Question

Given below are two statements: **Statement I:** The set of numbers $(7,8,9,a,b,10,8,7)$ has an arithmetic mean of 9 and mode (most frequently occurring number) as 8, then $a 	imes b = 120$. **Statement II:** Let $a$ and $b$ be two positive integers such that $a + b + a 	imes b = 84$, then $a + b = 20$. In the light of the above statements, choose the most appropriate answer from the options given below:

Accepted Answer

A. **Statement I:** Sum of numbers = $7 + 8 + 9 + a + b + 10 + 8 + 7 = 49 + a + b$. There are 8 elements, so mean = $\frac{49 + a + b}{8} = 9 \implies 49 + a + b = 72 \implies a + b = 23$. In the given list, 7 occurs twice and 8 occurs twice. For 8 to be the unique mode, at least one of $a$ or $b$ must be 8. If $a = 8$, then $b = 15$. Then $a 	imes b = 8 	imes 15 = 120$. So Statement I is correct. **Statement II:** $a + b + ab = 84 \implies (1+a)(1+b) = 85$. Since $a, b$ are positive integers, the factor pairs of 85 are $(1, 85)$ and $(5, 17)$. Since $a, b \ge 1$, we take $(1+a, 1+b) = (5, 17) \implies a=4, b=16$ (or vice versa). Then $a + b = 4 + 16 = 20$. So Statement II is correct. Both Statement I and Statement II are correct.

Mean and Mode Statement Analysis

Explanation

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