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Intern  Joined: 06 Feb 2013
Posts: 21
If a child is randomly selected from Columbus elementary sch  [#permalink]

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Difficulty:   45% (medium)

Question Stats: 66% (01:52) correct 34% (01:57) wrong based on 285 sessions

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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
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Joined: 02 Sep 2009
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Re: If a child is randomly selected from Columbus elementary sch  [#permalink]

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

$$P(b)=\frac{b}{b+g}=?$$

(1) If 25 boys are removed from the school, the probability of selecting a boy will be 0.75 --> $$\frac{b-25}{(b-25)+g}=\frac{3}{4}$$ --> $$b-3g=25$$. Not sufficient.

(2) There are 35 more boys than there are girls --> $$g=b-35$$. Not sufficient.

(1)+(2) We have two linear equation with two unknowns ($$b-3g=25$$ and $$g=b-35$$), thus we can solve for both and get the value of $$\frac{b}{b+g}$$. Sufficient.

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Intern  Joined: 06 Feb 2013
Posts: 21
Re: If a child is randomly selected from Columbus elementary sch  [#permalink]

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Bunuel wrote:

$$P(b)=\frac{b}{b+g}=?$$

Thanks, Bunuel.
Could you please clarify how the statement 1 "...the probability of selecting a boy will be 0.75" is different from the question itself "what is the probability that the child will be a boy". I'm stuck here because to me it looks like they provide the same information.
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Re: If a child is randomly selected from Columbus elementary sch  [#permalink]

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LinaNY wrote:
Bunuel wrote:

$$P(b)=\frac{b}{b+g}=?$$

Thanks, Bunuel.
Could you please clarify how the statement 1 "...the probability of selecting a boy will be 0.75" is different from the question itself "what is the probability that the child will be a boy". I'm stuck here because to me it looks like they provide the same information.

(1) says that "IF 25 boys are removed from the school, the probability of selecting a boy will be 0.75"
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Re: If a child is randomly selected from Columbus elementary sch  [#permalink]

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Bunuel wrote:
If a child is randomly selected from Columbus elementary school, what is the probability that the child will be a boy?

$$P(b)=\frac{b}{b+g}=?$$

(1) If 25 boys are removed from the school, the probability of selecting a boy will be 0.75 --> $$\frac{b-25}{(b-25)+g}=\frac{3}{4}$$ --> $$b-3g=25$$. Not sufficient.

(2) There are 35 more boys than there are girls --> $$g=b-35$$. Not sufficient.

(1)+(2) We have two linear equation with two unknowns ($$b-3g=25$$ and $$g=b-35$$), thus we can solve for both and get the value of $$\frac{b}{b+g}$$. Sufficient.

Hi Bunuel,

Can you please shed some light on why it would not be correct to state the following:

For the statement 1, p(girl)=(g/(b-25+g))=0.25.

Assuming this inference is correct, we can find the number of boys using a two equation,two unknowns approach.

Thank you!
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Joined: 02 Sep 2009
Posts: 55732
Re: If a child is randomly selected from Columbus elementary sch  [#permalink]

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Pmar2012 wrote:
Bunuel wrote:
If a child is randomly selected from Columbus elementary school, what is the probability that the child will be a boy?

$$P(b)=\frac{b}{b+g}=?$$

(1) If 25 boys are removed from the school, the probability of selecting a boy will be 0.75 --> $$\frac{b-25}{(b-25)+g}=\frac{3}{4}$$ --> $$b-3g=25$$. Not sufficient.

(2) There are 35 more boys than there are girls --> $$g=b-35$$. Not sufficient.

(1)+(2) We have two linear equation with two unknowns ($$b-3g=25$$ and $$g=b-35$$), thus we can solve for both and get the value of $$\frac{b}{b+g}$$. Sufficient.

Hi Bunuel,

Can you please shed some light on why it would not be correct to state the following:

For the statement 1, p(girl)=(g/(b-25+g))=0.25.

Assuming this inference is correct, we can find the number of boys using a two equation,two unknowns approach.

Thank you!

Yes, it's correct but if you simplify it you'd still get the same equation: $$b-3g=25$$. Thus you'd still have only one equation with two unknowns.

Hope it's clear.
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Posts: 7756
Re: child is randomly selected from Columbus elementary school  [#permalink]

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fitzpratik 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

Hi..
For getting a ratio here, you require all terms with variable.
But here by first look, you have a term without variable so none can be sufficient individually but let's see

1) $$\frac{b-25}{b+g-25}=0.75=\frac{3}{4}...... 4b-100=3b+3g-75....b=25+3g$$
Ratio can't be found
Insufficient
2) b=g+35
Again ratio cannot be found
Insuff

Combined..
You can find values of b and g and thus get ratio or PROBABILITY
35+g=3g+25.....2g=10...G=5
b=35+5=40..
Probability of picking boy is b/t=40/40+5=8/9..
Suff
C
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Re: If a child is randomly selected from Columbus elementary sch  [#permalink]

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LinaNY 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)?

The video posted below has tips on rephrasing the target question

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

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_________________ Re: If a child is randomly selected from Columbus elementary sch   [#permalink] 28 Nov 2018, 08:50
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