Two pipe can fill the cistern in 12 and 18 hr respectively. Both the pipes were opened at 10.00AM and by 6:00PM the cistern was full. what could be the min possible duration during which one of the pipe must have been closed during that interval?
@vbhvgupta said:Two pipe can fill the cistern in 12 and 18 hr respectively. Both the pipes were opened at 10.00AM and by 6:00PM the cistern was full. what could be the min possible duration during which one of the pipe must have been closed during that interval?
1/12(t)+1/18(t-x)=1
t=8
so x=2 hours = OA?
t=8
so x=2 hours = OA?
@vbhvgupta said:Two pipe can fill the cistern in 12 and 18 hr respectively. Both the pipes were opened at 10.00AM and by 6:00PM the cistern was full. what could be the min possible duration during which one of the pipe must have been closed during that interval?
let the amount of work =36W
so, effiiciency of P1=3 W/hr
P2=2 W/hr
for the time to be min, P2 has to be switched off
Now, total time taken=8hrs
let x be the no of hrs for which P2 was switched off
therefore:
(3*8)+[2(8-x)]=36
x=2 hrs
@vbhvgupta said:@Zedai OA 4/3 Hour
ok, wait i got it. min possible time to keep it closed, means max time to fill, so pipe two having lower efficency will run for more time.
so
t/12+(t-x)/18=1
aise mistakes 
@cynara said:ok, wait i got it. min possible time to keep it closed, means max time to fill, so pipe two having lower efficency will run for more time.so t/12+(t-x)/18=1aise mistakes
@cynara said:1/12(t)+1/18(t-x)=1t=8so x=2 hours = OA?
If we keep the 18 open for the full time we get 1hr 20 min?
(8-x)/12 + 8/18 = 1.
regards
scrabbler
(8-x)/12 + 8/18 = 1.
regards
scrabbler
@sbharadwaj said:In a school of 100 students, 70 play Hockey, 60 play Tennis and 55 play Football. What is the max number of sudents who play all 3.?And what is the minimum.? Is there any trick.? Plz explain
Plz help..!!
@sbharadwaj -yar Max can be found out very easily... but for Min, there is a short trick... i knew it..just slipped out of my mind..
Pl. help anyone...
find x ....
(271/x)^x(log271-logx-1)
(271/x)^x(log271-logx-1)=0
@sbharadwaj said:Plz help..!!
@sbharadwaj said:In a school of 100 students, 70 play Hockey, 60 play Tennis and 55 play Football.What is the max number of sudents who play all 3.?And what is the minimum.?Is there any trick.? Plz explain
Minimum will be 15.
Maximum will be 55.
"for maximum:
maximize the no. of people who play nothing..i.e. 100-70 = 30 nos.
start from 1 onwards...
Hockey : student 1 to student 70.
Tennis: Student 1 to student 60.
football: student 1 to student 55.
student who play all three will be 55.""
for minimum
minimize the no. of people who play nothing..i.e. 100-100 = 0 nos.
Tennis: Student 1 to student 60.
football: Student 46 to student 100.
football: Student 46 to student 100.
Hockey: Student 1 to Student 70.
Minimum will be 15 nos.
@leonidas. said:@sbharadwaj said:In a school of 100 students, 70 play Hockey, 60 play Tennis and 55 play Football.What is the max number of sudents who play all 3.?And what is the minimum.?Is there any trick.? Plz explainMinimum will be 15.Maximum will be 55. "for maximum: maximize the no. of people who play nothing..i.e. 100-70 = 30 nos. start from 1 onwards...Hockey : student 1 to student 70.Tennis: Student 1 to student 60.football: student 1 to student 55. student who play all three will be 55.""for minimumminimize the no. of people who play nothing..i.e. 100-100 = 0 nos. Tennis: Student 1 to student 60.football: Student 46 to student 100. Hockey: Student 1 to Student 70.Minimum will be 15 nos.
_/\_ Thanks bhai 😃 Looks simple n great..
Now what if in case, each student should play atleast one sport.? Any other trick for this case as well.??
Now what if in case, each student should play atleast one sport.? Any other trick for this case as well.??
@sbharadwaj @leotheking
@leonidas. So the trick goes like this:
For maximum: Find the game which is played by minimum no of players-this value would be maximum.
For minimum-Get sum of any two games and make a subtraction of total no of students.The minimum among those would be the answer
Eg-
(70+60)-100=30
(70+55)-100=25
(60+55)-100=15
So, miminum should be 15.
@sbharadwaj said:In a school of 100 students, 70 play Hockey, 60 play Tennis and 55 play Football. What is the max number of sudents who play all 3.?And what is the minimum.? Is there any trick.? Plz explain
Max will be when the smallest group is a subset of the others. So the 55 people who play football also play hockey and tennis. Hence 55.
For the minimum, note that 70 + 60 + 55 = 185. Since this is less than 200, it is possible to distribute all so that they play only 2 games. Hence the minimum playing all 3 should be 0.
If the total were to be more than 200, the excess people would be the minimum playing all 3. For example if the number of people in each game were 85, 87 and 90, then the maximum would be 85 and the minimum would be (85 + 87 + 90 - 200) = 62.
regards
scrabbler
For the minimum, note that 70 + 60 + 55 = 185. Since this is less than 200, it is possible to distribute all so that they play only 2 games. Hence the minimum playing all 3 should be 0.
If the total were to be more than 200, the excess people would be the minimum playing all 3. For example if the number of people in each game were 85, 87 and 90, then the maximum would be 85 and the minimum would be (85 + 87 + 90 - 200) = 62.
regards
scrabbler
@leonidas. said:minimize the no. of people who play nothing..i.e. 100-100 = 0 nos. Tennis: Student 1 to student 60.football: Student 46 to student 100. Hockey: Student 1 to Student 70.Minimum will be 15 nos.
What if Hockey is student 1 to 40 and 71-100?
@saurabhlumarrai said:@sbharadwaj@leotheking@leonidas. So the trick goes like this:(70+60)-100=30 (70+55)-100=25(60+55)-100=15So, miminum should be 15.
Look at the attached doc please.
regards
scrabbler
regards
scrabbler
@saurabhlumarrai said:@sbharadwaj@leotheking@leonidas. So the trick goes like this:For maximum: Find the game which is played by minimum no of players-this value would be maximum.For minimum-Get sum of any two games and make a subtraction of total no of students.The minimum among those would be the answerEg-(70+60)-100=30 (70+55)-100=25(60+55)-100=15So, miminum should be 15.
thanx very much!
@vbhvgupta said:A pipe can fill a tank in 4 hrs while a leak which is at 1/4th of thr height of the tank can empty upto that part in 2 hrs. If both r operated simultaneously and initiallly the tank is fulll, then when it will be 1/4th fullA.2 B.2 1/3 c. 1 1/2 D. 6hrs
incase anybody needs an explanation
clearly the filling pipe fills 1/4th of the tank every hour(i.e in 4hours it does W work then in one hour it does W/4 of the work) and the emptying pipe empties 3/8th of the tank in one hour(i.e it can empty only (1-1/4)=3/4th of the tank in 2 hours so in 1 hour it emties 3/8th of the tank) so the net work done is 1/4-3/8=-1/8
now it is given that the tank is full so in 2 hours 2*-1/8=-1/4 of the tank gets empty so in 6 hours 3/4th of the tank gets empty leaving 1/4th of the tank filled
clearly the filling pipe fills 1/4th of the tank every hour(i.e in 4hours it does W work then in one hour it does W/4 of the work) and the emptying pipe empties 3/8th of the tank in one hour(i.e it can empty only (1-1/4)=3/4th of the tank in 2 hours so in 1 hour it emties 3/8th of the tank) so the net work done is 1/4-3/8=-1/8
now it is given that the tank is full so in 2 hours 2*-1/8=-1/4 of the tank gets empty so in 6 hours 3/4th of the tank gets empty leaving 1/4th of the tank filled