Statement 1 is suff to answer the q.
statement 2 is insuff to ans: x=2 , y=-2 ,z=-5 or x=-4 y = 4 z = -1
So I will go with option A
i also picked A. but somehow OA is E. and its a GMATPREP question... there must be smthing we are missing here.
i also picked A. but somehow OA is E. and its a GMATPREP question... there must be smthing we are missing here.
MissionPGPX Saysi also picked A. but somehow OA is E. and its a GMATPREP question... there must be smthing we are missing here.
Interesting one,I tink the OA is correct.We overlooked the q.
x = 3^ 0.25
y = 2^0.25
z = 6^ 0.25
Then, x^2 + y^2 > z^2 but x^4 + y^4
So option E is the correct ans.
guys, have a look at this problem:
How many different 5 person teams can be formed from a group of x individuals.?
a) if there had been x+2 individuals in the group,exactly 126 - 5 different person teams could have been formed.
b) If there had been x+1 individuals in the group, exactly 56 different 3 -person teams could have been formed.
The answer to this is D as we can calculate x by either of the 2 options ,but in the explanation available to me ,combination has been used to calculate x and not permutation.
I feel this is a permutation problem and we have to find unique teams and we will use x+2! P 5 =126 and x+1 ! P 3 =56 to find x.
Please comment .
Are the integers A,B and C consequetive?
1) b-a = c-b
2) The average of {a,b,c} is equal to b.
Are the integers A,B and C consequetive?
1) b-a = c-b
2) The average of {a,b,c} is equal to b.
Statement 1 :
We can only say that b = a+c/2.This doesn't mean the 3 nos are consective. 1,3,5 are simple examples to prove this stuff..So this statement is insuff.
statement 2
This is again a repetition of statement 1.So this is insuff.
I would go with option E
Statement 1 :
We can only say that b = a+c/2.This doesn't mean the 3 nos are consective. 1,3,5 are simple examples to prove this stuff..So this statement is insuff.
statement 2
This is again a repetition of statement 1.So this is insuff.
I would go with option E
True. OA is E.
Hello puys, 
I was reading somewhere that since all of us focus all the attention towards the verbal section, we end up screwing our quant and DS sections, starting today i will be posting plenty of question in DS which i am unable to solve and hope the same from your end. Let's make this thread active like that for CR's and SC's ..
Set X has 5 integers: a, b, c, d, and e. If m is the mean and D, where D = sqrt ,is the standard deviation of the set X, is D>2 ?
(1) a, b, c, d and e are different integers
(2) m is an integer not equal to any elements of the set X
If x and y are integers, is x > |y?
(1) x| = |y + 1
(2) xy = x! + |y
Set X has 5 integers: a, b, c, d, and e. If m is the mean and D, where D = sqrt ,is the standard deviation of the set X, is D>2 ?
(1) a, b, c, d and e are different integers
(2) m is an integer not equal to any elements of the set X
If x and y are integers, is x > |y?
(1) x| = |y + 1
(2) xy = x! + |y
My take : C
smokinskull86 SaysMy take : C
Plz post explanation... without that its a question answer quiz session..nevertheless u r right...
If is an integer, what is the value of
?
1.
2.
Set X has 5 integers: a, b, c, d, and e. If m is the mean and D, where D
= sqrt ,is the standard deviation of the set X, is D>2 ?
(1) a, b, c, d and e are different integers
(2) m is an integer not equal to any elements of the set X
If x and y are integers, is x > |y?
(1) x| = |y + 1
(2) xy = x! + y
Consider 5 numbers : 3,4,5,6,7, the sum is 15 and the average is 5. The standard deviation is sqrt(14/5) 2, hence A alone is out, therefore A and D are insufficient
Consider 5 numbers : 1,1,1,3,4 the average is 2 and is an integer not included in the set, thus the standard deviation is sqrt{(1+1+1+1+4)/5 B is also insufficient.
Now, combining both the statements we get a set, in which all the numbers are unique and no number is equal to the mean of the 5 numbers.
Eg. {3,5,6,10,11} , the mean of this set is 7,
Therefore, m = 7
Also D = sqrt{(16+4+1+9+16)/5} = sqrt(36/5) = sqrt 7.2 > 2, Hence Option C is correct.
Would appreciate if someone could come up with a better solution.
MissionPGPX SaysPlz post explanation... without that its a question answer quiz session..nevertheless u r right...
dude, already got a couple of questions wrong today, didnt want to sound foolish with a wrong explainiation that's why didnt answer the question at the onset. 😁
Ifis an integer, what is the value of
1.
2.
Good question, i would go with option A in this one, A is sufficient
Reason : Trying to fit in random values here, only one value fits here,
consider p=1, thus p! =1 and (p!)^p = 1 = LHS
RHS = p! = 1
For all other values of p, this is not true, for negative numbers (eg. -1), the LHS will be negative whilst the RHS will be positive, hence rejected. Thus there is only one unique solution.
For option 2, we can have 2 solutions, p=1,p=2 ; both satisfy the equation, hence B is insufficient.
Post the OA, i am rather not confident about this one.
Thx
Good question, i would go with option A in this one, A is sufficient
Reason : Trying to fit in random values here, only one value fits here,
consider p=1, thus p! =1 and (p!)^p = 1 = LHS
RHS = p! = 1
For all other values of p, this is not true, for negative numbers (eg. -1), the LHS will be negative whilst the RHS will be positive, hence rejected. Thus there is only one unique solution.
For option 2, we can have 2 solutions, p=1,p=2 ; both satisfy the equation, hence B is insufficient.
Post the OA, i am rather not confident about this one.
Thx
O O O... how come for negative numbers LHS will be negative.
Statement 1 :
So S1 is not sufficient because of two values 1 and -1. So the answer is C.
Consider 5 numbers : 3,4,5,6,7, the sum is 15 and the average is 5. The standard deviation is sqrt(14/5) 2, hence A alone is out, therefore A and D are insufficient
Consider 5 numbers : 1,1,1,3,4 the average is 2 and is an integer not included in the set, thus the standard deviation is sqrt{(1+1+1+1+4)/5| B is also insufficient.
Now, combining both the statements we get a set, in which all the numbers are unique and no number is equal to the mean of the 5 numbers.
Eg. {3,5,6,10,11} , the mean of this set is 7,
Therefore, m = 7
Also D = sqrt{(16+4+1+9+16)/5} = sqrt(36/5) = sqrt 7.2 > 2, Hence Option C is correct.
Would appreciate if someone could come up with a better solution.
exactly, is putting values and doing calculations the only way to arrive at answer?
anyway how about second question?
O O O... how come for negative numbers LHS will be negative.
Statement 1 :take p=-1 so p!^p = -1!^(-1) = 1.
So S1 is not sufficient because of two values 1 and -1. So the answer is C.
Yes, you are right, i forgot to check out the modulus sign,:nono::nono:
and anyways even 0 is valid in the first statement, 0! is 1 and 1^0 = 1, so -1,0,1 are the answers from statement 1 and 1,2 from statement 2, hence Option C
1) If , is
(1) is less than
(2) is less than
1) If, is
(1)is less than
(2)is less than
I think answer is A because option A tells us that value of x is between 2.1 and 2.9 when y = 1. Other values of y do not satisfy both the equations ( x>2y and x-y
While there are infinite values for y that satisfy the equation in option B.
Thanks,
Anurag...
At a certain university, if 50 percent of the people who inquire about admission policies actually submit applications for admission, what percent of those who submit applications for admission enroll in classes at the university?
(1) Fifteen percent of those who submit applications for admission are accepted at the university.
(2) Eighty percent of those who are accepted send a deposit to the university.
please explain why the OA is E??
1) If you are accepted by the university, it does not necessarily mean you will also end up finally enrolling.
2) Similar explanation is valid for 2. Sending a deposit does not guarantee enrollment by the candidate at the classes.
Combining both, no conclusion also.
Hence E.