Question

A is twice as old as B five years ago his age was three times b age find their present age​

Answer 1

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

Answer 2

Answer:

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

Let the ade of A be a and B be b.a = 2b……….(1)

Let the ade of A be a and B be b.a = 2b……….(1)Five years ago

Let the ade of A be a and B be b.a = 2b……….(1)Five years agoA’s age was a-5 and B’s age was b-5

Let the ade of A be a and B be b.a = 2b……….(1)Five years agoA’s age was a-5 and B’s age was b-5a-5 = 3(b-5)

Let the ade of A be a and B be b.a = 2b……….(1)Five years agoA’s age was a-5 and B’s age was b-5a-5 = 3(b-5)a-5 = 3b -15………(2)

Let the ade of A be a and B be b.a = 2b……….(1)Five years agoA’s age was a-5 and B’s age was b-5a-5 = 3(b-5)a-5 = 3b -15………(2)We have to solve two linear equations (1) & (2)

Let the ade of A be a and B be b.a = 2b……….(1)Five years agoA’s age was a-5 and B’s age was b-5a-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)

Let the ade of A be a and B be b.a = 2b……….(1)Five years agoA’s age was a-5 and B’s age was b-5a-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

Let the ade of A be a and B be b.a = 2b……….(1)Five years agoA’s age was a-5 and B’s age was b-5a-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 - 15b = 10 & a = 20

Let the ade of A be a and B be b.a = 2b……….(1)Five years agoA’s age was a-5 and B’s age was b-5a-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 - 15b = 10 & a = 20So ages of A & B are 20 & 10 years respectively