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Baker's Dozen

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I'm posting the next set of medium/hard PS questions. I'll post OA's with detailed explanations after some discussion. Please, post your solutions along with the answers. Good luck!

1. A password on Mr. Wallace's briefcase consists of 5 digits. What is the probability that the password contains exactly three digit 6?

A. 860/90,000
B. 810/100,000
C. 858/100,000
D. 860/100,000
E. 1530/100,000

Solution: baker-s-dozen-128782-20.html#p1057502

2. If \(y=\frac{(3^5-3^2)^2}{(5^7-5^4)^{-2}}\), then y is NOT divisible by which of the following?
A. 6^4
B. 62^2
C. 65^2
D. 15^4
E. 52^4

Solution: baker-s-dozen-128782-20.html#p1057503

3. For the past k days the average (arithmetic mean) cupcakes per day that Liv baked was 55. Today Bibi joined and together with Liv they baked 100 cupcakes, which raises the average to 60 cupcakes per day. What is the value of k?
A. 6
B. 8
C. 9
D. 10
E. 12

Solution: baker-s-dozen-128782-20.html#p1057504

4. What is the smallest positive integer \(k\) such that \(126*\sqrt{k}\) is the square of a positive integer?
A. 14
B. 36
C. 144
D. 196
E. 441

Solution: baker-s-dozen-128782-20.html#p1057505

5. There are 7 red and 5 blue marbles in a jar. In how many ways 8 marbles can be selected from the jar so that at least one red marble and at least one blue marble to remain in the jar?
A. 460
B. 490
C. 493
D. 455
E. 445

Solution: baker-s-dozen-128782-20.html#p1057507

6. A pool has two water pumps A and B and one drain C. Pump A alone can fill the whole pool in x hours, and pump B alone can fill the whole pool in y hours. The drain can empty the whole pool in z hours, where z>x. With pumps A and B both running and the drain C unstopped till the pool is filled, which of the following represents the amount of water in terms of the fraction of the pool which pump A pumped into the pool?
A. \(\frac{yz}{x+y+z}\)

B. \(\frac{yz}{yz+xz-xy}\)

C. \(\frac{yz}{yz+xz+xy}\)

D. \(\frac{xyz}{yz+xz-xy}\)

E. \(\frac{yz+xz-xy}{yz}\)

Solution: baker-s-dozen-128782-20.html#p1057508

7. Metropolis Corporation has 4 shareholders: Fritz, Luis, Alfred and Werner. Number of shares that Fritz owns is 2/3 rd of number of the shares of the other three shareholders, number of the shares that Luis owns is 3/7 th of number of the shares of the other three shareholders and number of the shares that Alfred owns is 4/11 th of number of the shares of the other three shareholders. If dividends of $3,600,000 were distributed among the 4 shareholders, how much of this amount did Werner receive?
A. $60,000
B. $90,000
C. $100,000
D. $120,000
E. $180,000

Solution: baker-s-dozen-128782-20.html#p1057509

8. A set A consists of 7 consecutive odd integers. If the sum of 5 largest integers of set A is -185 what is the sum of the 5 smallest integers of set A?
A. -165
B. -175
C. -195
D. -205
E. -215

Solution: baker-s-dozen-128782-20.html#p1057512

9. If x and y are negative numbers, what is the value of \(\frac{\sqrt{x^2}}{x}-\sqrt{-y*|y|}\)?
A. 1+y
B. 1-y
C. -1-y
D. y-1
E. x-y

Solution: baker-s-dozen-128782-20.html#p1057514

10. If x^2<81 and y^2<25, what is the largest prime number that can be equal to x-2y?
A. 7
B. 11
C. 13
D. 17
E. 19

Solution: baker-s-dozen-128782-20.html#p1057515

11. In an infinite sequence 1, 3, 9, 27, ... each term after the first is three times the previous term. What is the difference between the sum of 13th and 15th terms and the sum of 12th and 14th terms of the sequence?
A. 10*3^11
B. 20*3^11
C. 10*3^12
D. 40*3^11
E. 20*3^12

Solution: baker-s-dozen-128782-40.html#p1057517

12. x, y and z are positive integers such that when x is divided by y the remainder is 3 and when y is divided by z the remainder is 8. What is the smallest possible value of x+y+z?
A. 12
B. 20
C. 24
D. 29
E. 33

Solution: baker-s-dozen-128782-40.html#p1057519

13. If \(x=\frac{(8!)^{10}-(8!)^6}{(8!)^{5}-(8!)^3}\), what is the product of the tens and the units digits of \(\frac{x}{(8!)^3}-39\)?
A. 0
B. 6
C. 7
D. 12
E. 14

Solution: baker-s-dozen-128782-40.html#p1057520
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New post 29 Jul 2014, 09:16
tyagigar wrote:
Bunuel wrote:
6. A pool has two water pumps A and B and one drain C. Pump A alone can fill the whole pool in x hours, and pump B alone can fill the whole pool in y hours. The drain can empty the whole pool in z hours, where z>x. With pumps A and B both running and the drain C unstopped till the pool is filled, which of the following represents the amount of water in terms of the fraction of the pool which pump A pumped into the pool?
A. \(\frac{yz}{x+y+z}\)

B. \(\frac{yz}{yz+xz-xy}\)

C. \(\frac{yz}{yz+xz+xy}\)

D. \(\frac{xyz}{yz+xz-xy}\)

E. \(\frac{yz+xz-xy}{yz}\)

With pumps A and B both running and the drain unstopped the pool will be filled in a rate \(\frac{1}{x}+\frac{1}{y}-\frac{1}{z}=\frac{yz+xz-zy}{xyz}\) pool/hour. So, the pool will be filled in \(\frac{xyz}{yz+xz-xy}\) hours (time is reciprocal of rate).

In \(\frac{xyz}{yz+xz-xy}\) hours A will pump \(\frac{1}{x}*\frac{xyz}{yz+xz-xy}=\frac{yz}{yz+xz-xy}\) amount of the water into the pool.

Answer: B.


can you please tell me if there is any difeerence in these two statements

amount of water in terms of the fraction of the pool which pump A pumped into the pool?

amount of water which pump A pumped into the pool?

and if there is a difference can you please explain?

i had solved the question but was stumped by the language, is it simply asking me the amount of water pumped by A

thanks


The first question asks about the fraction of the water pumped by A. For example, if the capacity of the pool is 100 gallons and A pumped 50, then the fraction would be 50/100.

The second question is about the amount. In this case the answer would be 50 gallons.
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Collection of Questions:
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DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: Baker's Dozen [#permalink]

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New post 06 Sep 2014, 09:15
With pumps A and B both running and the drain unstopped the pool will be filled in a rate \frac{1}{x}+\frac{1}{y}-\frac{1}{z}=\frac{yz+xz-zy}{xyz} pool/hour. So, there is a typo here,

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New post 07 Sep 2014, 09:30
Bunuel wrote:
SOLUTIONS:

1. A password on Mr. Wallace's briefcase consists of 5 digits. What is the probability that the password contains exactly three digit 6?

A. 860/90,000
B. 810/100,000
C. 858/100,000
D. 860/100,000
E. 1530/100,000

Total # of 5 digit codes is 10^5, notice that it's not 9*10^4, since in a code we can have zero as the first digit.

# of passwords with three digit 6 is \(9*9*C^3_5=810\): each out of two other digits (not 6) has 9 choices, thus we have 9*9 and \(C^3_5\) is ways to choose which 3 digits will be 6's out of 5 digits we have.

\(P=\frac{favorable}{total}=\frac{810}{10^5}\)

Answer: B.



Hello Bunuel.
My P&C area is kind of weak, and I am working on it. In the solution why is it important to pay attention to no. of ways three 6's can be used in the password? Can you please elaborate on why have you used the Combination formula at all in the solution?

Thanks.

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New post 07 Sep 2014, 09:39
p2bhokie wrote:
Bunuel wrote:
SOLUTIONS:

1. A password on Mr. Wallace's briefcase consists of 5 digits. What is the probability that the password contains exactly three digit 6?

A. 860/90,000
B. 810/100,000
C. 858/100,000
D. 860/100,000
E. 1530/100,000

Total # of 5 digit codes is 10^5, notice that it's not 9*10^4, since in a code we can have zero as the first digit.

# of passwords with three digit 6 is \(9*9*C^3_5=810\): each out of two other digits (not 6) has 9 choices, thus we have 9*9 and \(C^3_5\) is ways to choose which 3 digits will be 6's out of 5 digits we have.

\(P=\frac{favorable}{total}=\frac{810}{10^5}\)

Answer: B.



Hello Bunuel.
My P&C area is kind of weak, and I am working on it. In the solution why is it important to pay attention to no. of ways three 6's can be used in the password? Can you please elaborate on why have you used the Combination formula at all in the solution?

Thanks.


We have 5-digit password: ABCDE, out of which there are three 6's. These 6's can take any place: ABC, ABD, ABE, ... \(C^3_5\) gives the number of ways to choose which 3 digits (which 3 letters out of 5) will be 6's:
666DE
66C6E
...
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Collection of Questions:
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New post 11 Sep 2014, 10:49
Bunuel wrote:
SOLUTIONS:

1. A password on Mr. Wallace's briefcase consists of 5 digits. What is the probability that the password contains exactly three digit 6?

A. 860/90,000
B. 810/100,000
C. 858/100,000
D. 860/100,000
E. 1530/100,000

Total # of 5 digit codes is 10^5, notice that it's not 9*10^4, since in a code we can have zero as the first digit.

# of passwords with three digit 6 is \(9*9*C^3_5=810\): each out of two other digits (not 6) has 9 choices, thus we have 9*9 and \(C^3_5\) is ways to choose which 3 digits will be 6's out of 5 digits we have.

\(P=\frac{favorable}{total}=\frac{810}{10^5}\)

Answer: B.



Hi Bunuel,

I was unsure how to do this problem and when I saw your explanation, I was wondering how you got ( 9 * 9 * 3C5 ) ? Not sure where the 9s came from.

Thanks,

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DangerPenguin wrote:
Bunuel wrote:
SOLUTIONS:

1. A password on Mr. Wallace's briefcase consists of 5 digits. What is the probability that the password contains exactly three digit 6?

A. 860/90,000
B. 810/100,000
C. 858/100,000
D. 860/100,000
E. 1530/100,000

Total # of 5 digit codes is 10^5, notice that it's not 9*10^4, since in a code we can have zero as the first digit.

# of passwords with three digit 6 is \(9*9*C^3_5=810\): each out of two other digits (not 6) has 9 choices, thus we have 9*9 and \(C^3_5\) is ways to choose which 3 digits will be 6's out of 5 digits we have.

\(P=\frac{favorable}{total}=\frac{810}{10^5}\)

Answer: B.



Hi Bunuel,

I was unsure how to do this problem and when I saw your explanation, I was wondering how you got ( 9 * 9 * 3C5 ) ? Not sure where the 9s came from.

Thanks,


XX666

Each X can take any value from 0 to 9, inclusive but 6, so total of 9 values. Notice that the first digit can be zero here since the 5-digit password unlike 5-digit number can start with 0.
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: Baker's Dozen [#permalink]

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New post 06 Oct 2014, 10:02
Bunuel,

Whoaa!!Those were some PS!Tried to solve it by taking 2 mins/question.Completed 7 ques in almost 19+ mins and gave up after that.Got 4 wrong!
Could u please tell me the level of 1st 7 sums.Like 700-800 or 600-700,something like that.It'll help me to understand exactly how good/bad I am

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New post 06 Oct 2014, 10:05
DebWenger wrote:
Bunuel,

Whoaa!!Those were some PS!Tried to solve it by taking 2 mins/question.Completed 7 ques in almost 19+ mins and gave up after that.Got 4 wrong!
Could u please tell me the level of 1st 7 sums.Like 700-800 or 600-700,something like that.It'll help me to understand exactly how good/bad I am


Almost all of the questions here, as well as other questions from my signature below, are 700+.
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: Baker's Dozen [#permalink]

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New post 06 Oct 2014, 12:00
Thanks Bunuel
Got it!Could u please tell me if an accuracy of 3 out of 7 in 700+ PS(Around 43%) any good for achieving a 50 in QA.Also for 700+ questions what sort of accuracy should I aim for.
Thanks!!

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New post 16 Nov 2014, 13:28
Bunuel wrote:
2. If \(y=\frac{(3^5-3^2)^2}{(5^7-5^4)^{-2}}\), then y is NOT divisible by which of the following?
A. 6^4
B. 62^2
C. 65^2
D. 15^4
E. 52^4


\(y=\frac{(3^5-3^2)^2}{(5^7-5^4)^{-2}}=(3^5-3^2)^2*(5^7-5^4)^2=3^4*(3^3-1)^2*5^8*(5^3-1)^2=3^4*26^2*5^8*124^2=2^6*3^4*5^8*13^2*31^2\).

Now, if you analyze each option you'll see that only \(52^4=2^8*13^4\) is not a factor of \(y\), since the power of 13 in it is higher than the power of 13 in \(y\).

Answer: E.


Hi Brunel,
Great set of questions! Could you please explain the logic behind how you factorised this:
\(y=(3^5-3^2)^2*(5^7-5^4)^2=3^4*(3^3-1)^2*5^8*(5^3-1)^2\).
I am not too clear on how you pulled out factors of 3^4 and 5^8, and still have a 3^3 and a 5^3 left in the brackets.

Thanks!

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New post 16 Nov 2014, 13:35
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Madrigal wrote:
Bunuel wrote:
2. If \(y=\frac{(3^5-3^2)^2}{(5^7-5^4)^{-2}}\), then y is NOT divisible by which of the following?
A. 6^4
B. 62^2
C. 65^2
D. 15^4
E. 52^4


\(y=\frac{(3^5-3^2)^2}{(5^7-5^4)^{-2}}=(3^5-3^2)^2*(5^7-5^4)^2=3^4*(3^3-1)^2*5^8*(5^3-1)^2=3^4*26^2*5^8*124^2=2^6*3^4*5^8*13^2*31^2\).

Now, if you analyze each option you'll see that only \(52^4=2^8*13^4\) is not a factor of \(y\), since the power of 13 in it is higher than the power of 13 in \(y\).

Answer: E.


Hi Brunel,
Great set of questions! Could you please explain the logic behind how you factorised this:
\(y=(3^5-3^2)^2*(5^7-5^4)^2=3^4*(3^3-1)^2*5^8*(5^3-1)^2\).
I am not too clear on how you pulled out factors of 3^4 and 5^8, and still have a 3^3 and a 5^3 left in the brackets.

Thanks!


\((ab-ac)^2=(a(b-c))^2=a^2*(b-c)^2\). So, \((3^5-3^2)^2=(3^2*(3^3-1))^2=(3^2)^2*(3^3-1)^2=3^4*(3^3-1)^2\).

Similarly \((5^7-5^4)^2=(5^4(5^3-1))^2=5^8*(5^3-1)^2\).

Hope it's clear.
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Collection of Questions:
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Re: Baker's Dozen [#permalink]

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New post 20 Nov 2014, 09:30
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Bunuel wrote:
DangerPenguin wrote:
Bunuel wrote:
SOLUTIONS:

1. A password on Mr. Wallace's briefcase consists of 5 digits. What is the probability that the password contains exactly three digit 6?

A. 860/90,000
B. 810/100,000
C. 858/100,000
D. 860/100,000
E. 1530/100,000

Total # of 5 digit codes is 10^5, notice that it's not 9*10^4, since in a code we can have zero as the first digit.

# of passwords with three digit 6 is \(9*9*C^3_5=810\): each out of two other digits (not 6) has 9 choices, thus we have 9*9 and \(C^3_5\) is ways to choose which 3 digits will be 6's out of 5 digits we have.

\(P=\frac{favorable}{total}=\frac{810}{10^5}\)

Answer: B.



Hi Bunuel,

I was unsure how to do this problem and when I saw your explanation, I was wondering how you got ( 9 * 9 * 3C5 ) ? Not sure where the 9s came from.

Thanks,


XX666

Each X can take any value from 0 to 9, inclusive but 6, so total of 9 values. Notice that the first digit can be zero here since the 5-digit password unlike 5-digit number can start with 0.


Hi Bunuel

I have approached this a little differently.

= 9C1 x 9C1 x 1 [(5!/3!)+(5!/3!x2!)] / 10^5

Numerator:
9C1 is the choice of 9 digits from 0 to 9 except 6. So this is twice
1 is the way of choosing 6
Now I have chosen the numbers - three sixes and two other nos (these 2 could be same or different)

Now I form the permutations possible
5!/3! is the number of ways in which the two nos chosen are different
5!/3!x2! is the number of ways in which the two nos chosen are the same

By this I arrive at the total number of ways to get to this 5 digit nos

The universal set is 10^5

Have I misunderstood. What is the flaw in my approach?
Thanks!
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Re: Baker's Dozen [#permalink]

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New post 04 Dec 2014, 07:33
Bunuel wrote:
4. What is the smallest positive integer \(k\) such that \(126*\sqrt{k}\) is the square of a positive integer?
A. 14
B. 36
C. 144
D. 196
E. 441

\(126=2*3^2*7\), so in order \(126*\sqrt{k}\) to be a square of an integer \(\sqrt{k}\) must complete the powers of 2 and 7 to even number, so the least value of \(\sqrt{k}\) must equal to 2*7=14, which makes the leas value of \(k\) equal to 14^2=196.

Answer: D.


Would really appreciate if someone can explain this bit "\sqrt{k} must complete the powers of 2 and 7 to even number,"
Why does it have to be powers of 2*7

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Re: Baker's Dozen [#permalink]

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New post 04 Dec 2014, 07:57
saadis87 wrote:
Bunuel wrote:
4. What is the smallest positive integer \(k\) such that \(126*\sqrt{k}\) is the square of a positive integer?
A. 14
B. 36
C. 144
D. 196
E. 441

\(126=2*3^2*7\), so in order \(126*\sqrt{k}\) to be a square of an integer \(\sqrt{k}\) must complete the powers of 2 and 7 to even number, so the least value of \(\sqrt{k}\) must equal to 2*7=14, which makes the leas value of \(k\) equal to 14^2=196.

Answer: D.


Would really appreciate if someone can explain this bit "\sqrt{k} must complete the powers of 2 and 7 to even number,"
Why does it have to be powers of 2*7


Let me ask you: by what positive integer we should multiply \(126=2*3^2*7\) for the product to be the square of an integer?
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Baker's Dozen [#permalink]

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New post 04 Dec 2014, 08:28
Bunuel wrote:
saadis87 wrote:
Bunuel wrote:
4. What is the smallest positive integer \(k\) such that \(126*\sqrt{k}\) is the square of a positive integer?
A. 14
B. 36
C. 144
D. 196
E. 441

\(126=2*3^2*7\), so in order \(126*\sqrt{k}\) to be a square of an integer \(\sqrt{k}\) must complete the powers of 2 and 7 to even number, so the least value of \(\sqrt{k}\) must equal to 2*7=14, which makes the leas value of \(k\) equal to 14^2=196.

Answer: D.


Would really appreciate if someone can explain this bit "\sqrt{k} must complete the powers of 2 and 7 to even number,"
Why does it have to be powers of 2*7


Let me ask you: by what positive integer we should multiply \(126=2*3^2*7\) for the product to be the square of an integer?



I just started solving these again It made sense to me now.
Screen fatigue :P

and OMG dude amazing questions

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Re: Baker's Dozen [#permalink]

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New post 21 Dec 2014, 23:44
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Bunuel wrote:
6. A pool has two water pumps A and B and one drain C. Pump A alone can fill the whole pool in x hours, and pump B alone can fill the whole pool in y hours. The drain can empty the whole pool in z hours, where z>x. With pumps A and B both running and the drain C unstopped till the pool is filled, which of the following represents the amount of water in terms of the fraction of the pool which pump A pumped into the pool?
A. \(\frac{yz}{x+y+z}\)

B. \(\frac{yz}{yz+xz-xy}\)

C. \(\frac{yz}{yz+xz+xy}\)

D. \(\frac{xyz}{yz+xz-xy}\)

E. \(\frac{yz+xz-xy}{yz}\)

With pumps A and B both running and the drain unstopped the pool will be filled in a rate \(\frac{1}{x}+\frac{1}{y}-\frac{1}{z}=\frac{yz+xz-xy}{xyz}\) pool/hour. So, the pool will be filled in \(\frac{xyz}{yz+xz-xy}\) hours (time is reciprocal of rate).

In \(\frac{xyz}{yz+xz-xy}\) hours A will pump \(\frac{1}{x}*\frac{xyz}{yz+xz-xy}=\frac{yz}{yz+xz-xy}\) amount of the water into the pool.

Answer: B.

I did not get why 1/x is multiplied. Would you mind terribly, to explain?
Thanks

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New post 22 Dec 2014, 00:02
anuu wrote:
1) Ans B

5) For selecting 8 marbles from 12 marbles, a maximim of 4 marble can remain in
the jar . The scenarios for atleast 1 red marble and 1 blue marble to remain are:

1r1b 7c1*5c1 = 35
1r2b 7c1*5c2 = 70
2r1b 7c2*5c1 = 105
2r2b 7c2*5c2 = 210
3r1b 7c3*5c1 = 175
1r3b 7c1*5c3 = 70

Total 665 . Not sure if the logic is right..



I liked this approach. Tiny mistake though- In the first 3 cases the total of R and B is not 4, so when picking marbles we are not actually picking 8 of them. So I guess the requisites should be-
R>=1, B>=1 AND R+B=4.
Total the last 3 cases and the answer will be 455.
may help
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Re: Baker's Dozen [#permalink]

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New post 22 Dec 2014, 03:19
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Expert's post
deeuk wrote:
Bunuel wrote:
6. A pool has two water pumps A and B and one drain C. Pump A alone can fill the whole pool in x hours, and pump B alone can fill the whole pool in y hours. The drain can empty the whole pool in z hours, where z>x. With pumps A and B both running and the drain C unstopped till the pool is filled, which of the following represents the amount of water in terms of the fraction of the pool which pump A pumped into the pool?
A. \(\frac{yz}{x+y+z}\)

B. \(\frac{yz}{yz+xz-xy}\)

C. \(\frac{yz}{yz+xz+xy}\)

D. \(\frac{xyz}{yz+xz-xy}\)

E. \(\frac{yz+xz-xy}{yz}\)

With pumps A and B both running and the drain unstopped the pool will be filled in a rate \(\frac{1}{x}+\frac{1}{y}-\frac{1}{z}=\frac{yz+xz-xy}{xyz}\) pool/hour. So, the pool will be filled in \(\frac{xyz}{yz+xz-xy}\) hours (time is reciprocal of rate).

In \(\frac{xyz}{yz+xz-xy}\) hours A will pump \(\frac{1}{x}*\frac{xyz}{yz+xz-xy}=\frac{yz}{yz+xz-xy}\) amount of the water into the pool.

Answer: B.

I did not get why 1/x is multiplied. Would you mind terribly, to explain?
Thanks


The question asks to find the amount of water in terms of the fraction of the pool which pump A pumped into the pool. We know that the pool will be filled in \(\frac{xyz}{yz+xz-xy}\) hours and the rate of A is 1/x pool/hour. The job done by A in that time is (job) = (time)*(rate) = \(\frac{1}{x}*\frac{xyz}{yz+xz-xy}\).

Hope it's clear.
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New post 22 Dec 2014, 08:36
2. If y=\frac{(3^5-3^2)^2}{(5^7-5^4)^{-2}}, then y is NOT divisible by which of the following?
A. 6^4
B. 62^2
C. 65^2
D. 15^4
E. 52^4

Answer: E
Y = 3^4*(3-1)^2*(3^2+3+1)^2 * 5^8*(5-1)^4*(5^2+5+1)^4 = 2^10*3^4*5^8*13^2*31^4
where
6^4=2^4*3^4
62^2=2^2*31^2
65^2=5^2*13^2
15^4=3^4*5^4
52^4=2^8*13^4

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Re: Baker's Dozen   [#permalink] 22 Dec 2014, 08:36

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