Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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

LCM of x, 2^6 and (2*3)^5 = (2*3)^6 therefore x must at least have a 3^6 in it. Plus it can have other powers of 2 ranging from 2^0 to 2^6 that makes 7 possible values for X And C

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.

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 Kudos points for each correct solution!!!

----

2. Set S contains 7 different letters. How many subsets of set S, including an empty set, contain at most 3 letters?

Now in this Q, the Q for at most 3 letters in a subset that means it can have 0 or 1 or 2 or 3. Therefore, when the set contains 0, it can be represented as 7c0 which yields us 1. Second.... when the set contains 1, it can be represented as 7c1 which yields us 7. Third, .........when the set contains 2, it can be represented as 7c2 which yields us 21. Forth, .........when the set contains 3, it can be represented as 7c3 which yields us 35. Now as we know that these four conditions are joined together with the help of "or" so we must add them.... Therefore, 1+7+21+35=64. Hence, 64 answer....... E. _________________

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

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. _________________

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

Here, it is given that the LCM of x, 4^3 & 6^5 is 6^6. we can write is like this way. LCM (x,2^6, 2^5*3^5) is 2^6*3^6. Tha means that the LCM has 2^6*3^6 & we have 2^6, 2^5*3^5 therefore x must contain 3^6. alongwith 0,1,2,3,4,5,6. Therefore, x can be 7 different values. Therefore, 7. ........................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.

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 Kudos points for each correct solution!!!

--------------------------------------------------------------------------------------------------------------- Yeah, Here if we add 1/3 + 1/9 + 1/27 + 1/37 we will get the value as ::::: 0.50850850850850850850850850850851. it will keep on repeating itself in groups of three & @ 101th place it has a value. 0. Therefore, ......A. _________________

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.

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.

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 Kudos points for each correct solution!!!

----------------------------------------------------------------------------------------------------------------- The Q asks, How many of the subsets contains the five elements leaving behind 0. so we can calculate the # of subsets as 2^5 = 32. Therefore, 32 . .........D. _________________

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.

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 Kudos points for each correct solution!!!

------------------------------------------------------------------------------------------------------------- Here it is given that the GCF of two integers is 25 or 5^2. That means that they both share 5^2 only. Now we can depict the intergers as 25a & 25b sharing only 25 & a & b don't share anything b/w them. okay !! Now, as given in the question that sum of the integers is 350. therefore, 25a+25b=350 or 25(a+b)=350 or a+b =14. Now we look for those values of a & b those don't share anything b/w them. So, a+b= 3+11 or 1+13 or 5+9.........so Three values. Therefore, 3. C.......................... _________________

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

Question 1 Diagonals: 15x, 11x, 9x Area of S: 225.x^2/2 Area of R: 99.x^2/2

Difference: 63.x^2

For I to be possible: x has to be equal to 1 (ok!) For II to be possible: x has to be equal to sqrt(2) (NOT ok, since diagonals must be integers) For III to be possible: x has to be equal to 2 (ok!)

Question 5: Best solution is through elimination: 18! is divisible by 15, and 17. Therefore, 18! + 1 is not divisible by neither of these. 18! is also divisible by 33 as it has the factors 11 and 3. Therefore, 18! + 1 is not divisible by 33 18! is also divisible by 39 as it has the factors 13 and 3. Therefore, 18! + 1 is not divisible by 39

x + y = 350 x = 25*p, y = 25*q where p and q may not have any other factors in common

Substituting into the equation we have: p+q = 14 The only solutions for pair (p,q) in which p and q do not have common factors are: (1,13), (3,11),(5,9)

There are exactly 3 pairs: (25,325), (75, 275), (125, 225)

Hello everyone! Researching, networking, and understanding the “feel” for a school are all part of the essential journey to a top MBA. Wouldn’t it be great... ...

A few weeks ago, the following tweet popped up in my timeline. thanks @Uber_Mumbai for showing me what #daylightrobbery means!I know I have a choice not to use it...

“This elective will be most relevant to learn innovative methodologies in digital marketing in a place which is the origin for major marketing companies.” This was the crux...