Does and have to be distinct integers?

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

JEE MainMathematics

Reviewed by official_key

+4 marks−1 if incorrectSingle correctpyq

Statistics and Probability: Mathematics | JEE Main

Let

M

be the maximum value of the product of two positive integers when their sum is 66. Let the sample space

S = \left\{x \in \mathbb{Z} : x(66-x) \ge \frac{5}{9}M \right\}

and the event

A = \{x \in S : x \text{ 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.

\frac{7}{22}

\frac{1}{5}

\frac{15}{44}

\frac{1}{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 ·

\frac{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 $x(66-x) \ge \frac{5}{9}M$ first to determine the exact boundaries of the sample space $S$ before counting favorable outcomes.

B ·

\frac{1}{5}

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

C ·

\frac{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 ·

\frac{1}{3}

None. This is the correct option. None. 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 $ab$ . Sample space $S = \{x \in \mathbb{Z} : x(66 - x) \ge \frac{5}{9}M\}$ , and event $A = \{x \in S : x \text{ is a multiple of } 3\}$ .

02goal

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

03approach

Use the AM-GM inequality to maximize the product $ab$ 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, $\frac{a+b}{2} \ge \sqrt{ab} \implies \sqrt{ab} \le 33 \implies ab \le 33^2 = 1089$ . Since $a, b$ can be $33, 33$ , the maximum is $M = 33^2 = 1089$ .

05execute

Substitute $M$ into the condition for $S$ : $x(66-x) \ge \frac{5}{9}(33^2) = 5 \times 11^2 = 605$ . This rearranges to $x^2 - 66x + 605 \le 0$ . Factoring the quadratic: $(x - 11)(x - 55) \le 0$ . Thus, $x \in [11, 55]$ . Since $x \in \mathbb{Z}$ , $S = \{11, 12, \dots, 55\}$ , 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 $S$ ?
$x^2 - 66x + 605 \le 0$
What integer range corresponds to the solution of $x^2 - 66x + 605 \le 0$ ?
$11 \le x \le 55$
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
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
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: $\frac{7}{22}$ - 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 $x(66-x) \ge \frac{5}{9}M$ first to determine the exact boundaries of the sample space $S$ before counting favorable outcomes.
B: $\frac{1}{5}$ - Miscalculating the roots of the quadratic equation as 22 and 44 instead of 11 and 55. Solve $x^2 - 66x + 605 = 0$ using the quadratic formula: $x = \frac{66 \pm \sqrt{4356 - 2420}}{2} = \frac{66 \pm 44}{2}$ , which yields 11 and 55.
C: $\frac{15}{44}$ - 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: $\frac{1}{3}$ - None. This is the correct option. None. 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 $ab$ . Sample space $S = \{x \in \mathbb{Z} : x(66 - x) \ge \frac{5}{9}M\}$ , and event $A = \{x \in S : x \text{ is a multiple of } 3\}$ .
goal: Find the probability $P(A) = \frac{n(A)}{n(S)}$ .
approach: Use the AM-GM inequality to maximize the product $ab$ 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, $\frac{a+b}{2} \ge \sqrt{ab} \implies \sqrt{ab} \le 33 \implies ab \le 33^2 = 1089$ . Since $a, b$ can be $33, 33$ , the maximum is $a, b$ .
execute: Substitute $M$ into the condition for $S$ : $x(66-x) \ge \frac{5}{9}(33^2) = 5 \times 11^2 = 605$ . This rearranges to $x^2 - 66x + 605 \le 0$ . Factoring the quadratic: $(x - 11)(x - 55) \le 0$ . Thus, $x \in [11, 55]$ . Since $x \in \mathbb{Z}$ , $S = \{11, 12, \dots, 55\}$ , 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 $15/45 = 1/3$ .

Statistics and Probability: Mathematics | JEE Main

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

Let $g: R \rightarrow R$ be a non constant twice differentiable such that…

Let $\mu$ be the mean and $\sigma$ be the standard deviation of the distribution…

Students also ask

Statistics and Probability: Mathematics | JEE Main

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

Let g:R→Rg: R \rightarrow Rg:R→R be a non constant twice differentiable such that…

Let μ\muμ be the mean and σ\sigmaσ be the standard deviation of the distribution…

Students also ask

Let $g: R \rightarrow R$ be a non constant twice differentiable such that…

Let $\mu$ be the mean and $\sigma$ be the standard deviation of the distribution…