Five years ago,A's age was four times the age of B.Five years hence A's age will be twice of the age of B. Find their present ages.
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Question:
- Five years ago,A's age was four times the age of B. 5 years hence, A's age will be twice of the age of B. Find their present ages.
Given:
- Five years ago, A's age was 4 times the age of B.
- Five years hence A's age will be twice of the age of B.
To find:
- Age of A
- Age of B
Let :
- present age of A = x
- present age of B = y
Answer:
Five years ago:
- Age of A= x-5
- Age of B=y-5
As it's Given 5 years ago A's age was 4 times the age of B
.°. Equation formed as follows:
Now Let's solve this equation:
Five years hence:
- Age of A= x+5
- Age of B = y+5
As it's given 5 years hence A's age will be twice of the age of B.
.°. Equation formed as follows:
Now put value of y in Equation 2
multiply each digit by 4
we got value of x i.e.25
Put value of x in equation 1:
As we have supposed A's age as x
.°. A's age = 25
As we have supposed B's age as y
.°. B's age = 10
Now Let's Verify their ages:
Put age of A and B in Equation 1:
☆Hence Verified ☆
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And all we are done! ✔
:D
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