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If 2 different representatives are to be selected at random [#permalink]
31 Jan 2010, 23:19
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Difficulty:
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Question Stats:
62% (01:34) correct
38% (01:20) wrong based on 51 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
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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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