Question

Sunny father is 7 times bigger than his son .two years ago he is thirteen times bigger than his son.if sum of their ages is 32.find their present ages

Answer 1

Let the Present Age Of the Father and his son be x and y respectively

Two years ago their ages were x - 2 and y - 2 respectively
x - 2 = 13( y - 2)
x - 2 = 13y - 26
x - 13y = -24 ------------------1

Sum of their Present ages = 32
So,
x + y = 32 --------------------2

Solving simultaneously
x - 13y = -24
x + y = 32 Multiplying this equation by -1

So,
x - 13y = -24
-x - y = -32

-14y = -56
14y = 56
y = 56/14
y = 4 years

Substituting the value of y in equation 2 
x + y = 32
x + 4 = 32
x = 32 - 4
x = 28 years

Alternative Method And Easy Method
Sunny father is 7 times bigger than Sunny
So,
 
x = 7y -------------------1
x + y = 32 --------------2
Substituting the value of x in equation 2
x + y = 32 -------------Given
7y + y = 32
8y = 32
y = 32/8
y = 4 years

x = 7y
x= 7 * 4
x = 28 years

Ans = Age of the father ans the son is equal to 28 and 7 years respectively

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Answer 2

Let Sunny's age be= x
Therefore, His father's age = 7x

Since sum of their ages is= 32

Therefore, x+7x= 32
8x= 32
x= 32/8= 4

Therefore their present ages are

Sunny= 4 years
His Father= 28 years.