Question

The present ages of three persons is proportions 3:6:8. 4 years ago, the sum of their ages was 56. Find their ages.

Answer 1

Let the present age of 1st person be 3x, 2nd be 6x and 3rd be 8x.
Ages 4 years ago:
1st person = (3x - 4) years
2nd person = (6x - 4) years
3rd person = (8x - 4) years

3x - 4 + 6x - 4 + 8x - 4 = 56
17x - 12 = 56
17x = 56 + 12 = 68
$x = \frac{68}{17} = 4$
Present ages:
1st person = 3x = 3 × 4 = 12 years
2nd person = 6x = 6 × 4 = 24 years
3rd person = 8x = 8 × 4 = 32 years

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Answer 2

Let the age be6x 3x 8x
four year ago 56
After four year56+4=60
6x*3x*8x=60
126x^3=60
X^3=21/10
X=2.1
The first man age is 3*21=61
solve it by your self

I hope answer is right