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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

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 My answer C Squares less than 36 : 1,4,9,16,25 = 5 primes less than 36 : 2,3,5,7,11,13,17,19,23,29,31. =11 and sum =16. Hence the max value N has to be less than 37 as N =38 will increase the number of primes by 1 and squares by 1. And the sum will be 18 Minimum values of N has to be 32, as any value less than 32 will decrease the number of primes by 1.and the sum will be 15

A. 15 B. 17 C. 19 D. 33 E. 39 My answer is C. 15, 17,33 and 39 , perfectly divide into 18! Hence leave a reminder of 1, when they divide 18!+1. Hence can be eliminated , leaving 19 the answer.

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 My Answer C 66 = 26 *36 and 65 = 25 *35 . 4^3 = 26 Hence X should have 36 and It can have 20 till 26 that is 7 values from 20*36 till 26*36

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.

Hi Bunuel, Had the question been 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.

Please share your reasoning.

Thanks H

Hi himanshu. According to Wilson's Theorem, if P is a prime no. then the remainder when (p-1)! is divided by p is (p-1) Therefore, 18! on division by 19 will give 18 as a remainder. Now 18+1 is divisible by 19 therefore answer to your query is 19. add kudos if this helped you

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,

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¿??

[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.

Answer: D.

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.

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?
_________________

Any and all kudos are greatly appreciated. Thank you.

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.

Thanks..

Yes, an empty set is a subset of all sets.
_________________

If x = 31, then f(31) = 5 and g(31) = 10: f(x) + g(x) = 5 + 10 = 15.

Why g(31) = 10 is not a prime number.

Bunuel wrote:

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

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 used a choose method and got 31. With the binary method you used, there is a set which contains no numbers. This increases the number to 32. But the question doesn't say there can be a subset with no numbers because 0 is also not included...

Empty set is a subset of all non-empty sets.
_________________

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

From the given inequality, for any y and x=1, we would have x^y = 1. Also, for any x(and not equal to 0) and y = 0, we would again have the same inequality.
_________________

10. If x is not equal to 0 and x^y=1, then which of the following must be true?

I. x=1 False \(100^0=1\) II. x=1 and y=0 False \(2^0=1\) III. x=1 or y=0 True Infact there are two cases: every number with a 0 exponent equals 1, and 1 raised to any exp equals 1.

IMO C. III only _________________

It is beyond a doubt that all our knowledge that begins with experience.

lets try with the options. A. x between 30 and 36 f(n) = 5. No of squares = 5 (1,4,9,16 and 25) g(n) = 11 (Primes are 2,3,5,7,11,13,17,19,23,29,31) f(n) + g(n) = 16

Q9. reducing the equation makes it 508/999 when any number is divided by 999, the remainder after the decimal keeps repeating itself infinetely.

there the above fraction will be 0.508508508 ... and so on every 2, 5, 8th ... didgit will be 0 and every 3, 6, 9, 12th ... didgit will be 8 since 101 = 3*x + 2 101st digit in decimal will be 0 ans A

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.

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

Kudos points for each correct solution!!!

-----------

If x is not equal to 0 and x^y=1, then which of the following must be true?

The Q itself says that x is not equal to zero. Therefore, x can be +ve or -ve both okay. now, option 1 says...... x=1 it is false becoz it can't be must be true as x can be one but it can be some other number as well. for ex. this also satisfies the condition given. 2^0=1. therefore it must not be true.

Option second says ...... II. x=1 and y=0 , similar to the above one the given condition can be true but it is not must to be true as 2^0=1 again satisfies the condition given. Therefore False for me again. Option -III. it says either x=1 or y=0 now this option fulfills the criteria as if x =1 then no matter the power raise to it it will alwz remains 1 or if y will be equal to zero then no matter the base but the exp. remains in every case. Hence, III must be true ...... Therefore, C.
_________________

If you don’t make mistakes, you’re not working hard. And Now that’s a Huge mistake.

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.

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 Kudos points for each correct solution!!!

----------------------------------------------------------------------------------------------------------------- It is given that F(n) = # of perfect Squares < n & G(n) = # of primes less than n. & it is given that F(n)+G(n) = 16 so, the values in the two functions can be anything, so to keep the options limited, lets look at the answer choices. A. 30 < x < 36 ......... Perfect Squares < 30 = 1 4 9 16 25 i.e. 5. Prime Numbers < 36 = 2 3 5 7 11 13 17 19 23 19 31 i.e. 11. Therefore, this satisfies the condition given. Therefore,............. A.
_________________

If you don’t make mistakes, you’re not working hard. And Now that’s a Huge mistake.

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