SSC CGL Tier II 12 Apr 2015 Paper

Question : 79 of 100

Marks: +1, -0

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

Solution:

(3)

{3 a+4 b}/{2}>50

⇒3 a+4 b>100

⇒3 a+ {4 a}/{2}>100

[∵ a = 2b]

⇒3 a+2 a>100

⇒5 a+100

⇒a>20

∴ Minimum value of a = 21

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