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If 2 different representatives are to be selected at random [#permalink]
31 Jan 2010, 23:19

2

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D

E

Difficulty:

(N/A)

Question Stats:

64% (01:25) correct
36% (01:05) wrong based on 41 sessions

1. If 2 different representatives are to be selected at random from a group of 10 employees and if p is the probability that both representatives selected will be women, is p > 1/2

(1) More than 1/2 of the 10 employees are women. (2) The probability that both representatives selected will be men is less than 1/10

Answer : E

2. If x and y are positive, is x3 > y? (1) x > y (2) x > y Answer: E

3. What is the value of the integer k? (1) k + 3 > 0 (2) k4 ≤ 0 Answer: B

4. If the units digit of the three-digit positive integer k is nonzero, what is the tens digit of k? (1) The tens digit of k + 9 is 3. (2) The tens digit of k + 4 is 2. Answer: A _________________

Re: difficulty faced during test [#permalink]
01 Feb 2010, 05:22

3

This post received KUDOS

Expert's post

1. If 2 different representatives are to be selected at random from a group of 10 employees and if p is the probability that both representatives selected will be women, is p > 1/2

(1) More than 1/2 of the 10 employees are women. (2) The probability that both representatives selected will be men is less than 1/10

What is the probability of choosing 2 women out of 10 people \(\frac{w}{10}*\frac{w-1}{9}\) and this should be \(>1/2\). So we have \(\frac{w}{10}*\frac{w-1}{9}>\frac{1}{2}\) --> \(w(w-1)>45\) this is true only when \(w>7\). (w # of women \(<=10\))

So basically question asks is \(w>7\)?

(1) \(w>5\) not sufficient.

(2) \(\frac{10-w}{10}*\frac{10-w-1}{9}<\frac{1}{10}\) --> \((10-w)(9-w)<9\) --> \(w>6\), not sufficient

(1)+(2) \(w>5\), \(w>6\) not sufficient

Answer E.

You can use Combinations, to solve as well:

\(C^2_w\) # of selections of 2 women out of \(w\) employees;

\(C^2_{10}\) total # of selections of 2 representatives out of 10 employees.

Q is \(\frac{C^2_w}{C^2_{10}}>\frac{1}{2}\) --> \(\frac{\frac{w(w-1)}{2}}{45}>\frac{1}{2}\) --> --> \(w(w-1)>45\) --> \(w>7\)?

(1) \(w>5\), not sufficient.

(2) \(C^2_{(10-w)}\) # of selections of 2 men out of \(10-w=m\) employees --> \(\frac{C^2_{(10-w)}}{C^2_{10}}<\frac{1}{10}\) --> \(\frac{\frac{(10-w)(10-w-1)}{2}}{45}<\frac{1}{10}\) --> \((10-w)(9-w)<9\) --> \(w>6\), not sufficient

(1)+(2) \(w>5\), \(w>6\) not sufficient

Answer E.

3. What is the value of the integer k?

(1) k + 3 > 0 --> k>-3, not sufficient to determine single numerical value of k.

(2) k^4 ≤ 0, as number in even power is never negative k^4 can not be less than zero, thus k^4=0 --> k=0. sufficient.

Answer: B

4. If the units digit of the three-digit positive integer k is nonzero, what is the tens digit of k? (1) The tens digit of k + 9 is 3. (2) The tens digit of k + 4 is 2.

k=abc, c not zero, question b=?

(1) The tens digit of k + 9 is 3: abc +9 --- a3x

Tens digit of k+9, which is 3, gains 1 unit from c+9 (as c is not zero), hence b+1=3 --> b=2. Sufficient.

(2) The tens digit of k + 4 is 2 abc +4 --- a2x

Tens digit of k + 4, which is 2, may or may not gain 1 unit from c+4, hence either b+1=2 or b+0=2. Two different values for b. Not sufficient.

Answer: A.

Check the statements for question 2. _________________

Re: If 2 different representatives are to be selected at random [#permalink]
24 Jun 2012, 22:59

Hi Bunuel, Request you to clarify my below doubt-

In Choice B, can we flip the statement and say that : P(both Women) > 9/10 ; Since P(both men) < 1/10 (Given)

I know this wrong but not able to convince myself.

Just found out the reason that above mentioned statement is wrong because/ Can we say that - Since Total probability = 1 and Total Probability must be equal to P(2M) + P(2w) + P(M,W) + (W,M) = 1 Therefore, P(both Women) > 9/10 ---This is Wrong. Please evaluate my reasoning. If it is wrong, I would like to seek your inputs on the same.

Thanks H

Bunuel wrote:

1. If 2 different representatives are to be selected at random from a group of 10 employees and if p is the probability that both representatives selected will be women, is p > 1/2

(1) More than 1/2 of the 10 employees are women. (2) The probability that both representatives selected will be men is less than 1/10

What is the probability of choosing 2 women out of 10 people \(\frac{w}{10}*\frac{w-1}{9}\) and this should be \(>1/2\). So we have \(\frac{w}{10}*\frac{w-1}{9}>\frac{1}{2}\) --> \(w(w-1)>45\) this is true only when \(w>7\). (w # of women \(<=10\))

So basically question asks is \(w>7\)?

(1) \(w>5\) not sufficient.

(2) \(\frac{10-w}{10}*\frac{10-w-1}{9}<\frac{1}{10}\) --> \((10-w)(9-w)<9\) --> \(w>6\), not sufficient

Re: If 2 different representatives are to be selected at random [#permalink]
25 Jun 2012, 01:16

Expert's post

imhimanshu wrote:

Hi Bunuel, Request you to clarify my below doubt-

In Choice B, can we flip the statement and say that : P(both Women) > 9/10 ; Since P(both men) < 1/10 (Given)

I know this wrong but not able to convince myself.

Just found out the reason that above mentioned statement is wrong because/ Can we say that - Since Total probability = 1 and Total Probability must be equal to P(2M) + P(2w) + P(M,W) + (W,M) = 1 Therefore, P(both Women) > 9/10 ---This is Wrong. Please evaluate my reasoning. If it is wrong, I would like to seek your inputs on the same.

Thanks H

Bunuel wrote:

1. If 2 different representatives are to be selected at random from a group of 10 employees and if p is the probability that both representatives selected will be women, is p > 1/2

(1) More than 1/2 of the 10 employees are women. (2) The probability that both representatives selected will be men is less than 1/10

What is the probability of choosing 2 women out of 10 people \(\frac{w}{10}*\frac{w-1}{9}\) and this should be \(>1/2\). So we have \(\frac{w}{10}*\frac{w-1}{9}>\frac{1}{2}\) --> \(w(w-1)>45\) this is true only when \(w>7\). (w # of women \(<=10\))

So basically question asks is \(w>7\)?

(1) \(w>5\) not sufficient.

(2) \(\frac{10-w}{10}*\frac{10-w-1}{9}<\frac{1}{10}\) --> \((10-w)(9-w)<9\) --> \(w>6\), not sufficient

(1)+(2) \(w>5\), \(w>6\) not sufficient

Answer E.

When selecting two representatives only 3, mutually exclusive, cases are possible: 1. Both are men; 2. Both are women; 3. One is a man and another is a woman.

The sum of the probabilities of these cases must be 1. So, knowing that P(Both Men)<1/10 does not necessarily mean that P(Both Women)>9/10. _________________

Re: difficulty faced during test [#permalink]
26 Jul 2012, 21:51

Bunuel wrote:

1. If 2 different representatives are to be selected at random from a group of 10 employees and if p is the probability that both representatives selected will be women, is p > 1/2

(1) More than 1/2 of the 10 employees are women. (2) The probability that both representatives selected will be men is less than 1/10

What is the probability of choosing 2 women out of 10 people \(\frac{w}{10}*\frac{w-1}{9}\) and this should be \(>1/2\). So we have \(\frac{w}{10}*\frac{w-1}{9}>\frac{1}{2}\) --> \(w(w-1)>45\) this is true only when \(w>7\). (w # of women \(<=10\))

So basically question asks is \(w>7\)?

(1) \(w>5\) not sufficient.

(2) \(\frac{10-w}{10}*\frac{10-w-1}{9}<\frac{1}{10}\) --> \((10-w)(9-w)<9\) --> \(w>6\), not sufficient

(1)+(2) \(w>5\), \(w>6\) not sufficient

Answer E.

You can use Combinations, to solve as well:

\(C^2_w\) # of selections of 2 women out of \(w\) employees;

\(C^2_{10}\) total # of selections of 2 representatives out of 10 employees.

Q is \(\frac{C^2_w}{C^2_{10}}>\frac{1}{2}\) --> \(\frac{\frac{w(w-1)}{2}}{45}>\frac{1}{2}\) --> --> \(w(w-1)>45\) --> \(w>7\)?

(1) \(w>5\), not sufficient.

(2) \(C^2_{(10-w)}\) # of selections of 2 men out of \(10-w=m\) employees --> \(\frac{C^2_{(10-w)}}{C^2_{10}}<\frac{1}{10}\) --> \(\frac{\frac{(10-w)(10-w-1)}{2}}{45}<\frac{1}{10}\) --> \((10-w)(9-w)<9\) --> \(w>6\), not sufficient

(1)+(2) \(w>5\), \(w>6\) not sufficient

Answer E.

3. What is the value of the integer k?

(1) k + 3 > 0 --> k>-3, not sufficient to determine single numerical value of k.

(2) k^4 ≤ 0, as number in even power is never negative k^4 can not be less than zero, thus k^4=0 --> k=0. sufficient.

Answer: B

4. If the units digit of the three-digit positive integer k is nonzero, what is the tens digit of k? (1) The tens digit of k + 9 is 3. (2) The tens digit of k + 4 is 2.

k=abc, c not zero, question b=?

(1) The tens digit of k + 9 is 3: abc +9 --- a3x

Tens digit of k+9, which is 3, gains 1 unit from c+9 (as c is not zero), hence b+1=3 --> b=2. Sufficient.

(2) The tens digit of k + 4 is 2 abc +4 --- a2x

Tens digit of k + 4, which is 2, may or may not gain 1 unit from c+4, hence either b+1=2 or b+0=2. Two different values for b. Not sufficient.

Answer: A.

Check the statements for question 2.

Hi, regarding question 1, In statement two, how do we know that (10-w)(9-w)<9? Do we do it by trial and error?

Re: difficulty faced during test [#permalink]
21 Aug 2012, 14:09

reagan wrote:

Bunuel wrote:

1. If 2 different representatives are to be selected at random from a group of 10 employees and if p is the probability that both representatives selected will be women, is p > 1/2

(1) More than 1/2 of the 10 employees are women. (2) The probability that both representatives selected will be men is less than 1/10

What is the probability of choosing 2 women out of 10 people \(\frac{w}{10}*\frac{w-1}{9}\) and this should be \(>1/2\). So we have \(\frac{w}{10}*\frac{w-1}{9}>\frac{1}{2}\) --> \(w(w-1)>45\) this is true only when \(w>7\). (w # of women \(<=10\))

So basically question asks is \(w>7\)?

(1) \(w>5\) not sufficient.

(2) \(\frac{10-w}{10}*\frac{10-w-1}{9}<\frac{1}{10}\) --> \((10-w)(9-w)<9\) --> \(w>6\), not sufficient

(1)+(2) \(w>5\), \(w>6\) not sufficient

Answer E.

You can use Combinations, to solve as well:

\(C^2_w\) # of selections of 2 women out of \(w\) employees;

\(C^2_{10}\) total # of selections of 2 representatives out of 10 employees.

Q is \(\frac{C^2_w}{C^2_{10}}>\frac{1}{2}\) --> \(\frac{\frac{w(w-1)}{2}}{45}>\frac{1}{2}\) --> --> \(w(w-1)>45\) --> \(w>7\)?

(1) \(w>5\), not sufficient.

(2) \(C^2_{(10-w)}\) # of selections of 2 men out of \(10-w=m\) employees --> \(\frac{C^2_{(10-w)}}{C^2_{10}}<\frac{1}{10}\) --> \(\frac{\frac{(10-w)(10-w-1)}{2}}{45}<\frac{1}{10}\) --> \((10-w)(9-w)<9\) --> \(w>6\), not sufficient

(1)+(2) \(w>5\), \(w>6\) not sufficient

Answer E.

3. What is the value of the integer k?

(1) k + 3 > 0 --> k>-3, not sufficient to determine single numerical value of k.

(2) k^4 ≤ 0, as number in even power is never negative k^4 can not be less than zero, thus k^4=0 --> k=0. sufficient.

Answer: B

4. If the units digit of the three-digit positive integer k is nonzero, what is the tens digit of k? (1) The tens digit of k + 9 is 3. (2) The tens digit of k + 4 is 2.

k=abc, c not zero, question b=?

(1) The tens digit of k + 9 is 3: abc +9 --- a3x

Tens digit of k+9, which is 3, gains 1 unit from c+9 (as c is not zero), hence b+1=3 --> b=2. Sufficient.

(2) The tens digit of k + 4 is 2 abc +4 --- a2x

Tens digit of k + 4, which is 2, may or may not gain 1 unit from c+4, hence either b+1=2 or b+0=2. Two different values for b. Not sufficient.

Answer: A.

Check the statements for question 2.

Hi, regarding question 1, In statement two, how do we know that (10-w)(9-w)<9? Do we do it by trial and error?

Reagan

You cross-multiply LHS to RHS and divide by 10. So you get left with the numerators.

Re: If 2 different representatives are to be selected at random [#permalink]
13 Apr 2013, 15:17

eski wrote:

How is (10-w)(9-w) < 9 leads to w>6 ????

\(w^2-19w+81<0\) has two roots: \((19+-\sqrt{37})/2\) w1=(almost)\(\frac{19+6}{2}=\frac{25}{2}=12.5\) w2=(almost)\(\frac{19-6}{2}=\frac{13}{2}=6.5\) w1=12.5 w2=6.5 and the solution is 6.5<w<12.5 Remember that w is the number of women, so because we have 10 employees, the solution is \(6,5<w\leq{10}\) But because we cannot select 6 women and 1/2, the solution of the equation is women>6.

Hope it makes sense now _________________

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