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

Answer: (B)

$M=33 \times 33$
$ x(66-x) \geq \frac{5}{9} \times 33 \times 33 $
$ 11 \leq x \leq 55$
$ A:\{12,15,18, \ldots 54\} $
$ P(A)=\frac{15}{45}=\frac{1}{3}$