Three taps A, B and C can fill a tank in 12, 15 and 20 hours respectively. If A is open all the time and B and C are open for one hour each alternately, the tank will be full in:
A. 6 hours
B. $$6\frac{2}{3}$$ hours
C. 7 hours
D. $$7\frac{1}{2}$$ hours
Answer: Option C
Solution(By Examveda Team)
$$\eqalign{ & \left( {{\text{A + B}}} \right){\text{'s 1 hour work}} \cr & {\text{ = }} {\frac{1}{{12}} + \frac{1}{{15}}} = \frac{9}{{60}} = \frac{3}{{20}} \cr & \left( {{\text{A + C}}} \right){\text{'s 1 hour work}} \cr & {\text{ = }} {\frac{1}{{12}} + \frac{1}{{20}}} = \frac{8}{{60}} = \frac{2}{{15}} \cr & {\text{Part filled in 2 hrs}} \cr & {\text{ = }} {\frac{3}{{20}} + \frac{2}{{15}}} = \frac{{17}}{{60}} \cr & {\text{Part filled in 6 hrs}} \cr & {\text{ = }} {3 \times \frac{{17}}{{60}}} = \frac{{17}}{{20}} \cr & {\text{Remaining part}} \cr & {\text{ = }} {1 - \frac{{17}}{{20}}} = \frac{3}{{20}} \cr} $$Now it is the turn of A and B and
$$\frac{3}{{20}}$$ part is filled by A and B in 1 hour
∴ Total time taken to fill tank
= (6 + 1) hrs
= 7 hrs
This Question Belongs to Arithmetic Ability >> Pipes And Cistern
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Related Questions on Pipes and Cistern
A. $$\frac{5}{{11}}$$
B. $$\frac{6}{{11}}$$
C. $$\frac{7}{{11}}$$
D. $$\frac{8}{{11}}$$
A. $$1\frac{{13}}{{17}}$$ hours
B. $$2\frac{8}{{11}}$$ hours
C. $$3\frac{9}{{17}}$$ hours
D. $$4\frac{1}{2}$$ hours
A. $$4\frac{1}{3}$$ hours
B. 7 hours
C. 8 hours
D. 14 hours
17/60 comes as in 2 hours (B, C). 17 again in 2 hours and 17 again in another 2 hours. So totally 51/60 comes in 6 hours.
Remaining 51 + 9 = 60, part will fill in 7 th hour.
6hrs??????
I can't understand from where 6 hrs come? Could anyone help me?
tap 'A' can fill empty water tank in 12 hour while tap 'B' in 15 hour. If both the tap opened together and after 3 hours tap 'B' is closed find is what time tap 'A' will fill the remaining tank.