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:
Option 1: 20
Option 2: 18
Option 3: 21
Option 4: 19
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Correct Answer: 21
Solution :
According to the question,
$\frac{3a+4b}{2}>50$
$\Rightarrow3a+4b>100$
$\Rightarrow3a+\frac{4a}{2}>100 (\because a = 2b)$
$\Rightarrow3a+2a>100$
$\Rightarrow5a>100$
$\Rightarrow a>20$
$\therefore$ Minimum value of $a = 21$
Hence, the correct answer is 21.
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