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 (By Examveda Team)

$$\eqalign{ & {\text{According to the question,}} \cr & \frac{{3a + 4b}}{2} > 50 \cr & 3a + 4b > 100 \cr & a = 2b\,\,\left( {{\text{Given}}} \right) \cr & 6b + 4b > 100 \cr & 10b > 100 \cr & b > 10 \cr & {\text{If }}b = 10.1 \cr & \therefore a = 2 \times 10.1 \cr & a = 20.2 \cr & \therefore a = 21\left( {{\text{Approx}}} \right) \cr} $$