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Re: Fresh Meat!!!
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18 Apr 2013, 21:54
Q5) Which of the following is a factor of 18!+1?Answer is C.18! + 1 is only 1 more than the 18!, so none of the factors  2, 3, 4, 5, ...,15, 16, 17, 18 of 18! would be factors for 18! + 1
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Updated on: 19 Apr 2013, 13:32
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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Originally posted by nt2010 on 18 Apr 2013, 22:02.
Last edited by nt2010 on 19 Apr 2013, 13:32, edited 1 time in total.



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18 Apr 2013, 22:04
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: C
P.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, 22:36
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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19 Apr 2013, 01:02
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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Updated on: 19 Apr 2013, 01:51
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
Originally posted by rbansal6 on 19 Apr 2013, 01:14.
Last edited by rbansal6 on 19 Apr 2013, 01:51, edited 1 time in total.



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19 Apr 2013, 01:48
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, 01:53
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, 02:01
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, 04:48
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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19 Apr 2013, 05:29
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, 05:33
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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19 Apr 2013, 05:38
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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19 Apr 2013, 05:43
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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Updated on: 19 Apr 2013, 06:05
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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Originally posted by smyarga on 19 Apr 2013, 05:51.
Last edited by smyarga on 19 Apr 2013, 06:05, edited 1 time in total.



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19 Apr 2013, 05:56
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, 06:02
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, 09:18
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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19 Apr 2013, 09:18
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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19 Apr 2013, 09:18
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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