Question

The difference between the ages of two cousins is 10 years. 15 years ago, if the elder one was twice as old as younger one, find their present ages? from RS aggarwal..

Answer 1

let be the age of first boy is x
and the age of second boyis y

A TO Q,

x-y=10

x=y+10..........................................(1)

before 15 years,

(x-15)=2(y-15).............................…...(2)

putting the value of x in equation (2)

(y+10-15)= 2y-30

y-5=2y-30

-5+30 = 2y-y

y=25 years

putting the value of y in equation(1)

x=y+10

x=25+10

x=35years

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Answer 2

Here is the solution

Let the age of elder be x
And the age of the younger be y

A/q
X-y=10----------------equation 1

Again in II case
(X-15)= 2(y-15)
X-15=2y-30
X-2y= -15 - - - - - - - - equation 2

Subtracting equation 2 from 1 we get

X-y = 10
X-2y= -15
- + +
= y = 25

Now finding x, by putting the value of y in equation 1

X - 25 = 10
X = 10+ 25
X = 35
Therefore the age of elder one is 35 years and the age of the younger one is 25 years.

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