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Baker's Dozen

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Baker's Dozen [#permalink] New post 08 Mar 2012, 12:27
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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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Re: Baker's Dozen [#permalink] New post 10 Mar 2012, 23:51
Bunuel, could u please tell us which question is still unanswered or answered wrong by all of us? I want to think on such questions more, before u post solutions.

and btw, do u read our solutions or just only answers? I wonder whether my way of thinking was ok. in some cases I tried to use another method not to repeat others. I just wonder whether it worked, or it was just coincidence
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Re: Baker's Dozen [#permalink] New post 11 Mar 2012, 00:04
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LalaB wrote:
Bunuel, could u please tell us which question is still unanswered or answered wrong by all of us? I want to think on such questions more, before u post solutions.

and btw, do u read our solutions or just only answers? I wonder whether my way of thinking was ok. in some cases I tried to use another method not to repeat others. I just wonder whether it worked, or it was just coincidence


I do read all the solutions and award +1 Kudos if there is at least one correct explanation in the post.

As for your other question: I don't see a solution for #7, so you can try this one.
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Re: Baker's Dozen [#permalink] New post 11 Mar 2012, 00:43
Was it the wrong solution of #9. It seems to me that my post was removed :|
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Re: Baker's Dozen [#permalink] New post 11 Mar 2012, 00:50
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Re: Baker's Dozen [#permalink] New post 11 Mar 2012, 00:51
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Bunuel wrote:

As for your other question: I don't see a solution for #7, so you can try this one.

ok my answer to the q7 is D , because D is the leader of the day hehe :-D kidding
the answer is really D ,because-

2/3(L+A+W) =F
3/7(F+W+A)=L
4/11(F+W+L)=A
total=360 (I deleted the 0s just to make a number easy)
W=?

2/3(L+A+W) =F means F=2/3(360-F) =>(2/3)*360=5/3F=> F=144
3/7(F+W+A)=L means L=3/7(360-L) => L=108
4/11(F+W+L)=A means A=4/11(360-A)=>A=96

W=360- (L+A+F)=360-144-108-96=12


or W=120 000 (return the 0s which were deleted before)
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Re: Baker's Dozen [#permalink] New post 11 Mar 2012, 01:44
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utkarshlavania wrote:
[quote2]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

Correction/suggestion/advice/confirmation --- needed
Answer to 9th question If x and y are negative numbers, what is the value of \(\frac{\sqrt{x^2}}{x}-\sqrt{-y*|y|}\)?[/b]
Sqrtx^2/x-sqrt(-y*|y|) = |x|/x-|y|= -1-y


Responding to a pm:

[Reveal] Spoiler:
First of all, I would suggest you to put x = -2 and y = -3 and see what you get.

\(\frac{\sqrt{x^2}}{x}-\sqrt{-y*|y|}\)

\(\frac{\sqrt{(-2)^2}}{-2}-\sqrt{-(-3)*|-3|}\)

\(\frac{2}{-2}-\sqrt{9}\) (because square root will always be positive)

-1 - 3 = -1 + y (since y = -3)

Coming to your solution:

I see you have everything correct till the last step: |x|/x-|y|

|x|/x will be -1 which is fine.
What about |y|?
Do you remember how we define mods?
We say:
|x| = x if x >= 0
|x| = -x if x < 0

Why? If x >= 0 e.g. say x = 5,
|x| = x = 5
Instead, if x < 0 e.g. say x = -6,
is |x| = x? No! mods are never negative. So |x| = -x = -(-6) = 6 (Since x itself is negative, mod of x is negative of x which becomes positive.)

Therefore, if we know that y is negative, what is the value of |y|?
|y| = -y

Therefore, |x|/x-|y| = -1 -(-y) = -1 + y

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Re: Baker's Dozen [#permalink] New post 11 Mar 2012, 04:36
@karishma

then why is |x|/x =-1 , should not it be 1 as |x|=-x and denominator is negative as well
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Re: Baker's Dozen [#permalink] New post 11 Mar 2012, 04:59
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utkarshlavania wrote:
@karishma

then why is |x|/x =-1 , should not it be 1 as |x|=-x and denominator is negative as well


If \(x<0\) then \(|x|=-x\) and \(\frac{|x|}{x}=\frac{-x}{x}=-1\), (\(-\frac{x}{x}=-1\) no matter whether \(x\) is negative or positive).
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Re: Baker's Dozen [#permalink] New post 11 Mar 2012, 19:52
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utkarshlavania wrote:
@karishma

then why is |x|/x =-1 , should not it be 1 as |x|=-x and denominator is negative as well


You are right that |x|=-x since x is negative but tell me, what is -x, negative or positive? Negative of negative gives you positive, right? So -x must be positive. Now, if x is negative,
-x/x must be positive/negative giving you -1.

You are confusing yourself too much with negatives and positives. Just think of it this way:

|x|/x = -x/x (By definition, since |x| = -x when x < 0)
-x/x = -1 ( x and x get canceled here leaving you with -1)
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Re: Baker's Dozen [#permalink] New post 13 Mar 2012, 02:43
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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.
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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.
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Re: Baker's Dozen [#permalink] New post 13 Mar 2012, 02:46
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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

Total cupcakes for k days was 55k, which means that total cupcakes for k+1 days was 55k+100. The new average is (55k+100)/(k+1)=60 --> 55k+100=60k+60 --> k=8

Answer: B.
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Re: Baker's Dozen [#permalink] New post 13 Mar 2012, 02:48
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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.
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Re: Baker's Dozen [#permalink] New post 13 Mar 2012, 02:50
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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

Total ways to select 8 marbles out of 7+5=12 is \(C^8_{12}\);
Ways to select 8 marbles so that zero red marbles is left in the jar is \(C^7_7*C^1_5\);
Ways to select 8 marbles so that zero blue marbles is left in the jar is \(C^5_5*C^3_7\);

Hence ways to select 8 marbles so that at least one red marble and at least one blue marble to remain the jar is \(C^8_{12}-(C^7_7*C^1_5+C^5_5*C^3_7)=495-(5+35)=455\).

Answer: D.
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Baker's Dozen [#permalink] New post 13 Mar 2012, 02:52
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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.
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Re: Baker's Dozen [#permalink] New post 13 Mar 2012, 02:53
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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

Fritz owns is \(\frac{2}{3}\)rd of the shares of the other three shareholders --> Fritz owns \(\frac{2}{2+3}=\frac{2}{5}\)th of all shares;
Luis owns is \(\frac{3}{7}\)th of the shares of the other three shareholders --> Luis owns \(\frac{3}{3+7}=\frac{3}{10}\)th of all shares;
Alfred owns is \(\frac{4}{11}\)th of the shares of the other three shareholders --> Alfred owns \(\frac{4}{4+11}=\frac{4}{15}\)th of all shares;

Together those three own \(\frac{2}{5}+\frac{3}{10}+\frac{4}{15}=\frac{29}{30}\)th of all shares, which means that Werner owns \(1-\frac{29}{30}=\frac{1}{30}\). Hence from $3,600,000 Werner gets \($3,600,000*\frac{1}{30}=$120,000\).

Answer: D.
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Re: Baker's Dozen [#permalink] New post 13 Mar 2012, 02:56
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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

Say 7 consecutive odd integers are: \(x\), \(x+2\), \(x+4\), \(x+6\), \(x+8\), \(x+10\), \(x+12\).

Question: \(x+(x+2)+(x+4)+(x+6)+(x+8)=5x+20=?\)

Given: \((x+4)+(x+6)+(x+8)+(x+10)+(x+12)=-185\) --> \((x+4)+(x+6)+(x+8)+(x+10)+(x+12)=5x+40=-185\) --> \((5x+20)+20=-185\) --> \(5x+20=-205\)

Answer: D.
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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

Note that \(\sqrt{a^2}=|a|\). Next, since \(x<0\) and \(y<0\) then \(|x|=-x\) and \(|y|=-y\).

So, \(\frac{\sqrt{x^2}}{x}-\sqrt{-y*|y|}=\frac{|x|}{x}-\sqrt{(-y)*(-y)}=\frac{-x}{x}-\sqrt{y^2}=-1-|y|=-1+y\)

Answer: D.
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Re: Baker's Dozen [#permalink] New post 13 Mar 2012, 02:59
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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

Notice that we are not told that \(x\) and \(y\) are integers.

\(x^2<81\) means that \(-9<x<9\) and \(y^2<25\) means that \(-5<y<5\). Now, since the largest value of \(x\) is almost 9 and the largest value of \(-2y\) is almost 10 (for example if \(y=-4.9\)), then the largest value of \(x-2y\) is almost 9+10=19, so the actual value is less than 19, which means that the largest prime that can be equal to \(x-2y\) is 17. For example: \(x=8\) and \(y=-4.5\).

Answer: D.
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Re: Baker's Dozen [#permalink] New post 13 Mar 2012, 03:00
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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

You don't need to know geometric progression formula to solve this question. All you need is to find the pattern:

\(b_1=1=3^0\);
\(b_2=3=3^1\);
\(b_3=9=3^2\);
\(b_4=27=3^3\);
...
\(b_n=3^{n-1}\);

\(b_{13}+b_{15}-(b_{12}+b_{14})=3^{12}+3^{14}-3^{11}-3^{13}=3^{11}(3+3^3-1-3^2)=20*3^{11}\)

Answer: B.
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Re: Baker's Dozen   [#permalink] 13 Mar 2012, 03:00

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