Pipe A can fill a tank in 4 hours and pipe B can fill it 6 hours. If they are opened on alternate hours and if pipe A is opened first then in how many hours, the tank shall be full ?
A. $${\text{4}}\frac{1}{2}$$ hours
B. $${\text{4}}\frac{2}{3}$$ hours
C. $${\text{3}}\frac{1}{2}$$ hours
D. $${\text{3}}\frac{1}{4}$$ hours
Answer: Option B
Solution(By Examveda Team)
A → 4 hoursB → 6 hours
According to question,
⇒ For the first hour tap A is opened and B for second hour
⇒ Work done by both in 2 hours
$$\eqalign{ & \to \left( {3\,{\text{lit/h}} + 2\,{\text{lit/h}}} \right) \times 2 = 10\,{\text{units}} \cr & \,\,\,\,\,\,\,\,\mathop {\,\,|\,\,\, \times 2}\limits_{{\text{4 hours}}}^{{\text{2 hours}}} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\mathop {\,\,|\,\,\, \times 2}\limits_{{\text{10 litres}}}^{{\text{5 liters}}\,} \cr} $$
⇒ Remaining part
= 12 - 10 = 2 liters
⇒ Again 5th hour A will be opened Tap A will fill the 2 liters water with its efficiency = $$\frac{2}{3}$$ hours
⇒ Therefore tank will be filled in
= $$\left( {4 + \frac{2}{3}} \right)$$ hours
= $${\text{4}}\frac{2}{3}$$ hours
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
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