A is twice as old as B. After 8 years the sum of their ages will be 28 years. Find their
present ages
Answers
Answered by
2
Answer:
A = 28, B = 21
solution:
present age of A = x
present age of B = 49 - x ( since sum of their ages = 49)
difference between their ages = x - (49 -x) = 2x - 49 ----> (1)
when A's age was 49 - x
B's Age must be (A's age - their age difference)
B's past age = (49-x) - (2x - 49) = 98 - 3x
it is given that A' present age is twice the B's past age:
=> x = 2(98 - 3x)
7x = 196
value of x, A's current age = 28
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Answered by
0
Answer:
Age of B = x
Age of A = 2x
After 8 years,
( 2x + 8) + ( x + 8) = 28
2x + 8 + x + 8 = 28
3x + 16 = 28
3x = 28 - 16 = 12
x = 12/3 = 4
Present age of B = 4 and A = 8
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