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Baker's Dozen [#permalink]
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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: 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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Baker's Dozen [#permalink]
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13 Mar 2017, 09:01
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: http://gmatclub.com/forum/bakersdozen ... l#p1057509Let F represents Fritz, L for Luis, A for Alfred and W for Werner Fritz owns 2/3rd of the shares of the other three shareholders > F = 2/3 (3,600,000  F) > 3F = 7,200,000  2F > F = 1,440,000 Luis owns 3/7th of the shares of the other three shareholders > L = 3/7 (3,600,000  L) > 7L = 10,800,000  3L > L = 1,080,000 Alfred owns 4/11th of the shares of the other three shareholders > A = 4/11 (3,600,000  A) > 11A = 14,400,000  4A > A = 960,000 W= 3,600,000 (F + L + A) > 3,600,000  (1,440,000 + 1,080,000 + 960,000) = 120,000 Answer: D
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Re: Baker's Dozen [#permalink]
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27 Mar 2017, 18:07
Here's my solution for #2 (see visual below). It helps to use brackets!
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Baker's Dozen [#permalink]
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19 Dec 2017, 03:51
Bunuel niks18Quote: 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 Quote: Say 7 consecutive odd integers are: \(x\), \(x+2\), \(x+4\), \(x+6\), \(x+8\), \(x+10\), \(x+12\). I solved the same problem taking hint from number theory that a consecutive odd number is represented by: 2x+1 , 2x+3 My approach had slightly complex calculations but it ensured that my first no in series is always odd.Is this understanding correct? Or can I still take a random integer x as start and then add +2 (to get consecutive ones)
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Re: Baker's Dozen [#permalink]
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19 Dec 2017, 04:03
adkikani wrote: Bunuel niks18Quote: 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 Quote: Say 7 consecutive odd integers are: \(x\), \(x+2\), \(x+4\), \(x+6\), \(x+8\), \(x+10\), \(x+12\). I solved the same problem taking hint from number theory that a consecutive odd number is represented by: 2x+1 , 2x+3 My approach had slightly complex calculations but it ensured that my first no in series is always odd.Is this understanding correct? Or can I still take a random integer x as start and then add +2 (to get consecutive ones) In this question, as shown, you can take 7 consecutive odd integers to be \(x\), \(x+2\), \(x+4\), \(x+6\), \(x+8\), \(x+10\), and \(x+12\). For some number properties questions, you should represent odd integers, as 2k + 1, 2k + 3, ... Here you can also take the integers, to be 2k + 1, 2k + 3, ... but it's not necessary.
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Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
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Re: Baker's Dozen [#permalink]
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26 Mar 2018, 04:59
nahid78 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. I know this is how we solve this question. But can anyone please clarify why i don't get the same result if i work differently .... Suppose other two numbers are also same 6,6,6, and 1,1 or 6,6,6 and 2,2 so on... or other two are different. 6,6,6,1, and 2, or 6,6,6,2,and 4 You do get the same result in this case too. If the other 2 numbers are same, number of ways = 9C1 * 5!/3!*2! = 90 If the other 2 numbers are different, number of ways = 9C2 * 5!/3! = 720 Total = 90+720 = 810
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