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09 May 2017, 02:56
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
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62% (01:22) correct
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How many different teams of 3 boys can be made from the class of 10 children?
(1) Twenty-four teams of three can be made of only girls.
(2) The probability of randomly selecting a boy from the class is one half.
(1) Twenty-four teams of three can be made of only girls.
(2) The probability of randomly selecting a boy from the class is one half.
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09 May 2017, 06:34
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If Probability of selecting a boy is 1/2, then total boys must be 5/10=1/2. Now from 5 boys, 3 can be selected in 5C3 ways, which equals 10.
First statement doesnt look sufficient. Hence I would go with option B(second is enough to answer)
First statement doesnt look sufficient. Hence I would go with option B(second is enough to answer)
09 May 2017, 08:13
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Bunuel
How many different teams of 3 boys can be made from the class of 10 children?
(1) Twenty-four teams of three can be made of only girls.
(2) The probability of randomly selecting a boy from the class is one half.
(1) Twenty-four teams of three can be made of only girls.
(2) The probability of randomly selecting a boy from the class is one half.
(1) Call \(k\) is the number of girls in that class.
We have
\(C^3_k=24 \implies \frac{k!}{3!(k-3)!}=24 \implies \frac{k(k-1)(k-2)}{6}=24 \implies k(k-1)(k-2)=6\times 6 \times 4\)
Note that \(6\times 6=9 \times 4\), so the right side is divisible by 9. This means the left side must be divisible by 9.
Hence \(k \geq 9\).
However, \((k-1)(k-2) \geq 8 \times 7 > 4 \times 4 \implies k(k-1)(k-2)>6 \times 24\). This means that there is no \(k\) satisfies the equation. Insufficient.
(2) We have 50% students in that class are boy. Hence, there are 5 boys in that class. We can form \(5C3\) different teams of 3 boys. Sufficient.
The answer is B.
