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GMAT Ratio and Fraction Practice Questions

Ratios, fractions and decimals are the quiet workhorses of GMAT Focus Quant. They appear on their own, inside word problems and mixtures, and throughout Data Insights, where a ratio is often the only thing a table lets you compute.

Treat a ratio as parts of a whole. If boys and girls are in the ratio 3 : 5, write boys = 3k and girls = 5k. The total is 8k, so boys are 3/8 of the group, and the total must be a multiple of 8. That multiplier k is how you connect a ratio to real counts and how you handle a change: when 6 boys join, the new ratio is (3k + 6) : 5k, one equation with one unknown.

When two ratios share a term, such as a : b = 2 : 3 and b : c = 4 : 5, scale them so the shared term matches (8 : 12 : 15) before you combine them. For fractions, compare with cross-multiplication or benchmarks such as 1/2, and turn awkward decimals into fractions early.

The traps are part-to-part versus part-to-whole, and ratios that tell you nothing about size. A ratio alone never gives an actual count, which is a Data Sufficiency favourite: knowing that the ratio is 3 : 5 is not enough to find how many boys there are.

4 practice questions

Problem SolvingFractions/Ratios/DecimalsEasy

Question 1

In a chess club, the ratio of juniors to seniors is $5:3$. After $12$ new seniors join and no one leaves, the ratio of juniors to seniors becomes $5:4$. How many seniors are in the club after the new members join?

  1. A$48$
  2. B$52$
  3. C$60$
  4. D$64$
  5. E$72$
Show answer & explanation

Correct answer: A

Key Concept or Approach

Use a multiplier $k$: juniors $=5k$, seniors $=3k$. The number of juniors does not change.

Step-by-Step Solution

  • Step 1: Set up $\frac{5k}{3k+12}=\frac{5}{4}$.
  • Step 2: Cross-multiply: $20k = 15k + 60$, so $k = 12$.
  • Step 3: Seniors after joining: $3(12)+12 = 48$ (juniors $=60$, and $60:48 = 5:4$).

Final Answer

Answer: A

Common trap: 60 - that is the number of juniors, not seniors; 36 is the seniors count before the new members.

Data SufficiencyFractions/Ratios/DecimalsMedium

Question 2

A box contains only red marbles and blue marbles. How many red marbles are in the box?

(1) If 5 red marbles were added to the box, the numbers of red and blue marbles would be equal.

(2) Blue marbles make up 60% of the marbles in the box.

  1. AStatement (1) ALONE is sufficient, but statement (2) ALONE is NOT sufficient.
  2. BStatement (2) ALONE is sufficient, but statement (1) ALONE is NOT sufficient.
  3. CBOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
  4. DEACH statement ALONE is sufficient.
  5. EStatements (1) and (2) TOGETHER are NOT sufficient.
Show answer & explanation

Correct answer: C

Key Concept or Approach

Let $r$ = red and $b$ = blue. Each statement gives one relation between $r$ and $b$; two independent relations are needed.

Step-by-Step Solution

  • Step 1: Statement (1): $r + 5 = b$. $(r, b) = (1, 6)$ and $(10, 15)$ both work. Not sufficient.
  • Step 2: Statement (2): $b = 0.6(r + b)$, so $0.4b = 0.6r$ and $b = 1.5r$. $(r, b) = (2, 3)$ and $(10, 15)$ both work. Not sufficient.
  • Step 3: Together: $r + 5 = 1.5r$, so $0.5r = 5$ and $r = 10$. Sufficient.

Final Answer

Answer: C

Common trap: B: students treat the 60% as enough on its own, confusing a ratio with an actual count.

Problem SolvingSequencesFractions/Ratios/DecimalsHard

Question 3

For each positive integer $n$, $a_n = \frac{1}{n(n+1)}$. What is the least positive integer $k$ for which $a_1 + a_2 + \cdots + a_k > 0.98$?

  1. A$50$
  2. B$51$
  3. C$98$
  4. D$99$
  5. E$100$
Show answer & explanation

Correct answer: A

Key Concept or Approach

Split each term: $\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$. The sum telescopes.

Step-by-Step Solution

  • Step 1: $a_1 + \cdots + a_k = (1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + \cdots + (\frac{1}{k} - \frac{1}{k+1}) = 1 - \frac{1}{k+1}$.
  • Step 2: Need $1 - \frac{1}{k+1} > 0.98$, i.e. $\frac{1}{k+1} < 0.02 = \frac{1}{50}$, so $k + 1 > 50$.
  • Step 3: Least integer: $k = 50$ (at $k = 49$ the sum is exactly $0.98$, not greater).

Final Answer

Answer: A

Common trap: 51 - writing the partial sum as 1 - 1/k (an off-by-one in the telescoping), which pushes the answer to 51.

Problem SolvingFractions/Ratios/DecimalsSequencesHard

Question 4

What is the sum of the first $50$ digits after the decimal point in the decimal expansion of $\frac{3}{7}$?

  1. A$216$
  2. B$220$
  3. C$222$
  4. D$225$
  5. E$230$
Show answer & explanation

Correct answer: C

Key Concept or Approach

The decimal for $\frac{3}{7}$ repeats in a block of 6 digits. Count full blocks, then add the leftover digits.

Step-by-Step Solution

  • Step 1: Long division: $\frac{3}{7} = 0.428571428571\ldots$, repeating block $428571$ with digit sum $4 + 2 + 8 + 5 + 7 + 1 = 27$.
  • Step 2: $50 = 6 \cdot 8 + 2$: eight full blocks give $8 \cdot 27 = 216$.
  • Step 3: The next 2 digits are $4$ and $2$: total $216 + 6 = 222$.

Final Answer

Answer: C

Common trap: 225 - multiplying the block sum 27 by 50/6 (about 8.33) instead of adding the actual leftover digits 4 and 2.

More Quantitative Reasoning practice

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