yeah its 1. please explain how you got the focal points
@vijay_chandola said:Sample space:Choosing 1 student from N students, C(N, 1) ways.He can pass the baton to anyone of the remaining (N-1) students.again, the second student can pass the baton to anyone of the (N-1) students.So, total number of ways=C(N, 1)*(N-1)*(N-1)Favorable cases: Select any 2 random students who can pass the baton to each other,So, C(N, 2) cases.=> probability=C(N, 2)/{C(N, 1)*(N-1)*(N-1)}=1/2*(N-1)
question says "first students" it doesn't say any particular should get baton after 1 operation
I think we need to multiply nC2 by 2!
@anantn said:yeah its 1. please explain how you got the focal points
Because those are the points given there in the question and also that is the definition of an ellipse.. If the sum of point P 's distance from points A and B is constant, then the curve traced by such a point P is an ellipse with the focal points at A, B. (PA + PB = 10 is mentioned in the question)
@jain4444 said:
Let S be a set of positive integers, not necessarily distinct, in which the number 68 appears. The arithmetic mean of the numbers in S is 56. However, if 68 is removed, the arithmetic mean of the remaining numbers is 55. Determine the largest number that can appear in S.
(S + 68)/N = 56 => S = 56N - 68
(S)/(N-1) = 55 => S = 55N - 55
=> N = 13
S = 660
So, 12 numbers are there whose sum is 660 , to find maximum, I will take 11 numbers as 1
=> Max numbers = 660 - 11 = 649 ?
@anantn said:
Q2: There are 5 envelopes, 5 letter boxes and 5 cities.Each letter has only one correct letter box corresponding to it, and each of the letter boxes correspond to only one city. The letters are first posted in the letterboxes and then the letterboxes deliver mail to the cities. What is the probability of all the letters, as well as letterboxes reaching the wrong letterbox and city respectively?
Can the letters reach the right city ? Or it is not allowed ?
If allowed then,
P(letters reach the wrong letterbox) = Dearrangment(5)/ 5! = 44/120 = 11/30
P(letterboxes reaching wrong city) = 44/120 = 11/30
So, final prob = 121/900 ?
If not allowed, then I need to dig deeper :(
The average age of 10 members of a committee is the same as it was 4 years ago, because an old member has been replaced by a young member. Find how much younger is the new member?
@YouMadFellow : its like the letters are correspond only to mailboxes, not to cities. the mailboxes correspond to cities. so the letter can reach the right mailbox and the mailbox can deliver to the right city, multiple cases are possible.
On buying a camera, the shopkeeper gives three rolls of film free. On buying a cameraand six rolls of film, the shopkeeper gives additional four rolls of film free. If the equivalent discount is the same in both cases, then how many rolls will be equal in value to a camera? 12, 15 18 24
A :18. used logic
1C + 6R gives 4 R
but 1C gives 3R => 6R gives 1R
for value to b same as of Camera 3 * 6R= 18R.
if somewhere wrong please let me know.
@hiteshkhurana82 said:The average age of 10 members of a committee is the same as it was 4 years ago, because an old member has been replaced by a young member. Find how much younger is the new member?
40 years younger ? (EDITED)
(S - 36) + (O - 4) = S + Y
Y = O - 40 (EDITED)
(S - 36) + (O - 4) = S + Y
Y = O - 40 (EDITED)
@hiteshkhurana82 said:The average age of 10 members of a committee is the same as it was 4 years ago, because an old member has been replaced by a young member. Find how much younger is the new member?
S9 + Old = 10*A
S9 + 9*4 + New = 10*A
So, Old - New = 36, So the age difference now = 36 + 4 = 40 ?
my take was : total cases 5^2*4^2*3^2*2^2= 14400
total favourable cases= the letters have to be dearranged and then the mailboxes have to be derranged independently so:
(derrangement of 5)^2/1440 = 1936/14400 = 107/800
@catnov2013 said:On buying a camera, the shopkeeper gives three rolls of film free. On buying a cameraand six rolls of film, the shopkeeper gives additional four rolls of film free. If the equivalent discount is the same in both cases, then how many rolls will be equal in value to a camera? 12, 15 18 24
18 rolls should be the answer.
1C + 3R at the price of 1C
Discount = 3R/(1C + 3R)
1C + 10R at the price of 1C + 6R
Discount = 4R/(1C + 10R)
Equate the two:
3C + 30R = 4C + 12R
C = 18R
1C + 3R at the price of 1C
Discount = 3R/(1C + 3R)
1C + 10R at the price of 1C + 6R
Discount = 4R/(1C + 10R)
Equate the two:
3C + 30R = 4C + 12R
C = 18R
@hiteshkhurana82 said:The average age of 10 members of a committee is the same as it was 4 years ago, because an old member has been replaced by a young member. Find how much younger is the new member?
A=10(A+4)-N+X/10
N-X=40
40??
@hiteshkhurana82 said:The average age of 10 members of a committee is the same as it was 4 years ago, because an old member has been replaced by a young member. Find how much younger is the new member?
40??
@hiteshkhurana82 said:The average age of 10 members of a committee is the same as it was 4 years ago, because an old member has been replaced by a young member. Find how much younger is the new member?
Let the avg 4 years ago be "a"
It is the same today, hence
10 ( a+4) - O + Y /10 = a
From here, O - Y = 40
Hence, The new member is 40 years younger
@YouMadFellow said:Can the letters reach the right city ? Or it is not allowed ?If allowed then, P(letters reach the wrong letterbox) = Dearrangment(5)/ 5! = 44/120 = 11/30P(letterboxes reaching wrong city) = 44/120 = 11/30So, final prob = 121/900 ? If not allowed, then I need to dig deeper
Dearrangment(5)/ 5!
why are you dividing by 5!
cant understand !
why are you dividing by 5!
cant understand !
@hiteshkhurana82 said:The average age of 10 members of a committee is the same as it was 4 years ago, because an old member has been replaced by a young member. Find how much younger is the new member?
should be 40
@joyjitpal said:Dearrangment(5)/ 5! why are you dividing by 5! cant understand !
Total arrangements are 5! .. L1 to L5 can go to B1 to B5 in 5! ways, but we need only the ones where they are not going into the corresponding boxes
@anantn said:my take was : total cases 5^2*4^2*3^2*2^2= 14400total favourable cases= the letters have to be dearranged and then the mailboxes have to be derranged independently so:(derrangement of 5)^2/1440 = 1936/14400 = 107/800
I think your approach seems fine, but you have a calculation mistake in the last step. Also, tag the person in the post so that someone would know that you replied or EDIT the original post where you posted the question. :)
Determine the exact value:
The summation of 1/[n(n+1)]^3 as n takes integral values fr. [1,+inf).
The summation of 1/[n(n+1)]^3 as n takes integral values fr. [1,+inf).