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If a child is randomly selected from Columbus Elementary

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If a child is randomly selected from Columbus Elementary

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Magoosh

If a child is randomly selected from Columbus Elementary School, what is the probability that the child will be a boy?

1) If 25 boys are removed from the school, the probability of selecting a boy will be 0.75.
2) There are 35 more boys than there are girls.

OA C

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AAPL wrote:
If a child is randomly selected from Columbus Elementary School, what is the probability that the child will be a boy?
1) If 25 boys are removed from the school, the probability of selecting a boy will be 0.75.
2) There are 35 more boys than there are girls.
Target question: What is the probability that the child will be a boy?
This is a good candidate for rephrasing the target question.
Let G = # of girls in the school
Let B = # of boys in the school
So, G + B = total number of children in the school
So, P(selected child is a boy) = B/(G + B)
REPHRASED target question: What is the value of B/(G + B)?

Aside: Here’s a video with tips on rephrasing the target question: http://www.gmatprepnow.com/module/gmat-data-sufficiency?id=1100

Statement 1: If 25 boys are removed from the school, the probability of selecting a boy will be 0.75.
So, the number of boys = B - 25, and the total number of children = G + (B - 25)
We can write: (B - 25)/(G + B - 25) = 3/4
Since we have a linear equation with TWO variables, there's no way to solve this equation for B and G. So, statement 1 is NOT SUFFICIENT

If you're not convinced, consider these two CONFLICTING cases:
Case a: B = 28 and G = 1. After 25 boys leave, there are 3 boys and 1 girl. So, P(boy) = 3/4 = 0.75, which satisfies statement 1. In this case, the answer to the REPHRASED target question is B/(G + B) = 28/(1 + 28) = 28/29
Case b: B = 31 and G = 2. After 25 boys leave, there are 6 boys and 2 girls. So, P(boy) = 6/8 = 0.75, which satisfies statement 1. In this case, the answer to the REPHRASED target question is B/(G + B) = 31/(2 + 31) = 31/33
Since we cannot answer the REPHRASED target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: There are 35 more boys than there are girls.
There are several CONFLICTING cases that satisfy statement 2. Here are two:
Case a: B = 36 and G = 1. In this case, the answer to the REPHRASED target question is B/(G + B) = 36/(1 + 36) = 36/37
Case b: B = 37 and G = 2. In this case, the answer to the REPHRASED target question is B/(G + B) = 37/(2 + 37) = 37/39
Since we cannot answer the REPHRASED target question with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
From statement 1, we can write: (B - 25)/(G + B - 25) = 3/4
Cross multiply to get: 3(G + B - 25) = 4(B - 25)
Expand: 3G + 3B - 75 = 4B - 100
Rearrange to get: 3G - B = - 25

From statement 2, we can write: B = G + 35

At this point, we have two different linear equations with two variables. So, we COULD solve the system for B and G, which means we COULD answer the REPHRASED target question with certainty.
So, the combined statements are SUFFICIENT

Answer: C

Cheers,
Brent

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AAPL wrote:
Magoosh

If a child is randomly selected from Columbus Elementary School, what is the probability that the child will be a boy?

1) If 25 boys are removed from the school, the probability of selecting a boy will be 0.75.
2) There are 35 more boys than there are girls.
$$? = {b \over {b + {\rm{g}}}}$$
$$\left( 1 \right)\,\,{{b - 25} \over {\left( {b - 25} \right) + g}} = {3 \over 4}\,\,\,\left\{ \matrix{
\,{\rm{Take}}\,\,\left( {b,g} \right) = \left( {28,1} \right)\,\,\,\, \Rightarrow \,\,\,\,? = {{28} \over {29}} \hfill \cr
\,{\rm{Take}}\,\,\left( {b,g} \right) = \left( {31,2} \right)\,\,\,\, \Rightarrow \,\,\,\,? = {{31} \over {33}} \ne {{28} \over {29}} \hfill \cr} \right.$$
$$\left( 2 \right)\,\,b = g + 35\,\,\,\,\,\,\left\{ \matrix{
\,{\rm{Take}}\,\,\left( {b,g} \right) = \left( {36,1} \right)\,\,\,\, \Rightarrow \,\,\,\,? = {{36} \over {37}} \hfill \cr
\,{\rm{Take}}\,\,\left( {b,g} \right) = \left( {37,2} \right)\,\,\,\, \Rightarrow \,\,\,\,? = {{37} \over {39}} \ne {{36} \over {37}} \hfill \cr} \right.$$
$$\left( {1 + 2} \right)\,\,\,{{\left( {g + 35} \right) - 25} \over {\left[ {\left( {g + 35} \right) - 25} \right] + g}} = {3 \over 4}\,\,\,\,\mathop \Rightarrow \limits^{{1^{{\rm{st}}}}\,\,{\rm{degree}}} \,\,\,\,g\,\,{\rm{unique}}\,\,\,\, \Rightarrow \,\,\,\,b\,\,{\rm{unique}}\,\,\,\, \Rightarrow \,\,\,\,{\rm{SUFF}}.$$

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

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Fabio Skilnik :: https://GMATH.net (Math for the GMAT) or https://GMATH.com.br (Portuguese version)
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