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# Fresh Meat!!!

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Math Expert
Joined: 02 Sep 2009
Posts: 47977

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21 Apr 2013, 21:53
7
10
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.

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Posts: 47977

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21 Apr 2013, 21:58
4
5
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

1 empty set;
$$C^1_7=7$$ sets with one element;
$$C^2_7=21$$ sets with two elements;
$$C^3_7=35$$ sets with three element.

Total 1+7+21+35=64 sets.

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21 Apr 2013, 22:07
5
12
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.

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21 Apr 2013, 22:29
5
12
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.

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Posts: 47977

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21 Apr 2013, 22:38
8
16
5. Which of the following is a factor of 18!+1?

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

18! and 18!+1 are consecutive integers. Two consecutive integers are co-prime, which means that they don't share ANY common factor but 1. For example 20 and 21 are consecutive integers, thus only common factor they share is 1.

Now, since we can factor out each 15, 17, 33=3*11, and 39=3*13 out of 18!, then 15, 17, 33 and 39 ARE factors of 18! and are NOT factors of 18!+1. Therefore only 19 could be a factor of 18!+1.

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21 Apr 2013, 22:56
5
16
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.

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Joined: 02 Sep 2009
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21 Apr 2013, 23:27
3
6
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

Given: $$\frac{377,910 *x}{3,300}=integer$$.

Factorize the divisor: $$3,300=2^2*3*5^2*11$$.

Check 377,910 for divisibility by 2^2: 377,910 IS divisible by 2 and NOT divisible by 2^2=4 (since its last two digits, 10, is not divisible by 4). Thus x must have 2 as its factor (377,910 is divisible only by 2 so in order 377,910*x to be divisible by 2^2, x must have 2 as its factor);

Check 377,910 for divisibility by 3: 3+7+7+9+1+0=27, thus 377,910 IS divisible by 3.

Check 377,910 for divisibility by 5^2: 377,910 IS divisible by 5 and NOT divisible by 25 (in order a number to be divisible by 25 its last two digits must be 00, 25, 50, or 75, so 377,910 is NOT divisible by 25). Thus x must have 5 as its factor.

Check 377,910 for divisibility by 11: (7+9+0)-(3+7+1)=5, so 377,910 is NOT divisible by 11, thus x must have 11 as its factor.

Therefore the least value of x is $$2*5*11=110$$.

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21 Apr 2013, 23:45
1
23
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

$$\frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \frac{1}{37}=\frac{333}{999} + \frac{111}{999} + \frac{37}{999} + \frac{27}{999}=\frac{508}{999}=0.508508...$$.

102nd digit will be 8, thus 101st digit will be 0.

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21 Apr 2013, 23:48
8
10
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.

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Joined: 07 Sep 2010
Posts: 288

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25 Apr 2013, 19:42
Hi Bunuel,
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.

Thanks
H
Math Expert
Joined: 02 Sep 2009
Posts: 47977

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26 Apr 2013, 00:43
imhimanshu wrote:
Hi Bunuel,
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.

Thanks
H

Yes, we could get the correct answer with Wilson's theorem, but you don't need it for the GMAT.
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Joined: 23 Apr 2013
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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?
Math Expert
Joined: 02 Sep 2009
Posts: 47977

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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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Joined: 29 Aug 2012
Posts: 29
WE: General Management (Consulting)

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07 Jul 2013, 07:56
1
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¿??

Math Expert
Joined: 02 Sep 2009
Posts: 47977

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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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Joined: 06 Jun 2012
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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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Joined: 28 Jun 2013
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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
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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Re: Fresh Meat!!! &nbs [#permalink] 31 Aug 2013, 10:50

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