Aptitude - Pipes and Cistern - Discussion

A in one hour = 1/12 part,
B in one hour = 1/15 part,
C in one hour = 1/20 part,

Now, A is opened continuously and B and C one hour alternately. So, 1st hour A+B, then for the next hour A+C.
A+B in one hour= 1/12 + 1/15= 3/20 parts,
A+C in one hour= 1/12 +1/20= 2/15 parts,
Now, in the first 2 hours ( ie; A+B for first hour and A+C for second hour)= 3/20 + 2/15 parts= 17/60 parts.

For the tank to be filled the part must become 60/60.
We know, 17 * 3 = 51 (which is the multiple less than 60),
So, 2 * 3 hours = 6 hours.
In 6 hours, part filled is 51/60. Now, simplifying it we get 17/20.
17/20 part is filled, so the rest part to be filled is 1-17/20= 3/20.
A+B in first hour.
A+C in second hour.
A+B in third hour.
A+C in fourth hour.
A+B in fifth hour.
A+C in sixth hour.
Now, it is the turn for A+B.
A+B can fill the rest in 1 hour, since part filled by A+B in one hour is 3/20.
So, the total time taken is 6+1 hours= 7 hours.

Hopes all got it.

Please carefully read the question A is open for all times, B and C are opened alternatively for one hour, that means if B is opened for 1st hour that time C is closed(A and B filling the tank) For 2nd hour B is closed C is opened(A and C is Filling the tank)3rd hour B opened C closed and so on...

1st hour:

1 hour work of A and B is 1/12+1/15 = 3/20.

2nd hour:

1 hour work of A and C is 1/12+1/20 = 2/15.

3rd hour:

1 hour work of A and B is 1/12+1/15 = 3/20.

4th hour:

1 hour work of A and C is 1/12+1/20 = 2/15.

5th hour:

1 hour work of A and B is 1/12+1/15 = 3/20.

6th hour:

1 hour work of A and C is 1/12+1/20 = 2/15.

7th hour:

1 hour work of A and B is 1/12+1/15 = 3/20.

(Until you have to calculate 60/60).

If we add the whole we will get 1 is the answer that is capacity of tank.

3/20+2/15+3/20+2/15+3/20+2/15+3/20+2/15+3/20 = (9+8+9+8+9+8+9)/60 = 1.

Hence, time taken to fill the tank is 7 hours.

Consider the total capacity = X Litre
Amount filled in 1h by tank "A" = (X/12) Litre
Amount filled in 1h by tank "B" = (X/15) Litre
Amount filled in 1h by tank "C" = (X/20) Litre.

Now the problem says that the tank "A" is kept open and tanks "B" and "C" are opened alternatively for 1h until the tank gets full.

For the first 1 hour, the amount that both "A" and "B" can fill collectively = (X/12)+(X/15)=(27X/180)=(3X/20).

For the next 1hour the amount that both "A" and "C" can fill collectively = (X/12)+(X/20)= (32X/240)=(2X/15).

Total amount of water after 2 hours = {(3X/20)+(2X/15)} = 17X/60.
Now, this amount (17X/60), corresponding to 2hrs, repeats as a unit.
Since the max.amount (Capacity) = X Litre
(17X/60) * n = X
n = 3.5.
17X/60 Litres of 2hrs repeats for 3.5 times which means 3.5*2 = 7 hrs.

If you take more than 6hrs negative value so tank overflows since no leakage thats why we took upto 6hrs


Why 2 hours are taken first and then 6 is because of easy calculation it is not compulsory take like this since alternative filling


(A + B)'s 1 hour's work= 1/12+1/15=3/20

(A + C)'s hour's work =1/12+1/20=2/15


For 2hrs 17/60

so for 1hr take half from 2hrs i.e 17/60*1/2=17/120

now for 2hrs =17/120+17/120(2*17/120)

for 3hrs=17/120+17/120+120(3*17/120)

for 4hrs=4*17/120

for 5hrs=5*17/120

for 6hrs=6*17/120=17/20 right?

for 7hrs=7*17/120...................


Remaining part=(1-17/20)=3/20
suppose if you take for 7hrs gets negative value total time will be more than time required to fill the tank

Therefore from above part is filled by A and B in 1 hour is 3/20 and the ramaining part also 3/20 that means remaining part takes 1hr

Hence total time = 6hrs + 1hr = 7hrs.

I have solved it my way. First take the lcm of 12, 15 and 20 which is 60. So we assume the total capacity of tank to b 60 units.

That means A fills 5, B fills 4 and C fills 3 in 1 hr.
For the first hr A and B are opened so tank filled is 9 (5+4).
For the second hr A and C are opened so tank filled in 2nd hr is 8 (5+3).
So total tank filled in 2 hrs is 9+8=17. So in 6hrs 51 unit of tank is filled.

Now only 9 units of tank is to be filled which will b filled in next hr which is 7th hr and in this hr B will b opened. So in 7th hr 5 (By A) +4 (By B) =9unit will be filled which is sufficient to fill the remaining tank.

Hene total tank will be filled in 7 hrs. Hope you all got this.

For the first hour, part filled from the tank (1\12)+(1\15) = 3\20.

Second (1\12)+(3\20) = 2\15.

Third (1\12)+(1\15) = 3\20.

Fourth (1\12)+(1\15) = 2\15.

Fifth (1\12)+(1\15) = 3\20.

Lets see how much the tank full in 5 hours.

3*(3\20)+ 2*(2\15) = 43\60 So the tank is not full yet.

For the sixth hour ,part filled from the tank (1\12)+(3\20) = 2\15.

Let's see how much the tank full in 6 hour.

3*(3\20)+3*(1\15) = 17\20 so the tank is not full yet.

For the seventh hour, part filled from the tank(1\12)+(2\15) = 3\20.
Let's see how much the tank full in 7 hour.

4*(3\20)+3*(2\15) = 1 so the tank is completely full after 7 hour.

@All.

Just take a common multiple of these no. (12,15,20) ---> 60(and consider it as tanks total capacity).

Now, divide this capacity with each of these to get the rate of work of each tap
tap A ---> 60/12 --> 5 units /hr.
tap B ---> 60/15 --> 4 units /hr.
tap C ---> 60/20 ---> 3 units /hr.

Hence we can say --
A+B can fill 9 units/hr and A+C can fill 8 units /hr.

Since we are given that B and c are to be opened alternatively and all is opened for all the time.
we can do
(A+B)+(A+C)+(A+B)+........till we get their sum as 60(which is our tanks capacity)
9+ 8 + 9 + 8 +9 + 8 + 9 = > 60units.
i.e 7 hrs.

1 hr tank can fill 3/20 -> (1/12+1/15).

2nd hr tank can fill 3/20+2/15 = 17/60.

For 2 hr it fills 17/60 then for 1 hr it fills 17/120 -> (1/2*17/60).

Consider 17 table 17*6 = 102, so in six hours it it fills 102/120 and next hr B Pipe will be open and.

Fills as 102/120+3/20 = 1. So 7 hrs it will take.

Another way:

For 2 hrs it fills 17/60.

Consider next 2 hrs i.e. 4 hrs 2*17/60 = 34/60.

Consider next 2 hrs i.e 6 hrs 51/60 only 9 is less for 60 i.e to full.

So next 1 hr b pipe open 51/60+9/60 = 1.

I HAVE TRIED TO SOLVE IT IN MY OWN WAY,PZ REPLY WHETHER ITS IS CORRECT OR NOT

WORK DONE
A IN 1 HOUR = 1\12
B IN 1 HOUR = 1\15
C IN 1 HOUR = 1\20

SINCE A WORKS ALL THE TIME,B AND C WORKS ALTERNATELY
HOURS: 1 2 3 4 5 6 7
PERSONS (A,B) + (A,C) + (A,B) + (A,C)+ (A,B) + (A,C) + (A,B)
VALUE 3/20 2/15 3/20 2/15 3/20 2/ 15 3/20

WATCH CAREFULLY:WHEN U ADD THIS DENOMINATOR COMES TO 60 AND NUMERATOR ALSO TO 60 ie 1 ,SO TANK IS FULL IN 7 HOURS

Taking the lcm(12,15,20) = 60 liters (capacity of the tank).
In 1 hr : A can fill 5 liters(60/12), B = 4 liters C = 3 liters.
It's given that A is open all time and B & C alternately.
So in 1st hr : A+B = 9 liters.
2nd hr : A+C = 8 liters.
3rd hr : A+B = 9 liters.
4th hr : A+C = 8 liters.
5th hr : A+B = 9 liters.
6th hr : A+C = 8 liters.
7th hr : A+B = 9 liters.
Total = 60 liters (Capacity of the tank).
So, Ans = C. 7 hours.