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# Baker's Dozen

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Baker's Dozen [#permalink]  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]  10 Mar 2012, 08:03
Bunuel wrote:
utkarshlavania wrote:
-y = +y and |y| no matter what, comes out as positive hence Sqrt (-y*|y| ) comes out as possitive y and sqrtx^2 =|x|= positive x , of course from your tone i get that i'm going wrong some where but

What I meant is that if y<0 then |y|=-y and -1-|y|=-1-(-y)=-1+y.

isn't -1 -|y|=-1-y for the same reason |x|/x became -1 because numerator x was positive and denominator x was negative .
sqrt(negative -y *|y|) so negative -y =postive y and |y| = positive hence when y comes out it comes as positive y and because of the negative sign
-1 - y=-1-y
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Re: Baker's Dozen [#permalink]  10 Mar 2012, 08:12
utkarshlavania wrote:
Bunuel wrote:
utkarshlavania wrote:
-y = +y and |y| no matter what, comes out as positive hence Sqrt (-y*|y| ) comes out as possitive y and sqrtx^2 =|x|= positive x , of course from your tone i get that i'm going wrong some where but

What I meant is that if y<0 then |y|=-y and -1-|y|=-1-(-y)=-1+y.

isn't -1 -|y|=-1-y for the same reason |x|/x became -1 because numerator x was positive and denominator x was negative .
sqrt(negative -y *|y|) so negative -y =postive y and |y| = positive hence when y comes out it comes as positive y and because of the negative sign
-1 - y=-1-y

sqroot of (negative number ^2) = (negative number ^2)^1/2=negative number

for example sqroot (-5*|5|)=sqroot (-5^2)=((-5)^2)^1/2=-5
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Re: Baker's Dozen [#permalink]  10 Mar 2012, 10:55
LalaB wrote:
(

sqroot of (negative number ^2) = (negative number ^2)^1/2=negative number

for example sqroot (-5*|5|)=sqroot (-5^2)=((-5)^2)^1/2=-5[/quote]

agreed but here isn't the concept little different, as it is already mentioned that y is negative hence -y= positive y
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Re: Baker's Dozen [#permalink]  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]  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]  11 Mar 2012, 00:50
Expert's post
SergeNew wrote:
Was it the wrong solution of #9. It seems to me that my post was removed

Even if it were wrong I wouldn't remove it, it's just on the second page: baker-s-dozen-128782-20.html#p1056538
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Re: Baker's Dozen [#permalink]  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]  11 Mar 2012, 04:59
Expert's post
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]  11 Mar 2012, 19:52
Expert's post
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]  14 Mar 2012, 14:45
Bunuel wrote:
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, $$\frac{\sqrt{x^2}}{x}-\sqrt{-y*|y|}=\frac{|x|}{x}-\sqrt{(-y)*(-y)}=\frac{-x}{x}-\sqrt{y^2}=-1-|y|=-1+y$$

Hi Bunuel,

I didn't quite understand this ..

X and Y are odd then how can |x| = -x it has to be +ive.
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Re: Baker's Dozen [#permalink]  14 Mar 2012, 16:10
Expert's post
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onedayill wrote:
Bunuel wrote:
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, $$\frac{\sqrt{x^2}}{x}-\sqrt{-y*|y|}=\frac{|x|}{x}-\sqrt{(-y)*(-y)}=\frac{-x}{x}-\sqrt{y^2}=-1-|y|=-1+y$$

Hi Bunuel,

I didn't quite understand this ..

X and Y are odd then how can |x| = -x it has to be +ive.

x and y are negative numbers, not odd numbers.

If $$x\geq{0}$$ then $$|x|=x$$;
If $$x<{0}$$ then $$|x|=-x$$. So if $$x$$ is negative then $$|x|=-x=-negative=positive$$. For example, if $$x=-2$$ then $$|x|=|-2|=2=-x$$.

Check Absolute Values chapter of Math Book for more: math-absolute-value-modulus-86462.html

Hope it helps.
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Re: Baker's Dozen [#permalink]  14 Mar 2012, 22:43
Got 9 correct with @ 2.5 mins/Q ... how about others ? (2 Q left without going to deep as killed more than 4 mins on those Q4,7- this proves that not to waste time on something you don't get idea in 1.5-2 mins time frame and missed 2 narrowly- Q9, 10: 10 I din't read 2y but only y so...)..will post in details some explanations, esp. those which are little different from already posted by the very own legend Mr B!!
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Re: Baker's Dozen [#permalink]  16 Mar 2012, 02:47
Expert's post
chetan2u 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 fraction of the pool which pump A filled?
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 pump A will do $$\frac{1}{x}*\frac{xyz}{yz+xz-xy}=\frac{yz}{yz+xz-xy}$$ part of the job.

hi , i feel the ans to this Q is wrong. as we have to tell what fraction of the pool was filled up by pump A. it will not depend on the drain C as the filling was done only by pump A and B. ans shud be y/(x+y).
the given ans would be correct if we were to find "the amount of water pumped by A in that duration"

Actually that was the intended meaning of the question. I edited it so to avoid ambiguity: "which of the following represents the amount of water in terms of the fraction of the pool which pump A pumped into the pool?"

Hope now it's more precise.
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Re: Baker's Dozen [#permalink]  17 Mar 2012, 00:12
Expert's post
GMATD11 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 $$52^4=2^4*13^4$$ is not a factor of $$y$$, since the power of 13 in it is higher than the power of 13 in $$y$$.

Pls correct the Typo 2^4 to 2^8

Typo corrected, thanks.
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Re: Baker's Dozen [#permalink]  18 Mar 2012, 15:45
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.

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

Hello Bunnel....

Can you kindly clarify this question and elaborate on the solution!!

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Re: Baker's Dozen [#permalink]  19 Mar 2012, 03:10
Expert's post
balas 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.

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

Hello Bunnel....

Can you kindly clarify this question and elaborate on the solution!!

Can you please specify which part didn't you understand? Thanks.
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Re: Baker's Dozen [#permalink]  27 Mar 2012, 10:04
Awesome set of questions!
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Re: Baker's Dozen [#permalink]  28 Apr 2012, 02:41
Can any one please explain: why it is (9*9*1*1*1* 5C3) in Question no. 1 (password problem)?
Why not : 9*9*1*1*1* 5!/2! ??

Last edited by mofasser08 on 28 Apr 2012, 03:00, edited 1 time in total.
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Re: Baker's Dozen [#permalink]  28 Apr 2012, 18:47
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.

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

Hey Bunuel,

Sorry to trouble you. If you could, I would appreciate if you could clarify more about 5C3. So this is how I approached the problem:

We have three - 6's. With two slots that can be filled by 9 numbers.

666 9 9 . Now the question is how many ways can we rearrange these codes of three 6's and 2 non 6's? Well, have 5 elements 3 identical (because we have 3 identical 6's and two non-6's) Hence we have 5!/(3!2!) ways of rearranging the numbers. Is this why we have 5C3 in the problem?

But the two non-6's can be identical or different. For example 66611 66622 66633 66644 66655 ... 66699... etc all satisfy the condition of only three 6's. The number of ways to rearrange the letters {aaabb} is 5!/(3!2!). Moreover, we can also have the two non-6's be different, 66612 66654 66624, etc . the number of ways to rearrange the letters {aaabc} is 5!/(3!).

What do we do in this situation? Do we look at this problem as (#6)(#6)(#6)(not #6)(not #6) therefore 5!/(3!2!). Three repeating elements of #6 and two repeating elements of not #6.

OR

Do we look at the problem as (#6)(#6)(#6)(x)(y), where x and y two numbers that are not 6, that may or may not be equal, therefore 5!/3! (3! for the 3 repeating 6's but we cannot say anything about whether x and y are equal to each other)

Bunuel, as always, thank you so much!! And again, sorry to trouble you.
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Re: Baker's Dozen [#permalink]  28 Apr 2012, 18:51
mofasser08 wrote:
Can any one please explain: why it is (9*9*1*1*1* 5C3) in Question no. 1 (password problem)?
Why not : 9*9*1*1*1* 5!/2! ??

Mofasser,

We pretty much have the same question. But I think the reasons it's 5C3 is because we have

(#6)(#6)(#6)(not #6)(not #6)=aaabb.

The number of ways to rearrange aaabb is 5!/(3!2!). But we need to check with the Oracle, Bunuel.
Re: Baker's Dozen   [#permalink] 28 Apr 2012, 18:51

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