Aptitude - Pipes and Cistern - Discussion

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.

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.

This can be easily solved.
1/A + 1/B + 1/C = 1/6.
1/A + 1/B = 1/6 - 1/C.

Therefore, a+b+c worked for 2hrs + a+b worked for 7hrs ( a+b=1/6 - 1/c) = Total work i.e., 1
i.e., 2(1/6) + 7(1/6 - 1/C ) = 1.
2/ 6 - 7/6 -1 = 1/C.
c = 14.

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.

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.

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.

Part filled in 2 hours= 2/6.

This step is not clear. Please explain.

Part filled in 2 hours = 2/6 = 1/3. Remaining part = 2/3.
(A + B)'s 7 hours' work = 2/3.
So, (A+B)'s 1 hr work = 2/21,

Therefore C's 1 hour's work
= [(A+B+C)'s 1 hour]-[(A+B)'s 1 hr work].
=(1/6 - 2/21) = 1/14.
C alone can fill the tank in 14 hours.

A+B+C=6hr-------6ltr TC------1ltr/he fill tank.
A+B+C can run 2hr only - 2 LTR tank fill both.

Remaining part 6 - 2 = 4.

Now C is closed after 2he now remaining part fill A+B in 7hr.

So A+B = fill 4ltr ----7hr , so 1hr---4/7 effi, C efficiency is 3/7, So C can fill the tank Alone 6/3/7 = 14.