Let the sum of two positive integers be 24. If the probability that their product is not less than times their greatest possible product is where then equals

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Solution

Q.
Let the sum of two positive integers be 24. If the probability that their product is not less than $\frac{3}{4}$ times their greatest possible product is $\frac{m}{n},$ where $g c d (m, n) = 1,$ then $n - m$ equals [2024]

1 9

2 10

3 11

4 8

Ans.
(2)

Let the two positive integers be $x$ and $y$ .

Now, $x + y = 24,$ $x, y \in N$

Now, $\frac{x + y}{2} \geq \sqrt{x y}$    $[∵ A . M . \geq G . M .]$

$\Rightarrow x y \leq 144$

Since, $\frac{3}{4} \times 144 = 108$

So, $x y \geq 108$

$∴$   Favourable cases = ${(6, 18), (18, 6), (7, 17), (17, 7), (8, 16), (16, 8), (9, 15), (15, 9), (10, 14), (14, 10), (11, 13), (13, 11), (12, 12)}$

Total choices for $x + y = 24$ are 23

$∴$   Required probability $= \frac{13}{23} = \frac{m}{n}$

$∴$    $n - m = 23 - 13 = 10$

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