Seven years ago A was 4 times as old as B was then. Seven years hence,, A will be twice as old as B
will be then. Find the present age of each.
Answers
GIVEN:
- Seven years ago A was 4 times as old as B was then.
- Seven years hence,, A will be twice as old as B
TO FIND:
- What are the present ages of A and B ?
SOLUTION:
CASE :- 1)
Seven years ago A was 4 times as old as B was then
- A's age = (A –7)
- B's age = (B –7)
According to question:-
CASE :- 2)
Seven years hence, A will be twice as old as B
- A's age = (A+7)
- B's age = (B +7)
According to question:-
Put the value of A from equation 1) in equation 2)
Put the value of B in equation 2)
❝ Hence, the present ages of A and B are 35 and 14 ❞
______________________
Given :-
• Seven years ago A was 4 times as old as B was then.
• Seven years hence, A will be twice as old as B.
To Find :-
• Present age of each.
__________________________
Case 1)
Seven years ago A was 4 times as old as B then
• age of A => A - 7
• age of B => B - 7
According to question,
A - 7 = 4 ( B - 7 )
=> A - 7 = 4B - 28
=> A - 4B = - 28 + 7
=> A = - 21 + 4B ______1)
__________________________
Case 2)
Seven years hence, A will be twice as old as B.
• age of A = A + 7
• age of B = B + 7
According to question,
A + 7 = 2 ( B + 7 )
=> A + 7 = 2B + 14
=> A - 2B = 14 - 7
=> A - 2B = 7 ______2)
__________________________
Put the value of A in 2nd equation
Then,
- 21 + 4B - 2B = 7
=> - 21 + 2B = 7
=> 2B = 7 + 21
=> 2B = 28
=> B = 14
__________________________
Put the value of B in 2nd equation
Then,
A - 2(14) = 7
=> A - 28 = 7
=> A = 7 + 28
=> A = 35
Hence, the present ages of A and B are 35 and 14 respectively.