JEE MainMathematics

Answer verified against the official NTA key

+4 marks−1 if incorrectSingle correctPrevious-year question

Probability over an Integer Sample Space Defined by a Quadratic Inequality

Q: Does a and b have to be distinct integers?

No, the stem states 'two positive integers', which allows a = b = 33 unless explicitly stated as distinct.

Let M be the maximum value of the product of two positive integers when their sum is 66. Let the sample space S = {x ∈ ℤ: x(66-x) ≥ 5/9M } and the event A = {x ∈ S: x is a multiple of 3}. Then P(A) is equal to

Your answer stays private

What feels right?

No score. Commit to your first instinct. We’ll show what your mind noticed and what it missed.

A7/22B1/5C15/44D1/3

Step-by-step solutionView

Correct answer

The sample space contains 45 consecutive integers from 11 to 55, exactly 15 of which are multiples of 3, giving a probability of 1/3.

Option analysis

Why each option works or fails

A · 7/22

Believing the sample space consists of all positive integers summing to 66 (from 1 to 65 or 1 to 66) rather than the restricted set defined by the inequality. Solve the quadratic inequality $x (66 - x) \geq \frac{5}{9} M$ first to determine the exact boundaries of the sample space $S$ before counting favorable outcomes.

B · 1/5

Miscalculating the roots of the quadratic equation as 22 and 44 instead of 11 and 55. Solve $x^{2} - 66 x + 605 = 0$ using the quadratic formula: $x = \frac{66 \pm \sqrt{4356 - 2420}}{2} = \frac{66 \pm 44}{2}$ , which yields 11 and 55.

C · 15/44

Miscounting the total number of integers in the interval $[11, 55]$ by computing $55 - 11 = 44$ instead of $55 - 11 + 1 = 45$ . When counting consecutive integers in an inclusive range $[a, b]$ , use the formula $b - a + 1$ .

D · 1/3

This is the correct option. The sample space is $S = {11, 12, \dots, 55}$ containing $45$ integers, and the multiples of 3 are ${12, 15, \dots, 54}$ totaling $15$ integers, so $P (A) = \frac{15}{45} = \frac{1}{3}$ .

Reviewed route

Solution

StepWorking

01Given

Two positive integers $a, b$ have sum $a + b = 66$ . $M$ is the maximum value of their product $a b$ . Sample space $S = {x \in Z : x (66 - x) \geq \frac{5}{9} M}$ , and event $A = {x \in S : x is a multiple of 3}$ .

02Goal

Find the probability $P (A) = \frac{n (A)}{n (S)}$ .

03Approach

Use the AM-GM inequality to maximize the product $a b$ subject to $a + b = 66$ , finding $M$ . Then solve the resulting quadratic inequality for integer values of $x$ to identify $S$ , and count multiples of 3 in $S$ to find $n (A)$ .

04Execute

By AM-GM, $\frac{a + b}{2} \geq \sqrt{a b} ⟹ \sqrt{a b} \leq 33 ⟹ a b \leq 33^{2} = 1089$ . Since $a, b$ can be $33, 33$ , the maximum is $M = 33^{2} = 1089$ .

05Execute

Substitute $M$ into the condition for $S$ : $x (66 - x) \geq \frac{5}{9} (33^{2}) = 5 \times 11^{2} = 605$ . This rearranges to $x^{2} - 66 x + 605 \leq 0$ . Factoring the quadratic: $(x - 11) (x - 55) \leq 0$ . Thus, $x \in [11, 55]$ . Since $x \in Z$ , $S = {11, 12, \dots, 55}$ , so $n (S) = 55 - 11 + 1 = 45$ .

06Execute

The elements of $A$ are the multiples of 3 in $[11, 55]$ . The smallest multiple is 12 and the largest is 54. The number of terms is $\frac{54 - 12}{3} + 1 = 15$ . Therefore, $P (A) = \frac{n (A)}{n (S)} = \frac{15}{45} = \frac{1}{3}$ .

✓Verify

In any set of 45 consecutive integers, exactly $\frac{45}{3} = 15$ of them must be multiples of 3 because 45 is divisible by 3. Hence the ratio must be exactly $15 / 45 = 1 / 3$ .

Hints that build this answer step by step

What is the maximum value $M$ of the product of two positive integers whose sum is 66?
$M = 33 \times 33 = 1089$
What inequality defines the integers $x$ belonging to the sample space $S$ ?
$x^{2} - 66 x + 605 \leq 0$
What integer range corresponds to the solution of $x^{2} - 66 x + 605 \leq 0$ ?
$11 \leq x \leq 55$
How many integers are in $S$ , and how many of them are multiples of 3?
$∣ S ∣ = 45$ and $∣ A ∣ = 15$

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 8 September 2026

Quick checks

Students also ask

Does

a

and

b

have to be distinct integers?

No, the stem states 'two positive integers', which allows $a = b = 33$ unless explicitly stated as distinct.

Answer

The sample space contains 45 consecutive integers from 11 to 55, exactly 15 of which are multiples of 3, giving a probability of 1/3.

Why each option works or fails

A: 7/22 - Believing the sample space consists of all positive integers summing to 66 (from 1 to 65 or 1 to 66) rather than the restricted set defined by the inequality. Solve the quadratic inequality $x (66 - x) \geq \frac{5}{9} M$ first to determine the exact boundaries of the sample space $S$ before counting favorable outcomes.
B: 1/5 - Miscalculating the roots of the quadratic equation as 22 and 44 instead of 11 and 55. Solve $x^{2} - 66 x + 605 = 0$ using the quadratic formula: $x = \frac{66 \pm \sqrt{4356 - 2420}}{2} = \frac{66 \pm 44}{2}$ , which yields 11 and 55.
C: 15/44 - Miscounting the total number of integers in the interval $[11, 55]$ by computing $55 - 11 = 44$ instead of $55 - 11 + 1 = 45$ . When counting consecutive integers in an inclusive range $[a, b]$ , use the formula $b - a + 1$ .
D · correct: 1/3 - This is the correct option. The sample space is $S = {11, 12, \dots, 55}$ containing $45$ integers, and the multiples of 3 are ${12, 15, \dots, 54}$ totaling $15$ integers, so $P (A) = \frac{15}{45} = \frac{1}{3}$ .

Step-by-step solution

Given: Two positive integers $a, b$ have sum $a + b = 66$ . $M$ is the maximum value of their product $a b$ . Sample space $S = {x \in Z : x (66 - x) \geq \frac{5}{9} M}$ , and event $A = {x \in S : x is a multiple of 3}$ .
Goal: Find the probability $P (A) = \frac{n (A)}{n (S)}$ .
Approach: Use the AM-GM inequality to maximize the product $a b$ subject to $a + b = 66$ , finding $M$ . Then solve the resulting quadratic inequality for integer values of $x$ to identify $S$ , and count multiples of 3 in $S$ to find $n (A)$ .
Execute: By AM-GM, $\frac{a + b}{2} \geq \sqrt{a b} ⟹ \sqrt{a b} \leq 33 ⟹ a b \leq 33^{2} = 1089$ . Since $a, b$ can be $33, 33$ , the maximum is $M = 33^{2} = 1089$ .
Execute: Substitute $M$ into the condition for $S$ : $x (66 - x) \geq \frac{5}{9} (33^{2}) = 5 \times 11^{2} = 605$ . This rearranges to $x^{2} - 66 x + 605 \leq 0$ . Factoring the quadratic: $(x - 11) (x - 55) \leq 0$ . Thus, $x \in [11, 55]$ . Since $x \in Z$ , $S = {11, 12, \dots, 55}$ , so $n (S) = 55 - 11 + 1 = 45$ .
Execute: The elements of $A$ are the multiples of 3 in $[11, 55]$ . The smallest multiple is 12 and the largest is 54. The number of terms is $\frac{54 - 12}{3} + 1 = 15$ . Therefore, $P (A) = \frac{n (A)}{n (S)} = \frac{15}{45} = \frac{1}{3}$ .
Verify: In any set of 45 consecutive integers, exactly $\frac{45}{3} = 15$ of them must be multiples of 3 because 45 is divisible by 3. Hence the ratio must be exactly $15 / 45 = 1 / 3$ .

Probability over an Integer Sample Space Defined by a Quadratic Inequality

What feels right?

The sample space contains 45 consecutive integers from 11 to 55, exactly 15 of which are multiples of 3, giving a probability of 1/3.

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

05Execute

06Execute

✓Verify

We think you should solve this next.

25% of the population are smokers. A smoker has 27 times more chances to develop lung can…

The mean and variance of $7$ observations are $8$ and $16$ respectively. If one observati…

The same trap, in a different chapter — Sequences and Series, 2023

Students also ask

Probability over an Integer Sample Space Defined by a Quadratic Inequality

The sample space contains 45 consecutive integers from 11 to 55, exactly 15 of which are multiples of 3, giving a probability of 1/3.

01Given

02Goal

03Approach

04Execute

05Execute

06Execute

✓Verify

25% of the population are smokers. A smoker has 27 times more chances to develop lung can…

The mean and variance of 77 observations are 88 and 1616 respectively. If one observati…

The same trap, in a different chapter — Sequences and Series, 2023

Students also ask

The mean and variance of $7$ observations are $8$ and $16$ respectively. If one observati…