Aptitude - Pipes and Cistern - Discussion

Discussion Forum : Pipes and Cistern - General Questions (Q.No. 1)
1.
Three pipes A, B and C can fill a tank from empty to full in 30 minutes, 20 minutes, and 10 minutes respectively. When the tank is empty, all the three pipes are opened. A, B and C discharge chemical solutions P,Q and R respectively. What is the proportion of the solution R in the liquid in the tank after 3 minutes?
5
11
6
11
7
11
8
11
Answer: Option
Explanation:

Part filled by (A + B + C) in 3 minutes = 3 1 + 1 + 1 = 3 x 11 = 11 .
30 20 10 60 20

Part filled by C in 3 minutes = 3 .
10

Required ratio = 3 x 20 = 6 .
10 11 11

Discussion:
72 comments Page 3 of 8.

Sandhiya said:   1 decade ago
Pipe A fills the tank in 30 min
in one min pipe A fills 1/30 of the tank
pipe B fills the tank in 20 min
in one min pipe B fills 1/20 of the tank
pipe C fills the tank in 10 min
in one min pipe C fills 1/10 of the tank
part of the tank filled when all the three pipes are opened for one min is {(1/30)+(1/20)+(1/10)}=11/60
the part of the tank filled when all three pipes are opened for
three min is =3*(11/60)=33/60=11/20
now the chemical solution R present after 3 min=3*(part of the tank filled by C for one min)
=3*1/10=3/10
propotion of sol R in the tank after 3min=(chemical soln r in the tank after 3 min)/(total chemical soln in the tank)
=(3/10)/(11/20)=(3*20)/(11*10)=60/110=6/11
hence the ans is 6/11

Kiran said:   1 decade ago
@Lalitha

= 1/30+1/20+1/10
So L.C.M of 30,20,10=60
Then 1/30+1/20+1/10=2+3+6/60
=11/60
ok

Shrikant said:   1 decade ago
Don't make this type of questions complicated.

I am giving you simple solution which I gives in my class for this type of questions.

Just take LCM of all given values.

30, 20, 10 = 60 (LCM).

Assume that capacity of tank is 60 liters.

Now find the capacity of each pipe of filling the tank by.

* Total capacity of tank in liters / time taken by specific pipe.

1) Pipe A capacity 60 / 30 = 2 liters every min(solution P).
2) Pipe B capacity 60 / 20 = 3 liters every min(solution Q).
3) Pipe c capacity 60 / 10 = 6 liters every min(solution R).

Clearly if all works together than the capacity is 2+3+6 = 11.

Liters per minute. Now out of 11 pipe c (solution R) have 6 liters so in all solution in 1 minute solution are share should 6/11 (don't need to multiply by 3 minutes because the ratio will remain same for 1, 2, 3, 4. Any number of minutes) its very easy to solve any problem of pipe and cistern using this method, it looks lengthy in reading but if you practice I assure you, everyone of you will solve this type of questions in few seconds.
(3)

Niranjan kumar said:   1 decade ago
A - 30 min.
B - 20 min.
C - 10 min.
Now take LCM (30,20,10) = 60.

So,
A - 60/30 = 2 LTR.
B - 3LTR.
C - 6 LTR.

So TOTAL 2+3+6 = 11 LTR.
So proportion of C is 6/11.

So Simple.

Ankit jhunjhunwala said:   1 decade ago
At any time proportion of solution R is same with respect to other two solution,

In 1 minute A will drop =1/30.
" " "" " " B will drop =1/20.
" " " " " C will drop =1/10.

Multiplying 60 to each one,
A:B:C = 2:3:6.
Since solution R corresponds to C we have = 6/(6+3+2) = 6/11.

Shubha sahu said:   1 decade ago
Please tell me if there is any shortcut to solve this questions?

G.Nandhini said:   1 decade ago
Proportion of are in the tank = Amount of are filled by Tap C in 3 min/Total amount of P, Q, are filled by Taps A, B, C for 3 mins.

Sachin said:   1 decade ago
Why divided 3/10 by 11/60?

Anand said:   10 years ago
Percentage filled by A in 3 minute = 100x3/30.....1.

Percentage filled by B in 3 minute = 100x3/20.....2.

Percentage filled by C in 3 minute = 100x3/10.....3.

= 3/(1+2).

Xyz said:   10 years ago
Please give some simple explanation. Please tell me how to calculate LCM? I forgot it.


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