Question

Find the least number which when divided by $12$, leaves a remainder of $7$, when divided by $15$, leaves a remainder of $10$ and when divided by $16$, leaves a remainder of $11$

• A. $115$
• B. $235$
• C. $247$
• D. $475$

Answer 1

The correct option is B $235$

$12=2^2\times 315=3\times 516=2^{4}LCM=2^4\times 3\times 5=240$
Now, the number has a remainder of $7$ when divided by $12$. Similarly, number has remainder $10$ on dividing by $15$ and remainder is $11$ when divided by $16$. In each case, we are 5 short of getting a zero remainder. So the desired number must be $240-5=235$
Option B