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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
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: \(3^6\); \(2*3^6\); \(2^2*3^6\); \(2^3*3^6\); \(2^4*3^6\); \(2^5*3^6\); \(2^6*3^6\). _________________
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 want to convert any fraction with denominator 9, 99, 999, so on,to decimal form, then see what is the value of the fraction with 10, 100, 1000,so on, as denominator.
For eg, 457/999 = ?
See 457/1000 = 0.457
Then, 457/999 = 0.457457457457...
This knowledge comes very handy at times with complex fractions.
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
Notice that if x=-1 and y is any even number, then \((-1)^{even}=1\), thus none of the options must be true.
Answer: E.
Hi Bunuel,
As per the question which of the following must be true.
So as per the given choice B) II only is true right where 1^0 = 1 as X =1 and Y=0 given.
As ur explanation gives another chance as X coud be = -1 , and Y = any even.
Please clarify where i am wrong.
Thanks in Advance, Rrsnathan
x=1 and y=0 indeed satisfies x^y=1, but the question asks "which of the following must be true". So, this option is NOT necessarily true, because x can be -1 and y any even number. _________________
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.
Bunuel, question here. Could you help me out?
I am guessing x can assume 8 values and not 7. The values are \(3^6\), \(2^0\),\(2^1\),\(2^2\),\(2^3\),\(2^4\),\(2^5\) and \(2^6\). Anything wrong in my approach here?
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.
Bunuel, question here. Could you help me out?
I am guessing x can assume 8 values and not 7. The values are \(3^6\), \(2^0\),\(2^1\),\(2^2\),\(2^3\),\(2^4\),\(2^5\) and \(2^6\). Anything wrong in my approach here?
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.
Answer: D.
I would like to request some some help with this question..
Could you please elaborate on the theory behind 2^n? Isnt the null set considered to be a subset of this set? _________________
You've been walking the ocean's edge, holding up your robes to keep them dry. You must dive naked under, and deeper under, a thousand times deeper! - Rumi
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.
Answer: D.
I would like to request some some help with this question..
Could you please elaborate on the theory behind 2^n? Isnt the null set considered to be a subset of this set?
Consider simpler set: {a, b, c}. How many subsets does it have? Each of a, b, and c has two choices either to be included in subsets or not. Thus total of 2^38 subsets (including an empty set). The subsets are:
{a, b, c} --> each is included; {a, b} --> a and b are included; {a, c}; {b, c}; {a}; {b}; {c}; {} empty set, none is included.
Hi bunuel , Is following approach right ? As hcf is actually difference or multiple of difference of two numbers.. So a- b = 25k, a+b=350
Also k can be only an even number less than 12, that is a- b can at most be only 300, for an odd k anyway integer condition won't satisfy when we solve two equations, So a-b can be only 50, 100, 150, 200,250,300 , This give us three pairs of a and b
quote="Bunuel"]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.
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