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# M70-25

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Math Expert
Joined: 02 Sep 2009
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03 Sep 2018, 04:56
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Question Stats:

75% (00:29) correct 25% (00:00) wrong based on 4 sessions

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If each of the children who participated in a party put one card with his or her name on it into a hat, how many children participated in this party?

(1) The probability of pulling out a card with a girl’s name, when picking a card out of the hat at random, was $$\frac{2}{3}$$.

(2) There were 3 boys at the party.

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03 Sep 2018, 04:56
Official Solution:

We’ll go for LOGICAL since logic is the first option in Data Sufficiency.

(1) This tells us the ratio between girls and boys. But without any exact numbers to work with, we can’t tell how many children there are. (A) and (D) are eliminated.

(2) We know how many boys there were, but how many girls were there? (B) is eliminated.

Combining (1) and (2) gives us the number of boys, and the ratio of boys to girls – two pieces of information that are enough in order to find the number of girls, and thus the total number of children. (E) is eliminated.

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30 Sep 2018, 05:55
How do we know that the ratio is divided in what proportion... there can be 9 students and there can be 18 stufdents Answer should have been E?
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30 Sep 2018, 06:10
How do we know that the ratio is divided in what proportion... there can be 9 students and there can be 18 stufdents Answer should have been E?

If each of the children who participated in a party put one card with his or her name on it into a hat, how many children participated in this party?

(1) The probability of pulling out a card with a girl’s name, when picking a card out of the hat at random, was 2/3.

(2) There were 3 boys at the party.

No of Girls = G

No of Boys = B

St 1 : G/(B + G) = 2/3

So not sufficient to calculate B+G.

St 2 : B =3

again not sufficient to calculate B+G.

St 1 and St 2 together :

B =3 so 3G = 2B + 2G = 6 + 2G

Hence G = 6

B + G = 9

Sufficient.

Does this help?
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Re: M70-25   [#permalink] 30 Sep 2018, 06:10
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# M70-25

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