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A committee consists of n women and k men. In addition there

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A committee consists of n women and k men. In addition there  [#permalink]

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Updated on: 17 Feb 2012, 04:19
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A committee consists of n women and k men. In addition there are 4 alternates, 2 of whom are women. If one of the committee members is selected at random to be replaced by one of the alternates, also selected at random, what is the probability that the number of women on the committee will increase?

(1) n + k = 12
(2) k/n = 1/3

Originally posted by tom09b on 15 Feb 2012, 10:07.
Last edited by tom09b on 17 Feb 2012, 04:19, edited 2 times in total.
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Re: A committee consists of n women and k men. In addition there  [#permalink]

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15 Feb 2012, 11:20
4
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A committee consists of n women and k men. In addition there are 4 alternates, 2 of whom are women. If one of the committee members is selected at random to be replaced by one of the alternates, also selected at random, what is the probabilty that the number of women on the committee will increase?

Notice that the probability of selecting a woman from 4 alternates is 1/2.

Now, the number of women on the committee will increase if we replace a man from the committee with a woman from 4 alternates. Hence, the probabilty that the number of women on the committee will increase is k/(k+n)*1/2.

(1) n + k = 12. Not sufficient as we can not get the probability of selecting a man from the committee.
(2) k/n = 1/3 --> k/(k+n)=1/4. Sufficient.

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Re: A committee consists of n women and k men. In addition there  [#permalink]

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07 Jun 2013, 08:17
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tom09b wrote:
A committee consists of n women and k men. In addition there are 4 alternates, 2 of whom are women. If one of the committee members is selected at random to be replaced by one of the alternates, also selected at random, what is the probabilty that the number of women on the committee will increase?

(1) n + k = 12
(2) k/n = 1/3

Rewording of the question:
What is the probability that a man is chosen to be replaced and the alternate to replace him is a woman.

What you need is (probability of man chosen) x (probability of woman alternate)

$$\frac{k}{k+n} * \frac{2}{4} = \frac{k}{2(k+n)}$$

(1) n+k = 12
if n=1 and k=11, $$\frac{11}{2(12)} = \frac{11}{24}$$
if n=2 and k=10, $$\frac{10}{2(12)} = \frac{10}{24}$$
insufficient

(2) $$\frac{k}{n} = \frac{1}{3}$$
if k=1 and n=3, $$\frac{1}{2(4)} = \frac{1}{8}$$
if k=2 and n=6, $$\frac{2}{2(8)} = \frac{1}{8}$$; etc...
sufficient

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Re: A committee consists of n women and k men. In addition there  [#permalink]

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15 Feb 2012, 11:31
Bunuel wrote:
A committee consists of n women and k men. In addition there are 4 alternates, 2 of whom are women. If one of the committee members is selected at random to be replaced by one of the alternates, also selected at random, what is the probabilty that the number of women on the committee will increase?

Notice that the probability of selecting a woman from 4 alternates is 1/2.

Now, the number of women on the committee will increase if we replace a man from the committee with a woman from 4 alternates. Hence, the probabilty that the number of women on the committee will increase is k/(k+n)*1/2.

(1) n + k = 12. Not sufficient as we can not get the probability of selecting a man from the committee.
(2) k/n = 1/3 --> k/(k+n)=1/4. Sufficient.

My intuition drove me to B, as well but.. I couldn't find the way! Thank you!!
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Re: A committee consists of n women and k men. In addition there  [#permalink]

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03 Jul 2012, 08:40
1
Here is how I analyzed it if it helps:

The probability of selecting a woman from the alternates as given is - (2/4) = (1/2)
The probability of selecting a woman from the committee is - n/(n+k)

Now, we need to figure out the probability of pick a woman from the committee AND from the alternates [P(W&W)]. Therefore this is an AND problem.

1. n/12 Insufficient
2. k/n=1/3. Therefore n/(n+k)=3/4
Sufficient because P(W&W)=(3/4)*(1/2)
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Re: A committee consists of n women and k men. In addition there  [#permalink]

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07 Jun 2013, 05:12
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Bumping for review and further discussion*. Get a kudos point for an alternative solution!

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Re: A committee consists of n women and k men. In addition there  [#permalink]

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20 Mar 2014, 04:34
Bunuel wrote:
A committee consists of n women and k men. In addition there are 4 alternates, 2 of whom are women. If one of the committee members is selected at random to be replaced by one of the alternates, also selected at random, what is the probabilty that the number of women on the committee will increase?

Notice that the probability of selecting a woman from 4 alternates is 1/2.

Now, the number of women on the committee will increase if we replace a man from the committee with a woman from 4 alternates. Hence, the probabilty that the number of women on the committee will increase is k/(k+n)*1/2.

(1) n + k = 12. Not sufficient as we can not get the probability of selecting a man from the committee.
(2) k/n = 1/3 --> k/(k+n)=1/4. Sufficient.

. Hence, the probabilty that the number of women on the committee will increase is k/(k+n)*1/2.
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Re: A committee consists of n women and k men. In addition there  [#permalink]

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20 Mar 2014, 04:57
tusharGupta1 wrote:
Bunuel wrote:
A committee consists of n women and k men. In addition there are 4 alternates, 2 of whom are women. If one of the committee members is selected at random to be replaced by one of the alternates, also selected at random, what is the probabilty that the number of women on the committee will increase?

Notice that the probability of selecting a woman from 4 alternates is 1/2.

Now, the number of women on the committee will increase if we replace a man from the committee with a woman from 4 alternates. Hence, the probabilty that the number of women on the committee will increase is k/(k+n)*1/2.

(1) n + k = 12. Not sufficient as we can not get the probability of selecting a man from the committee.
(2) k/n = 1/3 --> k/(k+n)=1/4. Sufficient.

. Hence, the probabilty that the number of women on the committee will increase is k/(k+n)*1/2.

For the number of women on the committee to increase we must replace a man from the committee with a woman from 4 alternates.

The probability of selecting a man from the committee is (# of men in the committee)/(total # of people in the committee) = k/(k+n);
The probability of selecting a woman from 4 alternates is (# of women)/(total # of people) = 2/4 = 1/2.

Therefore, the probability that we select a man from the committee AND a woman from 4 alternates is k/(k+n)*1/2.

Hope it's clear.
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Re: A committee consists of n women and k men. In addition there  [#permalink]

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20 Mar 2014, 05:33
Bunuel wrote:
tusharGupta1 wrote:
Bunuel wrote:
A committee consists of n women and k men. In addition there are 4 alternates, 2 of whom are women. If one of the committee members is selected at random to be replaced by one of the alternates, also selected at random, what is the probabilty that the number of women on the committee will increase?

Notice that the probability of selecting a woman from 4 alternates is 1/2.

Now, the number of women on the committee will increase if we replace a man from the committee with a woman from 4 alternates. Hence, the probabilty that the number of women on the committee will increase is k/(k+n)*1/2.

(1) n + k = 12. Not sufficient as we can not get the probability of selecting a man from the committee.
(2) k/n = 1/3 --> k/(k+n)=1/4. Sufficient.

. Hence, the probabilty that the number of women on the committee will increase is k/(k+n)*1/2.

For the number of women on the committee to increase we must replace a man from the committee with a woman from 4 alternates.

The probability of selecting a man from the committee is (# of men in the committee)/(total # of people in the committee) = k/(k+n);
The probability of selecting a woman from 4 alternates is (# of women)/(total # of people) = 2/4 = 1/2.

Therefore, the probability that we select a man from the committee AND a woman from 4 alternates is k/(k+n)*1/2.

Hope it's clear.

Thankx a ton ............................................................................................................................................................
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Re: A committee consists of n women and k men. In addition there  [#permalink]

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12 May 2014, 04:12
Bunuel wrote:
tusharGupta1 wrote:
Bunuel wrote:
A committee consists of n women and k men. In addition there are 4 alternates, 2 of whom are women. If one of the committee members is selected at random to be replaced by one of the alternates, also selected at random, what is the probabilty that the number of women on the committee will increase?

Notice that the probability of selecting a woman from 4 alternates is 1/2.

Now, the number of women on the committee will increase if we replace a man from the committee with a woman from 4 alternates. Hence, the probabilty that the number of women on the committee will increase is k/(k+n)*1/2.

(1) n + k = 12. Not sufficient as we can not get the probability of selecting a man from the committee.
(2) k/n = 1/3 --> k/(k+n)=1/4. Sufficient.

. Hence, the probabilty that the number of women on the committee will increase is k/(k+n)*1/2.

For the number of women on the committee to increase we must replace a man from the committee with a woman from 4 alternates.

The probability of selecting a man from the committee is (# of men in the committee)/(total # of people in the committee) = k/(k+n);
The probability of selecting a woman from 4 alternates is (# of women)/(total # of people) = 2/4 = 1/2.

Therefore, the probability that we select a man from the committee AND a woman from 4 alternates is k/(k+n)*1/2.

Hope it's clear.

Dear Bunnel

Statement A: was clearly insufficient as we didnt know n & k individually.
statement B:
ratio of m:w = 1:3

so suppose, 1 m and 3 women r there in the committee

and if we replace the man with women, then the no. of women will increase.

now we have no man just 4 women..

probability of a new alternate as women in the committee: 3/4*1/2 = 3/8... lesser than 3/4

but if we replace the man with a man... there should be no change right as there will be 1 man and 3 women still
ans is diff therefore insufficient..

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Re: A committee consists of n women and k men. In addition there  [#permalink]

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12 May 2014, 07:53
nandinigaur wrote:

Dear Bunnel

Statement A: was clearly insufficient as we didnt know n & k individually.
statement B:
ratio of m:w = 1:3

so suppose, 1 m and 3 women r there in the committee

and if we replace the man with women, then the no. of women will increase.

now we have no man just 4 women..

probability of a new alternate as women in the committee: 3/4*1/2 = 3/8... lesser than 3/4

but if we replace the man with a man... there should be no change right as there will be 1 man and 3 women still
ans is diff therefore insufficient..

Don't understand what is your question...

The question asks what is the probability that the number of women on the committee will increase? The probability that the number of women on the committee will increase is k/(k+n)*1/2.

From (2) we get that k/(k+n)*1/2 = 1/4*1/2 = 1/8.
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Re: A committee consists of n women and k men. In addition there  [#permalink]

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01 Nov 2015, 12:56
unnecessary to get a number:

in order to increase the number of women, the only way is to replace a man with a woman

1. if a woman is replaced by a woman, the number will remain;

2. if a woman is replaced by a man, the number will decrease;

3. if a man is replaced by a man, the number will remain

so the possibility is:

(k/n+k)*(2/4), the key is the ratio of woman to man

(1) n+k=12, insufficient

(2)k/n=1/3, so k/n+k=1/4, sufficient

B
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Re: A committee consists of n women and k men. In addition there  [#permalink]

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17 Oct 2016, 07:01
Hi Bunuel,
Why does k/(k+n)*1/2 will represent the number of woman increase? How it relates to increment. Will you explain it?
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Re: A committee consists of n women and k men. In addition there  [#permalink]

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17 Oct 2016, 07:05
NaeemHasan wrote:
Hi Bunuel,
Why does k/(k+n)*1/2 will represent the number of woman increase? How it relates to increment. Will you explain it?

For the number of women on the committee to increase we must replace a man from the committee with a woman from 4 alternates.

The probability of selecting a man from the committee is (# of men in the committee)/(total # of people in the committee) = k/(k+n);
The probability of selecting a woman from 4 alternates is (# of women)/(total # of people) = 2/4 = 1/2.

Therefore, the probability that we select a man from the committee AND a woman from 4 alternates is k/(k+n)*1/2.

Hope it's clear.
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Re: A committee consists of n women and k men. In addition there  [#permalink]

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02 Jun 2018, 21:56
Given,
A committee of
n - women
k - men

Alternate choices, in case of replacement, available are 2 women & 2 men.

The probability of increasing the # of women, is by replacement of 1 man in the committee with 1 woman from the alternates.

Consider this as a selection of 2 people, one man from k men & one woman from the 2 alternates.

#of ways selecting 1 man from k men = k
#of way of selecting 1 woman from 2 women = 2

Required Probability = Probability of selecting 1 man from (n+k) members * Probability of selecting 1 woman from 4 alternates

Therefore required Probability is $${k/(n+k)}*{2/4}$$

Ok now lets check

Statement 1 : $$n+k = 12$$, clearly not sufficient, as n=10, k= 2 or n=7, k=5, or many other combinations.

Statement 2: $$k/n =1/3$$

which is $$n/k = 3/1$$

By adding 1 to each side can be converted to

$$(n+k)/k = 4/1$$

$$k/(n+k) = 1/4$$

Hence statement 2 is sufficient to find the required probability.
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Re: A committee consists of n women and k men. In addition there &nbs [#permalink] 02 Jun 2018, 21:56
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