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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Answered by
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
0
Answer:
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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