Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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

2

This post received KUDOS

00:00

A

B

C

D

E

Difficulty:

(N/A)

Question Stats:

60% (01:29) correct
40% (01:05) wrong based on 34 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 _________________

It is beyond a doubt that all our knowledge that begins with experience.