Question

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

• A. $\frac{15}{44}$
• B. $\frac{1}{3}$
• C. $\frac{7}{22}$
• D. $\frac{1}{5}$

Answer 1

The Correct Option is B

To solve this problem, we need to determine the maximum value of the product of two positive integers whose sum is 66. We then identify the sample space and event to calculate the required probability.

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Step 1: Determine the Maximum Product

Let the two integers be \(x\) and \(66-x\). Their product is given by:

\(P(x) = x(66-x) = 66x - x^2\)

To find the maximum product, we take the derivative of \(P(x)\) and set it to zero:

\(\frac{d}{dx}(66x - x^2) = 66 - 2x = 0\)

Solving this gives:

\(2x = 66 \Rightarrow x = 33\)

Thus, \(M = P(33) = 33 \times 33 = 1089\).

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Step 2: Determine the Sample Space \(S\)

The condition for the sample space is:

\(x(66-x) \geq \frac{5}{9} \times 1089\)

Calculating the right-hand side:

\(\frac{5}{9} \times 1089 = 605\)

So, the inequality becomes:

\(x(66-x) \geq 605\)

Solve for \(x\):

This inequality is a quadratic equation:

\(66x - x^2 \geq 605\)

Rearrange to solve:

\(x^2 - 66x + 605 \leq 0\)

Using the quadratic formula or factoring, we find the roots (approximating):

\(x = 11 \text{ and } x = 55\)

Thus, \(x\) should lie in the closed interval [11, 55].

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Step 3: Identify Event \(A\)

The event \(A\) consists of all integers in \(S\) that are multiples of 3.

From 11 to 55, the multiples of 3 are: 12, 15, 18, ..., 54.

To count them, we note the sequence is arithmetic with first term 12 and last term 54:

The number of terms is:

\(\frac{54-12}{3} + 1 = 15\)

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Step 4: Calculate Probability \(P(A)\)

The total number of integers in \(S\) from 11 to 55 is:

\(55 - 11 + 1 = 45\)

The probability \(P(A) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{15}{45} = \frac{1}{3}\)

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Therefore, the correct answer is \(\frac{1}{3}\).