Math, asked by advaithmora007, 9 months ago

A is twice old as B.five years ago A was 3 times as old as B .find their present ages​

Answers

Answered by snehamittal220
3

Answer:

Let age of B is x

Then age of A is 2x

5 years ago A is 3 times as old as B.

5 years ago the age of A was (2x-5)

Similarly 5 years ago age of B was( x-5)

According to the question, 2x-5=3(x-5)

2x-5=3x-15

x=10 yrs

So, present age of B is x i. e 10 yrs and present age of A is 2x=2×10=20 yrs

Answered by Anonymous
0

Answer:

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Let the age 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.

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