Two pipes can fill a tank in 20 and 24 minutes respectively and a waste pipe can empty 3 gallons per minute. All the three pipes working together can fill the tank in 15 minutes. The capacity of the tank is:
A. 60 gallons
B. 100 gallons
C. 120 gallons
D. 180 gallons
E. None of these
Answer: Option C
Solution(By Examveda Team)
Work done by the waste pipe in 1 minute$$\eqalign{ & = \frac{1}{{15}} - \left( {\frac{1}{{20}} + \frac{1}{{24}}} \right) \cr & = {\frac{1}{{15}} - \frac{{11}}{{120}}} \cr & = - \frac{1}{{40}}\,\,\,\,\,\left[ { - ve\,{\text{sign}}\,{\text{means}}\,{\text{emptying}}} \right] \cr & \therefore {\text{Volume}}\,{\text{of}}\,\frac{1}{{40}}{\text{part}} = 3\,{\text{gallons}} \cr & {\text{Volume}}\,{\text{of}}\,{\text{whole}} \cr & = \left( {3 \times 40} \right){\text{gallons}} \cr & = 120\,{\text{gallons}} \cr} $$
This Question Belongs to Arithmetic Ability >> Pipes And Cistern
Join The Discussion
Comments ( 3 )
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
Let, capacity is x gallon
(X/20+X/24)*15= 11x/8
Now,11x/8-x=15*3
3x/8=45
X= 120
Add up the work of two pipes i.e. 11/120. Let total work is x then work done by these two pipes in one minute is 11x/110. Now, equation will be
(11x/110)-3=x/15
I didn't understand this plz explain