Question

The sum of present ages of A, B and C is 93 years. Ten years ago the ratio of their ages was
2 : 3 : 4 respectively. What is the present age of C?

Answer 1

Let the present ages be a,b,c
a+b+c = 93
Their ages 10 years ago are a-10,b-10,c-10
ratio of ages is 2:3:4
sum of ages 10 years ago is a-10+b-10+c-10 = a+b+c - 30 = 93-30 = 63
Let 2x,3x,4x be the ages
2x+3x+4x = 63x = 7
C's age 10 years ago = 4(7) = 28 years
C's present age = 28+10 = 38 years

Answer 2

The correct answer is 38 years.

Given: The sum of present ages = 93.

Ratio of ages 10 years ago = 2 : 3 : 4.

To Find: The present age of C.

Solution:

Let the age of A 10 years ago = 2x

Then the age of A at present = 2x + 10

Let the age of B 10 years ago = 3x

Then the age of B at present = 3x + 10

Let the age of C 10 years ago = 4x

Then the age of C at present = 4x + 10

Sum of present ages = 2x+10 + 3x+10 + 4x+10 =93

9x + 30 = 93

9x = 63

x = 7

Present age of C = 4x+10 = 4(7) + 10 = 28+10 = 38

Hence, the present age of C is 38 years.

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