QJEEM24Ma1b8d: JEE Main 2024 Mathematics - Class 12 JEEM24Ma1b8d Solution: Let the sum of two positive integers be 24 . If the…

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QJEEM24Ma1b8d: JEE Main 2024 Mathematics - Class 12 JEEM24Ma1b8d Solution: Let the sum of two positive integers be 24 . If the…

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 $\gcd (m, n) = 1$ , then $n$ - $m$ equals

Multiple Choice Options

A 10
B 11
C 9
D 8

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Accepted Answer

Take two numbers as $a$ and $b$

$a + b = 24$

JEE Main 2024 (Online) 8th April Morning Shift Mathematics - Probability Question 20 English Explanation

For product to be maximum

$\begin{aligned} \frac{a + b}{2} \geq \sqrt{a b} \\ 144 > a b \end{aligned}$

Maximum product is 144

Now, $a b \geq \frac{3}{4} \cdot 144 = 108$

Sample space $= {(23, 1), (22, 2), \dots}$

Integer points on line in shaded region

$\begin{aligned} {(6, 18), (7, 17), (8, 16), \dots (18, 6)} \\ P (E) = \frac{n (E)}{n (S)} = \frac{13}{23} = \frac{m}{n} \Rightarrow n - m = 10 \end{aligned}$

QJEEM24Ma1b8d: JEE Main 2024 Mathematics - Class 12 JEEM24Ma1b8d Solution: Let the sum of two positive integers be 24 . If the…

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