\frac{3}{12}$. When two fractions have the same numerator, the fraction with the smaller denominator is larger, because the whole is divided into fewer, larger pieces. Since $10 < 12$, it follows that $\frac{3}{10} > \frac{3}{12}$. Therefore, the person who took slices from the pie cut into 10 slices got more pie.

Question 10

1. Chloe ate more pizza on Monday.
2. Chloe ate more pizza on Tuesday.
3. Chloe ate the same amount of pizza on both days.
4. It cannot be determined who ate more pizza. (correct answer)

Explanation: The fractions refer to different-sized wholes (a 'small pizza' versus a 'large pizza'). Because the total sizes of the pizzas are not the same and are not specified, we cannot compare the absolute amounts of pizza eaten. For example, $\frac{1}{3}$ of a very large pizza could be more than $\frac{1}{2}$ of a very small pizza. Without more information about the sizes of the pizzas, no definitive comparison can be made.

Question 11

1. $\frac{3}{8}$
2. $\frac{4}{7}$
3. $\frac{5}{9}$ (correct answer)
4. $\frac{7}{12}$

Explanation: To find which fraction is closest to $\frac{1}{2}$, we find the absolute difference between each fraction and $\frac{1}{2}$. A) $|\frac{3}{8} - \frac{4}{8}| = \frac{1}{8}$. B) $|\frac{4}{7} - \frac{1}{2}| = |\frac{8}{14} - \frac{7}{14}| = \frac{1}{14}$. C) $|\frac{5}{9} - \frac{1}{2}| = |\frac{10}{18} - \frac{9}{18}| = \frac{1}{18}$. D) $|\frac{7}{12} - \frac{1}{2}| = |\frac{7}{12} - \frac{6}{12}| = \frac{1}{12}$. Now we must find the smallest of these differences: $\frac{1}{8}, \frac{1}{14}, \frac{1}{18}, \frac{1}{12}$. When fractions have the same numerator (1 in this case), the one with the largest denominator is the smallest. The largest denominator is 18, so $\frac{1}{18}$ is the smallest difference. Therefore, $\frac{5}{9}$ is closest to $\frac{1}{2}$.

Question 12

1. Marco has more pizza left over.
2. Jada has more pizza left over. (correct answer)
3. They have the same amount of pizza left over.
4. There is not enough information to determine who has more left.

Explanation: First, determine the fraction of pizza each person has left. Marco has $1 - \frac{2}{5} = \frac{3}{5}$ of his pizza left. Jada has $1 - \frac{3}{8} = \frac{5}{8}$ of her pizza left. Next, compare these two fractions. To compare $\frac{3}{5}$ and $\frac{5}{8}$, find a common denominator, which is 40. Marco's remaining pizza is $\frac{3}{5} = \frac{24}{40}$. Jada's remaining pizza is $\frac{5}{8} = \frac{25}{40}$. Since $\frac{25}{40} > \frac{24}{40}$, Jada has more pizza left over.

Question 13

1. $\frac{1}{4}$ cup
2. $\frac{5}{12}$ cup (correct answer)
3. $\frac{2}{3}$ cup
4. $\frac{3}{5}$ cup

Explanation: To find a fraction between $\frac{1}{3}$ and $\frac{1}{2}$, convert them to fractions with a common denominator. A common denominator for all the fractions is 60. $\frac{1}{3} = \frac{20}{60}$ and $\frac{1}{2} = \frac{30}{60}$. We need a fraction between $\frac{20}{60}$ and $\frac{30}{60}$. Let's convert the answer choices: A) $\frac{1}{4} = \frac{15}{60}$ (too small). B) $\frac{5}{12} = \frac{25}{60}$ (this is between $\frac{20}{60}$ and $\frac{30}{60}$). C) $\frac{2}{3} = \frac{40}{60}$ (too large). D) $\frac{3}{5} = \frac{36}{60}$ (too large). Therefore, $\frac{5}{12}$ is the correct amount.

Question 14

1. The bookshelf that is $\frac{7}{9}$ full.
2. The bookshelf that is $\frac{3}{4}$ full. (correct answer)
3. They have the same amount of empty space.
4. The fuller bookshelf has more empty space.

Explanation: First, calculate the empty space for each bookshelf. The first bookshelf's empty space is $1 - \frac{7}{9} = \frac{2}{9}$. The second bookshelf's empty space is $1 - \frac{3}{4} = \frac{1}{4}$. Now, compare the fractions of empty space, $\frac{2}{9}$ and $\frac{1}{4}$. Using a common denominator of 36: $\frac{2}{9} = \frac{8}{36}$ and $\frac{1}{4} = \frac{9}{36}$. Since $\frac{9}{36} > \frac{8}{36}$, the bookshelf that is $\frac{3}{4}$ full has more empty space.

Question 15

1. $\frac{3}{4}$
2. $\frac{5}{7}$
3. $\frac{7}{10}$
4. $\frac{8}{13}$ (correct answer)

Explanation: We need to compare each fraction to $\frac{2}{3}$ to find the one that is smaller. We can use cross-multiplication. A) For $\frac{3}{4}$, $3 \times 3 = 9$ and $4 \times 2 = 8$. Since $9 > 8$, $\frac{3}{4} > \frac{2}{3}$. B) For $\frac{5}{7}$, $5 \times 3 = 15$ and $7 \times 2 = 14$. Since $15 > 14$, $\frac{5}{7} > \frac{2}{3}$. C) For $\frac{7}{10}$, $7 \times 3 = 21$ and $10 \times 2 = 20$. Since $21 > 20$, $\frac{7}{10} > \frac{2}{3}$. D) For $\frac{8}{13}$, $8 \times 3 = 24$ and $13 \times 2 = 26$. Since $24 < 26$, $\frac{8}{13} < \frac{2}{3}$. Thus, $\frac{8}{13}$ is the fraction that is not greater than $\frac{2}{3}$.

Question 16

1. David ran farther.
2. Sarah ran farther. (correct answer)
3. They ran the same distance.
4. Sarah ran exactly $\frac{1}{2}$ mile farther.

Explanation: To compare the mixed numbers $2\frac{3}{5}$ and $2\frac{2}{3}$, notice that the whole number part (2) is the same for both. Therefore, we only need to compare the fractional parts, $\frac{3}{5}$ and $\frac{2}{3}$. Find a common denominator, which is 15. Convert the fractions: $\frac{3}{5} = \frac{9}{15}$ and $\frac{2}{3} = \frac{10}{15}$. Since $\frac{10}{15} > \frac{9}{15}$, Sarah's fractional distance is greater. This means Sarah ran farther than David.

Question 17

1. The cake required more sugar. (correct answer)
2. The cookies required more sugar.
3. They required the same amount of sugar.
4. The cake required twice as much sugar as the cookies.

Explanation: To compare $\frac{11}{4}$ and $2\frac{1}{2}$, we should convert them to the same format. Let's convert the improper fraction $\frac{11}{4}$ to a mixed number. Divide 11 by 4, which is 2 with a remainder of 3. So, $\frac{11}{4} = 2\frac{3}{4}$. Now we compare $2\frac{3}{4}$ (for the cake) with $2\frac{1}{2}$ (for the cookies). Since the whole numbers are both 2, we compare the fractions $\frac{3}{4}$ and $\frac{1}{2}$. We know that $\frac{3}{4} > \frac{2}{4} = \frac{1}{2}$. Therefore, the cake required more sugar.

Question 18

1. The turtle is faster. (correct answer)
2. The snail is faster.
3. They have the same speed.
4. The turtle is twice as fast as the snail.

Explanation: The faster animal is the one that takes less time to travel the same distance. We need to compare the times $\frac{1}{5}$ hour and $\frac{2}{9}$ hour and find the smaller value. Using cross-multiplication to compare the fractions: for $\frac{1}{5}$, the product is $1 \times 9 = 9$; for $\frac{2}{9}$, the product is $2 \times 5 = 10$. Since $9 < 10$, it means $\frac{1}{5} < \frac{2}{9}$. The turtle takes less time, so the turtle is faster.

Question 19

1. There are more blue marbles than red marbles.
2. There are more red marbles than blue marbles. (correct answer)
3. There are equal numbers of red and blue marbles.
4. There are more green marbles than red marbles.

Explanation: The question asks to compare the number of red and blue marbles. This requires comparing the fractions $\frac{2}{7}$ (red) and $\frac{1}{4}$ (blue). We can use cross-multiplication: for $\frac{2}{7}$, the product is $2 \times 4 = 8$. For $\frac{1}{4}$, the product is $1 \times 7 = 7$. Since $8 > 7$, the fraction $\frac{2}{7}$ is greater than $\frac{1}{4}$. Therefore, there are more red marbles than blue marbles. The information about green marbles is extra and not needed to answer the question asked.

Question 20

1. The new fraction is smaller than $\frac{2}{5}$.
2. The new fraction is equal to $\frac{2}{5}$.
3. The new fraction is larger than $\frac{2}{5}$. (correct answer)
4. The result depends on the whole number that is added.

Explanation: Let's test this concept by adding a positive whole number, for example, 1. The new fraction becomes $\frac{2+1}{5+1} = \frac{3}{6} = \frac{1}{2}$. To compare $\frac{1}{2}$ with $\frac{2}{5}$, we use a common denominator of 10. $\frac{1}{2} = \frac{5}{10}$ and $\frac{2}{5} = \frac{4}{10}$. Since $\frac{5}{10} > \frac{4}{10}$, the new fraction is larger. This pattern holds true for any positive proper fraction; adding the same positive number to the numerator and denominator increases its value, bringing it closer to 1.