@bodhi_vriksha said:One more:A cylindrical hole of height 6cm is drilled through a sphere along its diameter. Find the volume of remaining part of the sphere.Team BV
36pi ?
jate jate ek aur :D
How many teams of at least one person can be formed for a mountaineering camp from the seven persons; Ala, Bia, Chu, Dim, Ela, Flu, Gini such that Ala and Dim don't like to be teamed together. Also Bia and Chu are not in good terms with each other.
{of course, team should be formed so that everyone is comfortable working in :) }
Team BV
@bodhi_vriksha said:jate jate ek aur How many teams of at least one person can be formed for a mountaineering camp from the seven persons; Ala, Bia, Chu, Dim, Ela, Flu, Gini such that Ala and Dim don't like to be teamed together. Also Bia and Chu are not in good terms with each other.{of course, team should be formed so that everyone is comfortable working in }Team BV
Main bhi jaate jaate....ek answer
(A or D or neither) = 3
(C or B or neither) = 3
Yes or no = 2 options each for the rest
So 3 * 3 * 2 * 2 *2 = 72 ways. One of these has 0 people.
Hence 71?
regards
scrabbler
@bodhi_vriksha said:Ooops...it became much easier I intended to ask: Find the largest positive integer which is divisible by all positive integers which are less than or equal to its cube root.e.g. 120 is divisible by all positive integers which are less than or equal to its cube root.Team BV
420.
Do we have to look for pattern only or there is some method to solve it?
:neutral:
@vijay_chandola said:420.Do we have to look for pattern only or there is some method to solve it?
"Pattern" is best method Vijay...provided it exists :P
@scrabbler said:Extremum case when the hole is of width 0 in a sphere of dia 6 =? r = 3 => V = 36pi.Since the question has no options and no other info I guess it will be the same whatever case we take? In which case 36pi?regardsscrabbler
How can you confirm Scrabbler that "It will be same whatever be the case"
@bodhi_vriksha said:How can you confirm Scrabbler that "It will be same whatever be the case"
Wording of the question + lack of options precludes multiple cases. It is basically either that or CBD (for which options are needed)....extreme cases are a good way of approaching such problems in general, as long as no CBD option exists.
Must run now...work :(
regards
scrabbler
Must run now...work :(
regards
scrabbler
@bodhi_vriksha said:How can you confirm Scrabbler that "It will be same whatever be the case"
Can solve it also btw (I think I see a Pyth approach) but take too long. And don't have pen/paper on me right now. So have to do orally.
regards
scrabbler
regards
scrabbler
@scrabbler said:9 cases, I had missed 1.If the number is abcdefg then a +5b +4c +6d +2e +3f +g should be divisible by 7. Also g = 5. hence a +5b +4c +6d +2e +3f = 7k+2. Also the variables are either 5 or 7. Solving, we get 9 cases:7555555 5557755 5755575 7755755 7575575 5777555 7757575 5577775 7775775 Edit :@ScareCrow28 , YMF and others have already got the same result I see...at least I guess the number are the same, 2-3 check kiye I'm late to the party regardsscrabbler
bhai yeh kya kiya hai? do you mind explaining your approach in detail..
@iLoveTorres said:bhai yeh kya kiya hai? do you mind explaining your approach in detail..
abcdefg
g+10*f+100*e............
now dividing this by 7
g + 3f +2e+.........
now variables are 5 and 7..........
@bodhi_vriksha said:jate jate ek aur How many teams of at least one person can be formed for a mountaineering camp from the seven persons; Ala, Bia, Chu, Dim, Ela, Flu, Gini such that Ala and Dim don't like to be teamed together. Also Bia and Chu are not in good terms with each other.{of course, team should be formed so that everyone is comfortable working in }Team BV
Total ways of forming a team = 2^7 - 1 = 127
When both A and D are there, no of ways = 2^5
When both B and C are there, no of ways = 2^5
When A, B, C, D all are there, no of ways = 2^3
So, required no of ways of forming a team = 127 - 32 - 32 + 8 = 81
@jain4444 said:Out of 200 in an aquarium, 99% are red. How many red fish must be removed in order to reduce the percentage of red fish to 98%?
100...
@bodhi_vriksha said:jate jate ek aur How many teams of at least one person can be formed for a mountaineering camp from the seven persons; Ala, Bia, Chu, Dim, Ela, Flu, Gini such that Ala and Dim don't like to be teamed together. Also Bia and Chu are not in good terms with each other.{of course, team should be formed so that everyone is comfortable working in }Team BV
7c1 + ( 7c2 - 2 ) + ( 7c3 - 2 * 5c1 ) + ( 7c4 - 2*5c2 +1 ) + (7c5 - 2*5c3 + 3c1 ) = 71 ;
@bodhi_vriksha said:Please find my solution in the attached pic. Area is 9+25 (root3)/4 sq. unitsAlternatively there is a direct relation between the distances from vertex and the side length of an equilateral triangle given by 3(x^4+y^4+z^4+a^4)=(x^2+y^2+z^2+a^2)^2. So you may also plug in the values to find a^2 and hence the area of thr triangle.
@ChirpiBird said:right!@chillfactor :you are also right!! @iLoveTorresOA is (36 + 25rt3)/4 ...
CAN YOU POST THE SOLUTION?
@bodhi_vriksha said:One more:A cylindrical hole of height 6cm is drilled through a sphere along its diameter. Find the volume of remaining part of the sphere.Team BV
Required volume
= Volume of sphere - volume of cylinder
Volume of sphere = 4pi*r^3/3
Volume of cylinder = pi(r^2 - 9)*6 + 2[Integration {pi(r^2 - x^2)*dx} from x = 3 to r]
= 6pi*r^2 - 54pi + 2(2pi*r^3/3 - 3pi*r^2 + 9pi)
= 4pi*r^3/3 - 36pi
=> Required volume = 36pi (it will be independent or r)
@iLoveTorres said:CAN YOU POST THE SOLUTION?
same way as @chillfactor did. please see his solution, if u dont get his solution, i'll draw n post mine. not in a mood to draw that figure. :P
@bodhi_vriksha said:jate jate ek aur How many teams of at least one person can be formed for a mountaineering camp from the seven persons; Ala, Bia, Chu, Dim, Ela, Flu, Gini such that Ala and Dim don't like to be teamed together. Also Bia and Chu are not in good terms with each other.{of course, team should be formed so that everyone is comfortable working in }Team BV
71 ?
when (A and B) or (C and D) or (A and C) or (B and D) or (only one of A,B,C,D) are there ---> (3c0 + 3c1 + 3c2 + 3c3)*8 = 64
when none of A,B,C,D are there ---> 3c1 + 3c2 + 3c3 = 7
total = 64 + 7 = 71...
@ChirpiBird said:same way as @chillfactor did. please see his solution, if u dont get his solution, i'll draw n post mine. not in a mood to draw that figure.
i need a simpler solution to the question.. the diagram by chillfactor is incomprehensible
@iLoveTorres said:bhai yeh kya kiya hai? do you mind explaining your approach in detail..
See, N = a*10^6 + b*10^5 + c*10^4 + d*10^3 + e*10^2 + f*10
Now take mod 7
N mod 7 = a(1) + b(5) + c(4) + d(6) + e(2) + f(3)
So Remainder is nothing but a+5b+4c+6d+2e+3f+g
The numbers in brackets are the numbers when powers of 10 are divided by 7. Hope that helps