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GMAT Ratio Practice Questions

Practice Ratio questions with worked explanations and timing guidance for Quantitative Reasoning.

Five-question preview. Answer 3 now without an account.
Question 1 of 3 free Easy

A certain galaxy is known to comprise approximately \( 4 \times 10^{11} \) stars. Of every 50 million of these stars, one is larger in mass than our sun. Approximately how many stars in this galaxy are larger than the sun?

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Correct answer: C

50 million can be represented in scientific notation as \( 5 \times 10^7 \). Restating this figure in scientific notation will enable us to simplify the division required to solve the problem. If one out of every \( 5 \times 10^7 \) stars is larger than the sun, we must divide the total number of stars by this figure to find the solution:

\( \frac{4 \times 10^{11}}{5 \times 10^7} \)

\( = \frac{4}{5} \times 10^{(11-7)} \)

\( = 0.8 \times 10^4 \)

The final step is to move the decimal point of 0.8 four places to the right, with a result of 8,000.

The correct answer is C.
Question 2 of 3 free Easy

In a certain pet shop, the ratio of dogs to cats to bunnies in stock is 3 : 5 : 7. If the shop carries 48 cats and bunnies total in stock, how many dogs are there?

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Correct answer: A

The ratio of dogs to cats to bunnies (Dogs: Cats: Bunnies) can be expressed as 3x : 5x : 7x. Here, x represents an "unknown multiplier." In order to solve the problem, we must determine the value of the unknown multiplier.

Cats + Bunnies = 48
5x + 7x = 48
12x = 48
\( x = 4 \)

Now that we know that the value of x (the unknown multiplier) is 4, we can determine the number of dogs.

Dogs = 3x = 3(4) = 12

The correct answer is A.
Question 3 of 3 free Medium

Which of the following fractions is at least twice as great as 11/50?

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Correct answer: E

First, let us rephrase the question. Since we need to find the fraction that is at least twice greater than \(\frac{11}{50}\), we are looking for a fraction that is equal to or greater than \(\frac{22}{50}\). Further, to facilitate our analysis, note that we can come up with an easy benchmark value for this fraction by doubling both the numerator and the denominator and thus expressing it as a percent: \(\frac{22}{50} = \frac{44}{100} = 44\%\). Thus, we can rephrase the question: "Which of the following is greater than or equal to \(44\%\)?"



Now, let's analyze each of the fractions in the answer choices using benchmark values:



\(\frac{2}{5}\): This fraction can be represented as \(40\%\), which is less than \(44\%\).



\(\frac{11}{34}\): This value is slightly less than \(\frac{11}{33}\) or \(\frac{1}{3}\). Therefore, it is smaller than \(44\%\).



\(\frac{43}{99}\): Note that the fraction \(\frac{43}{99}\) is smaller than \(\frac{44}{100}\), since fractions get smaller if the same number (in this case integer \(1\)) is subtracted from both the numerator and the denominator.



\(\frac{8}{21}\): We know that \(\frac{8}{21}\) is a little less than \(\frac{8}{20}\) or \(\frac{2}{5}\). Thus, \(\frac{8}{21}\) is less than \(44\%\).



\(\frac{9}{20}\): Finally, note that by multiplying the numerator and the denominator by \(5\), we can represent this fraction as \(\frac{45}{100}\), thus concluding that this fraction is greater than \(44\%\).



The correct answer is E.
Question 4 of 5 preview Medium

At the beginning of the year, the ratio of juniors to seniors in high school X was 3 to 4. During the year, 10 juniors and twice as many seniors transferred to another high school, while no new students joined high school X. If, at the end of the year, the ratio of juniors to seniors was 4 to 5, how many seniors were there in high school X at the beginning of the year?

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Question 5 of 5 preview Hard

Bag A contains red, white and blue marbles such that the red to white marble ratio is 1:3 and the white to blue marble ratio is 2:3. Bag B contains red and white marbles in the ratio of 1:4. Together, the two bags contain 30 white marbles. How many red marbles could be in bag A?

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