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Fresh Meat!!! [#permalink] New post 17 Apr 2013, 05:11
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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. 252


A. I only
B. II only
C. III only
D. I and III only
E. I, II and III

Solution: fresh-meat-151046-80.html#p1215318


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

Solution: fresh-meat-151046-100.html#p1215323

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

Solution: fresh-meat-151046-100.html#p1215329


4. 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: fresh-meat-151046-100.html#p1215335


5. Which of the following is a factor of 18!+1?

A. 15
B. 17
C. 19
D. 33
E. 39

Solution: fresh-meat-151046-100.html#p1215338


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

Solution: fresh-meat-151046-100.html#p1215345


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

Solution: fresh-meat-151046-100.html#p1215349


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

Solution: fresh-meat-151046-100.html#p1215359


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

Solution: fresh-meat-151046-100.html#p1215367


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

Solution: fresh-meat-151046-100.html#p1215370


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Re: Fresh Meat!!! [#permalink] New post 21 Apr 2013, 21:29
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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 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

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.
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Re: Fresh Meat!!! [#permalink] New post 21 Apr 2013, 21:56
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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.
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Re: Fresh Meat!!! [#permalink] New post 29 May 2013, 06:41
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prinkashar wrote:
1 3 +1 9 +1 27 +1 37 =333 999 +111 999 +37 999 +27 999 =508 999 =0.508508... .

would you please explain how 1/37 is 27/999


\frac{1}{37} =\frac{1*27}{37*27}=\frac{27}{999}.

Following link might help for this problem: math-number-theory-88376.html (check Converting Fractions chapter).
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Re: Fresh Meat!!! [#permalink] New post 30 May 2013, 08:13
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gultrage wrote:
imhimanshu wrote:
Hi Bunuel,
Had the question been
Which of the following is a factor of 18!+1?

A. 15
B. 17
C. 19
D. 33
E. None of These

Then, would it be possible to come at a conclusion that 19 will be the factor of 18!+1.
In the original question, we came to the answer by eliminating other choices.

Please share your reasoning.

Thanks
H

Hi himanshu.
According to Wilson's Theorem, if P is a prime no. then the remainder when (p-1)! is divided by p is (p-1)
Therefore, 18! on division by 19 will give 18 as a remainder. Now 18+1 is divisible by 19 therefore answer to your query is 19.
add kudos if this helped you


Check here: fresh-meat-151046-100.html#p1217213
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Re: Fresh Meat!!! [#permalink] New post 28 Aug 2013, 00:28
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Gagan1983 wrote:
Bunuel wrote:
SOLUTIONS:

[b]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?

Given that the ratio of the diagonal is d_s:d_1:d_2=15x:11x:9x, for some positive integer x (where d_s is the diagonal of square S and d_1 and d_2 are the diagonals of rhombus R).

area_{square}=\frac{d^2}{2} and area_{rhombus}=\frac{d_1*d_2}{2}.

The difference is area_{square}-area_{rhombus}=\frac{(15x)^2}{2}-\frac{11x*9x}{2}=63x^2.

If x=1, then the difference is 63;
If x=2, then the difference is 252;
In order the difference to be 126 x should be \sqrt{2}, which is not possible.

Answer: D.


Thank you. I have a question - Why cant x be [square_root]2. Why cant we have sides of lengths 5*[square_root]2, 11*[square_root]2 and 9*[square_root]2?


Firstly, these are not the sides of the given square and rhombus. They are diagonal values, where 15x corresponds to the square(where the diagonals are equal) and the 11x and 9x correspond to the rhombus(which has unequal diagonals). Also, it is mentioned that they are all integers, thus, if x = \sqrt{2}, then the value of the diagonal of the square/rhombus will no longer be an integer.

Hope this helps.
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Re: Fresh Meat!!! [#permalink] New post 31 Aug 2013, 21:23
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Cant we re write as 377910/3 X 11 X 10 X 10 ,

cancelling out 10, we get 37791 / 3 X 11 X 10

Eyeing 110 as one of the options, we check for divisibilty of 37791 for 3 and it is divisible.

which gives 12577 / 11 X 10, checked for 11, not divisible hence

minimum value of x is 11 X 10.

Please suggest Bunuel, if its wrong.
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Re: Fresh Meat!!! [#permalink] New post 18 Sep 2013, 02:34
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muzammil wrote:
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.


The point is that if the two integers are 175 and 175, then their greatest common divisor going to be 175, not 25 as given in the stem.

Does this make sense?
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Re: Fresh Meat!!! [#permalink] New post 03 Apr 2014, 07:25
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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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Re: Fresh Meat!!! [#permalink] New post 09 Apr 2014, 02:05
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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] New post 17 Apr 2013, 06:15
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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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Re: Fresh Meat!!! [#permalink] New post 17 Apr 2013, 06:20
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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Re: Fresh Meat!!! [#permalink] New post 17 Apr 2013, 09:24
Q4

f(n) = # perfect squares < n
g(n)= # primes < n

we need f(n) + g(n) = 16

lets try with the options.
A. x between 30 and 36
f(n) = 5. No of squares = 5 (1,4,9,16 and 25)
g(n) = 11 (Primes are 2,3,5,7,11,13,17,19,23,29,31)
f(n) + g(n) = 16

hence A
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Re: Fresh Meat!!! [#permalink] New post 17 Apr 2013, 09:54
Q9.
reducing the equation makes it 508/999
when any number is divided by 999, the remainder after the decimal keeps repeating itself infinetely.

there the above fraction will be 0.508508508 ... and so on
every 2, 5, 8th ... didgit will be 0 and
every 3, 6, 9, 12th ... didgit will be 8
since 101 = 3*x + 2
101st digit in decimal will be 0
ans A
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Re: Fresh Meat!!! [#permalink] New post 17 Apr 2013, 09:55
Bunuel wrote:
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.

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

Kudos points for each correct solution!!!

-----------

If x is not equal to 0 and x^y=1, then which of the following must be true?

The Q itself says that x is not equal to zero. Therefore, x can be +ve or -ve both okay.
now, option 1 says...... x=1 it is false becoz it can't be must be true as x can be one but it can be some other number as well. for ex. this also satisfies the condition given. 2^0=1. therefore it must not be true.

Option second says ...... II. x=1 and y=0 , similar to the above one the given condition can be true but it is not
must to be true as 2^0=1 again satisfies the condition given. Therefore False for me again.
Option -III. it says either x=1 or y=0 now this option fulfills the criteria as if x =1 then no matter the power raise to it it will alwz remains 1 or if y will be equal to zero then no matter the base but the exp. remains in every case. Hence,
III must be true ...... Therefore, C.
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Re: Fresh Meat!!! [#permalink] New post 17 Apr 2013, 11:17
Bunuel wrote:
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.

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
Kudos points for each correct solution!!!

-----------------------------------------------------------------------------------------------------------------
It is given that F(n) = # of perfect Squares < n &
G(n) = # of primes less than n. & it is given that F(n)+G(n) = 16 so, the values in the two functions can be anything,
so to keep the options limited, lets look at the answer choices.
A. 30 < x < 36 .........
Perfect Squares < 30 = 1 4 9 16 25 i.e. 5.
Prime Numbers < 36 = 2 3 5 7 11 13 17 19 23 19 31 i.e. 11.
Therefore, this satisfies the condition given. Therefore,............. A.
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Re: Fresh Meat!!! [#permalink] New post 17 Apr 2013, 15:00
Question 4:
Easier if you test numbers (considering that f(x) + g(x) is always crescent).

x = 30 -> f(30) = 5, g(30) = 10 -> f+g = 15
x = 31 -> f(31) = 5, g(31) = 10 -> f+g = 15
x = 32 -> f(32) = 5, g(32) = 11 -> f+g = 16
x = 33 -> f(33) = 5, g(33) = 11 -> f+g = 16
x = 34 -> f(34) = 5, g(34) = 11 -> f+g = 16
x = 35 -> f(35) = 5, g(35) = 11 -> f+g = 16
x = 36 -> f(36) = 5, g(36) = 11 -> f+g = 16
x = 37 -> f(37) = 6, g(37) = 11 -> f+g = 17

Therefore: 30 < x < 37

Answer: B
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Re: Fresh Meat!!! [#permalink] New post 17 Apr 2013, 15:01
Question 7:

x + y = 350
x = 25*p, y = 25*q where p and q may not have any other factors in common

Substituting into the equation we have: p+q = 14
The only solutions for pair (p,q) in which p and q do not have common factors are: (1,13), (3,11),(5,9)

There are exactly 3 pairs: (25,325), (75, 275), (125, 225)

Answer C

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Re: Fresh Meat!!! [#permalink] New post 17 Apr 2013, 15:02
Question 8
x.377910 = 3300k

=> x.(19.17.13.5.3^2.2) = 11.5^2.3.2^2.k
=> x. 19.17.13.3 = 11.5.2
=> x = 110.k/(19.17.13.3)

Therefore, the smallest number for x is 110 (when k = 19.17.13.3)

Answer: D

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Re: Fresh Meat!!! [#permalink] New post 17 Apr 2013, 15:03
Question 10

1. x = 1 ? Not necessary. Let x = 3, and y = 0, then x^y = 1
2. x = 1 and y = 0. Not necessary (see item 1)
3. x = 1 or y = 0. Not necessary. Let x = -1, y=2

Answer: E
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Last edited by caioguima on 18 Apr 2013, 03:48, edited 2 times in total.
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Re: Fresh Meat!!! [#permalink] New post 17 Apr 2013, 17:34
Question 5

I get that the answer is 19 that is C, but i am unable to find out how!
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Re: Fresh Meat!!!   [#permalink] 17 Apr 2013, 17:34
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