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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

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

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.

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. Please explain ! thanks
_________________

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.

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.

Answer: E.

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.

Thanks in Advance, 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.
_________________

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.

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?
_________________

You've been walking the ocean's edge, holding up your robes to keep them dry. You must dive naked under, and deeper under, a thousand times deeper! - Rumi

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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.

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.

Hi bunuel , Is following approach right ? As hcf is actually difference or multiple of difference of two numbers.. So a- b = 25k, a+b=350

Also k can be only an even number less than 12, that is a- b can at most be only 300, for an odd k anyway integer condition won't satisfy when we solve two equations, So a-b can be only 50, 100, 150, 200,250,300 , This give us three pairs of a and b

quote="Bunuel"]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

We are told that the greatest common factor of two integers is 25. So, these integers are \(25x\) and \(25y\), for some positive integers \(x\) and \(y\). Notice that \(x\) and \(y\) must not share any common factor but 1, because if they do, then GCF of \(25x\) and \(25y\) will be more that 25.

Next, we know that \(25x+25y=350\) --> \(x+y=14\) --> since \(x\) and \(y\) don't share any common factor but 1 then (x, y) can be only (1, 13), (3, 11) or (5, 9) (all other pairs (2, 12), (4, 10), (6, 8) and (7, 7) do share common factor greater than 1).

So, there are only three pairs of such numbers possible: 25*1=25 and 25*13=325; 25*3=75 and 25*11=275; 25*5=125 and 25*9=225.

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.

Answer: E.

Bunuel.. (II) x=1 andy=0 will give ans 1.. cant we say that this is the solution dat must give Ans "1"? 1^0=1?

The question asks which of the following MUST be true, not COULD be true. So, if if x=-1 and y is any even number, then \((-1)^{even}=1\), thus none of the options MUST be true.
_________________

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

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?

You are right. The question should read: f(n) is the number of positive perfect squares less than n. Edited.
_________________

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.

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.

Dear Bunnel

I didnt understand this.

y didnt we take the set {1,2,3,4,5} into consideration and solve like the above qs? where did we get 2^5 from?
_________________

Hope to clear it this time!! GMAT 1: 540 Preparing again

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.

Dear Bunnel

I didnt understand this.

y didnt we take the set {1,2,3,4,5} into consideration and solve like the above qs? where did we get 2^5 from?

What do you mean by the red part?

As for 2^5: the number of subsets of n-element set is 2^n, thus the number of subsets of 5-element set {1, 2, 3, 4, 5} is 2^5 (note that this includes an empty set as well as the original set {1, 2, 3, 4, 5}). Now, all subsets of {1, 2, 3, 4, 5} are the subsets of {0, 1, 2, 3, 4, 5} and does not include 0.

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.

Dear Bunnel

I didnt understand this.

y didnt we take the set {1,2,3,4,5} into consideration and solve like the above qs? where did we get 2^5 from?

What do you mean by the red part?

As for 2^5: the number of subsets of n-element set is 2^n, thus the number of subsets of 5-element set {1, 2, 3, 4, 5} is 2^5 (note that this includes an empty set as well as the original set {1, 2, 3, 4, 5}). Now, all subsets of {1, 2, 3, 4, 5} are the subsets of {0, 1, 2, 3, 4, 5} and does not include 0.

Does this make sense?

By red part i meant that y we havent solved it like we did the below qs:

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

y did 2^n come into qs 3 and not qs 2?
_________________

Hope to clear it this time!! GMAT 1: 540 Preparing again

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