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by bblast » Thu Jan 13, 2011 6:58 am
There are two schools in the village. The average age of pupils in the first school is 12.2 years; the average age of pupils in the second school is 13.1 years. What is the average age of all school pupils in the village?

1. There are 40 more pupils in the second school than there are in the first.
2. There are three times as many pupils in the second school as there are in the first.
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Source: — Data Sufficiency |
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by Night reader » Thu Jan 13, 2011 7:43 am
bblast wrote:There are two schools in the village. The average age of pupils in the first school is 12.2 years; the average age of pupils in the second school is 13.1 years. What is the average age of all school pupils in the village?

1. There are 40 more pupils in the second school than there are in the first.
2. There are three times as many pupils in the second school as there are in the first.
st(1) A=B-40, [A*12.2 + 13.1(A+40)] /(2A + 40) <=> (25.3A + 13.1*40)/(2A+40) Not Sufficient
st(2) 3A=B, (12.2A+3*13.1A)/4A Sufficient

B
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by bblast » Thu Jan 13, 2011 8:13 am
thats correct
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by MAAJ » Thu Jan 13, 2011 8:24 am
The above is right [spoiler](B)[/spoiler]

From the question steam:

(SUM1/n1) = 12.2 AND (SUM2/n2) = 13.1

What is (SUM1 + SUM2)/(n1+n2) ?

1) Not sufficient

2) (SUM1/n1) = 12.2 AND (SUM2/3n1) = 13.1

THUS SUM1 = 12.2(n1); SUM2 = 36.6(n1); (n1+n2) = 4n1

12.2(n1) + 39.3(n1) / 4(n1)
51.5(n1)/4(n1)
51.5/4
12.875 -> Sufficient
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