Question

7. A is twice old as B. Five years ago A was 3 times as old as B. Find their present ages.
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Answer 1

Let the ade of A be a and B be b.

a = 2b……….(1)

Five years ago

A’s age was a-5 and B’s age was b-5

a-5 = 3(b-5)

a-5 = 3b -15………(2)

We have to solve two linear equations (1) & (2)

Substituting value of a =2b in (2)

2b -5 = 3b - 15

b = 10 & a = 20

So ages of A & B are 20 & 10 years respectively.

I hope this helps you

Answer 2

Answer:

saumili 4.

okay.

Answer is correct.

It's easy method see.

let the present ages of A and B be x.

a = 2b.___1.

five years ago, I. e, back.

A's age was five years ago is a-5.

B's age was five years ago is b-5.

as per question, A is 3 times as old as B.

a-5= 3(b-5).

a-5 = 3b -15.___2.

now, only substitute equation 1 in 2.

we get,

2b -5 = 3b -15.

3b -2 b = 15- 5.

b = 10.

and put b in equation 1 .

we get, a = 2*10 = 20.

Now,

Present age of A = 20 years.

Present age of B = 10 years.

okay.

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