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# If 2 different representatives are to be selected at random

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If 2 different representatives are to be selected at random [#permalink]

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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

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

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

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.
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Re: difficulty faced during test [#permalink]

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01 Feb 2010, 06:22
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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

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

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.

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.

Check the statements for question 2.
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Re: difficulty faced during test [#permalink]

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02 Feb 2010, 09:45
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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 [#permalink]

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08 Feb 2010, 12:32
1
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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.

so E
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Re: difficulty faced during test [#permalink]

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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 [#permalink]

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10 Feb 2010, 10:48
1
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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.

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

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Re: difficulty faced during test [#permalink]

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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 [#permalink]

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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.

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Re: difficulty faced during test [#permalink]

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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.

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Re: If 2 different representatives are to be selected at random [#permalink]

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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 [#permalink]

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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 $$\frac{w}{10}*\frac{w-1}{9}$$ 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

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Re: If 2 different representatives are to be selected at random [#permalink]

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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 $$\frac{w}{10}*\frac{w-1}{9}$$ 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

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 [#permalink]

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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 $$\frac{w}{10}*\frac{w-1}{9}$$ 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

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

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.

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.

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?

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Re: difficulty faced during test [#permalink]

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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 $$\frac{w}{10}*\frac{w-1}{9}$$ 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

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

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.

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.

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.
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Re: If 2 different representatives are to be selected at random [#permalink]

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13 Apr 2013, 15:59
How is (10-w)(9-w) < 9 leads to w>6 ????
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Re: If 2 different representatives are to be selected at random [#permalink]

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13 Apr 2013, 16:17
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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Re: If 2 different representatives are to be selected at random [#permalink]

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13 Apr 2013, 16:24
Re: If 2 different representatives are to be selected at random   [#permalink] 13 Apr 2013, 16:24
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