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Fresh Meat!!!

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New post 18 Apr 2013, 22:54
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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New post Updated on: 19 Apr 2013, 14:32
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 D

1^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, 23:02.
Last edited by nt2010 on 19 Apr 2013, 14:32, edited 1 time in total.
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New post 18 Apr 2013, 23:04
1
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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New post 18 Apr 2013, 23:36
1
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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New post 19 Apr 2013, 02:02
1
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^2-99x^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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New post Updated on: 19 Apr 2013, 02:51
1
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 factor--1 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, 02:14.
Last edited by rbansal6 on 19 Apr 2013, 02:51, edited 1 time in total.
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New post 19 Apr 2013, 02:48
1
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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New post 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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New post 19 Apr 2013, 03:01
1
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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New post 19 Apr 2013, 05:48
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New post 19 Apr 2013, 06:29
1
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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New post 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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New post 19 Apr 2013, 06:38
1
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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New post 19 Apr 2013, 06:43
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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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New post Updated on: 19 Apr 2013, 07:05
1
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, 06:51.
Last edited by smyarga on 19 Apr 2013, 07:05, edited 1 time in total.
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New post 19 Apr 2013, 06:56
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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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New post 19 Apr 2013, 07:02
1
10. The answer is none. x could be -1. For example (-1)^2=1.
The correct answer is E.
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New post 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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New post 19 Apr 2013, 10:18
1
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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New post 19 Apr 2013, 10:18
1
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.
Re: Fresh Meat!!! &nbs [#permalink] 19 Apr 2013, 10:18

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