The smallest number, which when divided by 5, 10, 12 and 15, leaves remainder 2 in each case , but when divided by 7 leaves no remainder , is = ?

Answer: Option B

LCM of (5, 10, 12, 15) = 5 × 2 × 6 = 60
smallest number divided by (5, 10, 12, 15)
leaves remainder 2 and when divided by 7 leaves no remainder is
$$\begin{array}{rl}& \text{ = }\frac{60\text{K}+2}{7}=\frac{4\text{K}+2}{7}\\ & \text{At K = 3, }\frac{4\text{K}+2}{7}\\ & \Rightarrow \text{remainder = 0}\\ & \therefore \text{number = 60K + 2 }\\ & \text{ = 60}\times \text{3 + 2 }\\ & \text{ = 182}\end{array}$$