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Fresh Meat!!! [#permalink]
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17 Apr 2013, 06:11
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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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18 Apr 2013, 23:02
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 Answer is D1^y is always 1 and x is not equal to 0 x^0 is not always 1 especially when x is 0  undefined
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18 Apr 2013, 23:04
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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 Since X lies at most 38 from the options, Lets find the number of Perfect Squared below 40= 1,4,9,16,25,36 Lets find the number of Primes below 40= 2,3,5,7,11,13,17,19,23,29,31,37 Now, f(x) + g(x) = 16 Start with Option C: no of primes less than 31 = \(11\) no of p.s less than 31: \(5\) Total = 16 Answer: CP.S: The reason I started with option C is because when I analyzed choice A, the value of x is not fixed, and moving the value of A can shift the value of f(x)+g(x). Hence, I started looking for option where I can fix the value of "X"i.e. no prime number exists in that option.
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18 Apr 2013, 23:36
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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 A. I only B. II only C. III only D. I and III only E. I, II and III Given: Lengths of the diagonls are integers.Hence, multiplier must be integer. Lets have Ratio Mulitplier to be 1; Let Diagonal of Square: \(\sqrt{2a} = 15\) \(So, Area = 15*15/\sqrt{2}*\sqrt{2}\) Area of Rhombus is given by: \(1/2 d1*d2 = 1/2 *11*9\) Taking the difference of both values; we get :\(126/2 = 63\) Hence, 1st is True. Taking the Multiplier to be 2 and performing the same steps, we get difference: 252. Since, further taking the multiplier will only increase the difference, we shall stop here. Hence only 1 and 3 is correct. Answer: D
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1. The length of the diagonal of square S is 15x, the lengths of the diagonals of rhombus R are 11x and 9x, where x is a positive integer. The area of square S is d*d/2=(15x)^2/=225x^2/2 and the area of rhombus R is diagonal1*diagonal 2/2=11x*9x/2=99x^2/2. So, the difference between them is (225x^299x^2)/2=63x^2. Since x is an integer, the difference must be divisible by 63. Therefore, I is possible for x=1. II is not possible, since if 63x^2=126>x^2=2, which is not possible for an integer x. III is possible for x=2. The answer is D.
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19 Apr 2013, 02:14
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1. D 2. D 3. C 4. C 5. C 6. C 7. C 8. D 9. A 10.D
1. 15^2X^2  99X^2= 63X^2 2. 7C0 +7C1+7C2+7C3=63 3. {6C1+6C2++++6C6}{5C0+5C1+++5C5)=31..I have a question..Should we include 6C0?? 4. G(n)= 2, 3,5, 7, 11, 13, 17, 19, 23..parelly increasing F(n)= 1, 4, 9, 16, 25...G(n)+29, 31 total is 16 till x=36 5. only 19 is possible as all others are factors of 18! 6. 4^3=2^6 6^5=2^5.3^5 6^6=2^6.3^6 so X has to have 3^6 and can have any value from 2^0 to 2^6..so total=7
7. 25(a+b)=350...a+b=14..such that a and b has no common factor1 13, 3 11, 5 9...so 3
8. 377,910 is divisible by 3 & 10 but not 11..so x has to have 11 and another 10=110
9. .333333 + .1111111111 + .027027.. + .037037037  = .33333333333 + .11111111111 + .064064064  = .508508508 99th=8 100th =5 101th=0
10. D
Last edited by rbansal6 on 19 Apr 2013, 02:51, edited 1 time in total.



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19 Apr 2013, 02:48
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2. We have such cases for the sets: (1) Empty set. 1 (2) Contain only 1 element. There are 7 such subsets (3) Contain 2 elements. There are C_7^2=7!/(2!5!)=21 such subsets (4) Contain 3 elements. There are C_7^3=7!/(3!4!)=35 such subsets Therefore, total number of such subsets is 1+7+21+35=64 The answer is E.
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19 Apr 2013, 02:53
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
Here, should we include empty set?



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19 Apr 2013, 03:01
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3. To find different subsets that do not contain 0 is the same that find the number of different subsets of set {1,2,3,4,5}. The total number of such subsets is 2^5=32. The answer is D.
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19 Apr 2013, 05:48
rbansal6 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
Here, should we include empty set? Yes, empty set should be included.
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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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19 Apr 2013, 06:29
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4. Pick the numbers. Let's check the smallest possible: x=31. f(31)=5 (1,4,9,16,25) and g(31)=10 (2,3,5,7,11,13,17,19, 23,29). Therefore f(31)+g(31)=15, that is not equal to 16. But if we take x=32: f(32)=5 and g(32)=11 (10 previous and 31). So, x=32 is a possible value. Next perfect square of an integer is 36 and the next prime number is 37. So, for not changing the number 16, we can take x=33, 34, 35, 36. The correct answer is C.
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19 Apr 2013, 06:33
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5. The number 18!+1 is not divisible by any number from 2 to 18, because 18! is divisible by them. So, the answers A and B are not correct. Since 18!+1 is not divisible by 3, 18!+1 is not divisible by 33 and 39. So, answers В and E are wrong The correct answer is C.
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6. The problem states that lcm(x, 2^6, 2^5*3^5)=2^6*3^6. When we calculate lcm we have to take the highest powers in prime factorizations of numbers. If lcm contains 3^6 it must be in some number. So, x=3^6*y. y could be any factor of 2^6. The possible values of x: 3^6, 3^6*2, 3^6*2^2, 3^6*2^3, 3^6*2^4, 3^6*2^5, 3^6*2^6. The correct answer is C.
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7. Let two numbers be 25n and 25m, where gcd(n,m)=1. Then 25n+25m=350 or n+m=14. So, our goal is to find pairs of numbers (n,m) such that gcd(n,m)=1 and n+m=14: (1,13), (3,11), (5, 9). The correct answer is C.
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8. We know that x*377,910/3,300=some positive integer. Let's simplify this fraction: Divide numerator and denominator by 30: x*12,597/110. 12,597 is not divisible neither by 11 nor 2 nor 5. So, x must be divisble by 110. Then the least value of x is 110. The answer is D.
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Last edited by smyarga on 19 Apr 2013, 07:05, edited 1 time in total.



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9. Just add: 1/3+1/9+1/27+1/37=(333+111+37+27)/999=508/999=0.(508). 101st digit after decimal point is 0, because we have period with 3 digits and 101=3*33+2. The correct answer is A.
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19 Apr 2013, 07:02
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10. The answer is none. x could be 1. For example (1)^2=1. The correct answer is E.
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19 Apr 2013, 10:18
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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
A. I only B. II only C. III only D. I and III only E. I, II and III
Let’s say diagonal of square is 15x and diagonals of rhombus are 11x and 9x where x is an integer. Area of square = (1/2) * 15^2 * x^2 = (1/2) * 225 * x^2 Area of rhombus = (1/2) * 11*9 * x^2 = (1/2) * 99 * x^2 Difference of areas = (1/2) * 126 * x^2 = 63 * x^2 So, the different must be a multiple of 63.
I. Here x^2 = 1. Possible option. II. 126 = 63*2. Here x^2 = 2. So, x is not integer. This option is not possible. III. 252 = 63*4. Here x^2 = 4 > x = 2. Possible option
Correct answer is D.



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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
Number of subsets with 0 letter = 1 Number of subsets with 1 letter = 7C1 = 7 Number of subsets with 2 letters = 7C2 = 21 Number of subsets with 3 letters = 7C3 = 35 Total number of subsets with at most 3 letters = 1 + 7 + 21 + 35 = 64
Correct answer is E.



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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
Here we need to find out the possible subsets with the numbers {1,2,3,4,5}.
In the short way: Total number of subsets = 2^5 = 32
In the long way: Number of subsets with 0 element (null set) = 1 Number of subsets with 1 element = 5C1 = 5 Number of subsets with 2 elements = 5C2 = 10 Number of subsets with 3 elements = 5C3 = 10 Number of subsets with 4 elements = 5C4 = 5 Number of subsets with 5 elements = 5C5 = 1 Total number of subsets = 1+ 5 + 10 + 10 + 5 + 1 = 32
Correct answer is D.



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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
For x = 37, f(x) + g(x) = 5 + 12 = 17 For x = 36, f(x) + g(x) = 5 + 11 = 16 For x = 32, f(x) + g(x) = 5 + 11 = 16 For x = 31, f(x) + g(x) = 5 + 10 = 15 So, 31 < x < 37
Correct answer is C







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