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# Fresh Meat!!!

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Math Expert
Joined: 02 Sep 2009
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17 Apr 2013, 06:11
13
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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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01 Jul 2013, 10:12
Hello.. Can someone please explain why "x" ( 15x:11x:9x) needs to be an integer? Why not 1.5?

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01 Jul 2013, 10:23
CIyer wrote:
Hello.. Can someone please explain why "x" ( 15x:11x:9x) needs to be an integer? Why not 1.5?

We are told that the length of the diagonals are integers and their ratio is 15:11:9. This means that the lengths are multiples of 15, 11 and 9. If x=1.5, then the lengths won't be integers.

Hope it's clear.
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07 Jul 2013, 07:56
1
KUDOS
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 did this exercise as follows:

I eliminate the 0, so i have the following set: (1,2,3,4,5). Now, i use combinatorics.

Set containing 5 elements: 5C5=1
Set containing 4 elements: 4C5=5
Set containing 3 elements: 3C5=10
Set containing 2 elements: 2C5=10
Set containing 1 elements: 1C5=5

So, the total of posibilites are 31. What am I missing here¿??

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Posts: 42529

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07 Jul 2013, 08:03
jacg20 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 did this exercise as follows:

I eliminate the 0, so i have the following set: (1,2,3,4,5). Now, i use combinatorics.

Set containing 5 elements: 5C5=1
Set containing 4 elements: 4C5=5
Set containing 3 elements: 3C5=10
Set containing 2 elements: 2C5=10
Set containing 1 elements: 1C5=5

So, the total of posibilites are 31. What am I missing here¿??

You are missing 1 empty set, which is a subset of the original set and also does not contain 0.

Hope it's clear.
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08 Jul 2013, 09:03
1
KUDOS
1 useful formula: 2^n=nc0 + nc1+ ..... + ncn; here n=5, ans= 2^5= 32

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08 Jul 2013, 10:46
CIyer wrote:
1 useful formula: 2^n=nc0 + nc1+ ..... + ncn; here n=5, ans= 2^5= 32

That's correct. Another way of reaching the same formual is here: fresh-meat-151046-100.html#p1215329
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08 Jul 2013, 11:26
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.

Hi Bunuel,

This is probably a stupid question. But why can't x be \sqrt{2}?

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08 Jul 2013, 11:28
Aho92 wrote:
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.

Hi Bunuel,

This is probably a stupid question. But why can't x be \sqrt{2}?

Okay. I got it. Stupid me. They have to be integers

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09 Jul 2013, 00:40
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

*New project from GMAT Club!!! Check HERE
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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, 01:28
1
KUDOS
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.
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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, 10:55
2
KUDOS
Expert's post
2013gmat 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.

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

I don;t understand what you mean...

x can take the following 7 values:
$$3^6$$;
$$2*3^6$$;
$$2^2*3^6$$;
$$2^3*3^6$$;
$$2^4*3^6$$;
$$2^5*3^6$$;
$$2^6*3^6$$.
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31 Aug 2013, 22:23
1
KUDOS
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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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, 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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Re: Fresh Meat!!!   [#permalink] 03 Sep 2013, 00:53

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