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This is a pretty simple problem, however I was looking for purely algebraic way(s) to solve it. For some reason have trouble expressing it properly, please help to set it up.

Re: In a certain school, the ratio of boys to girls is 5 to 13.
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24 Sep 2012, 03:19

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ikokurin wrote:

This is a pretty simple problem, however I was looking for purely algebraic way(s) to solve it. For some reason have trouble expressing it properly, please help to set it up.

In a certain school, the ratio of boys to girls is 5 to 13. If there are 72 more girls than boys, how many boys are there? A. 27 B. 36 C. 45 D. 72 E. 117

Given: \(\frac{b}{g}=\frac{5x}{13x}\), for some positive integer \(x\). So, the number of boys must be a multiple of 5. Only answer choice C fits.

Answer: C.

Alternately you can write: \(5x+72=13x\) --> \(x=9\) --> \(boys=5x=5*9=45\).

WE: Information Technology (Internet and New Media)

Re: In a certain school, the ratio of boys to girls is 5 to 13.
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24 Sep 2012, 13:05

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The ratio of b to G is 5:13 and the other data point is G are more than boys by 72... Looking at the ratio we can say that the 8(13-5) extra parts caused this diff of 72. so 1 part corresponds to 72/8=9 and so 5 parts correspond to 5*9 = 45.

PS: always double check with the answer if u r using this approach.
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Re: In a certain school, the ratio of boys to girls is 5 to 13.
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06 May 2018, 04:52

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kalita wrote:

In a certain school, the ratio of boys to girls is 5 to 13. If there are 72 more girls than boys, how many boys are there?

A. 27 B. 36 C. 45 D. 72 E. 117

One approach:

GIVEN: the ratio of boys to girls is 5 to 13 There are several possible cases that meet this condition: - there are 5 boys and 13 girls - there are 10 boys and 26 girls - there are 15 boys and 39 girls . . . NOTE: We're also told that there are 72 more girls than boys. So, as we continue listing possible cases, we'll keep track of the DIFFERENCE in the number of boys and girls . . . - there are 20 boys and 52 girls (there are 32 more girls than boys) - there are 25 boys and 65 girls (there are 40 more girls than boys) - there are 30 boys and 78 girls (there are 48 more girls than boys) - there are 35 boys and 91 girls (there are 56 more girls than boys) - there are 40 boys and 104 girls (there are 64 more girls than boys) - there are 45 boys and 117 girls (there are 72 more girls than boys)

This is a pretty simple problem, however I was looking for purely algebraic way(s) to solve it. For some reason have trouble expressing it properly, please help to set it up.

Another approach: Let B = # of boys Let G = # of girls

The ratio of boys to girls is 5 to 13. We can write: B/G = 5/13 Cross multiply to get: 13B = 5G

There are 72 more girls than boys We can write G = B + 72 Alternatively, we can write: G - 72 = B

We now have two equations: 13B = 5G G - 72 = B

Take bottom equation and multiply both sides by 5 to get: 5G - 360 = 5B Now replace 5G with 13B[since the top equation tells us that 13B = 5G] We get: 13B - 360 = 5B Add 36 to both sides: 13B = 5B + 360 Subtract 5B from both sides: 8B = 360 Solve: B = 45

Re: In a certain school, the ratio of boys to girls is 5 to 13.
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20 Jun 2018, 08:57

we know that ratio of girls to boys is 5:13, and there are 72 more girls than boys. So we can interpret that (13-5) 8x=72 or x=9, since we know that boys are 5x, they number must be 45 (C)

gmatclubot

Re: In a certain school, the ratio of boys to girls is 5 to 13. &nbs
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20 Jun 2018, 08:57