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# Fresh Meat!!!

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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. 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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22 Jul 2013, 09:17
tricialin wrote:
hi Bunuel, thanks so much for your help - you're explanations to all kinds of problems have been invaluable to me in my studies thus far...

RE: the following question -

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:

Subset means selecting 0 or more from given set. since 0 should be excluded we have 5 numbers in set.

selecting 0 to n from given set is 2^n => 2^5= 32.

Ans: 32

Where do you get the 2 rom in 2*5?

OE's are in the original post here: fresh-meat-151046-100.html#p1213230. OE for this question is here: fresh-meat-151046-100.html#p1215329

Hope it helps.
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27 Aug 2013, 07:16
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.

Hi Bunuel,

I solved it as below and got the answer wrong. Can you let me know what i did wrong and please explain your approach in more detail.

Elements are {1, 2, 3, 4, 5}

Subset of 1: 5C1 = 5
Subset of 2: 5C2 = 10
Subset of 3: 5C3 = 10
Subset of 4: 5C4 = 5
Subset of 5: 5C5 = 1

Total of 31.
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27 Aug 2013, 07:22
summer101 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.

Hi Bunuel,

I solved it as below and got the answer wrong. Can you let me know what i did wrong and please explain your approach in more detail.

Elements are {1, 2, 3, 4, 5}

Subset of 1: 5C1 = 5
Subset of 2: 5C2 = 10
Subset of 3: 5C3 = 10
Subset of 4: 5C4 = 5
Subset of 5: 5C5 = 1

Total of 31.

All is fine except that you are forgetting an empty set which is also a subset and do not contain 0. As for my solution check this post it might help: how-many-subordinates-does-marcia-have-57169.html#p692676
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28 Aug 2013, 01:05
Bunuel wrote:
SOLUTIONS:

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 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.

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?
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28 Aug 2013, 10:42
Thanx bunnel for this set of question. I learned many good stuffs. Cheers
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28 Aug 2013, 11:55
mau5 wrote:
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.

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.

Thanks a bunch, Mau5, I did not read the given condition properly. Cheers.
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31 Aug 2013, 10:50
Bunuel wrote:
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.

Hi Bunuel,
x can take factor of 2 with power from 2 to 6 or no factor of 2. So the answer can be 6 too.
thanks
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31 Aug 2013, 22:30
$$\frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \frac{1}{37}=\frac{333}{999} + \frac{111}{999} + \frac{37}{999} + \frac{27}{999}=\frac{508}{999}=0.508508...$$.

How do we get these fractions with a common denominator?
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01 Sep 2013, 12:08
ygdrasil24 wrote:
$$\frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \frac{1}{37}=\frac{333}{999} + \frac{111}{999} + \frac{37}{999} + \frac{27}{999}=\frac{508}{999}=0.508508...$$.

How do we get these fractions with a common denominator?

$$\frac{1}{3} =\frac{1*333}{3*333}=\frac{27}{999}$$.

$$\frac{1}{9} =\frac{1*111}{9*111}=\frac{27}{999}$$.

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

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

Following link might help for this problem: math-number-theory-88376.html (check Converting Fractions chapter).
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02 Sep 2013, 19:52
manishuol wrote:
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.

Hi,

I have a small doubt

Is 38 is factor of 18! or 18!+1. Please explain.

Regards,
Rrsnathan.
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02 Sep 2013, 20:28
Bunuel wrote:
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

Notice that if x=-1 and y is any even number, then $$(-1)^{even}=1$$, thus none of the options must be true.

Hi Bunuel,

As per the question which of the following must be true.

So as per the given choice B) II only is true right
where 1^0 = 1 as X =1 and Y=0 given.

As ur explanation gives another chance as
X coud be = -1 , and Y = any even.

Please clarify where i am wrong.

Rrsnathan
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03 Sep 2013, 00:53

If you want to convert any fraction with denominator 9, 99, 999, so on,to decimal form, then see what is the value of the fraction with 10, 100, 1000,so on, as denominator.

For eg, 457/999 = ?

See 457/1000 = 0.457

Then, 457/999 = 0.457457457457...

This knowledge comes very handy at times with complex fractions.
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03 Sep 2013, 02:26
rrsnathan wrote:
Bunuel wrote:
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

Notice that if x=-1 and y is any even number, then $$(-1)^{even}=1$$, thus none of the options must be true.

Hi Bunuel,

As per the question which of the following must be true.

So as per the given choice B) II only is true right
where 1^0 = 1 as X =1 and Y=0 given.

As ur explanation gives another chance as
X coud be = -1 , and Y = any even.

Please clarify where i am wrong.

Rrsnathan

x=1 and y=0 indeed satisfies x^y=1, but the question asks "which of the following must be true". So, this option is NOT necessarily true, because x can be -1 and y any even number.
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07 Sep 2013, 09:32
How many different subsets of the set {0, 1, 2, 3, 4, 5} do not contain 0?

Will using combination concept be correct?Using this method i get 31 as the ans and not 32. Please guide!!

5c1+5c2+5c3+5c4+5c5=31
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07 Sep 2013, 09:35
Phoenix72 wrote:
How many different subsets of the set {0, 1, 2, 3, 4, 5} do not contain 0?

Will using combination concept be correct?Using this method i get 31 as the ans and not 32. Please guide!!

5c1+5c2+5c3+5c4+5c5=31

Check here: fresh-meat-151046-120.html#p1243696
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17 Sep 2013, 12:54
Bunuel wrote:
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.

Bunuel, question here. Could you help me out?

I am guessing x can assume 8 values and not 7. The values are $$3^6$$, $$2^0$$,$$2^1$$,$$2^2$$,$$2^3$$,$$2^4$$,$$2^5$$ and $$2^6$$. Anything wrong in my approach here?
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17 Sep 2013, 12:57
emailmkarthik wrote:
Bunuel wrote:
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.

Bunuel, question here. Could you help me out?

I am guessing x can assume 8 values and not 7. The values are $$3^6$$, $$2^0$$,$$2^1$$,$$2^2$$,$$2^3$$,$$2^4$$,$$2^5$$ and $$2^6$$. Anything wrong in my approach here?

Check here: fresh-meat-151046-140.html#p1262474
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18 Sep 2013, 02:57
Bunuel wrote:
25*1=25 and 25*13=325;
25*3=75 and 25*11=275;
25*5=125 and 25*9=225.

Hello, what about 175, 175 ? The question doesn't state they are distinct positive numbers.
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30 Sep 2013, 04:14
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.

I would like to request some some help with this question..

Could you please elaborate on the theory behind 2^n?
Isnt the null set considered to be a subset of this set?
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30 Sep 2013, 05:41
Transcendentalist 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.

I would like to request some some help with this question..

Could you please elaborate on the theory behind 2^n?
Isnt the null set considered to be a subset of this set?

Consider simpler set: {a, b, c}. How many subsets does it have? Each of a, b, and c has two choices either to be included in subsets or not. Thus total of 2^38 subsets (including an empty set). The subsets are:

{a, b, c} --> each is included;
{a, b} --> a and b are included;
{a, c};
{b, c};
{a};
{b};
{c};
{} empty set, none is included.

2^3=8 subsets.

Harder question to practice about the same concept: how-many-subordinates-does-marcia-have-57169.html#p692676

Hope it helps.
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