Arithmetic Word Problems and Quantity Distribution Previous Year Questions (PYQs) for CAT: 7+ Solved Questions with Step-by-Step Solutions

Solve 7+ Arithmetic Word Problems and Quantity Distribution previous year questions for CAT with answers and detailed solutions. Free sample questions below.

Chapter Roadmap: Arithmetic Word Problems and Quantity Distribution

CAT Quant • Arithmetic

Arithmetic Word Problems and Quantity Distribution

A chapter about distributing, selling, collecting, and splitting quantities without losing track.

chapter PYQs

💰

t1 — Money Distribution and Collections

Sequential shares, equal division changes, and cheque-count constraints.

Mastery: turn money stories into exact equations and integer cases.

3 PYQs

Selected

📦

t2 — Inventory, Sales and Remaining Quantities

Track what is sold, what remains, and how ratios change after selling.

4 PYQs

Highest here

🧮

t3 — Linear Constraints in Fees, Stocks and Scores

Convert multiple conditions into linear equations and inequalities.

3 PYQs

Moderate

💎

t4 — Proportional Value and Splitting

Handle values that change with proportional rules while splitting quantities.

1 PYQ

Selective

End goal: read a word problem and immediately decide: “Should I track shares, remaining amount, total count, or integer cases?”

Topic Hero: Money Distribution and Collections

Arithmetic → Quantity Distribution → t1 3 CAT PYQs

Money Distribution and Collections

The skill of converting money stories into clean equations and integer possibilities.

Sequential sharing

Equal division changes

Cheque collections

Integer constraints

One-line hook: In these questions, the answer is hidden in the phrase “of the total”, “of the remaining”, or “maximum possible”.

Arithmetic Word Problems and Quantity Distribution: Solved Questions with Step-by-Step Explanations (5 Problems)

Question 1 · Quantitative Ability NAT

In an examination, there were 75 questions. 3 marks were awarded for each correct answer, 1 mark was deducted for each wrong answer and 1 mark was awarded for each unattempted question. Rayan scored a total of 97 marks in the examination. If the number of unattempted questions was higher than the number of attempted questions, then the maximum number of correct answers that Rayan could have given in the examination is

Correct Answer:

Step-by-Step Solution

Key idea: this is a linear-constraint scoring question. The trigger is that three types of questions exist: correct, wrong, and unattempted, each with a different mark. We need a count equation and a value equation.

Step 1: Define variables.

Let $C$ be correct answers, $W$ wrong answers, and $U$ unattempted questions.

Step 2: Write the count equation.

Total questions = 75, so:

$C + W + U = 75$ .

Step 3: Write the score equation.

Correct gives +3, wrong gives -1, unattempted gives +1.

So:

$3 C - W + U = 97$ .

Step 4: Use the condition on unattempted questions.

Attempted questions = $C + W$ .

Unattempted is higher than attempted:

$U > C + W$ .

Since $C + W + U = 75$ , let attempted $A = C + W$ . Then $U = 75 - A$ .

The condition becomes:

$75 - A > A$

$75 > 2 A$

$A < 37.5$ .

Since $A$ is an integer, $A \leq 37$ .

Step 5: Eliminate $U$ from the score equation.

From the count equation, $U = 75 - C - W$ .

Substitute into the score equation:

$3 C - W + (75 - C - W) = 97$

$2 C - 2 W + 75 = 97$

$2 C - 2 W = 22$

$C - W = 11$ .

Step 6: Maximise $C$ under the attempted limit.

From $C - W = 11$ , we get $C = W + 11$ .

Attempted questions:

$C + W = (W + 11) + W = 2 W + 11$ .

Since attempted must be at most 37:

$2 W + 11 \leq 37$

$2 W \leq 26$

$W \leq 13$ .

To maximise $C$ , take the largest possible $W$ , which is 13.

Then $C = 13 + 11 = 24$ .

Step 7: Verify.

If $C = 24$ , $W = 13$ , then attempted = 37 and $U = 75 - 37 = 38$ .

Score = $3 (24) - 13 + 38 = 72 - 13 + 38 = 97$ .

Also, unattempted 38 is higher than attempted 37.

Answer: 24.

Question 2 · Quantitative Ability NAT

If a certain amount of money is divided equally among n persons, each one receives Rs 352. However, if two persons receive Rs 506 each and the remaining amount is divided equally among the other persons, each of them receive less than or equal to Rs 330. Then, the maximum possible value of n is

Correct Answer:

Step-by-Step Solution

This is an "Equal Division with Changed Payments" question, specifically the maximum-value variant. You recognise it because a total amount is first divided equally, then the same total is redistributed differently, and you're asked for the largest possible number of people under an inequality ("less than or equal to").

Step 1: Write the total using the original equal division.

If each of $n$ persons gets Rs 352, the total amount is:

$Total = 352 n$

Step 2: Account for the two special persons.

Two persons now receive Rs 506 each, using up:

$2 \times 506 = 1012$

The amount left for the remaining $(n - 2)$ persons is:

$352 n - 1012$

Step 3: Apply the inequality.

This remaining amount is shared equally among $(n - 2)$ persons, and each share is at most Rs 330:

$\frac{352 n - 1012}{n - 2} \leq 330$

Step 4: Solve the inequality for $n$ .

Since $n > 2$ , multiply both sides by $(n - 2)$ (positive, so inequality direction is unchanged):

$352 n - 1012 \leq 330 (n - 2)$

$352 n - 1012 \leq 330 n - 660$

$352 n - 330 n \leq 1012 - 660$

$22 n \leq 352$

$n \leq 16$

Step 5: Check that n = 16 actually works (division must be valid).

Total $= 352 \times 16 = 5632$ . Remaining $= 5632 - 1012 = 4620$ . Persons left $= 14$ .

Each remaining person gets $4620/14 = 330$ , which satisfies "less than or equal to 330" exactly.

Answer: the maximum possible value of $n$ is 16.

Common trap: flipping the inequality sign, or forgetting to check that the boundary value actually gives a valid (non-negative, sensible) share. Here $n = 16$ hits the boundary exactly at 330, confirming it is achievable and is indeed the maximum.

Question 3 · Quantitative Ability NAT

A donation box can receive only cheques of ₹100, ₹250, and ₹500. On one good day, the donation box was found to contain exactly 100 cheques amounting to a total sum of ₹15250. Then, the maximum possible number of cheques of ₹500 that the donation box may have contained, is

Correct Answer:

Step-by-Step Solution

Key idea: this is a fixed-denomination collection question. We know the total number of cheques and the total value, so we form two equations and then use the fact that cheque counts must be integers.

Step 1: Define variables.

Let $x$ be the number of ₹100 cheques, $y$ the number of ₹250 cheques, and $z$ the number of ₹500 cheques.

Step 2: Write the count equation.

There are 100 cheques in total:

$x + y + z = 100.$

Step 3: Write the value equation.

The total amount is ₹15250:

$100 x + 250 y + 500 z = 15250.$

Step 4: Simplify the value equation.

Divide by 50:

$2 x + 5 y + 10 z = 305.$

Step 5: Eliminate $x$ using $x = 100 - y - z$ .

Substitute:

$2 (100 - y - z) + 5 y + 10 z = 305.$

Expand:

$200 - 2 y - 2 z + 5 y + 10 z = 305.$

Combine like terms:

$200 + 3 y + 8 z = 305.$

So:

$3 y + 8 z = 105.$

Step 6: Maximise $z$ with integer constraints.

We need $y \geq 0$ and integer, so:

$3 y = 105 - 8 z .$

First, $105 - 8 z \geq 0$ , giving:

$z \leq 13.125.$

Second, $105 - 8 z$ must be divisible by 3. Since $105$ is divisible by 3 and $8 \equiv 2 (mod 3)$ , we need:

$2 z \equiv 0 (mod 3),$

so $z$ must be a multiple of 3.

Step 7: Choose the largest possible multiple of 3 below 13.125.

The largest such value is $z = 12$ .

Check:

$3 y = 105 - 8 (12) = 9,$

so $y = 3$ . Then $x = 100 - 3 - 12 = 85$ , which is valid.

Answer: 12.

Common trap: taking $z = 13$ just because $105/8 = 13.125$ . But $z = 13$ makes $y = 1/3$ , which is not an integer.

Question 4 · Quantitative Ability NAT

A fruit seller has a stock of mangoes, bananas and apples with at least one fruit of each type. At the beginning of a day, the number of mangoes make up 40% of his stock. That day, he sells half of the mangoes, 96 bananas and 40% of the apples. At the end of the day, he ends up selling 50% of the fruits. The smallest possible total number of fruits in the stock at the beginning of the day is

Correct Answer:

340

Step-by-Step Solution

Key idea: This is a smallest possible stock with divisibility constraints question, recognisable because percentages must produce whole fruits and we need to minimize the total subject to integer constraints.

Why this method: We express all quantities in terms of the total T, use the sales equation to link them, then find the smallest T satisfying all divisibility and positivity constraints.

Step-by-step working:

Define variables:

Total fruits = $T$
Mangoes = $0.4 T = \frac{2 T}{5}$
Bananas = $B$ , Apples = $A$
$B + A = 0.6 T$

Write the sales equation:

Sold: $\frac{1}{2} (0.4 T) + 96 + 0.4 A = 0.5 T$
$0.2 T + 96 + 0.4 A = 0.5 T$
$0.4 A = 0.3 T - 96$
$A = 0.75 T - 240 = \frac{3 T}{4} - 240$

Find B:

$B = 0.6 T - A = 0.6 T - 0.75 T + 240 = 240 - 0.15 T = 240 - \frac{3 T}{20}$

Apply integer constraints:

Mangoes = $\frac{2 T}{5}$ must be integer: $T$ divisible by 5
Apples = $\frac{3 T}{4} - 240$ must be integer: $T$ divisible by 4
Apples sold = $0.4 A = \frac{2 A}{5}$ must be integer: $A$ divisible by 5
Bananas = $240 - \frac{3 T}{20}$ must be integer: $T$ divisible by 20
Combined: $T$ must be divisible by 20

Apply positivity constraints:

$A \geq 1$ : $\frac{3 T}{4} \geq 241$ , so $T \geq 322$ (rounded up)
$B \geq 1$ : $240 - \frac{3 T}{20} \geq 1$ , so $T \leq 1593$
Mangoes $\geq 1$ : $T \geq 3$ (trivially satisfied)

Find smallest valid T:

$T \geq 322$ and $T$ divisible by 20
Smallest multiple of 20 that is $\geq 322$ is $340$

Verify T = 340:

Mangoes = 136, Apples = $255 - 240 = 15$ , Bananas = $240 - 51 = 189$
Total: $136 + 15 + 189 = 340$ ✓
Sold: $68 + 96 + 6 = 170 = 50%$ of 340 ✓
All quantities $\geq 1$ ✓

Answer: The smallest possible total number of fruits is 340.

Question 5 · Quantitative Ability NAT

In a school with 1500 students, each student chooses any one of the streams out of science, arts, and commerce, by paying a fee of Rs 1100, Rs 1000, and Rs 800, respectively. The total fee paid by all the students is Rs 15,50,000. If the number of science students is not more than the number of arts students, then the maximum possible number of science students in the school is

Correct Answer:

700

Step-by-Step Solution

Key idea: This is a linear constraint with inequality maximization question, recognisable because we have a count equation, a value equation, and an inequality constraint, and we need to maximize one variable.

Why this method: Two equations (count and fee total) reduce three unknowns to one free variable. The inequality constraint then bounds that variable.

Step-by-step working:

Define variables:

Let $S$ = science students, $A$ = arts students, $C$ = commerce students

Write the count equation:

$S + A + C = 1500$ ... (1)

Write the value equation:

$1100 S + 1000 A + 800 C = 1550000$ ... (2)

Eliminate C:

From (1): $C = 1500 - S - A$
Substitute into (2): $1100 S + 1000 A + 800 (1500 - S - A) = 1550000$
$300 S + 200 A + 1200000 = 1550000$
$300 S + 200 A = 350000$
$3 S + 2 A = 3500$ ... (3)

Express A in terms of S:

$A = \frac{3500 - 3 S}{2}$
For $A$ to be a non-negative integer: $S$ must be even, and $S \leq 1166$

Express C in terms of S:

$C = 1500 - S - \frac{3500 - 3 S}{2} = \frac{3000 - 2 S - 3500 + 3 S}{2} = \frac{S - 500}{2}$
For $C \geq 0$ : $S \geq 500$

Apply the inequality constraint:

$S \leq A = \frac{3500 - 3 S}{2}$
$2 S \leq 3500 - 3 S$
$5 S \leq 3500$
$S \leq 700$

Find maximum S:

$S$ must be even, $500 \leq S \leq 700$
Maximum $S = 700$

Verify:

$S = 700$ , $A = 700$ , $C = 100$ . Total = 1500 ✓
Fee: $770000 + 700000 + 80000 = 1550000$ ✓
$S \leq A$ : $700 \leq 700$ ✓

Answer: The maximum possible number of science students is 700.

Arithmetic Word Problems and Quantity Distribution Previous Year Questions (PYQs) for CAT: 7+ Solved Questions with Step-by-Step Solutions

Chapter Roadmap: Arithmetic Word Problems and Quantity Distribution

Arithmetic Word Problems and Quantity Distribution

Topic Hero: Money Distribution and Collections

Money Distribution and Collections

Arithmetic Word Problems and Quantity Distribution: 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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