Discussion of Question with ID = 088 under Problems-on-Ages

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Question

The sum of the ages of 10 calves of a whale born at a gap of 6 years is 390. What is the age of the youngest calf?

A

12 years.

B

13 years.

C

11 years.

D

14 years.

Soln.
Ans: a

The ages of the calves are in an AP with d = 6, n = 10, and sum S = 390. We have to find the first term a. We know S = ${n/2} × (2a + (n-1)d)$ Putting the values 390 = ${10/2} × (2a + (10-1)×6)$. Solving, get a = 12 years.


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