Aptitude - Problems on Ages - Discussion

Discussion Forum : Problems on Ages - General Questions (Q.No. 4)

Problems on Ages

A is two years older than B who is twice as old as C. If the total of the ages of A, B and C be 27, then how old is B?

Answer: Option

Explanation:

Let C's age be x years. Then, B's age = 2x years. A's age = (2x + 2) years.

(2x + 2) + 2x + x = 27

5x = 25

x = 5.

Hence, B's age = 2x = 10 years.

Discussion:

79 comments Page 5 of 8.

Annn said: 8 months ago

A + B + C = 27.
B = A + 2.
C = 2B.
A + (A+2) + 2(A+2) = 27,
4A + 6 = 27,
A = 5.
B = 7.

(7)

Vinayak said: 10 years ago

A = B+2.

C = 2B.

A+B+C = 27.

B+2+B++2B = 27.

4B = 25.

B = 6.25.

I got this. :).

Mourya said: 1 decade ago

B is twice of C so C is half of B so you have to replace c = x/2 you will get answer.

Sweta said: 1 decade ago

We can also solve as
A's age be x yrs. then B's age is x-2 and C's age is(x-2)/2.

Dharumaraj S said: 2 decades ago

(b-2)+ b + .5b =27

which is an alternate method to solve the same problem.....

Rakesh said: 4 years ago

A = 2+B.
B = 2C,
A+B+C = 27.

2 + 2C + 2C + C = 27
5C = 27 - 2,
5C = 25,
C = 5.

(8)

Karan vekariya said: 1 decade ago

Here, b = 2c and a = (b-2) then,
(b-2)+b+b/2 = 27.

=>5b = 50.
So, b = 10.

Pavani said: 8 years ago

If we do option verification, option C(ans 9) is satisfying.

i.e. 11+9+7=27.

(1)

Ganesh said: 1 decade ago

A=2+B,
B=2C

A+B+C=27
so (2+B) + B + B/2 = 27
2B + B/2 = 25
5B/2 = 25
B=10

Nikunj said: 5 years ago

Simply, the solution is
(x+2) + x + 0.5x = 27.
x + x + 0.5x = 25.
x = 10.

(1)

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