Arithmetic Word Problems and Quantity Distribution Notes for CAT: Concepts, Formulas, Worked Examples & Practice

Arithmetic Word Problems and Quantity Distribution notes for CAT: 60 study cards covering concepts, formulas, shortcuts and exam traps, plus solved practice questions.

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”.

The Real Topic: Bookkeeping, Not Calculation

Think in buckets

Money distribution questions become easy when every person or cheque type gets its own bucket.

👤

People

Pinu, Meena, Rinu, Seema

₹

Amounts

shares, fixed payments, remaining money

🧾

Counts

number of cheques or people

Core habit: before solving, name the buckets and write what each bucket receives.

Total = Sum of All Parts

The master equation

Total amount = part_{1} + part_{2} + part_{3} + \dots

For four people:

T = P + M + R + S

This is especially useful when one amount is not directly given. You find it as the leftover.

Arithmetic Word Problems and Quantity Distribution: Solved Questions with Step-by-Step Explanations (2 Problems)

Question 1 · Quantitative Ability NAT

In a school, $120$ students each choose exactly one of Science, Arts or Commerce. The fees are ₹ $1400$ , ₹ $1000$ and ₹ $700$ respectively. The total fee collected is ₹ $119000$ . If the number of Science students is fewer than the number of Arts students, what is the maximum possible number of Science students?

Correct Answer:

Step-by-Step Solution

Key idea: this is a fee count-value maximization question. The triggers are total students, total fee, and a comparison condition asking for a maximum.

Step 1: Let $S$ , $A$ and $C$ be the numbers of Science, Arts and Commerce students.

Step 2: Write the count equation.

$S + A + C = 120$

Step 3: Write the value equation.

$1400 S + 1000 A + 700 C = 119000$

Step 4: Eliminate $C$ using $C = 120 - S - A$ .

$1400 S + 1000 A + 700 (120 - S - A) = 119000$

$84000 + 700 S + 300 A = 119000$

$700 S + 300 A = 35000$

Divide by $100$ :

$7 S + 3 A = 350$

Step 5: Express $A$ in terms of $S$ .

$A = \frac{350 - 7 S}{3}$

For $A$ to be an integer, $350 - 7 S$ must be divisible by $3$ .

Since $350 \equiv 2 (mod 3)$ and $7 \equiv 1 (mod 3)$ , we need

$2 - S \equiv 0 (mod 3)$

$S \equiv 2 (mod 3)$

Step 6: Apply the strict condition $S < A$ .

$S < \frac{350 - 7 S}{3}$

$3 S < 350 - 7 S$

$10 S < 350$

$S < 35$

Step 7: Find the largest integer less than $35$ that satisfies $S \equiv 2 (mod 3)$ .

The candidates below $35$ are $32, 29, 26, \dots$ , so the largest is $32$ .

Step 8: Check feasibility.

If $S = 32$ , then

$A = \frac{350 - 224}{3} = 42$

and

$C = 120 - 32 - 42 = 46$

All counts are non-negative, and $32 < 42$ .

Answer: $32$ .

Common trap: reading “fewer than” as “not more than” gives the false boundary value $S = 35$ , where $S = A$ . Another trap is choosing $34$ just because it is below $35$ , without checking that $A$ must be an integer.

Question 2 · Quantitative Ability NAT

A donation box contains only cheques of ₹100, ₹200 and ₹500. The box has exactly 50 cheques with a total value of ₹13,000. If the box contains at least one cheque of each denomination, what is the maximum possible number of ₹500 cheques?

Correct Answer:

Step-by-Step Solution

Key idea: This is a fixed-denomination collection question, recognisable because only three cheque values, a total count, and a total value are given. The added layer is the condition that each denomination appears at least once.

Step 1: Let x, y, z be the numbers of ₹100, ₹200 and ₹500 cheques.

Step 2: Write the count equation.

x + y + z = 50

Step 3: Write the value equation.

100x + 200y + 500z = 13000

Divide by 100:

x + 2y + 5z = 130

Step 4: Eliminate x using x = 50 - y - z.

50 - y - z + 2y + 5z = 130

y + 4z = 80

Step 5: Express y and x in terms of z.

y = 80 - 4z

x = 50 - y - z = 50 - (80 - 4z) - z = 3z - 30

Step 6: Apply the hidden condition: at least one cheque of each denomination.

So x >= 1, y >= 1, z >= 1.

From y >= 1:

80 - 4z >= 1

4z <= 79

z <= 19.75

Hence z <= 19.

From x >= 1:

3z - 30 >= 1

3z >= 31

z >= 31/3

Hence z >= 11.

Step 7: The maximum possible integer value is z = 19.

Check: z = 19 gives y = 80 - 76 = 4 and x = 57 - 30 = 27.

Count: 27 + 4 + 19 = 50.

Value: 2700 + 800 + 9500 = 13000.

Answer: 19.

Common trap: If you ignore “at least one of each”, z = 20 looks possible because y = 0. But y = 0 violates the condition.

More notes in this unit

chapter

Arithmetic Word Problems and Quantity Distribution Notes for CAT: Concepts, Formulas, Worked Examples & Practice

Arithmetic Word Problems and Quantity Distribution notes for CAT: 60 study cards covering concepts, formulas, shortcuts and exam traps, plus solved practice q

A question from this chapter

Question 1

In a school, 120 students each choose exactly one of Science, Arts or Commerce. The fees are ₹ 1400, ₹ 1000 and ₹ 700 respectively. The total fee collected is ₹ 119000 . If the number of Science students is fewer than the number of Arts students, what is the maximum possible number of Science students?

Question 2

A donation box contains only cheques of ₹100, ₹200 and ₹500. The box has exactly 50 cheques with a total value of ₹13,000. If the box contains at least one cheque of each denomination, what is the maximum possible number of ₹500 cheques?

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