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# Baker's Dozen

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Math Expert
Joined: 02 Sep 2009
Posts: 59561
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13 Mar 2012, 04:02
12
52
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.

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13 Mar 2012, 04:05
18
38
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

Apply $$a^2-b^2=(a-b)(a+b)$$: $$x=\frac{(8!)^{10}-(8!)^6}{(8!)^{5}-(8!)^3}=\frac{((8!)^{5}-(8!)^3)((8!)^{5}+(8!)^3)}{(8!)^{5}-(8!)^3}=(8!)^{5}+(8!)^3$$.

Next, $$\frac{x}{(8!)^3}-39=\frac{(8!)^{5}+(8!)^3}{(8!)^3}=\frac{(8!)^{5}}{(8!)^3}+\frac{(8!)^{3}}{(8!)^3}-39=(8!)^2+1-39=(8!)^2-38$$.

Now, since $$8!$$ has 2 and 5 as its multiples, then it will have 0 as the units digit, so $$(8!)^2$$ will have two zeros in the end, which means that $$(8!)^2-38$$ will have 00-38=62 as the last digits: 6*2=12.

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Re: Baker's Dozen  [#permalink]

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14 Mar 2012, 11:37
2
3
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.

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

Did this way- prob of choosing 6 will be 1/10 on each of he three times (and these choices can be in $$5C3$$=10 ways) and the rest two digits can be any thing from set 0-9 (except 6) so in total we have 9 choices (i..e a prob of 9/10 for the rest of the 2 positions). So total prob will be$$10*(1/10*1/10*1/10*9/10*9/10)= 810/10^5$$(B)
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14 Mar 2012, 15:45
1
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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Posts: 59561
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14 Mar 2012, 17:10
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]

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14 Mar 2012, 23:43
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]

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15 Mar 2012, 03:36
3
2
Bunuel wrote:
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

Apply $$a^2-b^2=(a-b)(a+b)$$: $$x=\frac{(8!)^{10}-(8!)^6}{(8!)^{5}-(8!)^3}=\frac{((8!)^{5}-(8!)^3)((8!)^{5}+(8!)^3)}{(8!)^{5}-(8!)^3}=(8!)^{5}+(8!)^3$$.

Next, $$\frac{x}{(8!)^3}-39=\frac{(8!)^{5}+(8!)^3}{(8!)^3}=\frac{(8!)^{5}}{(8!)^3}+\frac{(8!)^{3}}{(8!)^3}-39=(8!)^2+1-39=(8!)^2-38$$.

Now, since $$8!$$ has 2 and 5 as its multiples, then it will have 0 as the units digit, so $$(8!)^2$$ will have two zeros in the end, which means that $$(8!)^2-38$$ will have 00-38=62 as the last digits: 6*2=12.

Although there is hardly anything left after your's explanations but still I'll post my approach, which might help few people out here (esp. on the # trailing 0's as you people have kept in the Math book)--

first I simplified the whole x value as it looks so clumsy/heavy: given
$$x = (8!)^6 [(8!)^4 - 1]/ (8!)^3[(8!)^2 - 1] = (8!)^3[(8!)^2 + 1][(8!)^2 - 1]/[(8!)^2 - 1] so, (x/ (8!)^3) - 39 = [(8!)^2 + 1] -39 = (8!)^2 -38$$
Now as per the Q we need to know unit & tenth digit of the above number- as we know factorials of numbers > 4 have trailing zero's so here we can determine how many trailing 0's in 8!. Since $$8 < 5^2$$ so that means we have only 1 trailing 0 in here (alternately you can have manual calc in this case as well because it's small number). So $$(8!)^2$$ has 2 trailing 0's. Now it does not matter what hundredth position is becasue we need to subtract only 38- this would result in 6 in tenth pos and 2 in unit pos. so product is = 12...

P.S.- just for refrence # trailing 0's in factorial N is determined by- $$N/5^1+N/5^2+....+N/5^k (where 5^k < N)$$for each division we take integer part (quotient) and then add them up. this should give # trailing 0's
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Re: Baker's Dozen  [#permalink]

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15 Mar 2012, 05:28
1
Bunuel wrote:
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 $$C^8_{12}-(C^7_7*C^1_5+C^5_5*C^3_7)=495-(5+35)=455$$.

I tried feagure out this way first but thought to calculate little differently-

Since there are a toatl of 12 Marbels and after picking the required # marbels we are left with 4 marbels in the Jar- so our task is determine out of these 4 in how many ways we can pick R, B marbels so that there exists at least 1 marbel of each type- (R-1,B-3), (R-2,B-2),(R-3,B-1)

So the required # ways = $$7C1*5C3 + 7C2*5C2 + 7C3*5C1$$
= $$(7 * 10) + (21 * 10)+ (35 * 5)$$
= $$70 + 210 + 175$$
= 455 (D)
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16 Mar 2012, 07:01
1
1
Hi Bunuel,
Please help me in understanding the below text. I think I am lacking this concept.I always try to solve by taking LCM in order to get rid of the fractions. However, I found your approach short n simple. If you could shed some light on this, would be really helpful

Fritz owns $$\frac{2}{2+3}=\frac{2}{5}$$th of all shares

Thanks
H

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. Math Expert Joined: 02 Sep 2009 Posts: 59561 Re: Baker's Dozen [#permalink] ### Show Tags 16 Mar 2012, 07:12 4 3 imhimanshu wrote: Hi Bunuel, Please help me in understanding the below text. I think I am lacking this concept.I always try to solve by taking LCM in order to get rid of the fractions. However, I found your approach short n simple. If you could shed some light on this, would be really helpful Fritz owns $$\frac{2}{2+3}=\frac{2}{5}$$th of all shares Thanks H 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 $$3,600,000*\frac{1}{30}=120,000$$.

It's quite simple: A has $2 and B has 3$ --> A has 2/3rd of B's amount and also 2/(2+3)=2/5th of total amount of $5. Hope it's clear. Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 9848 Location: Pune, India Re: Baker's Dozen [#permalink] ### Show Tags 19 Mar 2012, 04:37 4 2 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. $$\frac{yz+xz-xy}{yz}$$ Solution: baker-s-dozen-128782-20.html#p1057508 Reading the discussion on this question made me realize how I end up retracting on the suggested strategies again and again. Since we are talking about Quant, you would expect to have a definite set of rules but no sir, you don't. I repeat this quite often: When you have a question using variables and the answer is in terms of those variables, your life is easier than if you get a question using numbers. The reason - you can assign your own 'cool' values to the variables for which the relations hold. Then just go on and check the option which gives you the right answer. I retract it with the following - But if the number of variables is high, say 3 or 4 or more, it might be too cumbersome to plug in values for each variable and keep in mind what stands for what. This could lead to a lot of confusion and errors. I retract it again - But if you can give some very convenient values to the variables, go ahead and plug it in. This question has variables and the answer is in terms of the variables so plugging in values is a good option. But the number of variables is 3 which could make it cumbersome. But, you can give the variables such values that you get your answer quickly. I need to find the rate of A. There are no constraints on the values x, y, and z can take except z > x (drain C empties slower than pipe A fille) Let's say, x = 4, y = 8, z = 8 What did I do here? I made the rate of B same as the rate of C. This means, whenever both of them are working together, drain C cancels out the work of pipe B. So the entire pool will be filled by pipe A and the amount of water pumped in by A will be the entire pool. Hence, if y = z, pipe A fills the entire pool i.e. the amount of water in terms of fraction of the pool pumped by A is 1. In the options, put y = z and see which option gives you 1. Only options (B) and (E) do. Now let's say, x = 8, y = 4, z = 8.00001 ( z should be greater than x but let's assume it is infinitesimally greater than x such that we can approximate it to 8 only) Rate of work of C is half the rate of work of B. Rate of work of C is same as rate of work of A. All the work done by pipe A is removed by drain C. So if pipe B fills the pool, drain C empties half of it in that time (this is the water pumped by pipe A) Put x = z in the options B and E. You get y/z = 4/8 = 1/2 in option (B). Answer (B) Even if you end up feeling that this method is complicated, try and wrap your head around it. It might give you some ideas of logical solutions in some other questions. _________________ Karishma Veritas Prep GMAT Instructor Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options > Intern Joined: 14 Apr 2012 Posts: 5 Re: Baker's Dozen [#permalink] ### Show Tags Updated on: 28 Apr 2012, 04:00 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! ?? Originally posted by mofasser08 on 28 Apr 2012, 03:41. Last edited by mofasser08 on 28 Apr 2012, 04:00, edited 1 time in total. Math Expert Joined: 02 Sep 2009 Posts: 59561 Re: Baker's Dozen [#permalink] ### Show Tags 28 Apr 2012, 03:59 3 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. Director Joined: 22 Mar 2011 Posts: 584 WE: Science (Education) Re: Baker's Dozen [#permalink] ### Show Tags 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 Answer B 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 Answer E $$y=\frac{(3^5-3^2)^2}{(5^7-5^4)^{-2}}=(3^5-3^2)^2(5^7-5^4)^2=3^4*5^8(3^2-1)^2*(5^3-1)^2=3^4*5^8*8^2*124^2$$. 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 Answer B 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 Answer D 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 Answer D 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. $$\frac{yz+xz-xy}{yz}$$ Answer B 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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Re: Baker's Dozen  [#permalink]

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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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Re: Baker's Dozen  [#permalink]

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25 Aug 2012, 11:07
1
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]

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02 Sep 2012, 12:55
Can you explain how you got from (3^5-3^2)^2 to 3^4*(3^3-1)^2? When extracting 3^2 from (3^5-3^2), do we also square the 3^2? I know it may be simple algebra but its a bit confusing, I got 3^2(3^3-1)^2. I know thats wrong but can you explain why? Thanks.

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 $$y$$.

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Re: Baker's Dozen  [#permalink]

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03 Sep 2012, 05:33
2
1
ctiger100 wrote:
Can you explain how you got from (3^5-3^2)^2 to 3^4*(3^3-1)^2? When extracting 3^2 from (3^5-3^2), do we also square the 3^2? I know it may be simple algebra but its a bit confusing, I got 3^2(3^3-1)^2. I know thats wrong but can you explain why? Thanks.

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 $$y$$.

Consider this: $$(ab-ac)^2=(a(b-c))^2=a^2*(b-c)^2$$, so $$(3^5-3^2)^2=(3^2*(3^3-1))^2=(3^2)^2*(3^3-1)^2=3^4*(3^3-1)^2$$.

Hope it helps.
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Re: Baker's Dozen  [#permalink]

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01 Oct 2012, 09:52
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.

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

Bunuel,
I did a mistake while calculating 10^5 - I calculated 9*10^4. I have a doubt : when the question says "5 digit" code, doesn't it mean that numbers start from 10000? I have seen some GMATClub questions were three digit codes start from 100. Can you please help?
Thanks
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Posts: 59561
Re: Baker's Dozen  [#permalink]

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01 Oct 2012, 09:57
2
voodoochild 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.

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

Bunuel,
I did a mistake while calculating 10^5 - I calculated 9*10^4. I have a doubt : when the question says "5 digit" code, doesn't it mean that numbers start from 10000? I have seen some GMATClub questions were three digit codes start from 100. Can you please help?
Thanks

Since we are talking about a 5-digit password and not a 5-digit number, then it can start with zero. For example a password can be 00123 or 00001.

Hope it's clear.
Re: Baker's Dozen   [#permalink] 01 Oct 2012, 09:57

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