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

Question

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]

Smart Achievers · Accepted Answer

(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 $= \frac{13}{23} = \frac{m}{n}$

$∴$ $n - m = 23 - 13 = 10$

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

Want Us to Email you About Special Offers & Updates?

Solution

Q. Let the sum of two positive integers be 24. If the probability that their product is not less than 34 times their greatest possible product is mn, where gcd(m,n)=1, then n-m equals [2024]

1 9

2 10

3 11

4 8

Want Us to Email you About Special Offers & Updates?

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]