I didn't get how you have solved it. Can you explain me this all step by step?
Shaunak said:
(Sun, Nov 28, 2010 01:11:18 AM)
It's simple direct and inverse ratio theory. You can apply unitary method,but that'll take a hell lot of time.
Sivakumar said:
(Tue, Nov 30, 2010 11:04:02 AM)
a:b::c:d=>bc=ad
Math Master said:
(Fri, Dec 31, 2010 06:15:32 AM)
Better you do like this:
3 pumps take 16 hrs total (8 Hrs a day)
If 1 pump will be working then, it will need 16*3=48 hrs
1 pump need 48 Hrs
If I contribute 4 pumps then
48/4=12 hrs.
Anil said:
(Tue, Jan 18, 2011 06:52:38 AM)
That's the better way of explaining.
Mehar said:
(Wed, Feb 2, 2011 04:26:16 AM)
D P H
2 3 8 ==> Fill the tank
1 4 x ==> Have to do empty same tank ,
so work is same then
2*3*8=1*4*x => x=12.
Vishal said:
(Wed, Feb 2, 2011 12:58:10 PM)
3 pumb work in 2 days = 16 hours for empty the tank
1 pump work in 2 days = 16*3 = 48 hours for empty the tank
so 4 pump ...........= (16*3)/4 = 12..
Uttam said:
(Sat, Feb 5, 2011 12:45:24 AM)
@ Math Master
Your approach is good, but it dint use the case that pumps should empty the tank within 1 day, is it doesn't make any difference?
Miss Mathphobia said:
(Wed, Mar 30, 2011 09:47:01 AM)
I liked MathMaster's method...quite simple and easy 2 understand...
If 38 men , working 6 hours a day ,can do a piece of work in 12 days , find the number of days in which 57 men working 8 hours a day, can do twice that piece of work, supposing that two men of the first set do as much work in 1 hour as 3 men of the secand set do in 3/2 hours.
Umesh said:
(Thu, Jul 28, 2011 03:59:56 PM)
Total work (Time to empty the tank by one pump) = 3(pumps)* 8(hours)*2(days)
= 42hours
When 4 pumps are using it will take 42/4= 12 hours
Anamika said:
(Mon, Aug 8, 2011 06:52:36 PM)
Math Master's solution is really easy to uderstand
Cauverypithan said:
(Mon, Aug 8, 2011 10:30:21 PM)
Math master method is wrong if 3 pumps works a day for 16 hour and one pump will 16/3 not 16*3.
Naresh said:
(Mon, Aug 15, 2011 07:59:29 PM)
Thanks mehar.
Hiscan said:
(Sun, Aug 28, 2011 06:48:03 PM)
@Cauverypithan
Math master is rite....coz 3 pumps will empty in 16hrs....
if they r using one pump,then work load ll increase,then 48hrs is rite....
still not clear...
then c...3 pump ll empty one tank in 16hr...so more the no of pump faster it gets emptied,
lesser the no of pump slower it gets emptied...
Deepika said:
(Sat, Sep 17, 2011 07:43:18 PM)
Thanks for all for there short cuts.
Imraz said:
(Sun, Sep 25, 2011 11:44:22 PM)
3 pumps 8 hours 2 days =work
3*8*2=work----->eqn1
now to complete the same work with 4 pumps in 1 day how many hours are required? lets take the hours required as x. then we can write the same work as
4 pump x hours 1 day =work
4*x*1=work ---->eqn2
equating both equation
3*8*2=4*x*1
x=48/4
x=12 hours
Raja said:
(Wed, Oct 5, 2011 05:26:43 PM)
@maher. I am just asking where is the concept of chain rule (indirect propprtion) approach.
Sudheer said:
(Wed, Oct 5, 2011 07:45:29 PM)
Imraz said well.
Yogendra said:
(Thu, Dec 15, 2011 12:28:17 PM)
Thanks imraz your method is very easy to understand.
Maths Lover said:
(Thu, Dec 22, 2011 10:41:38 PM)
@math master and all.
Your approach is right because fortunately question is asking of 1 day. If not. Then we have to divide the answer by number of days. So people take care of that also.
Kuntal Paul said:
(Wed, Dec 28, 2011 12:16:28 PM)
Mathmaster method is so simple and effective.
Ram Chowdary M said:
(Mon, Feb 6, 2012 06:10:09 PM)
Math master really thank you sir your said easy and simple method.
Hari said:
(Fri, Mar 9, 2012 12:44:28 AM)
Yes it's very easy to understand solved by Mathmaster. ThanQ.
I can't understand your question Mr. Akhilesh sahu.
Sanjit said:
(Sun, Mar 18, 2012 10:48:51 AM)
As a lay man I liked the DPH formula of meher.
Thanks.
Sandeep said:
(Sun, Apr 1, 2012 12:08:21 PM)
GIVEN: 3 pumps' 1 hour's work= 1/16
SOLUTION : 1 pump's 1 hour's work= 1/48
4 pumps' 1 hour's work= (1/48)*4 = 1/12
hence 4 pumps will finish the work in 12 hours
Kartik said:
(Wed, May 23, 2012 03:21:13 PM)
Thanks to mathmaster.
Ravi Teja P said:
(Sun, Jun 17, 2012 12:07:01 AM)
Let pumps - 'p', no.of hrs/day - 'h', no.of days - 'd'.
For this type of problems, first know what we have to find? here we have to find hrs/day i.e. 'h'.
Now, find out the relation of 'h' with other parameters 'p' and 'd'.
If no.of pumps increases, no.of hrs/day decreases (inverse proportion).
p & (1/h). (1) **Assume &-proportionality symbol**.
If no.of days increases, no.of hrs/day decreases (inverse proportion).
d & (1/h). (2).
From (1) & (2).
pd & (1/h). (3).
=> pdh=constant.
=> p1*d1*h1=p2*d2*h2.
=> 3*2*8=4*1*x.
=> x=12.
Aara said:
(Thu, Oct 11, 2012 11:30:42 PM)
Why this formula given below does not work with this question but works with other such similar questions?
(Machine (1) x Time) /Work = (Machine (2) x Time) /Work.
Ashish And Amartya said:
(Sat, Oct 13, 2012 05:58:48 PM)
We found a new method.
step 1- calc the total time for 3 pumps ie(2*8)
2- now for 1 pump time=(2*8)*3
3- " " 4 pump time = ((2*8)*3)/4..ans.
Simple said:
(Sun, Oct 21, 2012 05:01:11 AM)
(Machine (1) x Time) /Work = (Machine (2) x Time) /Work.
This formula works here also
3 * 8(hrs)* 2(day) /1 =4 * x(hrs) * 1(day)/1
==> x= 12
Here work done is 1 since full tank is to be emptied.
Hanumanth said:
(Sat, Dec 15, 2012 11:10:17 AM)
pumps*days*hours=pumps*days*hours.
4*1*x=3*8*2.
x=12.
That's simple
Naveen Moorthy said:
(Wed, Feb 27, 2013 03:46:21 PM)
4 x 1 x x = 3 x 2 x 8
x =