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 = ?

A. 189

B. 182

C. 175

D. 91

Answer: Option B

Solution(By Examveda Team)

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
$$\eqalign{ & {\text{ = }}\frac{{60{\text{K}} + 2}}{7} = \frac{{4{\text{K}} + 2}}{7} \cr & {\text{At K = 3, }}\frac{{4{\text{K}} + 2}}{7} \cr & \Rightarrow {\text{remainder = 0}} \cr & \therefore {\text{number = 60K + 2 }} \cr & {\text{ = 60}} \times {\text{3 + 2 }} \cr & {\text{ = 182}} \cr} $$