Number Theory Word Problems and Diophantine Conditions Practice Questions for CAT: 17+ Solved Questions with Step-by-Step Solutions

Solve 17+ Number Theory Word Problems and Diophantine Conditions practice questions for CAT with answers and detailed solutions. Free sample questions below.

Chapter Roadmap: Number Theory Word Problems and Diophantine Conditions

🧩

CAT QA • Number System

Chapter Roadmap

Step 1 • Selected Topic • PYQ count: 1 • Importance: Moderate

① Diophantine Equations and Natural Number Constraints

You will learn how to solve equations where variables must be natural numbers, distinct, positive, or minimum/maximum satisfying conditions.

Mastery target: turn a scary equation like

3 a c = 8 (a + b)

into divisibility cases.

Step 2 • Context Topic • PYQ count: 1 • Importance: Moderate

② Ratio-Based Divisibility Word Problems

You will later use ratio, multiples, and divisibility conditions in word problems. This is nearby, but not taught in this selected topic.

Boundary note: today we focus only on Diophantine equations with natural-number restrictions.

Why this chapter exists: CAT sometimes hides number theory inside simple-looking equations. The trick is not heavy algebra; it is respecting integer conditions.

Diophantine Equations and Natural Number Constraints

Topic Hero

Diophantine Equations and Natural Number Constraints

The equation is only half the story. The real CAT move is: which integer values are actually allowed?

🎯

What you master

Convert equations into divisibility cases and find the smallest valid expression.

⚠️

Main trap

Treating $a, b, c$ like real numbers and forgetting natural-number restrictions.

CAT signal: words like natural numbers, distinct, integer, least possible, smallest value, number of solutions.

Number Theory Word Problems and Diophantine Conditions: Solved Questions with Step-by-Step Explanations (5 Problems)

Question 1 · Quantitative Ability MCQ

The daily outputs of two machines A and B are in the ratio $3 : 8$ , so they can be written as $3 k$ and $8 k$ for a natural number $k$ . The total output of A in 4 days is a multiple of 7, and the total output of B in 3 days is a multiple of 5. If $k$ and a natural number $t$ satisfy

$2 k + 3 t = 300,$

what is the maximum possible daily output of machine B?

A.
560
B.
840
C.
1120
D.
1400

Correct Answer:

Step-by-Step Solution

Key idea: this is a chapter-synthesis maximum problem. The ratio gives a hidden multiplier, the divisibility conditions restrict $k$ , and the extra Diophantine equation filters the remaining candidates.

Step 1: Write the daily outputs.

A produces $3 k$ per day and B produces $8 k$ per day.

Step 2: Use A's divisibility condition.

A's 4-day total is $12 k$ , and it is a multiple of 7.

Since $g cd (12, 7) = 1$ , $k$ must be a multiple of 7.

Step 3: Use B's divisibility condition.

B's 3-day total is $24 k$ , and it is a multiple of 5.

Since $g cd (24, 5) = 1$ , $k$ must be a multiple of 5.

Step 4: Combine the ratio-divisibility conditions.

$k$ must be a multiple of both 7 and 5, so $k$ is a multiple of

$lcm (7, 5) = 35.$

Step 5: Use the Diophantine equation.

$2 k + 3 t = 300$

Since $t$ is natural, $300 - 2 k$ must be a positive multiple of 3. Also $2 k < 300$ , so

$k < 150.$

Step 6: List multiples of 35 below 150.

$35, 70, 105, 140$

Now require $300 - 2 k$ to be divisible by 3.

Since $300 \equiv 0 (mod 3)$ , we need

$2 k \equiv 0 (mod 3) .$

Because 2 is invertible modulo 3, this means

$k \equiv 0 (mod 3) .$

Among $35, 70, 105, 140$ , only $105$ is divisible by 3.

Step 7: Confirm $t$ .

$t = \frac{300 - 2 ( 105 )}{3} = \frac{90}{3} = 30,$

which is natural.

Step 8: Compute B's daily output.

$8 k = 8 \cdot 105 = 840.$

Answer: B

Question 2 · Quantitative Ability MCQ

Two weekly collections are in the ratio $2 : 9$ . If $k$ is the common natural-number multiplier, which pair correctly represents the first collection and the second collection?

A.
$2 + k$ and $9 + k$
B.
$2 k$ and $9 k$
C.
$9 k$ and $2 k$
D.
$k /2$ and $k /9$

Correct Answer:

Step-by-Step Solution

Key idea: this is a ratio-to-multiplier representation question, recognisable because the problem gives a ratio and asks for the actual form of the quantities.

Step 1: The ratio $2 : 9$ gives the shape of the two quantities. The first quantity corresponds to $2$ , and the second corresponds to $9$ .

Step 2: A ratio does not mean the actual values are exactly $2$ and $9$ . It means both values are obtained by multiplying the ratio terms by the same common multiplier.

Step 3: Let that common multiplier be $k$ . Then the first collection is $2 k$ and the second collection is $9 k$ .

Step 4: Keep the order of the ratio. Since the first term is $2$ , the first quantity is $2 k$ , not $9 k$ .

Answer: $2 k$ and $9 k$ .

Question 3 · Quantitative Ability MSQ

The daily outputs of two machines P and Q are in the ratio $3 : 4$ , so they can be written as $3 k$ and $4 k$ for a natural number $k$ . The total output of P in 2 days is a multiple of 5, and the total output of Q in 2 days is a multiple of 16. Suppose $k$ and a natural number $m$ satisfy

$3 k + 4 m = 220.$

Which of the following statements must be true? Select all that apply.

A.
k is a multiple of 20
B.
m is a multiple of 5
C.
k can be 70
D.
The minimum possible value of m is 10

Correct Answer:

["A","B","D"]

Step-by-Step Solution

Key idea: this is a necessary-truth question. We must find all valid values of $k$ after combining the ratio-divisibility conditions with the Diophantine equation, then test each statement against every valid case.

Step 1: Write the daily outputs.

P produces $3 k$ per day and Q produces $4 k$ per day.

Step 2: Use P's condition.

P's 2-day total is $6 k$ , a multiple of 5.

Since $g cd (6, 5) = 1$ , $k$ must be a multiple of 5.

Step 3: Use Q's condition.

Q's 2-day total is $8 k$ , a multiple of 16.

$\frac{8 k}{16} = \frac{k}{2}$

For this to be an integer, $k$ must be even.

Step 4: Combine the ratio-divisibility conditions.

$k$ is a multiple of 5 and even, so $k$ is a multiple of 10.

Step 5: Use the Diophantine equation.

$3 k + 4 m = 220$

$m = \frac{220 - 3 k}{4}$

For $m$ to be a natural number, $220 - 3 k$ must be positive and divisible by 4. Positivity gives

$3 k < 220 \Rightarrow k < 73 \frac{1}{3} .$

Step 6: List multiples of 10 below $73 \frac{1}{3}$ .

$10, 20, 30, 40, 50, 60, 70$

Now impose divisibility by 4. Since $220$ is divisible by 4, we need

$3 k \equiv 0 (mod 4) .$

Because $g cd (3, 4) = 1$ , this means

$k \equiv 0 (mod 4) .$

Among the multiples of 10, the valid values are

$20, 40, 60.$

Step 7: Compute the corresponding $m$ values.

For $k = 20$ :

$m = \frac{220 - 60}{4} = 40.$

For $k = 40$ :

$m = \frac{220 - 120}{4} = 25.$

For $k = 60$ :

$m = \frac{220 - 180}{4} = 10.$

Step 8: Check the statements.

A: All valid $k$ values are $20, 40, 60$ , all multiples of 20. True.

B: The valid $m$ values are $40, 25, 10$ , all multiples of 5. True.

C: $k = 70$ is a multiple of 10, but it is not divisible by 4, so $m$ is not an integer. False.

D: The valid $m$ values are $40, 25, 10$ , so the minimum is 10. True.

Answer: A, B, D

Question 4 · Quantitative Ability MCQ

Two factories P and Q produce pens in a week in the ratio $5 : 6$ . In 2 weeks, P's total production is a multiple of 13. In 3 weeks, Q's total production is a multiple of 20. If the common natural-number multiplier in the weekly ratio is less than $500$ , how many values can it take?

A.
1
B.
2
C.
3
D.
4

Correct Answer:

Step-by-Step Solution

Key idea: this is the standard ratio-plus-multiple setup, but instead of asking for the minimum, it asks for the number of valid multipliers below a bound.

Step 1: Write weekly production using the multiplier.

Let P's weekly production be $5 k$ and Q's weekly production be $6 k$ , where $k$ is a natural number.

Step 2: Translate P's condition.

P's 2-week total is $2 \times 5 k = 10 k$ .

This is a multiple of 13, so $13$ divides $10 k$ .

Since $g cd (10, 13) = 1$ , $k$ must be a multiple of 13.

Step 3: Translate Q's condition.

Q's 3-week total is $3 \times 6 k = 18 k$ .

This is a multiple of 20, so $20$ divides $18 k$ .

Remove the common factor 2 from 18 and 20. We need $10$ to divide $9 k$ .

Since $g cd (9, 10) = 1$ , $k$ must be a multiple of 10.

Step 4: Combine the conditions.

$k$ must be a multiple of both 13 and 10. Thus $k$ must be a multiple of $lcm (13, 10) = 130$ .

Step 5: Count natural multiples below 500.

The positive multiples of 130 below 500 are $130, 260, 390$ .

There are 3 such values.

Answer: 3

Question 5 · Quantitative Ability NAT

If $p$ and $q$ are natural numbers satisfying $5 p + 8 q = 200$ , what is the minimum possible value of $p + q$ ?

Correct Answer:

Step-by-Step Solution

Key idea: this is a natural-number Diophantine minimization question. The equation does not have one solution; it has several valid natural-number pairs, and we must compare $p + q$ across them.

Step 1: Isolate $p$ .

$5 p = 200 - 8 q \Rightarrow p = \frac{200 - 8 q}{5}$

Step 2: Use divisibility by 5.

For $p$ to be an integer, $200 - 8 q$ must be divisible by 5. Since $200$ is divisible by 5, we need $8 q$ divisible by 5. Because $8$ and $5$ are coprime, $q$ must be a multiple of 5.

Step 3: Use positivity of $p$ .

We need $200 - 8 q > 0$ , so $q < 25$ . The positive multiples of 5 below 25 are $q = 5, 10, 15, 20$ .

Step 4: Compute the valid pairs and sums.

If $q = 5$ , $p = (200 - 40) /5 = 32$ , sum $= 37$ .

If $q = 10$ , $p = (200 - 80) /5 = 24$ , sum $= 34$ .

If $q = 15$ , $p = (200 - 120) /5 = 16$ , sum $= 31$ .

If $q = 20$ , $p = (200 - 160) /5 = 8$ , sum $= 28$ .

Step 5: Check the boundary.

$q = 25$ would give $p = 0$ , but natural numbers start from 1, so $q = 25$ is not allowed.

Answer: 28

Number Theory Word Problems and Diophantine Conditions Practice Questions for CAT: 17+ Solved Questions with Step-by-Step Solutions

Chapter Roadmap: Number Theory Word Problems and Diophantine Conditions

Chapter Roadmap

① Diophantine Equations and Natural Number Constraints

② Ratio-Based Divisibility Word Problems

Diophantine Equations and Natural Number Constraints

Diophantine Equations and Natural Number Constraints

Number Theory Word Problems and Diophantine Conditions: 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

More practice questions in this unit

A question from this chapter

Login to view the complete practice questions and solutions

Personalised first. High quality throughout.

Built around you, not around a syllabus PDF

Revision that hits your weak spots

Questions calibrated to the real exam

Notes written for recall, not for volume

One place for everything

Honest progress

Unlock the whole course