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# Baker's Dozen

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08 Mar 2012, 13:27
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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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28 Apr 2012, 03:59
2
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Expert's post
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*5P3 ??

That's a tough 700+ problem.

We are interested 666XX numbers. Now, two this can be arranged in 10 ways;
666XX;
66X6X;
6X66X;
X666X;

66XX6;
6X6X6;
X66X6;

6XX66;
XX666;

X6X66.

So, basically with $$C^3_5=10$$ we are choosing which 3 places out of 5 will be occupied by X's. Or we could do the other way around: $$C^2_510$$ choosing which 2 places out of 5 will be occupied by 6's.

Next, 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.

Total $$9*9*C^3_5=810$$.

Hope it's clear.
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28 Apr 2012, 19:47
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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28 Apr 2012, 19:51
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.
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29 Apr 2012, 06:16
1
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Expert's post
alphabeta1234 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.

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

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.

In this post there are all cases listed: baker-s-dozen-128782-60.html#p1079496

Please tell me what part of this post needs farther clarification.
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01 May 2012, 23:52
1
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Bunuel wrote:
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!

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

my solution: x^2<81 ==> Mod(x)<9 ==> -9 < x< 9----------------------------------#
y^2<25 ==> Mod(y)<5 ==> -5 < y<5 so, -10<2y<10 and (-)-10<(-)2y<(-)10 {multiplying - to all. after multiplication, sign reverses.}
so, it will be
-10<-2y<10 ------------------------------------------@
-19<x-2y<19
largest prime no is 17.
ANS : D

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

my solution : Difference = [(a13 + a15)- (a12 + a14)] : here series is in GP with common ratio r= 3 and first term a=1
= [ar^12 +ar^14 - ar^11- ar^13]
= [ar^11 (r+r^3-1-r^2]
= ar^11 ( 3+27-1-9)
= 1.3^11 .20
= 20*3^11
Ans : B

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

My solution :
x = y@ + 3
y= Z# +8
min value of z is 1 (given x, y and z are positive integers )
and Min value of # is 0
so, y = 0.1+8 =8 & z=1
min value of @ is 0
x=0+3 =3
X+Y+Z = 3+8+1 =12
.Ans :A

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

My solution :

$$x=\frac{(8!)^{10}-(8!)^6}{(8!)^{5}-(8!)^3}$$,
= (8!)^6 \frac{(8!)^{4}-1}{(8!)^3{(8!)^{2}-1}}[/m],

{(8!)^{4}-1} = {(8!)^{2}-1}{(8!)^{2}+1}

then
x = (8!)^(6-3) [(8!)^2+1]
x / (8!)^3 = [(8!)^2+1]
we have to calculate $$\frac{x}{(8!)^3}-39$$?[/b]
so,
$$\frac{x}{(8!)^3}-39$$?[/b] = [(8!)^2+1] -39
= (8!)^2-38

now we have top find the product of the tens and the units digits :
8! consist one 5 so, its unit digit must be zero.
on squaring the 8!, there will be two zero at the end i.e one at unit digit and tens digit.

so, xxxxxxx00-38 = yyyyyy62

Ans : 12 (D)
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02 May 2012, 10:59
Question 1 3 number of 6's can be arranged among themselves in 1 way. Number of ways we can select three position from 5= 5C3. For the remaining two position we can choose any digit but 6. So number of possible patterns=(5C3)*9*9
Hence required probability=(5C3)*9*9/10^5
I hope those who have doubt can understand this explanation
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08 May 2012, 18:46
Bunuel wrote:
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. Could some one explain this to me how does one arrive at 2/2+3 = 2/5 and so on? Math Expert Joined: 02 Sep 2009 Posts: 35240 Followers: 6619 Kudos [?]: 85310 [0], given: 10236 Re: Baker's Dozen [#permalink] ### Show Tags 09 May 2012, 01:10 mbafall2011 wrote: Bunuel wrote: 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 $$3,600,000*\frac{1}{30}=120,000$$.

Could some one explain this to me how does one arrive at 2/2+3 = 2/5 and so on?

Explained here: baker-s-dozen-128782-40.html#p1059585
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04 Jun 2012, 11:16
Bunuel wrote:
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*5P3 ??

That's a tough 700+ problem.

We are interested 666XX numbers. Now, two this can be arranged in 10 ways;
666XX;
66X6X;
6X66X;
X666X;

66XX6;
6X6X6;
X66X6;

6XX66;
XX666;

X6X66.

So, basically with $$C^3_5=10$$ we are choosing which 3 places out of 5 will be occupied by X's. Or we could do the other way around: $$C^2_510$$ choosing which 2 places out of 5 will be occupied by 6's.

Next, 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.

Total $$9*9*C^3_5=810$$.

Hope it's clear.

Hi Bunuel..can you tell me if I am thinking correct.

1/10 * 1/10 * 1/10 * 9/10 * 9/10 this gives me 3 chances of getting 6 and rest 2 for getting any other number.
now the above scenario can be arranged in 5C3 ways so

1/10 * 1/10 * 1/10 * 9/10 * 9/10 * 5C3..

I hope I am correct in this way..?

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12 Jun 2012, 18:39
Hi,
Thanks!
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13 Jun 2012, 02:52
sharmila79 wrote:
Hi,
Thanks!

Switch view mode of the topic from "Best Reply" to "Oldest" and the links from the initial post will lead you to the posts with solutions. The question about the shareholders is answered here: baker-s-dozen-128782-40.html#p1059585

Hope it helps.
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13 Jun 2012, 06:11
Thank you very much! I got it.
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16 Jun 2012, 06:11
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.

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

Hi Bunnel,

I didn't understand how you got 5C3, can you please explain more this...
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16 Jun 2012, 06:15
kotela 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.

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

Hi Bunnel,

I didn't understand how you got 5C3, can you please explain more this...

Check this: baker-s-dozen-128782-60.html#p1079496

Hope it helps.
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22 Jul 2012, 09:35
Bunuel wrote:
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: $$(x+4)+(x+6)+(x+8)+(x+10)+(x+12)=-185$$ --> $$(x+4)+(x+6)+(x+8)+(x+10)+(x+12)=5x+40=-185$$ --> $$(5x+20)+20=-185$$ --> $$5x+20=-205$$

Another easy way.

In 7 such numbers the middle three would remain same. So we are basically replacing 6th and 7th by 1st and 2nd.
Since these are consecutive odd numbers => 7th and 2nd differ by 10..... similarly 6th and 1st by 10...total difference = 20
so our answer should be = sum - 20 = -185-20 = -205
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24 Jul 2012, 16:44
Bunuel wrote:
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

Given $$x=qy+3$$, where $$q$$ is a quotient, an integer $$\geq0$$. Which means that the least value of $$x$$ is when $$q=0$$, in that case $$x=3$$. This basically means that $$x$$ is less than $$y$$. For example 3 divided by 4 yields remainder of 3.

Thus we have that:
$$x$$ is divided by $$y$$ the remainder is 3 --> minimum value of $$x$$ is 3;
$$y$$ is divided by $$z$$ the remainder is 8 --> minimum value of $$y$$ is 8 and minimum value of $$z$$ is one more than 8, so 9 (8 divided by 9 yields the remainder of 8);

So, the smallest possible value of $$x+y+z$$ is 3+8+9=20.

Hey Bunuel,

Thank you for the explanation. I am trying to understand why we can impose the condition that $$z>8$$ be greater than its remainded, but that we don't impose the same condition on $$y>3$$
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06 Aug 2012, 01:58
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.

I did a silly mistake here....after coming to the digit as 14. saw the option and marked A..!!! but didnt think that its sqrt of K...
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06 Aug 2012, 03:39
Bunuel wrote:
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

Number of codes with exactly 3 digits of 6 is 5C3*9*9= 810. Total number of possible codes $$10^5.$$
Requested probability $$\frac{810}{10^5}$$

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

$$y=\frac{(3^5-3^2)^2}{(5^7-5^4)^{-2}}=(3^5-3^2)^2(5^7-5^4)^2=3^4*5^8(3^2-1)^2*(5^3-1)^2=3^4*5^8*8^2*124^2$$. The number is not divisible by 13, which is a factor of 52.

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

55k+100 = 60(k+1), solve for k, and get k = 8.

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

126 = 2*9*7=9*14. To get a perfect square $$\sqrt{k}$$ must be equal to 14, so k =196.

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

The total number of possibilities to chose 8 marbles is 12C8 = $$\frac{12*11*10*9}{2*3*4}=\frac{990}{2}=495.$$. Count the number of possibilities when no red marble is left, so 7 red and 1 blue were selected, which is 5 possibilities.
Count the number of possibilities that no blue marble is left, so 3 red and 5 blue were selected, which is 7C3=$$\frac{7*6*5}{2*3}=35$$.
Therefore, the required number is 495 - 5 - 35 = 455.

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. $$\frac{yz+xz-xy}{yz}$$

Total time it takes to fill the whole pool is $$T=\frac{1}{\frac{1}{x}+\frac{1}{y}-\frac{1}{z}}=\frac{xyz}{yz+xz-xy}$$. Rate of pump A is 1/x, and T/x gives exactly the expression in B.

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

2/3 + 1 =5/3, 3/7+1=10/7, 4/11+1=15/11, therefore consider the total amount a multiple of 30, so 30N.
From the given information we can deduce that Fritz owns 12N, Luis 9N, and Alfred 8N. So, there is N left for Werner, which is 3,600,000/30 = 120,000.

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

We can deduce that the fifth number in the sequence is -185/5 = -37. Then, the third number is -37 - 2 -2 = -41.
The required sum is 5*(-41) = -205.

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

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

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

We can deduce that $$-9<x<9$$ and $$-5<y<5$$ or $$-10<-2y<10$$, and altogether $$-19<x-2y<19$$.
Therefore, the largest prime number which fulfills the above requirement is 17.

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

Let's denote by $$a_1=1, a_2=3,...$$ the terms of the given sequence.
$$a_{13}+a_{15}-(a_{12}+a_{14})=a_{13}+9a_{13}-(a_{12}+9a_{12})=10a_{13}-10a_{12}=30a_{12}-10a_{12}=20a_{12}=20*3^{11}$$

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

We can write $$x = ay+3$$, where $$a$$ is a non-negative integer. We can deduce that $$x\geq3$$ and $$y>3$$.
Also, $$y=bz+8$$, for some non-negative integer $$b$$. It follows from here that $$y\geq8$$ and $$z>8$$.
The minimum sum for $$x+y+z$$ is obtained for $$x=3, y=8, z=9$$.

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

$$x=\frac{(8!)^{10}-(8!)^6}{(8!)^{5}-(8!)^3}=\frac{(8!)^6((8!)^4-1)}{(8!)^3((8!)^2-1)}$$.
Since $$(8!)^4-1=((8!)^2+1)((8!)^2-1)$$, $$\frac{x}{(8!)^3}=(8!)^2+1$$, which is an integer which ends in 1 and has more than 2 zeros before the 1.
*001-39 ends in 62. The product of the last two digits is 6*2=12.

Solution: baker-s-dozen-128782-40.html#p1057520

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25 Aug 2012, 10:52
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, $$\frac{\sqrt{x^2}}{x}-\sqrt{-y*|y|}=\frac{|x|}{x}-\sqrt{(-y)*(-y)}=\frac{-x}{x}-\sqrt{y^2}=-1-|y|=-1+y$$

Bunuel, i am confused with the answer...

i have followed the method...

{sq-rt[(-x)sq]/(-x) } - {sq-rt[-(-y)*mod(-y)]}

{sq-rt(x2)/(-x)} - sq-rt(y2)

[x/(-x)] - y

-1-y

since we had used the negative values before, the final answer would have to be -(y+1) which is C. Can you please explain me why we have again substituted y with -y just before the final step to give -1+y.

Thanks
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Harsha

Note: Give me kudos if my approach is right , else help me understand where i am missing.. I want to bell the GMAT Cat

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25 Aug 2012, 11:07
1
KUDOS
harshavmrg 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, $$\frac{\sqrt{x^2}}{x}-\sqrt{-y*|y|}=\frac{|x|}{x}-\sqrt{(-y)*(-y)}=\frac{-x}{x}-\sqrt{y^2}=-1-|y|=-1+y$$

Bunuel, i am confused with the answer...

i have followed the method...

{sq-rt[(-x)sq]/(-x) } - {sq-rt[-(-y)*mod(-y)]}

{sq-rt(x2)/(-x)} - sq-rt(y2)

[x/(-x)] - y

-1-y

since we had used the negative values before, the final answer would have to be -(y+1) which is C. Can you please explain me why we have again substituted y with -y just before the final step to give -1+y.

Thanks

$$\sqrt{y^2}=|y|$$ and $$|y|=-y,$$ because $$y<0.$$ So you should have $$-(-y)=y.$$
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Re: Baker's Dozen   [#permalink] 25 Aug 2012, 11:07

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