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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

18! and 18!+1 are consecutive integers. Two consecutive integers are co-prime, which means that they don't share ANY common factor but 1. For example 20 and 21 are consecutive integers, thus only common factor they share is 1.

Now, since we can factor out each 15, 17, 33=3*11, and 39=3*13 out of 18!, then 15, 17, 33 and 39 ARE factors of 18! and are NOT factors of 18!+1. Therefore only 19 could be a factor of 18!+1.

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 that the ratio of the diagonal is \(d_s:d_1:d_2=15x:11x:9x\), for some positive integer x (where \(d_s\) is the diagonal of square S and \(d_1\) and \(d_2\) are the diagonals of rhombus R).

\(area_{square}=\frac{d^2}{2}\) and \(area_{rhombus}=\frac{d_1*d_2}{2}\).

The difference is \(area_{square}-area_{rhombus}=\frac{(15x)^2}{2}-\frac{11x*9x}{2}=63x^2\).

If x=1, then the difference is 63; If x=2, then the difference is 252; In order the difference to be 126 x should be \(\sqrt{2}\), which is not possible.

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

Consider the set without 0: {1, 2, 3, 4, 5}. Each out of 5 elements of the set {1, 2, 3, 4, 5} has TWO options: either to be included in the subset or not, so total number of subsets of this set is 2^5=32. Now, each such set will be a subset of {0, 1, 2, 3, 4, 5} and won't include 0.

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

Side square = 15x \(AreaS = \frac{15^2}{2}x^2\) Diagonals= 9x, 11x\(AreaR = \frac{11*9*x^2}{2}\) Difference = \(\frac{15^2x^2-11*9x^2}{2}= \frac{126x^2}{2}= 63x^2\) \(63=3*3*7\) if x=1 diff = 63 possible and easy to see \(126=2*3*3*7\) x sould be \(\sqrt{2}\) => no integer \(252=2*2*3*3*7\) x=2 possible

IMO D. I and III only _________________

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

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.

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

We are given that \(6^6=2^{6}*3^{6}\) is the least common multiple of the following three numbers:

x; \(4^3=2^6\); \(6^5 = 2^{5}*3^5\);

First notice that \(x\) cannot have any other primes other than 2 or/and 3, because LCM contains only these primes.

Now, since the power of 3 in LCM is higher than the powers of 3 in either the second number or in the third, than \(x\) must have \(3^{6}\) as its multiple (else how \(3^{6}\) would appear in LCM?).

Next, \(x\) can have 2 as its prime in ANY power ranging from 0 to 6, inclusive (it cannot have higher power of 2 since LCM limits the power of 2 to 6).

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

We are told that the greatest common factor of two integers is 25. So, these integers are \(25x\) and \(25y\), for some positive integers \(x\) and \(y\). Notice that \(x\) and \(y\) must not share any common factor but 1, because if they do, then GCF of \(25x\) and \(25y\) will be more that 25.

Next, we know that \(25x+25y=350\) --> \(x+y=14\) --> since \(x\) and \(y\) don't share any common factor but 1 then (x, y) can be only (1, 13), (3, 11) or (5, 9) (all other pairs (2, 12), (4, 10), (6, 8) and (7, 7) do share common factor greater than 1).

So, there are only three pairs of such numbers possible: 25*1=25 and 25*13=325; 25*3=75 and 25*11=275; 25*5=125 and 25*9=225.

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

The two numbers can be represented as 25a and 25b, where a and b are co-prime.Also, 25(a+b) = 350 --> (a+b) = 14 Thus, a=1,b=13 or a=3,b=11 or a=9,b=5.

9. What is the 101st digit after the decimal point in the decimal representation of 1/3 + 1/9 + 1/27 + 1/37?

1/3=0.333 1/9=0.333/3=0.111 1/27=0.037 and then repeats 1/37=0.027 and then repeats We can work on the first 3 digits: 0.111+0.333+0.027+0.037=0.508 After the 0 we have at first place a 5, second a 0, third an 8; and so on 4th=5, 5th=0, 6th=8. Every 10th position we have a "change" 10th=5 20th=0 30th=8 and so on 100th=5 and finally 101st=0

IMO A.0

Thanks for the set Bunuel! _________________

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

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.

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

Numbers are: x, 2^6, and (2^5)*(3^5) LCM of the numbers = (2^6)*(3^6) As 3^6 is not part of second and third numbers, it must be part of x. So, lowest and highest values of x can be 3^6 and 6^6 {=(3^6)*(2^6)}. So the values that x can take are: (3^6)*(2^0), (3^6)*(2^1), (3^6)*(2^2), (3^6)*(2^3), (3^6)*(2^4), (3^6)*(2^5), and (3^6)*(2^6).

1/3 + 1/9 + 1/27 + 1/37 =0.508508…………….. (recurring 508) --> 8 will be in every 3rd position --> 8 will be in 99th position --> 0 will be in 101st position

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

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

We are given that \(6^6=2^{6}*3^{6}\) is the least common multiple of the following three numbers:

x; \(4^3=2^6\); \(6^5 = 2^{5}*3^5\);

First notice that \(x\) cannot have any other primes other than 2 or/and 3, because LCM contains only these primes.

Now, since the power of 3 in LCM is higher than the powers of 3 in either the second number or in the third, than \(x\) must have \(3^{6}\) as its multiple (else how \(3^{6}\) would appear in LCM?).

Next, \(x\) can have 2 as its prime in ANY power ranging from 0 to 6, inclusive (it cannot have higher power of 2 since LCM limits the power of 2 to 6).

Thus, \(x\) could take total of 7 values.

Answer: C.

Hi Bunuel, x can take factor of 2 with power from 2 to 6 or no factor of 2. So the answer can be 6 too. Please explain ! thanks

I don;t understand what you mean...

x can take the following 7 values: \(3^6\); \(2*3^6\); \(2^2*3^6\); \(2^3*3^6\); \(2^4*3^6\); \(2^5*3^6\); \(2^6*3^6\). _________________

MBA Acceptance Rate by Country Most top American business schools brag about how internationally diverse they are. Although American business schools try to make sure they have students from...

McCombs Acceptance Rate Analysis McCombs School of Business is a top MBA program and part of University of Texas Austin. The full-time program is small; the class of 2017...