math - age

[Electrons]

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The sum of the ages of the brother and sister is 27 years. When the brother is twice as old as he is now, the sister will be three times as old as she is now. How old (in years) is the sister now?

a) 9
b) 12
c) 15
d) 18
e) 21

Answer is A. I got 10.5...which is not exactly 9 but luckily guessed it as A. What did I miss here in the setup that doesn't get 9 exactly?

my setup:
2x=3(27-x)
x=16.5 , therefore 27-x=10.5

 

[chessxwizard]

You made the assumption that at time t (When the brother is twice as old as he is now, the sister will be three times as old as she is now), both the sister and the brother will be the same age. The best way to solve this problem I think is to work backwards with the answer choices given.

 

[eleanor_rigby]

I solved it this way:

(1) b + s = 27
2b = (27-s) + x ----> x = 2b -(27-s)
3s = (27-b) + x ----> x = 3s - (27-b)

Therefore,

2b - (27-s) = 3s - (27-b) (substitute from equation 1)
2(27-s) - (27-s) = 3s - (27 - (27 -s)
27 - s = 3s - s
27 -s = 2s
27 = 3s
s = 9


Where's Streetwolf? He might derive an easier way of solving the problem.

 

[Dion]

Brother + Sister = 27 years old

so I quickly started to plug numbers in:

A. If brother was 9 years old and is twice that age now, it'll make him 18. So plug that into the brother + sister = 27 formula. So it will be 18+s=27. s=9 years old.

 

[Streetwolf]

b + s = 27
When b = 2b, s = 3s

2b = b + b

At that point, s(future age) = s(present age) + b

So s + b = 3s (she will be 3x her age)
2s = b

Then plug in the first equation

2s + s = 27
3s = 27
s = 9
b = 18