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boys in the community

Expert replies
by sanju09 » Wed Apr 15, 2009 4:40 am
In a community comprising of only boys and girls, the sum of ages of girls is 17 more than the sum of that of boys. What is the minimum possible number of boys in the community?

(1) The average (arithmetic mean) age of boys is 11 more than that of girls.

(2) The average (arithmetic mean) age of boys is 16.



MBM
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Sanjeev K Saxena
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The Princeton Review - Manya Abroad
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Source: — Data Sufficiency |

Re: boys in the community

by Vemuri » Wed Apr 15, 2009 9:51 am
Let G be the sum of ages of girls & B be the sum of ages of boys. The question statment says that G-B=17

Stmt 1: B/x - G/y = 11 (where 'x' is the number of boys & 'y' is the number of girls). Does not help.

Stmt 2: B/x = 16. Is not sufficient by itself.

Combining Stmt 1 & 2, G/y=5. Does not help either.

My answer is E.
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by vittalgmat » Wed Apr 15, 2009 12:45 pm
Very interesting one!!!!.

Great Job Sanju. keep it coming!!!>

I got C as the answer.
number of boys = 3 and number of girls = 13.

If this is correct, I will post the xplanation.
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by mals24 » Wed Apr 15, 2009 2:18 pm
Gosh really cool question Sanju :)

One more vote for C.

G: sum of ages of girls
g: number of girls
B: sum of ages of boys
b: number of boys

Given: G = 17+B

St1. B/b = G/g+11
St2. B/b = 16

Combining 1 and 2

16 = G/g + 11
G/g = 5
G =5g

B = 16b

5g = 17 + 16b


Now we need to find a minimum value.
Plug in values for b

When b = 0, 1 and 2
g will not be an integer.

Only when b = 3, g will be 13.

So the minimum value for b=3
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by vittalgmat » Wed Apr 15, 2009 9:13 pm
Thanks Mals for the excellent explanation.
I went thru the same path!!!, and took more than 2 mins :-( to
arrive at the answer.

I think this is a 700 level question.
What do u guys think ?

rgds
-V
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