Aptitude - Chain Rule - Discussion
Discussion Forum : Chain Rule - General Questions (Q.No. 1)
1.
3 pumps, working 8 hours a day, can empty a tank in 2 days. How many hours a day must 4 pumps work to empty the tank in 1 day?
Answer: Option
Explanation:
Let the required number of working hours per day be x.
More pumps, Less working hours per day (Indirect Proportion)
Less days, More working hours per day (Indirect Proportion)
| Pumps | 4 | : | 3 |  |
:: 8 : x |
| Days | 1 | : | 2 |
4 x 1 x x = 3 x 2 x 8
 x = |
(3 x 2 x 8) |
| (4) |
x = 12.
Discussion:
69 comments Page 4 of 7.
Vivek said:
1 decade ago
One pump requires 16 *3 = 48 hrs to empty the tank.
4 pump requires 48/4 = 12 hrs.
4 pump requires 48/4 = 12 hrs.
Lakshmi said:
1 decade ago
3 pumps -> 8h -> 2 days.
4 pumps -> ? -> 1 day.
3*8*2/4*1 = 12.
4 pumps -> ? -> 1 day.
3*8*2/4*1 = 12.
Shubhank said:
1 decade ago
Can anyone just explain the chain rule with direct and indirect proportion?
Naveen moorthy said:
1 decade ago
As tank are Same.
So 3x8=24 hours required for 1 day for that tank.
So hrs will be same.
For 4 tanks, 4x6=24 for 2 days.
If it 1 day, 6x2=12 hours.
So 3x8=24 hours required for 1 day for that tank.
So hrs will be same.
For 4 tanks, 4x6=24 for 2 days.
If it 1 day, 6x2=12 hours.
Hanumanth said:
1 decade ago
pumps*days*hours=pumps*days*hours.
4*1*x=3*8*2.
x=12.
That's simple
4*1*x=3*8*2.
x=12.
That's simple
Simple said:
1 decade ago
(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.
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.
Ashish and Amartya said:
1 decade ago
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.
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.
Aara said:
1 decade ago
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.
(Machine (1) x Time) /Work = (Machine (2) x Time) /Work.
Ravi teja P said:
1 decade ago
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.
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.
(1)
Kartik said:
1 decade ago
Thanks to mathmaster.
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