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If 2 different representatives are to be selected at random
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01 Feb 2010, 00:19
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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 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 threedigit 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
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Re: difficulty faced during test
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01 Feb 2010, 06:22
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 \(\frac{w}{10}*\frac{w1}{9}\) and this should be \(>1/2\). So we have \(\frac{w}{10}*\frac{w1}{9}>\frac{1}{2}\) > \(w(w1)>45\) this is true only when \(w>7\). (w # of women \(<=10\)) So basically question asks is \(w>7\)? (1) \(w>5\) not sufficient. (2) \(\frac{10w}{10}*\frac{10w1}{9}<\frac{1}{10}\) > \((10w)(9w)<9\) > \(w>6\), 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 \(\frac{C^2_w}{C^2_{10}}>\frac{1}{2}\) > \(\frac{\frac{w(w1)}{2}}{45}>\frac{1}{2}\) > > \(w(w1)>45\) > \(w>7\)? (1) \(w>5\), not sufficient. (2) \(C^2_{(10w)}\) # of selections of 2 men out of \(10w=m\) employees > \(\frac{C^2_{(10w)}}{C^2_{10}}<\frac{1}{10}\) > \(\frac{\frac{(10w)(10w1)}{2}}{45}<\frac{1}{10}\) > \((10w)(9w)<9\) > \(w>6\), 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 threedigit 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.
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Re: difficulty faced during test
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02 Feb 2010, 09:45
thanks for claering my doubts. Q) If x and y are positive, is x^3 > y? (1) \sqrt{x} > y (/ is formating error; read it sqare root of X is greater than Y) (2) x > y
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Re: difficulty faced during test
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08 Feb 2010, 12:32
For your #2: x & y positive. x^3 > y? (1) x^(1/2) > y i) take x = 1/4, and y = 1/3 so x^(1/2) = 1/2 > 1/3... but you can see that (1/4)^3 is NOT > 1/3... ii) and x = 4 and y = 1 so x^(1/2) = 2 > 1... and 4^3 is obviously > 1. we answered x^3 > y both NO and YES, so contradiction. Statement (1) not sufficient (2) x > y i) simply take fractions (x = 1/2 and y = 1/4... x^3 = 1/8) and you'll see that x^3 NOT > y ii) take numbers > 1 and you'll get x^3 > y. contradiction again... so E
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Re: difficulty faced during test
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10 Feb 2010, 04:37
WHY THIS GUY NEVER USE SPOILER ? Either give OA after some discussion or use spoiler



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Re: difficulty faced during test
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10 Feb 2010, 10:48
adalfu wrote: For your #2:
x & y positive. x^3 > y?
(1) x^(1/2) > y i) take x = 1/4, and y = 1/3 so x^(1/2) = 1/2 > 1/3... but you can see that (1/4)^3 is NOT > 1/3...
ii) and x = 4 and y = 1 so x^(1/2) = 2 > 1... and 4^3 is obviously > 1.
we answered x^3 > y both NO and YES, so contradiction. Statement (1) not sufficient
(2) x > y i) simply take fractions (x = 1/2 and y = 1/4... x^3 = 1/8) and you'll see that x^3 NOT > y
ii) take numbers > 1 and you'll get x^3 > y.
contradiction again...
so E doh, i forgot the (together): i.e., taking both statements together... same story: (1) x>y and (2) x^(1/2) > y take x = 1/4 and y = 1/5 satisfying both (1) and (2) above. you can see that x^3 > y => Not true take x = 4 and y = 1 implies x^3 > y => True contradiction
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Re: difficulty faced during test
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21 Mar 2011, 12:58
What is the source of these questions?



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Re: difficulty faced during test
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21 Mar 2011, 17:34
4.
1. Sufficient to know that there was a carry over to tens place. enough to answer the question.
2. Not sufficient . as we don't know whether there was a carry over from units place or not.
Answer A.



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Re: difficulty faced during test
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21 Mar 2011, 17:40
3.
1. Not sufficient K+3 >0 and k could be any positive integer,
2. Sufficient. (K^4) <=0
k cannot be ve or +ve , as even power of any positive integer is greater than 0 , the above condition will never be met.
k =0 satisfies the condition and enough to answer the question.
Answer B.



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Re: If 2 different representatives are to be selected at random
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19 Nov 2011, 02:45
Clearly E, after plugging in the numbers, there is an insufficiency in both the statements.
Eg: 1) Try x= 25, y= 4 (YES) & x= .04 & y =.3 (NO) 2) Try x= 25, y= 4 (YES) & x= .2, y= .1 (NO) (Some of the values that i randomly tried)



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Re: If 2 different representatives are to be selected at random
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24 Jun 2012, 23:59
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 \(\frac{w}{10}*\frac{w1}{9}\) and this should be \(>1/2\). So we have \(\frac{w}{10}*\frac{w1}{9}>\frac{1}{2}\) > \(w(w1)>45\) this is true only when \(w>7\). (w # of women \(<=10\))
So basically question asks is \(w>7\)?
(1) \(w>5\) not sufficient.
(2) \(\frac{10w}{10}*\frac{10w1}{9}<\frac{1}{10}\) > \((10w)(9w)<9\) > \(w>6\), not sufficient
(1)+(2) \(w>5\), \(w>6\) not sufficient
Answer E.



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Re: If 2 different representatives are to be selected at random
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25 Jun 2012, 02:16
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 \(\frac{w}{10}*\frac{w1}{9}\) and this should be \(>1/2\). So we have \(\frac{w}{10}*\frac{w1}{9}>\frac{1}{2}\) > \(w(w1)>45\) this is true only when \(w>7\). (w # of women \(<=10\))
So basically question asks is \(w>7\)?
(1) \(w>5\) not sufficient.
(2) \(\frac{10w}{10}*\frac{10w1}{9}<\frac{1}{10}\) > \((10w)(9w)<9\) > \(w>6\), 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.
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Re: difficulty faced during test
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26 Jul 2012, 22:51
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 \(\frac{w}{10}*\frac{w1}{9}\) and this should be \(>1/2\). So we have \(\frac{w}{10}*\frac{w1}{9}>\frac{1}{2}\) > \(w(w1)>45\) this is true only when \(w>7\). (w # of women \(<=10\))
So basically question asks is \(w>7\)?
(1) \(w>5\) not sufficient.
(2) \(\frac{10w}{10}*\frac{10w1}{9}<\frac{1}{10}\) > \((10w)(9w)<9\) > \(w>6\), 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 \(\frac{C^2_w}{C^2_{10}}>\frac{1}{2}\) > \(\frac{\frac{w(w1)}{2}}{45}>\frac{1}{2}\) > > \(w(w1)>45\) > \(w>7\)?
(1) \(w>5\), not sufficient.
(2) \(C^2_{(10w)}\) # of selections of 2 men out of \(10w=m\) employees > \(\frac{C^2_{(10w)}}{C^2_{10}}<\frac{1}{10}\) > \(\frac{\frac{(10w)(10w1)}{2}}{45}<\frac{1}{10}\) > \((10w)(9w)<9\) > \(w>6\), 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 threedigit 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 (10w)(9w)<9? Do we do it by trial and error? Reagan



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Re: difficulty faced during test
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21 Aug 2012, 15:09
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 \(\frac{w}{10}*\frac{w1}{9}\) and this should be \(>1/2\). So we have \(\frac{w}{10}*\frac{w1}{9}>\frac{1}{2}\) > \(w(w1)>45\) this is true only when \(w>7\). (w # of women \(<=10\))
So basically question asks is \(w>7\)?
(1) \(w>5\) not sufficient.
(2) \(\frac{10w}{10}*\frac{10w1}{9}<\frac{1}{10}\) > \((10w)(9w)<9\) > \(w>6\), 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 \(\frac{C^2_w}{C^2_{10}}>\frac{1}{2}\) > \(\frac{\frac{w(w1)}{2}}{45}>\frac{1}{2}\) > > \(w(w1)>45\) > \(w>7\)?
(1) \(w>5\), not sufficient.
(2) \(C^2_{(10w)}\) # of selections of 2 men out of \(10w=m\) employees > \(\frac{C^2_{(10w)}}{C^2_{10}}<\frac{1}{10}\) > \(\frac{\frac{(10w)(10w1)}{2}}{45}<\frac{1}{10}\) > \((10w)(9w)<9\) > \(w>6\), 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 threedigit 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 (10w)(9w)<9? Do we do it by trial and error? Reagan You crossmultiply LHS to RHS and divide by 10. So you get left with the numerators.



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Re: If 2 different representatives are to be selected at random
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13 Apr 2013, 15:59
How is (10w)(9w) < 9 leads to w>6 ????
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Re: If 2 different representatives are to be selected at random
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13 Apr 2013, 16:17
eski wrote: How is (10w)(9w) < 9 leads to w>6 ???? \(w^219w+81<0\) has two roots: \((19+\sqrt{37})/2\) w1=(almost)\(\frac{19+6}{2}=\frac{25}{2}=12.5\) w2=(almost)\(\frac{196}{2}=\frac{13}{2}=6.5\) 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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