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

Hello Dear!!!

Here's your answer....

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

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

Ten years ago,

we have to subtract 30 years from present age.

Because there are 3 persons.. so 3(10 years)

So 30 years...

Ten years ago.... The sum of their ages was

93-30

63

Their ratio is 2:3:4

2x + 3x + 4x = 63

9x = 63

x = 63/9

x = 7

Age of C (Before 10 years) = 4x = 4(7)
= 28

Age of C(present) = 28+10 = 38

Present age of C is 38.

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Hope this helps you....

Answer 2

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 93 - 30 = 63

Let 2x,3x,4x be the ages

2x+3x+4x = 63
x = 7

C's age 10 years ago = 4(7) = 28 years

C's present age = 28+10 = 38 years