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 9 M } and the event A = { x S : x is a multiple of 3 } . Then P ( A ) is equal to [2023]

Question

Let $M$ be the maximum value of the product of two positive integers when their sum is 66. Let the sample space $S = {x \in Z : x (66 - x) \geq \frac{5}{9} M}$ and the event $A = {x \in S : x is a multiple of 3} .$ Then $P (A)$ is equal to [2023]

Smart Achievers · Accepted Answer

(2)

$M = 33^{2} (Using A.M. and G.M. inequality)$

$x (66 - x) \geq \frac{5}{9} \times 33^{2}$

$\Rightarrow x^{2} - 66 x + \frac{5}{9} \times 33^{2} \leq 0 \Rightarrow x^{2} - 66 x + 605 \leq 0$ ...(i)

$\Rightarrow (x - 55) (x - 11) \leq 0;$ $11 \leq x \leq 55$

$∴$ $S = {11, 12, 13, \dots, 55} \Rightarrow n (S) = 45$

$Elements of S which are multiples of 3 are$

$12 + (n - 1) 3 = 54 \Rightarrow 3 (n - 1) = 42 \Rightarrow n = 15$

$n (A) = 15 \Rightarrow P (A) = \frac{15}{45} = \frac{1}{3}$

Solution

Q.
Let $M$ be the maximum value of the product of two positive integers when their sum is 66. Let the sample space $S = {x \in Z : x (66 - x) \geq \frac{5}{9} M}$ and the event $A = {x \in S : x is a multiple of 3} .$ Then $P (A)$ is equal to [2023]

1 $\frac{15}{44}$

2 $\frac{1}{3}$

3 $\frac{7}{22}$

4 $\frac{1}{5}$

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Solution

Q. 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)≥59M} and the event A={x∈S:x is a multiple of 3}. Then P(A) is equal to [2023]

1 1544

2 13

3 722

4 15