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Answer (Detailed Solution Below) Option 4 : 1 hour 15 minutes
10 Qs. 10 Marks 7 Mins
Two pipes can fill a cistern in 2 hours and 3 hours, while a third pipe can drain the cistern empty in 6 hours, When three pipes opened then their 1 hour's work = ½ + 1/3 - 1/6 = 2/3 Total cistern full = 1/6 ⇒ Remaining part to filled = 1 - 1/6 = 5/6 ⇒ time taken by three to fill 5/6th of the cistern is = (5/6) / (3/2) = 1.25 ∴ It will take 1 hour and 15 minutes to fill the cistern.India’s #1 Learning Platform Start Complete Exam Preparation
Video Lessons & PDF Notes Trusted by 3,00,16,304+ Students Answer: Option B Explanation: Solution 1Pipe A alone can fill the cistern in $37\dfrac{1}{2}=\dfrac{75}{2}$ minutes. Since it was open for $30$ minutes, part of the cistern filled by pipe A$=\dfrac{2}{75}×30=\dfrac{4}{5}$So the remaining $\dfrac{1}{5}$ part is filled by pipe B.Pipe B can fill the cistern in 45 minutes. So, time required to fill $\dfrac{1}{5}$ part$=\dfrac{45}{5}=9$ minutes. i.e., pipe B is turned off after 9 minutes. Solution 2Part filled by pipe A in 1 minute $=\dfrac{2}{75}$Part filled by pipe B in 1 minute $=\dfrac{1}{45}$Suppose pipe B is closed after $x$ minutes. Then, $\dfrac{2}{75}×30+\dfrac{1}{45}×x=1\\\dfrac{4}{5}+\dfrac{x}{45}=1\\x=9$ Solution 3LCM$\left(37\dfrac{1}{2},45\right)=225$Let capacity of the cistern $=225$ litre.Quantity filled by pipe A in $1$ min $=\dfrac{225}{37.5}=6$ litre.Quantity filled by pipe B in $1$ min $=\dfrac{225}{45}=5$ litre.In $30$ minute, pipe A fills $6×30=180$ litre.Remaining quantity of $225-180=45$ litre is filled by pipe B.Time taken for this $=\dfrac{45}{5}=9$ minutes. Therefore, pipe B is turned off after 9 minutes. Solution 4Pipe A can fill the cistern in $37\dfrac{1}{2}$ minutes $=\dfrac{75}{2}$ minutes.=> Part filled by pipe A in 1 minute $=\dfrac{2}{75}$Pipe B can fill the cistern in 45 minutes=> Part filled by pipe B in 1 minute $=\dfrac{1}{45}$Part filled by Pipe A and B together in 1 minute$=\dfrac{2}{75}+\dfrac{1}{45}=\dfrac{6+5}{225}=\dfrac{11}{225}$Assume that B is turned off after $x$ minutes. i.e., for $x$ minutes, both pipe A and B were open.Part filled by Pipe A and B together in $x$ minutes$=x×\dfrac{11}{225}=\dfrac{11x}{225}$Now, the cistern must be filled in $(30-x)$ minutes by pipe A alone.Part filled in $(30-x)$ minutes by pipe A alone$=(30-x)× \dfrac{2}{75}=\dfrac{2(30-x)}{75}$ $\dfrac{11x}{225}+\dfrac{2(30-x)}{75}=1\\\Rightarrow 11x+6(30-x)=225\\\Rightarrow 11x+180-6x=225\\\Rightarrow 5x=45\\x=9$
Exercise :: Pipes and Cistern - General Questions
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Exercise :: Pipes and Cistern - General Questions
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Two pipes can fill a cistern in 19 and 8 minutes respectively
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