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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: bakersdozen12878220.html#p10575022. If \(y=\frac{(3^53^2)^2}{(5^75^4)^{2}}\), 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: bakersdozen12878220.html#p10575033. 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: bakersdozen12878220.html#p10575044. 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: bakersdozen12878220.html#p10575055. 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: bakersdozen12878220.html#p10575076. 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. \(\frac{yz}{yz+xzxy}\) C. \(\frac{yz}{yz+xz+xy}\) D. \(\frac{xyz}{yz+xzxy}\) E. \(\frac{yz+xzxy}{yz}\) Solution: bakersdozen12878220.html#p10575087. 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: bakersdozen12878220.html#p10575098. 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: bakersdozen12878220.html#p10575129. If x and y are negative numbers, what is the value of \(\frac{\sqrt{x^2}}{x}\sqrt{y*y}\)?A. 1+y B. 1y C. 1y D. y1 E. xy Solution: bakersdozen12878220.html#p105751410. If x^2<81 and y^2<25, what is the largest prime number that can be equal to x2y?A. 7 B. 11 C. 13 D. 17 E. 19 Solution: bakersdozen12878220.html#p105751511. 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: bakersdozen12878240.html#p105751712. 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: bakersdozen12878240.html#p105751913. 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: bakersdozen12878240.html#p1057520
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Re: Baker's Dozen [#permalink]
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18 Sep 2013, 06:00
Bunuel wrote: 2. If \(y=\frac{(3^53^2)^2}{(5^75^4)^{2}}\), 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
\(y=\frac{(3^53^2)^2}{(5^75^4)^{2}}=(3^53^2)^2*(5^75^4)^2=3^4*(3^31)^2*5^8*(5^31)^2=3^4*26^2*5^8*124^2=2^6*3^4*5^8*13^2*31^2\).
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. Can someone please explain how the move from the second part of the equation to the third part was?



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Re: Baker's Dozen [#permalink]
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18 Sep 2013, 07:19
ronr34 wrote: Bunuel wrote: 2. If \(y=\frac{(3^53^2)^2}{(5^75^4)^{2}}\), 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
\(y=\frac{(3^53^2)^2}{(5^75^4)^{2}}=(3^53^2)^2*(5^75^4)^2=3^4*(3^31)^2*5^8*(5^31)^2=3^4*26^2*5^8*124^2=2^6*3^4*5^8*13^2*31^2\).
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. Can someone please explain how the move from the second part of the equation to the third part was? Consider this: \((abac)^2=(a(bc))^2=a^2*(bc)^2\), so \((3^53^2)^2=(3^2*(3^31))^2=(3^2)^2*(3^31)^2=3^4*(3^31)^2\). Hope it helps.
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Re: Baker's Dozen [#permalink]
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19 Sep 2013, 04:34
I have a question about the quick calculation with regards to the number of shares that each one holds.... What is the logic behind the fraction of 2/2+3 ? 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.



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Re: Baker's Dozen [#permalink]
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19 Sep 2013, 04:52
ronr34 wrote: I have a question about the quick calculation with regards to the number of shares that each one holds.... What is the logic behind the fraction of 2/2+3 ? 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. Check here: bakersdozen12878240.html#p1059585
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Re: Baker's Dozen [#permalink]
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08 Oct 2013, 07:32
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Hi Bunuel,
On Mr. Wallace's Briefcase problem, I was wondering if you can clarify the step where you multiple by 5 choose 3? Can I think of it as SSSNN ["S" representing the number 6 and "N" representing a non6 digit] and the number of ways to arrange this would be 5!/[3!*2!] since there are 3 repeats for S and 2 repeats for N. Is my reasoning the a logical way to approach this problem as well? Thnks!



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17 Oct 2013, 10:12
Bunuel wrote: SOLUTIONS:
1. A password on Mr. Wallace's briefcase consists of 5 digits. What is the probability that the password contains exactly three digit 6?
A. 860/90,000 B. 810/100,000 C. 858/100,000 D. 860/100,000 E. 1530/100,000
Total # of 5 digit codes is 10^5, notice that it's not 9*10^4, since in a code we can have zero as the first digit.
# of passwords with three digit 6 is \(9*9*C^3_5=810\): each out of two other digits (not 6) has 9 choices, thus we have 9*9 and \(C^3_5\) is ways to choose which 3 digits will be 6's out of 5 digits we have.
\(P=\frac{favorable}{total}=\frac{810}{10^5}\)
Answer: B. what's the 9*9 portion of that come from?



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17 Oct 2013, 10:15
AccipiterQ wrote: Bunuel wrote: SOLUTIONS:
1. A password on Mr. Wallace's briefcase consists of 5 digits. What is the probability that the password contains exactly three digit 6?
A. 860/90,000 B. 810/100,000 C. 858/100,000 D. 860/100,000 E. 1530/100,000
Total # of 5 digit codes is 10^5, notice that it's not 9*10^4, since in a code we can have zero as the first digit.
# of passwords with three digit 6 is \(9*9*C^3_5=810\): each out of two other digits (not 6) has 9 choices, thus we have 9*9 and \(C^3_5\) is ways to choose which 3 digits will be 6's out of 5 digits we have.
\(P=\frac{favorable}{total}=\frac{810}{10^5}\)
Answer: B. what's the 9*9 portion of that come from? There are 2 digits different from 6. Each can take 9 values. Total 9*9=81 values.
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Re: Baker's Dozen [#permalink]
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19 Oct 2013, 15:40
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 possibilities = 100,000
No of possibilities with exactly three 6s' = 6 6 6 X X
X can be any digit among 0,1,2,3,4,5,7,8,9
First case: XX can be same digits, so XX can be 00,11,22,33,44,55,77,88,99
No of ways to arrange this: (5!/2! * 3 !) * 9 = 90
second case: XX can be different digits, so XX can 01,02,04,05 etc.. likewise 72 possibilities.
No of ways to arrange this combination: (5!/3!) * 72 = 1440
So total no ways = 1530
I chose answer E but I know it is wrong. Can someone tell me where I am going wrong?



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25 Oct 2013, 14:30
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Hello Bunuel,
In the last problem: 8! does not end in 00 rather it is 40320. So the solution would be 8!38=40282
And the product of the tens and units digits would be : 8*2= 16 (Though not in option)
Can you please correct me if i am worng



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25 Oct 2013, 14:32



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01 Dec 2013, 03:32
Hello Bunuel, Since we have all negative numbers in the set, shouldn't we add up the first 5 numbers as the largest? \(x + (x+2) + (x+4) + (x+6) + (x+8) = 5x +20 = 185\) \(x = 41\) Then we take 5 smallest integers, which are the latter in our set \((x+4) + (x+6) + (x+8) + (x+10) + (x+12) = 5x +40\) \(x=  41\) \(41*5 + 40=165\) answer A? Please correct me if I'm wrong. 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: \((x+4)+(x+6)+(x+8)+(x+10)+(x+12)=185\) > \((x+4)+(x+6)+(x+8)+(x+10)+(x+12)=5x+40=185\) > \((5x+20)+20=185\) > \(5x+20=205\)
Answer: D.



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Re: Baker's Dozen [#permalink]
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01 Dec 2013, 06:54
trailrunner wrote: Hello Bunuel, Since we have all negative numbers in the set, shouldn't we add up the first 5 numbers as the largest? \(x + (x+2) + (x+4) + (x+6) + (x+8) = 5x +20 = 185\) \(x = 41\) Then we take 5 smallest integers, which are the latter in our set \((x+4) + (x+6) + (x+8) + (x+10) + (x+12) = 5x +40\) \(x=  41\) \(41*5 + 40=165\) answer A? Please correct me if I'm wrong. 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: \((x+4)+(x+6)+(x+8)+(x+10)+(x+12)=185\) > \((x+4)+(x+6)+(x+8)+(x+10)+(x+12)=5x+40=185\) > \((5x+20)+20=185\) > \(5x+20=205\)
Answer: D. Five largest integers from {\(x\), \(x+2\), \(x+4\), \(x+6\), \(x+8\), \(x+10\), \(x+12\)} are {\(x+4\), \(x+6\), \(x+8\), \(x+10\), \(x+12\)} no matter whether x is negative or positive.
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Re: Baker's Dozen [#permalink]
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18 Jan 2014, 21:23
Bunuel wrote: SOLUTIONS:
1. A password on Mr. Wallace's briefcase consists of 5 digits. What is the probability that the password contains exactly three digit 6?
A. 860/90,000 B. 810/100,000 C. 858/100,000 D. 860/100,000 E. 1530/100,000
Total # of 5 digit codes is 10^5, notice that it's not 9*10^4, since in a code we can have zero as the first digit.
# of passwords with three digit 6 is \(9*9*C^3_5=810\): each out of two other digits (not 6) has 9 choices, thus we have 9*9 and \(C^3_5\) is ways to choose which 3 digits will be 6's out of 5 digits we have.
\(P=\frac{favorable}{total}=\frac{810}{10^5}\)
Answer: B. Hi Bunuel, 6 6 6 The 6 smilefaces stand for the 6 possible position for the other 2 digit (not 6). If I am right, I think the number of passwords with 3 digit 6 is C(6, 2)P(9,1)P(9,1). But if so, there is no correct answer in the choice. Would you please advise me where I am wrong? Thank you.



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14 Feb 2014, 01:39
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. 1y C. 1y D. y1 E. xy
Note that \(\sqrt{a^2}=a\). Next, since \(x<0\) and \(y<0\) then \(x=x\) and \(y=y\).
So, \(\frac{\sqrt{x^2}}{x}\sqrt{y*y}=\frac{x}{x}\sqrt{(y)*(y)}=\frac{x}{x}\sqrt{y^2}=1y=1+y\)
Answer: D. Shouldn't 1y be 1y? since y will be positive always?



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17 Feb 2014, 07:50
dnawap123 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. 1y C. 1y D. y1 E. xy
Note that \(\sqrt{a^2}=a\). Next, since \(x<0\) and \(y<0\) then \(x=x\) and \(y=y\).
So, \(\frac{\sqrt{x^2}}{x}\sqrt{y*y}=\frac{x}{x}\sqrt{(y)*(y)}=\frac{x}{x}\sqrt{y^2}=1y=1+y\)
Answer: D. Shouldn't 1y be 1y? since y will be positive always? No. When \(x\leq{0}\) then \(x=x\), or more generally when \(some \ expression\leq{0}\) then \(some \ expression={(some \ expression)}\). When \(x\geq{0}\) then \(x=x\), or more generally when \(some \ expression\geq{0}\) then \(some \ expression={some \ expression}\). Since given that y is a negative number (y<0), then \(y = y\), hence \(1y=1(y)=1+y\). Hope it helps.
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Re: Baker's Dozen [#permalink]
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15 Jun 2014, 22:51
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
Sol.
Total outcomes = 10^5 desired = _ _ _ _ _ (three 6s and two any digits but not 6) i.e. number of ways to select digit that is not 6 = 9 ways to select 6 = 1 way therefore total ways of desired outcome = 9*9*5!(arrangement of 5 digits)/3!(arrangement of 6s) i.e. equal to 1620/10^5
Bunuel, in your solution the total ways of desired selection are 9*9*(5!/3!*2!) can you explain from where do we get 2! in denominator ? Dividing by 2! would mean that the other two digits are also same. ex. 55666 or 77666. But 57666 is also possible. Please correct where I am getting it wrong.



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Re: Baker's Dozen [#permalink]
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03 Jul 2014, 00:52
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 \(a^2b^2=(ab)(a+b)\): \(x=\frac{(8!)^{10}(8!)^6}{(8!)^{5}(8!)^3}=\frac{((8!)^{5}(8!)^3)((8!)^{5}+(8!)^3)}{(8!)^{5}(8!)^3}=(8!)^{5}+(8!)^3\).
Next, \(\frac{x}{(8!)^3}39=\frac{(8!)^{5}+(8!)^3}{(8!)^3}=\frac{(8!)^{5}}{(8!)^3}+\frac{(8!)^{3}}{(8!)^3}39=(8!)^2+139=(8!)^238\).
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 \((8!)^238\) will have 0038=62 as the last digits: 6*2=12.
Answer: D. Hi Bunnel, I have a doubt here I did it in a following way ((8!)^6 ((8!)^41))/((8!)^3 ((8!)^21)) then I will get (8!)^5 as i can ignore 1. (8!)^41 = (8!)^4. I have seen this kind of solution on some of the problems where you ignore 1 . I did the same here but getting different result. Please suggest.



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Re: Baker's Dozen [#permalink]
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03 Jul 2014, 06:09
PathFinder007 wrote: 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 \(a^2b^2=(ab)(a+b)\): \(x=\frac{(8!)^{10}(8!)^6}{(8!)^{5}(8!)^3}=\frac{((8!)^{5}(8!)^3)((8!)^{5}+(8!)^3)}{(8!)^{5}(8!)^3}=(8!)^{5}+(8!)^3\).
Next, \(\frac{x}{(8!)^3}39=\frac{(8!)^{5}+(8!)^3}{(8!)^3}=\frac{(8!)^{5}}{(8!)^3}+\frac{(8!)^{3}}{(8!)^3}39=(8!)^2+139=(8!)^238\).
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 \((8!)^238\) will have 0038=62 as the last digits: 6*2=12.
Answer: D. Hi Bunnel, I have a doubt here I did it in a following way ((8!)^6 ((8!)^41))/((8!)^3 ((8!)^21)) then I will get (8!)^5 as i can ignore 1. (8!)^41 = (8!)^4. I have seen this kind of solution on some of the problems where you ignore 1 . I did the same here but getting different result. Please suggest. That's totally wrong. We need to find the tens and units digits of some expression, ignoring anything there will change the result. Where did I say that we can ignore something when we want to find units/tens digit of an expression???
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Baker's Dozen [#permalink]
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29 Jul 2014, 05:40
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. \(\frac{yz}{yz+xzxy}\)
C. \(\frac{yz}{yz+xz+xy}\)
D. \(\frac{xyz}{yz+xzxy}\)
E. \(\frac{yz+xzxy}{yz}\)
With pumps A and B both running and the drain unstopped the pool will be filled in a rate \(\frac{1}{x}+\frac{1}{y}\frac{1}{z}=\frac{yz+xzzy}{xyz}\) pool/hour. So, the pool will be filled in \(\frac{xyz}{yz+xzxy}\) hours (time is reciprocal of rate).
In \(\frac{xyz}{yz+xzxy}\) hours A will pump \(\frac{1}{x}*\frac{xyz}{yz+xzxy}=\frac{yz}{yz+xzxy}\) amount of the water into the pool.
Answer: B. can you please tell me if there is any difeerence in these two statements amount of water in terms of the fraction of the pool which pump A pumped into the pool? amount of water which pump A pumped into the pool? and if there is a difference can you please explain? i had solved the question but was stumped by the language, is it simply asking me the amount of water pumped by A thanks







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