Question
If the arithmetic mean of 3a and 4b is greater than 50, and a is twice b, then the smallest possible integer value of a is
Answer: Option A Answer: (a)$\text"3a+4b"/2$ > 50⇒ 3a + 4b > 100 ⇒ 3a + ${4a}/2$ > 100 [Since a = 2b] ⇒ 3a + 2a > 100 ⇒ 5a > 100 ⇒ a > 20 ∴ Minimum value of a = 21
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