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# Fresh Meat!!!

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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

Kudos points for each correct solution!!!
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17 Apr 2013, 05:47
1
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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

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, 05:49
1
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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:

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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17 Apr 2013, 05:58
1
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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?

$$2^6, 2^5*3^5,2^6*3^6$$ 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, 06:01
1
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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, 06:04
1
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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, 06:11
1
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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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VP
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17 Apr 2013, 06:23
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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, 08:28
1
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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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Kudos [?]: 18 [1] , given: 17

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17 Apr 2013, 09:15
1
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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, 09:20
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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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17 Apr 2013, 09:36
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Q6.

LCM of x, 2^6 and (2*3)^5 = (2*3)^6
therefore x must at least have a 3^6 in it. Plus it can have other powers of 2 ranging from 2^0 to 2^6
that makes 7 possible values for X
And C
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17 Apr 2013, 09:43
1
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Q7,
gcf = 25 = 5^2
Sum of 2 integers = 350

25 , 325 -> GCF = 25, SUM = 350 (OK)
50 , 300 -> GCF = 50, SUM = 350 (NOT OK)
75 , 275 -> GCF = 25, SUM = 350 (OK)
100 , 250 -> GCF = 50, SUM = 350 (NOT OK)
125 , 225 -> GCF = 25, SUM = 350 (OK)
150 , 200 -> GCF = 50, SUM = 350 (NOT OK)
175 , 175 -> GCF = 175, SUM = 350 (NOT OK)

Hence there will be 3 such pairs of integers.
Ans C
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17 Apr 2013, 09:48
1
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Q8

x * 2 * (3^2) * 5 * 13 * 17 * 19 / (2^2 * 3 * 5^2 * 11)
after simplification we are left with:

x * 3 * 13 * 17 * 19 / 2 * 5 * 11

therefore x must at least be divisible by 110 (2*5*11)

Ans D
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17 Apr 2013, 10:12
1
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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.

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
Kudos points for each correct solution!!!

----

2. Set S contains 7 different letters. How many subsets of set S, including an empty set, contain at most 3 letters?

Now in this Q, the Q for at most 3 letters in a subset that means it can have 0 or 1 or 2 or 3.
Therefore, when the set contains 0, it can be represented as 7c0 which yields us 1.
Second.... when the set contains 1, it can be represented as 7c1 which yields us 7.
Third, .........when the set contains 2, it can be represented as 7c2 which yields us 21.
Forth, .........when the set contains 3, it can be represented as 7c3 which yields us 35.
Now as we know that these four conditions are joined together with the help of "or"
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17 Apr 2013, 10:25
1
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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.

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

A. 15
B. 17
C. 19
D. 33
E. 39
Kudos points for each correct solution!!!

------------------------------------------------------------------------------------------------------------

First of all, before solving this problem, we must remember that two consecutive integers have a GCF of 1. That means 2 consecutive integers only share one factor that is 1. They never share any other factor. therefore, the factors of 18! can never be the factors of 18!+1. Okay!!!

Now, lets take a look at the answer choices.
1. 15, it is a factor of 18! as 18! contains both 5 & 3. Therefore, Eliminated.
2. 17, it is a factor of 18! as 18! contains 17. Therefore, Eliminated.
3. 19, it is not a factor of 18! as 18! can't contain 19 in it. Therefore, Chosen... Correct.
4. 33, it is a factor of 18! as 18! contains both 11 & 3. Therefore, Eliminated.
5. 39, it is a factor of 18! as 18! contains both 13 & 3. Therefore, Eliminated.
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17 Apr 2013, 10:37
1
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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.

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
Kudos points for each correct solution!!!

--------------------------------------------------------------------------------------------------------------

In this Q. we have to find the least value of X ????

Now for this first of all we have prime factorize 377,910 i.e.;

377,910 = 4199*3^3*2*5
3300 = 2*3*5*11*2*5.

Now as given in the Question that, The product of a positive integer x and 377,910 is divisible by 3,300.
Therefore, 4199*3^3*2*5*x / 2*3*5*11*2*5.

After cancelling the required. we get X= 2*5*11 =110.
Therefore 110........................ D.
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Kudos [?]: 83 [1] , given: 11

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17 Apr 2013, 10:48
1
KUDOS
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.

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
Kudos points for each correct solution!!!

-------------------------------------------------------------------------------------------------------------------

Here, it is given that the LCM of x, 4^3 & 6^5 is 6^6.
we can write is like this way. LCM (x,2^6, 2^5*3^5) is 2^6*3^6.
Tha means that the LCM has 2^6*3^6 & we have 2^6, 2^5*3^5 therefore x must contain 3^6. alongwith 0,1,2,3,4,5,6. Therefore, x can be 7 different values. Therefore, 7. ........................C.
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17 Apr 2013, 11:05
1
KUDOS
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.

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
Kudos points for each correct solution!!!

---------------------------------------------------------------------------------------------------------------
Yeah, Here if we add 1/3 + 1/9 + 1/27 + 1/37 we will get the value as ::::: 0.50850850850850850850850850850851. it will keep on repeating itself in groups of three & @ 101th place it has a value. 0. Therefore, ......A.
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17 Apr 2013, 11:21
1
KUDOS
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.

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
Kudos points for each correct solution!!!

-----------------------------------------------------------------------------------------------------------------
The Q asks, How many of the subsets contains the five elements leaving behind 0.
so we can calculate the # of subsets as 2^5 = 32. Therefore, 32 . .........D.
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17 Apr 2013, 11:30
1
KUDOS
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.

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
Kudos points for each correct solution!!!

-------------------------------------------------------------------------------------------------------------
Here it is given that the GCF of two integers is 25 or 5^2. That means that they both share 5^2 only.
Now we can depict the intergers as 25a & 25b sharing only 25 & a & b don't share anything b/w them. okay !!
Now, as given in the question that sum of the integers is 350. therefore, 25a+25b=350 or 25(a+b)=350
or a+b =14. Now we look for those values of a & b those don't share anything b/w them. So,
a+b= 3+11 or 1+13 or 5+9.........so Three values. Therefore, 3. C..........................
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Re: Fresh Meat!!!   [#permalink] 17 Apr 2013, 11:30

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