Ratios, Percentages, Averages and Population Previous Year Questions (PYQs) for CAT: 22+ Solved Questions with Step-by-Step Solutions

Solve 22+ Ratios, Percentages, Averages and Population previous year questions for CAT with answers and detailed solutions. Free sample questions below.

Chapter Roadmap: Ratios, Percentages, Averages and Population

Chapter Roadmap

Ratios → Percentages → Averages → Classification

A clean journey from comparison to CAT-level arithmetic modelling.

⚖️

1. Ratios, Proportions and Sharing

10 PYQs • Medium-high

You master part-to-part comparison, shifting, sharing, age ratios, salary ratios, and mixture-style split logic.

📈

2. Percentage Change and Population Growth

5 PYQs • Moderate

You learn increase, decrease, successive change, and population-style growth logic.

🎯

3. Averages, Means and Weighted Averages

18 PYQs • Highest

You convert averages into totals and use weighted average thinking for groups and constraints.

🧩

4. Counting with Percentages and Classification

2 PYQs • Support

You combine category counts with percentages, ratios, and table-based thinking.

End goal: convert any arithmetic story into clean variables, ratios, totals, and equations without getting trapped by words.

Topic Hero: Ratios Are Comparison Machines

Topic Hero

Ratios, Proportions and Sharing

The art of converting comparison stories into equations.

3 : 2

Means the quantities are $3 x$ and $2 x$ , not necessarily 3 and 2.

x

The multiplier is found from extra information: total, difference, shift, or final ratio.

CAT skill: Do not calculate early. Represent first. Solve only after the relation becomes clear.

Ratios, Percentages, Averages and Population: Solved Questions with Step-by-Step Explanations (5 Problems)

Question 1 · Quantitative Ability NAT

In an examination, the average marks of students in sections A and B are 32 and 60, respectively. The number of students in section A is 10 less than that in section B. If the average marks of all the students across both the sections combined is an integer, then the difference between the maximum and minimum possible number of students in section A is

Correct Answer:

Step-by-Step Solution

Key idea: This is a combined average must be integer question, recognisable because two groups have different averages, different sizes linked by a fixed difference, and the combined average has an integrality constraint.

Why this method: The combined average is a weighted average. When we express it algebraically, the integrality condition becomes a divisibility condition, which we solve by finding divisors.

Step-by-step working:

Define variables:

Let students in section A = $a$
Students in section B = $a + 10$ (since A has 10 fewer than B)

Write the combined average:

Total marks = $32 a + 60 (a + 10) = 92 a + 600$
Total students = $a + (a + 10) = 2 a + 10$
Combined average = $\frac{92 a + 600}{2 a + 10}$

Simplify using polynomial division:

$92 a + 600 = 46 (2 a + 10) + 140$
So average = $46 + \frac{140}{2 a + 10} = 46 + \frac{70}{a + 5}$

Apply the integer constraint:

For the average to be an integer, $(a + 5)$ must divide $70$
Divisors of $70$ : $1, 2, 5, 7, 10, 14, 35, 70$
So $a + 5 \in {1, 2, 5, 7, 10, 14, 35, 70}$
Giving $a \in {- 4, - 3, 0, 2, 5, 9, 30, 65}$

Filter for valid values:

Since $a$ represents students in a section, $a \geq 1$
Valid values: $a \in {2, 5, 9, 30, 65}$

Verify the average stays between 32 and 60:

$a = 2$ : average = $46 + 70/7 = 56$ ✓
$a = 65$ : average = $46 + 70/70 = 47$ ✓
All intermediate values also give averages between 32 and 60.

Calculate the answer:

Maximum $a = 65$ , Minimum $a = 2$
Difference = $65 - 2 = 63$

Answer: The difference between maximum and minimum possible students in section A is 63.

Question 2 · Quantitative Ability NAT

The average of three distinct real numbers is 28. If the smallest number is increased by 7 and the largest number is reduced by 10, the order of the numbers remains unchanged, and the new arithmetic mean becomes 2 more than the middle number, while the difference between the largest and the smallest numbers becomes 64. Then, the largest number in the original set of three numbers is

Correct Answer:

Step-by-Step Solution

This is a "Pattern 4: Average After Changing Extreme Values" question. You recognise it because the average is given for three unknowns, then only the smallest and the largest are changed while the middle stays fixed — that is the exact trigger for this pattern.

Step 1: Convert average to total.

Let the three distinct numbers be $a < b < c$ (smallest, middle, largest).

Average is 28, so:

$a + b + c = 3 \times 28 = 84$

Step 2: Write the new numbers and new total.

Smallest increased by 7: $a + 7$ . Largest reduced by 10: $c - 10$ . Middle $b$ is untouched.

New sum $= (a + 7) + b + (c - 10) = (a + b + c) - 3 = 84 - 3 = 81$

New mean $= \frac{81}{3} = 27$

Step 3: Use "new mean is 2 more than the middle number."

$27 = b + 2 ⟹ b = 25$

Step 4: Use the sum to get $a + c$ .

Since $a + b + c = 84$ and $b = 25$ :

$a + c = 84 - 25 = 59$

Step 5: Use the new difference condition.

"Difference between largest and smallest becomes 64" refers to the new numbers:

$(c - 10) - (a + 7) = 64 ⟹ c - a - 17 = 64 ⟹ c - a = 81$

Step 6: Solve the pair of equations.

$a + c = 59, c - a = 81$

Adding: $2 c = 140 ⟹ c = 70$

Then $a = 59 - 70 = - 11$

Step 7: Verify order is preserved (the question insists on this).

Original: $- 11 < 25 < 70$ ✓

New: $a + 7 = - 4, b = 25, c - 10 = 60$ , so $- 4 < 25 < 60$ ✓ — order unchanged, exactly as stated.

Check the total: $- 11 + 25 + 70 = 84$ , average $= 28$ ✓. New mean $= (- 4 + 25 + 60) /3 = 81/3 = 27 = 25 + 2$ ✓. New difference $= 60 - (- 4) = 64$ ✓.

Answer: the largest number in the original set is 70.

Common trap: students apply the "increase by 7 / decrease by 10" change to the wrong variable, or use the new difference condition on the original numbers instead of the new ones. Reading "the new arithmetic mean becomes..." and "the difference... becomes 64" tells you both conditions apply to the changed numbers, not the original ones.

Question 3 · Quantitative Ability NAT

The arithmetic mean of scores of 25 students in an examination is 50. Five of these students top the examination with the same score. If the scores of the other students are distinct integers with the lowest being 30, then the maximum possible score of the toppers is

Correct Answer:

Step-by-Step Solution

Pattern recognition: This is a "maximize an extreme value subject to a fixed total and distinctness constraints" question. The trigger is: a fixed average (hence a fixed total), a group of toppers who share one score, and the rest of the group having distinct integer scores with a known minimum — and we're asked to maximize the topper's score.

Why this method: To make the toppers' score as large as possible, we must make the total left for everyone else as small as possible, since the total marks of all 25 students is fixed.

Step 1 — Find the total marks.

Average = 50 over 25 students, so:

$Total = 50 \times 25 = 1250$

Step 2 — Identify the two groups.

5 toppers share one common score (call it $T$ ). The other 20 students have distinct integer scores, with the lowest being 30.

Step 3 — Minimize the "other 20" total.

To maximize $T$ , we must minimize the sum of the other 20 scores. Since they must be distinct integers with the lowest being 30, the smallest possible sum uses 20 consecutive integers starting at 30:

$30, 31, 32, \dots, 49$

Sum of these 20 numbers:

$Sum = \frac{( 30 + 49 ) \times 20}{2} = \frac{79 \times 20}{2} = 790$

Step 4 — Find the toppers' total and individual score.

$Toppers’ total = 1250 - 790 = 460$

$T = \frac{460}{5} = 92$

Step 5 — Sanity check.

The topper score 92 is indeed higher than the highest of the other 20 scores (49), so it's consistent with them being the "toppers." All 20 other scores are distinct integers ≥ 30. Everything checks out.

Answer: 92.

Common trap: Students sometimes try to minimize the average of the other 20 instead of the sum, or forget that using consecutive integers starting exactly at the minimum (30) is what minimizes the sum — using non-consecutive numbers or a higher starting point wastes total and gives a lower, wrong answer for $T$ .

Question 4 · Quantitative Ability MCQ

Consider six distinct natural numbers such that the average of the two smallest numbers is 14, and the average of the two largest numbers is 28. Then, the maximum possible value of the average of these six numbers is

A.
23
B.
24
C.
23.5
D.
22.5

Correct Answer:

Step-by-Step Solution

Key idea: this is an ordered distinct-numbers average maximisation question. The trigger is that six numbers are distinct natural numbers, the two smallest have a fixed average, the two largest have a fixed average, and we need the maximum possible overall average.

Step 1: Sort the numbers.

Let the six distinct natural numbers be:

$a_{1} < a_{2} < a_{3} < a_{4} < a_{5} < a_{6}$ .

Step 2: Use the given pair averages.

Average of the two smallest numbers is 14:

$\frac{a _{1} + a _{2}}{2} = 14$ ,

so $a_{1} + a_{2} = 28$ .

Average of the two largest numbers is 28:

$\frac{a _{5} + a _{6}}{2} = 28$ ,

so $a_{5} + a_{6} = 56$ .

Step 3: Understand what can be maximised.

The total sum is:

$(a_{1} + a_{2}) + (a_{3} + a_{4}) + (a_{5} + a_{6})$ .

The first and last pair sums are fixed:

$28 + (a_{3} + a_{4}) + 56$ .

To maximise the average, we must maximise $a_{3} + a_{4}$ .

Step 4: Choose the largest pair to allow the biggest middle numbers.

Since $a_{5}$ and $a_{6}$ are distinct natural numbers and $a_{5} < a_{6}$ with sum 56, the largest possible value of $a_{5}$ is 27.

If $a_{5} = 27$ , then $a_{6} = 29$ .

We cannot take $a_{5} = 28$ , because then $a_{6}$ would also have to be 28, violating distinctness.

Step 5: Maximise the middle numbers.

With $a_{5} = 27$ , the middle numbers must satisfy:

$a_{2} < a_{3} < a_{4} < 27$ .

The largest possible distinct natural values below 27 are:

$a_{3} = 25$ and $a_{4} = 26$ .

Step 6: Check that the smallest pair can coexist with this.

We need $a_{1} + a_{2} = 28$ , $a_{1} < a_{2}$ , and $a_{2} < 25$ so that $a_{3} = 25$ is allowed.

One valid choice is $a_{1} = 4$ and $a_{2} = 24$ .

Then the sequence begins $4, 24, 25, 26, 27, 29$ , which satisfies all conditions.

Step 7: Compute the maximum total and average.

Maximum total:

$28 + 25 + 26 + 56 = 135$ .

Maximum average:

$\frac{135}{6} = 22.5$ .

Answer: Option D, 22.5.

Question 5 · Quantitative Ability NAT

The average of a non-decreasing sequence of N numbers $a_{1}, a_{2}, ..., a_{N}$ is 300. If $a_{1}$ , is replaced by $6 a_{1}$ , the new average becomes 400. Then, the number of possible values of $a_{1}$ , is

Correct Answer:

Step-by-Step Solution

Key idea: this is a replacement-in-a-sequence question. The trigger is that one term is replaced and the average changes. The average change gives a total change equation, and the non-decreasing condition gives bounds.

Step 1: Write the original total.

The average of $N$ numbers is 300.

So the original sum is:

$300 N$ .

Step 2: Write the new total after replacing $a_{1}$ by $6 a_{1}$ .

The sum increases by:

$6 a_{1} - a_{1} = 5 a_{1}$ .

The new average is 400, so the new sum is:

$400 N$ .

Therefore:

$300 N + 5 a_{1} = 400 N$ .

So:

$5 a_{1} = 100 N$ ,

$a_{1} = 20 N$ .

Step 3: Use the non-decreasing condition.

Since the sequence is non-decreasing, every term is at least $a_{1}$ .

Therefore the average must be at least $a_{1}$ .

The average is 300, so:

$a_{1} \leq 300$ .

Substitute $a_{1} = 20 N$ :

$20 N \leq 300$ ,

$N \leq 15$ .

Step 4: Check which values of $N$ are actually possible.

$N$ must be a positive integer.

But $N = 1$ is impossible. If $N = 1$ , the only number must itself have average 300, so $a_{1} = 300$ . But the replacement equation gives $a_{1} = 20$ , a contradiction.

For every $N$ from 2 to 15, a valid non-decreasing sequence can be formed.

One construction is:

take $a_{1} = a_{2} = \dots = a_{N - 1} = 20 N$ ,

and choose the last term to make the total $300 N$ .

The last term becomes:

$300 N - 20 N (N - 1) = 20 N (16 - N)$ .

For $N \leq 15$ , this last term is at least $20 N$ , so the sequence remains non-decreasing.

Step 5: Count possible values.

Valid $N$ values are $2, 3, \dots, 15$ .

Number of values = $15 - 2 + 1 = 14$ .

Since $a_{1} = 20 N$ , each valid $N$ gives a different $a_{1}$ .

Answer: 14.

Ratios, Percentages, Averages and Population Previous Year Questions (PYQs) for CAT: 22+ Solved Questions with Step-by-Step Solutions

Chapter Roadmap: Ratios, Percentages, Averages and Population

Ratios → Percentages → Averages → Classification

1. Ratios, Proportions and Sharing

2. Percentage Change and Population Growth

3. Averages, Means and Weighted Averages

4. Counting with Percentages and Classification

Topic Hero: Ratios Are Comparison Machines

Ratios, Proportions and Sharing

Ratios, Percentages, Averages and Population: Solved Questions with Step-by-Step Explanations (5 Problems)

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

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