Question

A boy is twice as old as his sister. Four years hence the product of their ages will be 160.Find their present ages, 5yrs before ages.

Answer 1

Answer:

Step-by-step explanation:

Present age of the boy be 2x

Her sister age be X

(2x+4) (X+4)= 160

2x^2+12x-144=0

Splitting the middle term

2x^2+24x-12x-144=0

2x(X+12)-12(X+12)

X+12=0. 2x-12=0

X=-12. 2x=12

X=6

The present age will be 6

Answer 2

hey mate

the answer to your question is :

let sister's age be x

so, boy's age will be 2x

a.t.q.

(4 + x) (4 + 2x) = 160

16 + 8x + 4x + 2x^2 - 160 = 0

2x^2 + 12 x - 144 = 0

dividing 2 both sides :

x^2 + 6x - 72 = 0

x^2 + 12x - 6x -72 = 0

x ( x + 12 ) -6 ( x +12) = 0

( x + 12) (x - 6) = 0

x + 12 =0 or x - 6 = 0

x = -12 or x = 6

we know that x can't be negative

so, x = 6

so, sisters age =6 yrs

and boys age = 12 yrs

hope it helps you