Asked by AoPS
Each of Alice and Beatrice has their birthday on the same day. In 8 years' time, Alice will be twice as old as Beatrice. Ten years ago, the sum of their ages was 21. How old is Alice now?
Answers
Answered by
Reiny
present time:
Alice --- a
Beatrice --- b
10 years ago:
a-10 + b-10 = 21
a+b = 41 **
in 8 year's time:
a+8 = 2(b+8)
a+8 = 2b + 16
a - 2b = 8 ***
subtract *** from **
3b = 33
b = 11 , then a = 30
State your conclusions
Alice --- a
Beatrice --- b
10 years ago:
a-10 + b-10 = 21
a+b = 41 **
in 8 year's time:
a+8 = 2(b+8)
a+8 = 2b + 16
a - 2b = 8 ***
subtract *** from **
3b = 33
b = 11 , then a = 30
State your conclusions
Answered by
oobleck
so put the words into math.
a+8 = 2(b+8)
a-10 + b-10 = 21
so, a=30
a+8 = 2(b+8)
a-10 + b-10 = 21
so, a=30
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