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Rato question

by Brent@GMATPrepNow » Sat Dec 13, 2008 11:07 am
Here's one for you:

The current ratio of boys to girls at a certain school is 2 to 5. If 12 additional boys were added to the school, the new ratio of boys to girls would be 4 to 9. How many boys currently attend the school?

(A) 27
(B) 48
(C) 54
(D) 72
(E) 108
Source: — Problem Solving |

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by dmateer25 » Sat Dec 13, 2008 11:52 am
Let x = Total students
Let y = current boys

2x/7 = y
x = 7y/2..............(1)

(4x + 48)/13 = y + 12
4x + 48 = 13y + 156.......................(2)


4(7y/2) + 48 = 13y + 156
28y/2 + 48 = 13 y + 156
14y = 13y + 108
y = 108

E

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by Brent@GMATPrepNow » Sat Dec 13, 2008 11:55 am
That's pretty good, but the answer can be found faster by using just one variable.
Anyone?

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by dmateer25 » Sat Dec 13, 2008 12:50 pm
With 1 variable:

Let x= # of original boys

(x/1) / (2/7) + 12 = ((x+12)/1) / (4/13)

7x/2 + 12 = (13x+156)/4

(7x + 24)/2 = (13x+156)/4

(14x + 48)/4 - (13x + 156)/4

x - 108 = 0

x = 108

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by Brent@GMATPrepNow » Sat Dec 13, 2008 5:27 pm
The OA is E.

Here's my method:
Let 2x be the number of boys currently attending the school.
This means that the number of girls is 5x

If we add 12 boys (to the existing 2x boys), the new ratio becomes 4/9.
This gives us the equation (2x+12)/5x = 4/9

Solve for 2x (since there are currently 2x boys) and you get 2x=108


Another method (if you're stuck) is to start plugging in answer choices.

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Re: Rato question

by logitech » Sat Dec 13, 2008 6:17 pm
Okay, the girls numbers dont change so only the TOP part of the B/G fraction changes

LCM(5:9)=45 so simply multiply the first boys with 9 and second boys with 5

18B+12=20B , so B is 6 and 18B is 108
LGTCH
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