JEE MAIN - Mathematics (2024 - 8th April Morning Shift - No. 10)
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 $$\operatorname{gcd}(m, n)=1$$, then $$n$$-$$m$$ equals
10
11
9
8
Explanation
Take two numbers as $$a$$ and $$b$$
$$a+b=24$$
_8th_April_Morning_Shift_en_10_1.png)
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 $$=\{(23,1),(22,2), \ldots\}$$
Integer points on line in shaded region
$$\begin{aligned} & \{(6,18),(7,17),(8,16), \ldots(18,6)\} \\ & P(E)=\frac{n(E)}{n(S)}=\frac{13}{23}=\frac{m}{n} \Rightarrow n-m=10 \end{aligned}$$
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