Question

Q8. Eight years ago, a father's age was thrice the age of his son and two
years later the father's age will be twice the age of his son. Find their present
ages.​

Answer 1

Answer:

Let the present age of the father be x and the present age of the son be y.

It is given that five years ago, the age of a father was thrice the age of his son that is:

x−5=3(y−5)

x−5=3y−15

x−3y=−15+5

x−3y=−10..........(1)

Also, in five years time, the age of the father will be twice the age of his son at that time that is:

x+5=2(y+5)

x+5=2y+10

x−2y=10−5

x−2y=5..........(2)

Now subtract equation 1 from equation 2 to eliminate x, because the coefficients of x are same. So, we get

(x−x)+(−2y+3y)=5+10

i.e. y=15

Substituting this value of y in (2), we get

x−30=5

i.e. x=5+30=35

Hence, the present age of the father is 35 years and the present age of the son is 15 years.

Step-by-step explanation:

Answer 2

Let father's age be X

and son's age be Y

Eight years ago father's age was thrice to his son

=> X-8/Y-8 = 3/1

=> 3Y-24 = X-8

=> X-3Y+16 = 0 ----------(1)

Two years hence, father is twice to son

=> X+2/Y+2 = 2/1

=> 2Y+4 = X+2

=>X-2Y-2 = 0 -----------(2)

do (1) -(2)

Then, Y=18 -----------(3)

Substitute (3) in (2) then

=> X-2*18 -2 = 0

=> X-36-2 = 0

=> X= 38

Therefore, Father's age is 38

Son's is age is 18

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