Aptitude - Pipes and Cistern - Discussion

A+B+C worked together for 2 hrs.

In 1hr they filled up to 1/6 parts thus for 2hrs they fill up to (1/6)*2 i.e.,2/6 parts. Which is 1/3.

Then we need to calculate remaining parts so total-filled parts give remaining part.

i.e., 1-1/3 = 2/3 part are yet remaining.

According to the question now A and B alone opened thus it works for 7 hrs to fill the remaining 2/3 part.

So (A+B)together fill (2/3)*(1/7)parts in 1hr i.e it together fill 2/21 parts in 1hr.

Now we need to calculate the time of C ;

So (A+B+C)'s 1hr work-(A+B)'s 1hr work will give C's 1hr work.

Therefore C's work in 1 HOUR = 1/6 - 2/21 = 1/14 parts.

So time taken by C to fill the tank is 14 hours.

first understand the question properly.

A, B, C all opened can fill in 6 hours.
ie 1/A+1/B+1/C=1/6________(x)
ie 1/6th of tank is filled if all the three operate for an hour.
So if all 3 operate for 2 hours 2/6th of the tank is filled.
So remaining is 2/3rd.(1-2/6).

After 2 hours, C is closed.
to fill the 2/3rd by A & B it takes 7 hours.
[to find the time by c alone we need to find 1/A+1/B to substitute in (x)..]
So A and B in 1 hour can complete 2/(3*7).
ie 1/A+1/B=2/21. sub in x.
1/C=1/6-2/21.
We get 1/C =1/14.
So, C takes 14 hours to fill the tank.

Let's take tank capacity as 42 litres.
Given A+B+C finishes in 6 hours.

Per hour, A+B+C= 42/6 = 7 litres/hour.
But, They worked together for only 2 hours.

For 2 hours,
A+B+C= 7 litres => 2 hours=14litres/2hrs.
Remaining= 28 litres.

(A+B) worked to finish it in 7 hours.
So, (A+B)= 28/7 = 4 litres/hour.

Normally, (A+B)+C= 42/6 = 7 litres/hr.
In this one hour, (A+B) will fill 4 litres.

Remaining 3 litres will be filled by C in 1 hour.
So, C alone = 42/3= 14 hours.

Three pipes can fill a pipe in 6 hours,
(A+B+C) * 6 -----> (1)

Again, 3 workers work for 2 hours altogether so,
(A+B+C) * 2 , But The rest of the work done by A+B together, (A+B) *7
so. (A+B+C) *2 + (A+B) * 7 -----> (2)

ATQ,
(A+B+C) *6 = (A+B+C)*2 + (A+B) * 7.

Solving the equation we will get = C : A+B :: 3 : 4 (put this value in 1 or 2).

putting the value in 1 we will get,

(4+3)*7 = 42, this is the total work.
C = 42/3 = 14 Hours (ans)

It can be done very easily by the LCM method. Let the capacity of cistern be 42 units LCM of 6, 2 , 7. Then according to question.

A+B+C fills 7 units per hour(42/6). hence in 2 hours of operation, they fill 14 units.

Remaining units 42"14 = 28, which A+B fills in 7 hours means A+B fills 4 units per hour. hence it is clear that C fills 3 units per hour. So, C will fill the cisterns i.e. 42 units in 42/3 = 14 Hours.

It can be done very easily by the LCM method. Let the capacity of cistern be 42 units LCM of 6, 2 , 7. Then according to question.

A+B+C fills 7 units per hour(42/6). hence in 2 hours of operation, they fill 14 units.

Remaining units 42"14 = 28, which A+B fills in 7 hours means A+B fills 4 units per hour. hence it is clear that C fills 3 units per hour. So, C will fill the cisterns i.e. 42 units in 42/3 = 14 Hours.

A .+ B + C = 6 hrs
2 hrs = 1/3 work,
remaining = 2/3 work.
A+B, 2/3 work= 7 hrs,
Total work 3/3 i.e 1 = 7 hrs + 7/2 hrs = 21/2 hrs
Now consider LCM of A + B + C & A+B.
i.e 6 & 21/2 is 42
Taking efficiency: A + B + C = 7units
A+B = 4 units.

So C's efficiency= A+ B+C -( A+ B).
= 7-4 = 3.
Total work is LCM i.e 42 & efficiency of C is 3.
So total time= 42/3 = 14.

@Sahil sharma see the options in the answer. No one options are below 6 hours so first we need to find 6 hours work.

We have 2 hours work. Now we need to find out 6 hours work.

We know 2*3=6. So multiply by 3 on RHS to find out 6 hours work. If we wanted to know 8 hours work then we need to do 2*4=8. If 10 hours then 2*5=10 and so on.

In this case, take LCM of 6 and 7, that is 42 and assume total manpower day is 42, the efficiency of A+B+C is 42/6 = 7.

A, B, C did for 2 days that is 7 * 2 = 14 MD complete, remaining 42-14=28 MD done by A+B in 7hr, So, A+B efficiency is 28/7 = 4.

So, C's efficiency is 7 - 4 = 3.

C can complete total work by 42/3=14 days.

Take,total hrs =total work.

A+B+C=6hrs(total hrs)=6 work(total wrk).
A+B +C work for 2hrs so 2work completed.
Remaining work (6-2) is 4.

Remaining wrk completed by A+B in 7hrs.
A+B=4/7 work/hrs.

Therefore, C's efficiency 1-(4/7)=3/7 wrk/hrs.

C-----alone---total work/efficiency.
6/(3/7) = 14 hrs.