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

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

Since X lies at most 38 from the options,

Lets find the number of Perfect Squared below 40= 1,4,9,16,25,36 Lets find the number of Primes below 40= 2,3,5,7,11,13,17,19,23,29,31,37

Now, f(x) + g(x) = 16 Start with Option C: no of primes less than 31 = \(11\) no of p.s less than 31: \(5\) Total = 16 Answer: C

P.S: The reason I started with option C is because when I analyzed choice A, the value of x is not fixed, and moving the value of A can shift the value of f(x)+g(x). Hence, I started looking for option where I can fix the value of "X"i.e. no prime number exists in that option.
_________________

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: Lengths of the diagonls are integers.Hence, multiplier must be integer.

Lets have Ratio Mulitplier to be 1; Let Diagonal of Square: \(\sqrt{2a} = 15\)

\(So, Area = 15*15/\sqrt{2}*\sqrt{2}\)

Area of Rhombus is given by: \(1/2 d1*d2 = 1/2 *11*9\)

Taking the difference of both values; we get :\(126/2 = 63\) Hence, 1st is True.

Taking the Multiplier to be 2

and performing the same steps, we get difference: 252.

Since, further taking the multiplier will only increase the difference, we shall stop here. Hence only 1 and 3 is correct. Answer: D _________________

1. The length of the diagonal of square S is 15x, the lengths of the diagonals of rhombus R are 11x and 9x, where x is a positive integer.

The area of square S is d*d/2=(15x)^2/=225x^2/2 and the area of rhombus R is diagonal1*diagonal 2/2=11x*9x/2=99x^2/2.

So, the difference between them is (225x^2-99x^2)/2=63x^2. Since x is an integer, the difference must be divisible by 63. Therefore, I is possible for x=1. II is not possible, since if 63x^2=126--->x^2=2, which is not possible for an integer x. III is possible for x=2.

The answer is D.
_________________

I'm happy, if I make math for you slightly clearer And yes, I like kudos:)

1. 15^2X^2 - 99X^2= 63X^2 2. 7C0 +7C1+7C2+7C3=63 3. {6C1+6C2++++6C6}-{5C0+5C1+++5C5)=31..I have a question..Should we include 6C0?? 4. G(n)= 2, 3,5, 7, 11, 13, 17, 19, 23..parelly increasing F(n)= 1, 4, 9, 16, 25...G(n)+29, 31|| total is 16 till x=36 5. only 19 is possible as all others are factors of 18! 6. 4^3=2^6 6^5=2^5.3^5 6^6=2^6.3^6 so X has to have 3^6 and can have any value from 2^0 to 2^6..so total=7

7. 25(a+b)=350...a+b=14..such that a and b has no common factor--1 13, 3 11, 5 9...so 3

8. 377,910 is divisible by 3 & 10 but not 11..so x has to have 11 and another 10=110

2. We have such cases for the sets: (1) Empty set. 1 (2) Contain only 1 element. There are 7 such subsets (3) Contain 2 elements. There are C_7^2=7!/(2!5!)=21 such subsets (4) Contain 3 elements. There are C_7^3=7!/(3!4!)=35 such subsets Therefore, total number of such subsets is 1+7+21+35=64

The answer is E.
_________________

I'm happy, if I make math for you slightly clearer And yes, I like kudos:)

3. To find different subsets that do not contain 0 is the same that find the number of different subsets of set {1,2,3,4,5}. The total number of such subsets is 2^5=32.

The answer is D.
_________________

I'm happy, if I make math for you slightly clearer And yes, I like kudos:)

4. Pick the numbers. Let's check the smallest possible: x=31. f(31)=5 (1,4,9,16,25) and g(31)=10 (2,3,5,7,11,13,17,19, 23,29). Therefore f(31)+g(31)=15, that is not equal to 16.

But if we take x=32: f(32)=5 and g(32)=11 (10 previous and 31). So, x=32 is a possible value. Next perfect square of an integer is 36 and the next prime number is 37. So, for not changing the number 16, we can take x=33, 34, 35, 36.

The correct answer is C.
_________________

I'm happy, if I make math for you slightly clearer And yes, I like kudos:)

5. The number 18!+1 is not divisible by any number from 2 to 18, because 18! is divisible by them. So, the answers A and B are not correct. Since 18!+1 is not divisible by 3, 18!+1 is not divisible by 33 and 39. So, answers В and E are wrong

The correct answer is C.
_________________

I'm happy, if I make math for you slightly clearer And yes, I like kudos:)

6. The problem states that lcm(x, 2^6, 2^5*3^5)=2^6*3^6. When we calculate lcm we have to take the highest powers in prime factorizations of numbers. If lcm contains 3^6 it must be in some number. So, x=3^6*y. y could be any factor of 2^6. The possible values of x: 3^6, 3^6*2, 3^6*2^2, 3^6*2^3, 3^6*2^4, 3^6*2^5, 3^6*2^6.

The correct answer is C.
_________________

I'm happy, if I make math for you slightly clearer And yes, I like kudos:)

7. Let two numbers be 25n and 25m, where gcd(n,m)=1. Then 25n+25m=350 or n+m=14. So, our goal is to find pairs of numbers (n,m) such that gcd(n,m)=1 and n+m=14: (1,13), (3,11), (5, 9).

The correct answer is C.
_________________

I'm happy, if I make math for you slightly clearer And yes, I like kudos:)

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

Let’s say diagonal of square is 15x and diagonals of rhombus are 11x and 9x where x is an integer. Area of square = (1/2) * 15^2 * x^2 = (1/2) * 225 * x^2 Area of rhombus = (1/2) * 11*9 * x^2 = (1/2) * 99 * x^2 Difference of areas = (1/2) * 126 * x^2 = 63 * x^2 So, the different must be a multiple of 63.

I. Here x^2 = 1. Possible option. II. 126 = 63*2. Here x^2 = 2. So, x is not integer. This option is not possible. III. 252 = 63*4. Here x^2 = 4 --> x = 2. Possible option

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

Number of subsets with 0 letter = 1 Number of subsets with 1 letter = 7C1 = 7 Number of subsets with 2 letters = 7C2 = 21 Number of subsets with 3 letters = 7C3 = 35 Total number of subsets with at most 3 letters = 1 + 7 + 21 + 35 = 64

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

Here we need to find out the possible subsets with the numbers {1,2,3,4,5}.

In the short way: Total number of subsets = 2^5 = 32

In the long way: Number of subsets with 0 element (null set) = 1 Number of subsets with 1 element = 5C1 = 5 Number of subsets with 2 elements = 5C2 = 10 Number of subsets with 3 elements = 5C3 = 10 Number of subsets with 4 elements = 5C4 = 5 Number of subsets with 5 elements = 5C5 = 1 Total number of subsets = 1+ 5 + 10 + 10 + 5 + 1 = 32

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

For x = 37, f(x) + g(x) = 5 + 12 = 17 For x = 36, f(x) + g(x) = 5 + 11 = 16 For x = 32, f(x) + g(x) = 5 + 11 = 16 For x = 31, f(x) + g(x) = 5 + 10 = 15 So, 31 < x < 37

After days of waiting, sharing the tension with other applicants in forums, coming up with different theories about invites patterns, and, overall, refreshing my inbox every five minutes to...

I was totally freaking out. Apparently, most of the HBS invites were already sent and I didn’t get one. However, there are still some to come out on...

In early 2012, when I was working as a biomedical researcher at the National Institutes of Health , I decided that I wanted to get an MBA and make the...