If the arithmetic mean of 3a and 4b is greater than 50,

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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

1. 20
2. 18
3. 21
4. 19

Correct Option: C

Here , The arithmetic mean of 3a and 4b is greater than 50

3a + 4b	> 50
2

⇒ 3a + 4b > 100

⇒ 3a +	4a	> 100 [∴ a = 2b]
	2

⇒ 3a + 2a > 100
⇒ 5a > 100
⇒ a > 20
∴ Minimum value of a = 21