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?
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:
76 comments Page 8 of 8.
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.
Shubha sahu said:
1 decade ago
Please tell me if there is any shortcut to solve this questions?
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.
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.
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.
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.
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
= 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
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
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
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