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.

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

8. The product of a positive integer x and 377,910 is divisible by 3,300, then the least value of x is:

3,300=3*11*2*5*2*5 377,910=37791*2*5 377,910*x/3300=\frac{37791*2*5*x}{3*11*2*5*2*5} Now x must contain 2*5, because 37791 is divisible by 3 x must not contain a 3, because 37791 is not divisible by 11 x must have it. x=2*5*11=110

IMO D. 110 _________________

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

With 6 different elements in a set, total number of subsets = 2^6 = 64 With 5 different elements in a set, total number of subsets = 2^5 = 32 Hence from set of 6, if we do not include 0 in any set, that would be equivalent to considering just 5 elements out of 6 sets and how many subsets can be obtained from 5 elements. that should be 2^5 = 32 sub sets. And D

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

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.

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