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17 Apr 2013, 06:11
The next set of PS questions. I'll post OA's with detailed explanations after some discussion. Please, post your solutions along with the answers.1. The length of the diagonal of square S, as well as the lengths of the diagonals of rhombus R are integers. The ratio of the lengths of the diagonals is 15:11:9, respectively. Which of the following could be the difference between the area of square S and the area of rhombus R?
I. 63 II. 126 III. 252A. I only B. II only C. III only D. I and III only E. I, II and III Solution: freshmeat15104680.html#p12153182. Set S contains 7 different letters. How many subsets of set S, including an empty set, contain at most 3 letters?A. 29 B. 56 C. 57 D. 63 E. 64 Solution: freshmeat151046100.html#p12153233. How many different subsets of the set {0, 1, 2, 3, 4, 5} do not contain 0?A. 16 B. 27 C. 31 D. 32 E. 64 Solution: freshmeat151046100.html#p12153294. The functions f and g are defined for all the positive integers n by the following rule: f(n) is the number of positive perfect squares less than n and g(n) is the number of primes numbers less than n. If f(x) + g(x) = 16, then x is in the range:A. 30 < x < 36 B. 30 < x < 37 C. 31 < x < 37 D. 31 < x < 38 E. 32 < x < 38 Solution: freshmeat151046100.html#p12153355. Which of the following is a factor of 18!+1?A. 15 B. 17 C. 19 D. 33 E. 39 Solution: freshmeat151046100.html#p12153386. If the least common multiple of a positive integer x, 4^3 and 6^5 is 6^6. Then x can take how many values?A. 1 B. 6 C. 7 D. 30 E. 36 Solution: freshmeat151046100.html#p12153457. The greatest common divisor of two positive integers is 25. If the sum of the integers is 350, then how many such pairs are possible?A. 1 B. 2 C. 3 D. 4 E. 5 Solution: freshmeat151046100.html#p12153498. The product of a positive integer x and 377,910 is divisible by 3,300, then the least value of x is:A. 10 B. 11 C. 55 D. 110 E. 330 Solution: freshmeat151046100.html#p12153599. What is the 101st digit after the decimal point in the decimal representation of 1/3 + 1/9 + 1/27 + 1/37?A. 0 B. 1 C. 5 D. 7 E. 8 Solution: freshmeat151046100.html#p121536710. If x is not equal to 0 and x^y=1, then which of the following must be true?
I. x=1 II. x=1 and y=0 III. x=1 or y=0A. I only B. II only C. III only D. I and III only E. None Solution: freshmeat151046100.html#p1215370Kudos points for each correct solution!!!
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31 Aug 2013, 10:55
2013gmat wrote: Bunuel wrote: 6. If the least common multiple of a positive integer x, 4^3 and 6^5 is 6^6. Then x can take how many values?
A. 1 B. 6 C. 7 D. 30 E. 36
We are given that \(6^6=2^{6}*3^{6}\) is the least common multiple of the following three numbers:
x; \(4^3=2^6\); \(6^5 = 2^{5}*3^5\);
First notice that \(x\) cannot have any other primes other than 2 or/and 3, because LCM contains only these primes.
Now, since the power of 3 in LCM is higher than the powers of 3 in either the second number or in the third, than \(x\) must have \(3^{6}\) as its multiple (else how \(3^{6}\) would appear in LCM?).
Next, \(x\) can have 2 as its prime in ANY power ranging from 0 to 6, inclusive (it cannot have higher power of 2 since LCM limits the power of 2 to 6).
Thus, \(x\) could take total of 7 values.
Answer: C. Hi Bunuel, x can take factor of 2 with power from 2 to 6 or no factor of 2. So the answer can be 6 too. Please explain ! thanks I don;t understand what you mean... x can take the following 7 values: \(3^6\); \(2*3^6\); \(2^2*3^6\); \(2^3*3^6\); \(2^4*3^6\); \(2^5*3^6\); \(2^6*3^6\).
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31 Aug 2013, 22:30
\(\frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \frac{1}{37}=\frac{333}{999} + \frac{111}{999} + \frac{37}{999} + \frac{27}{999}=\frac{508}{999}=0.508508...\).
How do we get these fractions with a common denominator?



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01 Sep 2013, 12:08



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02 Sep 2013, 20:28
Bunuel wrote: 10. If x is not equal to 0 and x^y=1, then which of the following must be true?
I. x=1 II. x=1 and y=0 III. x=1 or y=0
A. I only B. II only C. III only D. I and III only E. None
Notice that if x=1 and y is any even number, then \((1)^{even}=1\), thus none of the options must be true.
Answer: E. Hi Bunuel, As per the question which of the following must be true. So as per the given choice B) II only is true right where 1^0 = 1 as X =1 and Y=0 given. As ur explanation gives another chance as X coud be = 1 , and Y = any even. Please clarify where i am wrong. Thanks in Advance, Rrsnathan



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03 Sep 2013, 02:26
rrsnathan wrote: Bunuel wrote: 10. If x is not equal to 0 and x^y=1, then which of the following must be true?
I. x=1 II. x=1 and y=0 III. x=1 or y=0
A. I only B. II only C. III only D. I and III only E. None
Notice that if x=1 and y is any even number, then \((1)^{even}=1\), thus none of the options must be true.
Answer: E. Hi Bunuel, As per the question which of the following must be true. So as per the given choice B) II only is true right where 1^0 = 1 as X =1 and Y=0 given. As ur explanation gives another chance as X coud be = 1 , and Y = any even. Please clarify where i am wrong. Thanks in Advance, Rrsnathan x=1 and y=0 indeed satisfies x^y=1, but the question asks "which of the following must be true". So, this option is NOT necessarily true, because x can be 1 and y any even number.
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18 Sep 2013, 02:57
Bunuel wrote: 25*1=25 and 25*13=325; 25*3=75 and 25*11=275; 25*5=125 and 25*9=225.
Answer: C. Hello, what about 175, 175 ? The question doesn't state they are distinct positive numbers.



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30 Sep 2013, 04:14
Bunuel wrote: 3. How many different subsets of the set {0, 1, 2, 3, 4, 5} do not contain 0?
A. 16 B. 27 C. 31 D. 32 E. 64
Consider the set without 0: {1, 2, 3, 4, 5}. Each out of 5 elements of the set {1, 2, 3, 4, 5} has TWO options: either to be included in the subset or not, so total number of subsets of this set is 2^5=32. Now, each such set will be a subset of {0, 1, 2, 3, 4, 5} and won't include 0.
Answer: D. I would like to request some some help with this question.. Could you please elaborate on the theory behind 2^n? Isnt the null set considered to be a subset of this set?
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30 Sep 2013, 05:41
Transcendentalist wrote: Bunuel wrote: 3. How many different subsets of the set {0, 1, 2, 3, 4, 5} do not contain 0?
A. 16 B. 27 C. 31 D. 32 E. 64
Consider the set without 0: {1, 2, 3, 4, 5}. Each out of 5 elements of the set {1, 2, 3, 4, 5} has TWO options: either to be included in the subset or not, so total number of subsets of this set is 2^5=32. Now, each such set will be a subset of {0, 1, 2, 3, 4, 5} and won't include 0.
Answer: D. I would like to request some some help with this question.. Could you please elaborate on the theory behind 2^n? Isnt the null set considered to be a subset of this set? Consider simpler set: {a, b, c}. How many subsets does it have? Each of a, b, and c has two choices either to be included in subsets or not. Thus total of 2^38 subsets (including an empty set). The subsets are: {a, b, c} > each is included; {a, b} > a and b are included; {a, c}; {b, c}; {a}; {b}; {c}; {} empty set, none is included. 2^3=8 subsets. Harder question to practice about the same concept: howmanysubordinatesdoesmarciahave57169.html#p692676Hope it helps.
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12 Jan 2014, 22:15
Bunuel wrote: 9. What is the 101st digit after the decimal point in the decimal representation of 1/3 + 1/9 + 1/27 + 1/37?
A. 0 B. 1 C. 5 D. 7 E. 8
\(\frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \frac{1}{37}=\frac{333}{999} + \frac{111}{999} + \frac{37}{999} + \frac{27}{999}=\frac{508}{999}=0.508508...\).
102nd digit will be 8, thus 101st digit will be 0.
Answer: A. Great questions Bunuel! Is there a trick to convert 1/37 > 27/999?
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13 Jan 2014, 00:17
m3equals333 wrote: Bunuel wrote: 9. What is the 101st digit after the decimal point in the decimal representation of 1/3 + 1/9 + 1/27 + 1/37?
A. 0 B. 1 C. 5 D. 7 E. 8
\(\frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \frac{1}{37}=\frac{333}{999} + \frac{111}{999} + \frac{37}{999} + \frac{27}{999}=\frac{508}{999}=0.508508...\).
102nd digit will be 8, thus 101st digit will be 0.
Answer: A. Great questions Bunuel! Is there a trick to convert 1/37 > 27/999? \(\frac{1}{3} =\frac{1*333}{3*333}=\frac{27}{999}\). \(\frac{1}{9} =\frac{1*111}{9*111}=\frac{27}{999}\). \(\frac{1}{27} =\frac{1*37}{27*37}=\frac{27}{999}\). \(\frac{1}{37} =\frac{1*27}{37*27}=\frac{27}{999}\). Following link might help for this problem: mathnumbertheory88376.html (check Converting Fractions chapter).
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27 Feb 2014, 07:22
Bunuel wrote: 10. If x is not equal to 0 and x^y=1, then which of the following must be true?
I. x=1 II. x=1 and y=0 III. x=1 or y=0
A. I only B. II only C. III only D. I and III only E. None
Notice that if x=1 and y is any even number, then \((1)^{even}=1\), thus none of the options must be true.
Answer: E. Bunuel.. (II) x=1 andy=0 will give ans 1.. cant we say that this is the solution dat must give Ans "1"? 1^0=1?
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27 Feb 2014, 07:35
sanjoo wrote: Bunuel wrote: 10. If x is not equal to 0 and x^y=1, then which of the following must be true?
I. x=1 II. x=1 and y=0 III. x=1 or y=0
A. I only B. II only C. III only D. I and III only E. None
Notice that if x=1 and y is any even number, then \((1)^{even}=1\), thus none of the options must be true.
Answer: E. Bunuel.. (II) x=1 andy=0 will give ans 1.. cant we say that this is the solution dat must give Ans "1"? 1^0=1? The question asks which of the following MUST be true, not COULD be true. So, if if x=1 and y is any even number, then \((1)^{even}=1\), thus none of the options MUST be true.
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03 Apr 2014, 08:25
Bunuel wrote: 4. The functions f and g are defined for all the positive integers n by the following rule: f(n) is the number of perfect squares less than n and g(n) is the number of primes numbers less than n. If f(x) + g(x) = 16, then x is in the range:
A. 30 < x < 36 B. 30 < x < 37 C. 31 < x < 37 D. 31 < x < 38 E. 32 < x < 38
Perfect squares: 1, 4, 9, 16, 25, 36, .., Prime numbers: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, ...
If x = 31, then f(31) = 5 and g(31) = 10: f(x) + g(x) = 5 + 10 = 15. If x = 32, then f(32) = 5 and g(32) = 11: f(x) + g(x) = 5 + 11 = 16. ... If x = 36, then f(36) = 5 and g(36) = 11: f(x) + g(x) = 5 + 11 = 16. If x = 37, then f(37) = 6 and g(37) = 11: f(x) + g(x) = 6 + 11 = 17.
Thus x could be 32, 33, 34, 35 or 36: 31<x<37.
Answer: C. Why is zero not being considered in the count of perfect squares? I realize that it states that n is positive, but if f(n) is defined as perfect squares less than n and zero is less than, why do we not count it?
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03 Apr 2014, 08:44
MrWallSt wrote: Bunuel wrote: 4. The functions f and g are defined for all the positive integers n by the following rule: f(n) is the number of perfect squares less than n and g(n) is the number of primes numbers less than n. If f(x) + g(x) = 16, then x is in the range:
A. 30 < x < 36 B. 30 < x < 37 C. 31 < x < 37 D. 31 < x < 38 E. 32 < x < 38
Perfect squares: 1, 4, 9, 16, 25, 36, .., Prime numbers: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, ...
If x = 31, then f(31) = 5 and g(31) = 10: f(x) + g(x) = 5 + 10 = 15. If x = 32, then f(32) = 5 and g(32) = 11: f(x) + g(x) = 5 + 11 = 16. ... If x = 36, then f(36) = 5 and g(36) = 11: f(x) + g(x) = 5 + 11 = 16. If x = 37, then f(37) = 6 and g(37) = 11: f(x) + g(x) = 6 + 11 = 17.
Thus x could be 32, 33, 34, 35 or 36: 31<x<37.
Answer: C. Why is zero not being considered in the count of perfect squares? I realize that it states that n is positive, but if f(n) is defined as perfect squares less than n and zero is less than, why do we not count it? You are right. The question should read: f(n) is the number of positive perfect squares less than n. Edited.
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09 Apr 2014, 01:53
Bunuel wrote: 3. How many different subsets of the set {0, 1, 2, 3, 4, 5} do not contain 0?
A. 16 B. 27 C. 31 D. 32 E. 64
Consider the set without 0: {1, 2, 3, 4, 5}. Each out of 5 elements of the set {1, 2, 3, 4, 5} has TWO options: either to be included in the subset or not, so total number of subsets of this set is 2^5=32. Now, each such set will be a subset of {0, 1, 2, 3, 4, 5} and won't include 0.
Answer: D. Hi Bunuel, Is {NULL} a subset of {1,2,3,4,5}? Because 2^5 also contains {NULL} as one possibility. Thanks..



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09 Apr 2014, 03:05
riskietech wrote: Bunuel wrote: 3. How many different subsets of the set {0, 1, 2, 3, 4, 5} do not contain 0?
A. 16 B. 27 C. 31 D. 32 E. 64
Consider the set without 0: {1, 2, 3, 4, 5}. Each out of 5 elements of the set {1, 2, 3, 4, 5} has TWO options: either to be included in the subset or not, so total number of subsets of this set is 2^5=32. Now, each such set will be a subset of {0, 1, 2, 3, 4, 5} and won't include 0.
Answer: D. Hi Bunuel, Is {NULL} a subset of {1,2,3,4,5}? Because 2^5 also contains {NULL} as one possibility. Thanks.. Yes, an empty set is a subset of all sets.
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Re: Fresh Meat!!! [#permalink]
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17 May 2014, 07:44
Bunuel wrote: 3. How many different subsets of the set {0, 1, 2, 3, 4, 5} do not contain 0?
A. 16 B. 27 C. 31 D. 32 E. 64
Consider the set without 0: {1, 2, 3, 4, 5}. Each out of 5 elements of the set {1, 2, 3, 4, 5} has TWO options: either to be included in the subset or not, so total number of subsets of this set is 2^5=32. Now, each such set will be a subset of {0, 1, 2, 3, 4, 5} and won't include 0.
Answer: D. Dear Bunnel I didnt understand this. y didnt we take the set {1,2,3,4,5} into consideration and solve like the above qs? where did we get 2^5 from?
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Re: Fresh Meat!!! [#permalink]
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17 May 2014, 07:51
NGGMAT wrote: Bunuel wrote: 3. How many different subsets of the set {0, 1, 2, 3, 4, 5} do not contain 0?
A. 16 B. 27 C. 31 D. 32 E. 64
Consider the set without 0: {1, 2, 3, 4, 5}. Each out of 5 elements of the set {1, 2, 3, 4, 5} has TWO options: either to be included in the subset or not, so total number of subsets of this set is 2^5=32. Now, each such set will be a subset of {0, 1, 2, 3, 4, 5} and won't include 0.
Answer: D. Dear Bunnel I didnt understand this. y didnt we take the set {1,2,3,4,5} into consideration and solve like the above qs? where did we get 2^5 from? What do you mean by the red part? As for 2^5: the number of subsets of nelement set is 2^n, thus the number of subsets of 5element set {1, 2, 3, 4, 5} is 2^5 (note that this includes an empty set as well as the original set {1, 2, 3, 4, 5}). Now, all subsets of {1, 2, 3, 4, 5} are the subsets of {0, 1, 2, 3, 4, 5} and does not include 0. Does this make sense?
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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: Fresh Meat!!! [#permalink]
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17 May 2014, 08:20
Bunuel wrote: NGGMAT wrote: Bunuel wrote: 3. How many different subsets of the set {0, 1, 2, 3, 4, 5} do not contain 0?
A. 16 B. 27 C. 31 D. 32 E. 64
Consider the set without 0: {1, 2, 3, 4, 5}. Each out of 5 elements of the set {1, 2, 3, 4, 5} has TWO options: either to be included in the subset or not, so total number of subsets of this set is 2^5=32. Now, each such set will be a subset of {0, 1, 2, 3, 4, 5} and won't include 0.
Answer: D. Dear Bunnel I didnt understand this. y didnt we take the set {1,2,3,4,5} into consideration and solve like the above qs? where did we get 2^5 from? What do you mean by the red part? As for 2^5: the number of subsets of nelement set is 2^n, thus the number of subsets of 5element set {1, 2, 3, 4, 5} is 2^5 (note that this includes an empty set as well as the original set {1, 2, 3, 4, 5}). Now, all subsets of {1, 2, 3, 4, 5} are the subsets of {0, 1, 2, 3, 4, 5} and does not include 0. Does this make sense? By red part i meant that y we havent solved it like we did the below qs: 2. Set S contains 7 different letters. How many subsets of set S, including an empty set, contain at most 3 letters?
A. 29 B. 56 C. 57 D. 63 E. 64
1 empty set; C^1_7=7 sets with one element; C^2_7=21 sets with two elements; C^3_7=35 sets with three element.
Total 1+7+21+35=64 setsy did 2^n come into qs 3 and not qs 2?
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