Question

The total of the ages of a, b and c is 93 years. 10 years ago, the ratio of their ages was 2:3:4. What is the present age of c?

Answer 1

Present age of C is 38 years old

Solution:

Sum of present ages of A, B, C is 93 years

A + B + C = 93

Their ages 10 years ago are A - 10, B - 10, C - 10

Sum of ages ten years ago = A -10 + B - 10 + C - 10 = A + B + C - 30 = 93 - 30 = 63

Ten years ago their ages were in the ratio 2:3:4

Let 2x, 3x, 4x be the ages of A , B, C respectively 10 years ago

Therefore,

2x + 3x + 4x = 63

9x = 63

x = 7

Thus age of C ten years ago  = 4x = 4(7) = 28

Present age = 28 + 10 = 38

Thus present age of C is 38 years old

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Answer 2

The present age of C is 38 years.

Step-by-step explanation:

Since we have given that

Sum of ages of a,b and c = 93 years

10 years ago,

Ratio of their ages = 2:3:4

So, according to question, it becomes.,

$2x+3x+4x+30=93\\\\9x+30=93\\\\9x=93-30\\\\9x=63\\\\x=\dfrac{63}{9}\\\\x=7$

So, the present age of C is given by

$4x+10=4(7)\times 10=28+10=38\ years$

Hence, the present age of C is 38 years.

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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?

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