One year ago, a man was 8 times as old as his son. Now his age is equal to the
square of his son’s age. Their present ages are
(a) 7 years, 49 years
(b) 5 years, 25 years
(c) 1 years, 50 years
(d) 6 years, 49 years
Answers
Answered by
3
Step-by-step explanation:
let present age of his son x
present age of man is y
1 year ago
son's age = x-1
man's age = y-1
y-1 = 8(x-1)
y-1 = 8x-8
8x-y = 7.............1
now
y = x^2..............2
place the value of 2 in 1
8x-x^2=7
taking minus common
x^2-8x+7=0
x^2-(7+1)x+7=0
x^2-7x-x+7=0
x(x-7)-1(x-7)=0
(x-1)(x-7)=0
x=7years. x=1 not possible
y = x^2
y = 7^2
y =49 years
so option a is correct
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