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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

1. x = 1 ? Not necessary. Let x = 3, and y = 0, then x^y = 1 2. x = 1 and y = 0. Not necessary (see item 1) 3. x = 1 or y = 0. Not necessary. Let x = -1, y=2

Answer: E _________________

Please press "kudo" if this helped you!

Last edited by caioguima on 18 Apr 2013, 03:48, edited 2 times in total.

The WONDERFUL thing is that 37*27 = 999, therefore, we can write the sum as: 1/3 + 1/9 + 1/27 + 1/37 = 333/999 + 111/999 + 37/999 + 27/999 = 508/999 = 0.508508508508508...

Note that the decimals repeat themselves in period of 3. Since 101 divided by 3 gives remainder 2, we're looking for the position #2 in the repeated set 508.

I. X=1 ; x could be any no. if y=o II. Y=0; x could become 1 therefore negating this statement III. x=1 or y=0; well x could very well be -1; so not necessary _________________

When you feel like giving up, remember why you held on for so long in the first place.

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.”

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

Sol: We need to find condition which will be true under all conditions

St 1 when x =1 then X^ (any value of y) will always give us 1. So A, D are possible answers St 2 X=1 and y=0, Here and means both conditions simultaneously. But Y need not be zero as long as x= 1. Y can take any value as long as x=1 and hence B is ruled out

St 3 x=1 or y=0 here "or" means any one of the condition if true then we get x^y=1 which is true. Consider x=1 and y= 32 ----> x^y = 1, consider x= 1, y= -3, then x^y= 1 Similarly X= 2 but y=0 then 2^0 =1 Option C and D can be the possible answer

Ans Option D......

PS: I think there is a catch since St 1 covers only value of x whereas St 3 covers for all possible cases of (x,y) and I was tempted to go for option C alone as the answer. Let's see _________________

“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.”

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

Sol: 1/3 + 1/9 + 1/27 + 1/37

1/3= 0.333333 1/9= 1/(3*3)= 0.3333333/3-----> 0.1111111 1/27= 1(3^3)= 0.037037037 1/37= 0.027027027 So if we add up 0.333 0.111 0.037 0.027 Sum is ( .508508508)...99th Digit will be 8,100th digit will be 5 and 101st digit will be 0...

I guessed it under timed condition......Would like to have faster way.

Ans should A _________________

“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, 2^6is 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.

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

All are Could be true questions, but none of them is MUST be TRUE\ 1- X can be any other value than 1 2- NoT Necessarily 3- x=2 and y =0; x=1 and and y=1 can satisfy Hence, E _________________