DS question..
if x
1. x 2. y
I think answer is Statement 2 alone is sufficient to answer the question.
try this..
in how many ways can 1,2,3,4 and 5 be arranged so that the even numbers are not adjacent?
a. 108
b. 96
c. 72
d. 60
e. 48
I think the answer is c.
In the numbers 1 to 5, there are 3 odd numbers. These 3 odd numbers create 4 gaps. In these 4 gaps the two even numbers can be places in 4P2 ways. Again the 3 odd numbers can be permuted in 3! ways . So the answer is 4P2X3! = 72.
Correct me if I am wrong.
I agree. Statement 2 alone is sufficient.
1) xxy
because though for xxy
but when x=0, then x^2=xy
MBAKNOCKS SaysI think answer is Statement 2 alone is sufficient to answer the question.
1 more DS question.
1. if x is not equal to 1, is x^2/(x-1)>x?
1. x>0
2. x
2. if the probability that event A will occur is 0.5. what is the probability that event B will occur?
1. the probability that both event A and B will occur is 0.3.
2. the probability that at least one of the events will occur is 0.8.
1 more DS question.
1. if x is not equal to 1, is x^2/(x-1)>x?
1. x>0
2. x
2. if the probability that event A will occur is 0.5. what is the probability that event B will occur?
1. the probability that both event A and B will occur is 0.3.
2. the probability that at least one of the events will occur is 0.8.
I think both the statement togather is not sufficient enough to answer the question. Take x= .5 and x=1.5 to prove it.
Correct me if I am wrong.
I think both the statement togather is not sufficient enough to answer the question. Take x= .5 and x=1.5 to prove it.
Correct me if I am wrong.
:wow: if u've solved within 1.5 to 2 minutes then 'well done'.
If points A and C both lie on the circle with center B and the measurement of angle ABC is not a multiple of 30, what is the ratio of the area of the circle centered at point B to the area of triangle ABC?
A) 2B) 2(AB)2/(BC)2C) 4D) (BC)2/.5(BC)(AB)E) None of the Above
OA is E. its a good one...
If points A and C both lie on the circle with center B and the measurement of angle ABC is not a multiple of 30, what is the ratio of the area of the circle centered at point B to the area of triangle ABC?
A) 2B) 2(AB)2/(BC)2C) 4D) (BC)2/.5(BC)(AB)E) None of the Above
OA is E. its a good one...
area of circle = pie r^2
area of traingel = 1/2 r^2 sin@
required ratio = (2pie)/sin@
checking with options, none fits.
this question fits more 4 a DS problem.
Divide 43 by 2 till u get a quotient less than 2 and add all the quotients
43/2 = 21
21/2 = 10
10/2 = 5
5/2 = 2
2/2 = 1
So, 21+10+5+2+1 = 39
Hope it helps
Thanks Deepak
If there are 85 students in a statistics class and we assume that there are 365 days in a year, what is the probability that at least two students in the class have the same birthday (assuming birthdays are distributed independently)?
A)
B)
C)
D)
E)
A)
B)
C)
D)
E)
If there are 85 students in a statistics class and we assume that there are 365 days in a year, what is the probability that at least two students in the class have the same birthday (assuming birthdays are distributed independently)?
A)
B)
C)
D)
E)
probability that none of the students have b'day on same day = 365P85/365^85
so required probability(at least two) = 1 - 365P85/365^85 (option D)
probability that none of the students have b'day on same day = 365P85/365^85
so required probability(at least two) = 1 - 365P85/365^85 (option D)
ur answer is right, but u r missing the true justification. look at your solution and the answer choice D. Dont u think u r missing a term 85! in choice D. OA is D, so there must be some other way to solve it, it was 700+ level question.
@missionPGPX
i didnt solve it fully.. did not expand the permutation fully coz answer was directly visible from the options after solving half. i think its always better to use options to save time. juz expand the permutation if u want to solve it fully. 😃
this method is the shortest possible.
if u try to solve it taking case by case then it'll take more than half an hour to solve this question.
@PGPX
forgot 2 mention..
u wont get 85! after expanding that permutation..
@PGPX
forgot 2 mention..
u wont get 85! after expanding that permutation..
Hey dare2,
sorry abt that. I guess i need more practice on such questions. you were right. Its a permutation i wont get 85!. But can you please elaborate on that permutation and probability part? probability that none of the students have b'day on same day = 365P85/365^85.
Hey dare2,
sorry abt that. I guess i need more practice on such questions. you were right. Its a permutation i wont get 85!. But can you please elaborate on that permutation and probability part? probability that none of the students have b'day on same day = 365P85/365^85.
look.. for no two students to have b'day on same day, each of the 365 days can have only 1 student whose b'day falls on that day. or we can say selecting 85 days from 365 days.(if we have 85 days then only we can place each student to have unique b'days. so permutation will be 365P85 = (365!)/(365-85)! = (365!)/(280!)
for finding total number of possibilities..
1st student can have his b'day in any of the 365 days ie 365 possibilities.
2nd --------------------------------------------------365 possibilities.
3rd ---------------------------------------------------365 possibilites.
.
.
.
.
and so on..
so total possibilities = 365x365x365x...85 times or 365^85
hence probability for unique b'days = (365!/280!)/365^85.
we have to subtract this probability from the probability 1 so that we get the probability in which students dont have unique b'days.
i dont know whether i was able to explain conveniently.. plz revert back if u have ne more doubts on this..
wrong thread
Set A = {1, 2, 3, 4, 5, 6, y}
Which of the following possible values for y would cause Set A to have the smallest standard deviation?
(A) 1
(B) 2.5
(C) 3
(D) 3.5
(E) 7
Set A = {1, 2, 3, 4, 5, 6, y}
Which of the following possible values for y would cause Set A to have the smallest standard deviation?
(A) 1
(B) 2.5
(C) 3
(D) 3.5
(E) 7
will go for option D.
the less the data dispersed from the mean, the less the SD is..