# The sums of ages of two of them at a time are 4 years, 6 years and 8 years

There are three brothers. The sums of ages of two of them at a time are 4 years, 6 years and 8 years. The age difference between the eldest and the youngest is

(a) 3 years

(b) 4 years

(c) 5 years

(d)( 6 years

**Answer: (b) 4 years**

**Solution:-**

Lat the three brothers be A, B and C

The sum of two of them at a time are 4 years, 6 years and 8 years.

Then, A + B = 4 ……………………….. (1)

B + C = 6 ……………………………………… (2)

C + A = 8 …………………………………..(3)

From equation (1),

A = 4 – B

Put the value of A = 4 – B in equation (3),

Then, C + 4 – B = 8

C – B = 8 – 4

C – B = 4 ………………………………(4)

Add, equation (2) & (4),

B + C + C – B = 6 + 4

2 C = 10

C = 10/2**C = 5 years**

Put the value of C in equation (2),

Then, B +5 = 8

B = 8 – 5**B = 3 years**

Put the value of A in equation (1),

Then, 3 + A = 4

A = 4 – 3** A = 1 years**

Thus, the ages of the three brothers are 1 years, 3 years and 5 years.

So, the age difference between the eldest and smallest = 5 – 1

= 4 years.

Hence, the correct answer is option (b) 4 years.