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Fresh Meat!!!
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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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21 Apr 2013, 23:04
7. 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 We are told that the greatest common factor of two integers is 25. So, these integers are \(25x\) and \(25y\), for some positive integers \(x\) and \(y\). Notice that \(x\) and \(y\) must not share any common factor but 1, because if they do, then GCF of \(25x\) and \(25y\) will be more that 25. Next, we know that \(25x+25y=350\) > \(x+y=14\) > since \(x\) and \(y\) don't share any common factor but 1 then (x, y) can be only (1, 13), (3, 11) or (5, 9) (all other pairs (2, 12), (4, 10), (6, 8) and (7, 7) do share common factor greater than 1). So, there are only three pairs of such numbers possible: 25*1=25 and 25*13=325; 25*3=75 and 25*11=275; 25*5=125 and 25*9=225. Answer: C.
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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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17 Apr 2013, 06:28
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. 252
Side square = 15x \(AreaS = \frac{15^2}{2}x^2\) Diagonals= 9x, 11x\(AreaR = \frac{11*9*x^2}{2}\) Difference = \(\frac{15^2x^211*9x^2}{2}= \frac{126x^2}{2}= 63x^2\) \(63=3*3*7\) if x=1 diff = 63 possible and easy to see \(126=2*3*3*7\) x sould be \(\sqrt{2}\) => no integer \(252=2*2*3*3*7\) x=2 possible IMO D. I and III only
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17 Apr 2013, 06:31
2. Set S contains 7 different letters. How many subsets of set S, including an empty set, contain at most 3 letters?At most 3 letters = 0 letters or 1 letter or 2 letters or 3 letters 0=1 1=7C1=7 2=7C2=21 3=7C3=35 \(1+7+21+35=64\) IMO E. 64
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17 Apr 2013, 06:34
3. How many different subsets of the set {0, 1, 2, 3, 4, 5} do not contain 0?It's like asking how many subsets has {1,2,3,4,5} \(2^5=32\) IMO D. 32
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17 Apr 2013, 06:35
5. Which of the following is a factor of 18!+1? A. 15 B. 17 C. 19 D. 33 E. 39 18! and 18!+1 are consecutive integers, thus coprime. All options apart from C are present in 18!. Thus 19 is the only factor present in 18!+1. C.
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17 Apr 2013, 06:47
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 The no of primes less than 30 = 10 primes. Also,the number of perfect squares less than 30 = 1,4,9,16,25 = 5. Thus, for 31<x<37, the total sum is 16. C.
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17 Apr 2013, 06:49
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:perfect squares = 1 4 9 16 25 36 prime numbers = 2 3 5 7 11 13 17 19 23 19 31 37 The first number that makes f(x) + g(x) = 16 is 32 and the last is 36 IMO C. 31 < x < 37
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Updated on: 18 Apr 2013, 02:11
7. 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 The two numbers can be represented as 25a and 25b, where a and b are coprime.Also, 25(a+b) = 350 > (a+b) = 14 Thus, a=1,b=13 or a=3,b=11 or a=9,b=5. C.
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Originally posted by mau5 on 17 Apr 2013, 06:55.
Last edited by mau5 on 18 Apr 2013, 02:11, edited 2 times in total.



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17 Apr 2013, 06:58
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?\(2^6, 2^5*3^5,2^6*3^6\) and \(x\) \(x\) MUST have a \(3^6\) and can have any \(2^n\) with \(0\leq{n}\leq{6}\). So x can have \(7\) values IMO C. 7
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17 Apr 2013, 07:01
8. 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 We know that 377910 is not divisible by 11. Also, 3300 = 3*11*5^2*2^2. Now, as 377910 is divisible by 30, we are left with 11,5,2.Thus, the least value of x = 11*5*2 = 110. D.
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17 Apr 2013, 07:04
7. 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?\(GMD = 5^2\) the numbers are \(5^2k\) and \(5^2q\) where q and k do not share any factor \(25k+25q=350\) \(25(k+q)=350\) \(k+q=14\) The possible numbers that summed give us 14 are: 1+13, 2+12,... those however must have no factor in common and those pairs are:1+13,3+11,5+9 IMO C.3
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17 Apr 2013, 07:11
8. The product of a positive integer x and 377,910 is divisible by 3,300, then the least value of x is:\(3,300=3*11*2*5*2*5\) \(377,910=37791*2*5\) \(377,910*x/3300=\frac{37791*2*5*x}{3*11*2*5*2*5}\) Now x must contain 2*5, because 37791 is divisible by 3 x must not contain a 3, because 37791 is not divisible by 11 x must have it. \(x=2*5*11=110\) IMO D. 110
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17 Apr 2013, 07:15
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 From the given inequality, for any y and x=1, we would have x^y = 1. Also, for any x(and not equal to 0) and y = 0, we would again have the same inequality.
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17 Apr 2013, 07:20
10. If x is not equal to 0 and x^y=1, then which of the following must be true?I. x=1 False \(100^0=1\) II. x=1 and y=0 False \(2^0=1\) III. x=1 or y=0 True Infact there are two cases: every number with a 0 exponent equals 1, and 1 raised to any exp equals 1. IMO C. III only
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17 Apr 2013, 07:23
5. Which of the following is a factor of 18!+1?\(18!\) and \(18!+1\) are consecutives so they do not share any factor (except 1). A,B,D and E are factors of \(18!\) (ie:\(33=3*11\) both contained in \(18!\)) IMO C.19
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17 Apr 2013, 07:38
9. What is the 101st digit after the decimal point in the decimal representation of 1/3 + 1/9 + 1/27 + 1/37?1/3=0.333 1/9=0.333/3=0.111 1/27=0.037 and then repeats 1/37=0.027 and then repeats We can work on the first 3 digits: 0.111+0.333+0.027+0.037=0.508 After the 0 we have at first place a 5, second a 0, third an 8; and so on 4th=5, 5th=0, 6th=8. Every 10th position we have a "change" 10th=5 20th=0 30th=8 and so on 100th=5 and finally 101st=0 IMO A.0 Thanks for the set Bunuel!
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17 Apr 2013, 09:28
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 1/37 = 27/999= 0.027(recurring) 1/27 = 37/999 = 0.037(recurring) 1/9 = 0.111(recurring) 1/3 = 0.333(recurring) The total = 0.508(recurring) Thus the next 2 digits after the 99th digit = 5 then 0. A.
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17 Apr 2013, 10:15
Q2.
# Empty Set = 1 # Sets with 1 element = 7C1 = 7 # Sets with 2 elements = 7C2 = 21 # Sets with 3 elements = 7C3 = 35
# Subsets containing at most 3 letters = 1 + 7 + 21 + 35 = 64 Ans E



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17 Apr 2013, 10:20
Q3:
With 6 different elements in a set, total number of subsets = 2^6 = 64 With 5 different elements in a set, total number of subsets = 2^5 = 32 Hence from set of 6, if we do not include 0 in any set, that would be equivalent to considering just 5 elements out of 6 sets and how many subsets can be obtained from 5 elements. that should be 2^5 = 32 sub sets. And D




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