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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

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

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

A. 10 B. 11 C. 55 D. 110 E. 330

Sol: Given that (x* 377910/ 3*11*100)= Integer and therefore we need to least value of x

On simplifying we get (x* 12597/11*10). Now 12597 is not divisible by 11 and 100. Therefore for the expression to be to result an integer, least value of X should be 110

Ans D _________________

“If you can't fly then run, if you can't run then walk, if you can't walk then crawl, but whatever you do you have to keep moving forward.”

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

Sol: Given 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:

Best thing will be to assume value of X and check in which of the above range it satisfies for all values of X

Start with option C Let us take x - 32 so we have f(n) = 5 (1,4,9,16 and 25) and G(n) = 11 (2,3,5,7,11,13,17,19,23,29 and 31) Clearly f(n) +g(n)= 16. Note that for any value of X in the range for option C, we get the value of F(n)+ G(n)= 16 and Hence C should be the Answer.

To cross check, lets look at option A and Assume X = 32 then f(n)= 5 and g(n) = 11. F(n)+G(n)= 16 but if take X= 31, f(n)=5 and g(n)=10 and therefore their sum is not equal to 16

Consider option E if x= 33 then f(n) =5 and g(n)= 11 and f(n)+ g(n)= 16 but if X =37 then f(n)= 6 (1,,4,9,16,25 and 36) and g(n)= 11 and f(n)+g(n)= 17

Therefore Ans should be C _________________

“If you can't fly then run, if you can't run then walk, if you can't walk then crawl, but whatever you do you have to keep moving forward.”

We can say that 18! and 18! +1 are consecutive nos and therefore both the nos are co-prime. Therefore any number which is a factor of 18! will not be factor 18!+1

18!= 18*17*16*15*14*.....*1 Looking at option Choices we have 15,17, 33 (11*3) and 39 (3*13) as factors of 18!.

Therefore answer should be C i.e 19 _________________

“If you can't fly then run, if you can't run then walk, if you can't walk then crawl, but whatever you do you have to keep moving forward.”

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

Sol: Greatest Common divisors of 2 nos (n1,n2) is 25 and n1+n2= 350 Think of any 2 nos whose GCD will be 25 i.e (25,50), (25,25), (25,100), (25,125), (25,150)... Now we are bounded by the condition that n1 +n2=350

Consider n1 =25 then n2= 325 , GCD is 25 Consider n1 =50, n2=300, GCD is 50 (Cannot be the pair) Consider n1=75 and n2= (275), GCD is 25 Consider n1=100, n2=250, GCD is 50 n1=125, n2=225,GCD =25 n1=150, n2=200, GCD =50 n1=175, n2=175, GCD is 175

So Ans should be 3.....Option C _________________

“If you can't fly then run, if you can't run then walk, if you can't walk then crawl, but whatever you do you have to keep moving forward.”

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

Theory for solving such questions: 1- Perform the Prime Factorization of the Number 2- Find the Highest Index among the given numbers of all the Prime No's 3- L.C.M is the product of all these prime no's with the respective highest indexes

\(4^3 = 2^6\) \(6^5 = 2^5 * 3^5\) Since LCM is 2^6* 3^6

For \(3^6\) to be there, We require that to come from "X" Since, \(2^6\)is already there,

X may contain \(2^0, 2^1, 2^2,2^3,2^4,2^5, and 2^6\)

So, X can be any of the values - \(2^0 *3^6 2^1 *3^6 , 2^2*3^6, 2^3*3^6, 2^4*3^6, 2^5,*3^6 and 2^6*3^6\)

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

A. 10 B. 11 C. 55 D. 110 E. 330

Lets Break into Prime Nos First

\(3300 = 3* 2^2* 5^2 *11\)

\(377,910 = 2*3*5*7*2371\)

\(2*3*5*7*2371 * X / 3*2^2*5^2*11\)

2371 is not divisible by 2; Hence X must contain one 2(atleast), for the number to be divisible by 3300 Similarly, X should contain 01 Five and 01 eleven for the number to be divisible by 3300 Hence X must be at-least 2*5*11=110

Since these digits will be repeated, and for finding 101 digits, we can ignore the part till 99th digit as it is cyclical. Hence - Adding .333+.111+.037+.027 = .508. Hence, 101st digit will be 0

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

Since G.C.D of two integers X and Y is 25 So, X can be written as \(X= 25 a\) and \(Y = 25 b,\)where both a and b are co-primes.

Since sum of integers is 350, we can write as \(25a+25b = 350 25(a+b) = 25*14\)

Now, a+b = 14 such that a and b are co-prime to each other -

Since these are numbers not digits, there is no restriction. And since ths um of two numbers is even, we can ignore the even numbers, and shal concentrate only on ODD numbers.

\(a=1;b=13 a=3;b=11 a=5;b=9\) Hence, 3 such numbers.

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:)

Back to hometown after a short trip to New Delhi for my visa appointment. Whoever tells you that the toughest part gets over once you get an admit is...