Question

Sum of present ages of a, b and c is 92 years. If 4 years ago, the ratio of their ages were 1:2:3 respectively, find as present age.

Answer 1

The present ages of a, b and c are taken as a, b, c respectively too. So we have,

a + b + c = 72

(Sum being 92 doesn't make sense about their ages, so I took it as 72. Follow similar steps if sum is given any other number).

4 years ago, their ages will be a - 4, b - 4 and c - 4 respectively. Since these ages are in the ratio 1 : 2 : 3, let,

a - 4 = x → (1)

b - 4 = 2x → (2)

c - 4 = 3x → (3)

Adding (1), (2) and (3),

(a - 4) + (b - 4) + (c - 4) = x + 2x + 3x

a + b + c - 12 = 6x

72 - 12 = 6x

60 = 6x

x = 10

Then, in (1),

a = x + 4

a = 14

and in (2),

b = 2x + 4

b = 24

and, finally, in (3),

c = 3x + 4

c = 34

Hence the present ages are 14, 24 and 34 respectively.

#answerwithquality

#BAL

Answer 2

Answer:

A's present age = 18.33 years

Step-by-step explanation:

Present:

A + B + C = 92

4 years ago:

A - 4 : B - 4 : C - 4 = 1:2:3

$\frac{A-4}{B-4}$ = 1/2
$\frac{B-4}{C-4}$ = 2/3

A - 4 = $\frac{B-4}{2}$

A = $\frac{B}{2}$ - 2 + 4

A = $\frac{B}{2}$ + 2

C - 4 = $\frac{3(B-4)}{2}$

C = $\frac{3B}{2}$ - 6 + 4

C = $\frac{3B}{2}$ - 2

A + B + C = 92
$\frac{B}{2}$ + 2 + B + $\frac{3B}{2}$ - 2 = 92

$\frac{B + 3B}{2}$ + B = 92

3B = 92

B = 92/3

A = B/2 + 2

A = $\frac{92}{3*2}$ + 2

A = $\frac{49}{3}$ + 2

A = 16.33 + 2

A = 18.33 years