Questions and Answers Type: | MCQ (Multiple Choice Questions). |

Main Topic: | Quantitative Aptitude. |

Quantitative Aptitude Sub-topic: | Time and Work Aptitude Questions and Answers. |

Number of Questions: | 10 Questions with Solutions. |

- A pipe can fill the tank in \(10\) hours and pipe B can empty the tank in \(15\) hours. Find the time required to fill the tank, if both pipes are running simultaneously?
- \(10 \ hours\)
- \(20 \ hours\)
- \(30 \ hours\)
- \(40 \ hours\)

Answer: (c) \(30 \ hours\)

Solution: Part of the tank filled in one hour $$ = \frac{1}{10} - \frac{1}{15} $$ $$ = \frac{1}{30} $$ Hence, the tank will be filled in \(30\) hours

Solution: Part of the tank filled in one hour $$ = \frac{1}{10} - \frac{1}{15} $$ $$ = \frac{1}{30} $$ Hence, the tank will be filled in \(30\) hours

- A cistern is normally filled in \(10\) hours, but it takes \(12\) hours, when there is a leak in its bottom. If cistern is full, in what time shall the leak empty the cistern?
- \(30 \ hours\)
- \(40 \ hours\)
- \(50 \ hours\)
- \(60 \ hours\)

Answer: (d) \(60 \ hours\)

Solution: Let leak can empty the cistern in \(x\) hours, then $$ \frac{1}{10} - \frac{1}{x} = \frac{1}{12} $$ $$ \frac{1}{10} - \frac{1}{12} = \frac{1}{x} $$ $$ \frac{1}{60} = \frac{1}{x} $$ Hence, leak can empty the tank in \(60\) hours.

Solution: Let leak can empty the cistern in \(x\) hours, then $$ \frac{1}{10} - \frac{1}{x} = \frac{1}{12} $$ $$ \frac{1}{10} - \frac{1}{12} = \frac{1}{x} $$ $$ \frac{1}{60} = \frac{1}{x} $$ Hence, leak can empty the tank in \(60\) hours.

- If pipe M can fill the tank in \(6\) hours and pipe N can empty the tank in \(8\) hours, then find how many hours will it take to fill a half empty tank?
- \(10 \ hours\)
- \(12 \ hours\)
- \(15 \ hours\)
- \(18 \ hours\)

Answer: (b) \(12 \ hours\)

Solution: part of the tank filled in one hour $$ = \frac{1}{6} - \frac{1}{8} $$ $$ = \frac{1}{24} $$ Hence, full tank will be filled in \(24\) hours and half tank will be filled in \(12\) hours.

Solution: part of the tank filled in one hour $$ = \frac{1}{6} - \frac{1}{8} $$ $$ = \frac{1}{24} $$ Hence, full tank will be filled in \(24\) hours and half tank will be filled in \(12\) hours.

- Two taps P and Q can fill a tank in \(5\) hours and \(10\) hours, respectively. If both taps are opened together, then find the time required to fill the tank?
- \(\frac{10}{3} \ hours\)
- \(\frac{3}{10} \ hours\)
- \(\frac{12}{5} \ hours\)
- \(\frac{5}{12} \ hours\)

Answer: (a) \(\frac{10}{3} \ hours\)

Solution: net part of the tank filled in one hour $$ = \frac{1}{5} + \frac{1}{10} $$ $$ = \frac{3}{10} $$ Hence, tank will be filled in \(\frac{10}{3}\) hours.

Solution: net part of the tank filled in one hour $$ = \frac{1}{5} + \frac{1}{10} $$ $$ = \frac{3}{10} $$ Hence, tank will be filled in \(\frac{10}{3}\) hours.

- One tap can fill a tank in \(5\) hours and another tap can empty the tank in \(6\) hours. How long will they take to fill the tank, if both the taps are opened?
- \(10 \ hours\)
- \(20 \ hours\)
- \(30 \ hours\)
- \(40 \ hours\)

Answer: (c) \(30 \ hours\)

Solution: net part of the tank filled in one hour $$ = \frac{1}{5} - \frac{1}{6} $$ $$ = \frac{1}{30} $$ Hence, tank will be filled in \(30\) hours.

Solution: net part of the tank filled in one hour $$ = \frac{1}{5} - \frac{1}{6} $$ $$ = \frac{1}{30} $$ Hence, tank will be filled in \(30\) hours.

- Two inlet pipes can fill the tank in \(5\) hours and \(10\) hours respectively, while two outlet pipes can empty the full tank in \(15\) and \(20\) hours, respectively. If pipes are running simultaneously, then find the time required to fill the tank?
- \(\frac{60}{11} \ hours\)
- \(\frac{11}{60} \ hours\)
- \(\frac{65}{12} \ hours\)
- \(\frac{12}{65} \ hours\)

Answer: (a) \(\frac{60}{11} \ hours\)

Solution: Given, \(x_1 = 5\), \(x_2 = 10\), \(y_1 = 15\) and \(y_2 = 20\)

then net part of the tank filled in one hour $$ = \left(\frac{1}{5} + \frac{1}{10}\right) - \left(\frac{1}{15} + \frac{1}{20}\right) $$ $$ = \frac{3}{10} - \frac{7}{60} $$ $$ = \frac{11}{60} $$ Hence, the tank will be filled in \(\frac{60}{11}\) hours.

Solution: Given, \(x_1 = 5\), \(x_2 = 10\), \(y_1 = 15\) and \(y_2 = 20\)

then net part of the tank filled in one hour $$ = \left(\frac{1}{5} + \frac{1}{10}\right) - \left(\frac{1}{15} + \frac{1}{20}\right) $$ $$ = \frac{3}{10} - \frac{7}{60} $$ $$ = \frac{11}{60} $$ Hence, the tank will be filled in \(\frac{60}{11}\) hours.

- A cistern is normally filled in \(12\) hours but takes \(3\) hours more to fill because of a leak in its bottom. If the cistern is full, then how many hours the leak will take to empty the tank?
- \(30 \ hours\)
- \(40 \ hours\)
- \(50 \ hours\)
- \(60 \ hours\)

Answer: (d) \(60 \ hours\)

Solution: Let the leak can empty the tank in \(x\) hours, then $$ \frac{1}{12} - \frac{1}{x} = \frac{1}{15} $$ $$ = \frac{1}{12} - \frac{1}{15} = \frac{1}{x} $$ $$ \frac{1}{60} = \frac{1}{x} $$ $$ x = 60 \ hours $$

Solution: Let the leak can empty the tank in \(x\) hours, then $$ \frac{1}{12} - \frac{1}{x} = \frac{1}{15} $$ $$ = \frac{1}{12} - \frac{1}{15} = \frac{1}{x} $$ $$ \frac{1}{60} = \frac{1}{x} $$ $$ x = 60 \ hours $$

- Two pipes can fill a tank in \(4\) hours and \(6\) hours respectively, while a third pipe empties the full tank in \(10\) hours. If all the three pipes operate simultaneously, then how much time it will take to fill the tank?
- \(2.5 \ hours\)
- \(3.1 \ hours\)
- \(4.2 \ hours\)
- \(5.3 \ hours\)

Answer: (b) \(3.1 \ hours\)

Solution: part of the tank filled in one hour $$ = \frac{1}{4} + \frac{1}{6} - \frac{1}{10} $$ $$ = \frac{19}{60} $$ Hence, the tank will be filled in \(\frac{60}{19}\) hours or \(3.1\) hours.

Solution: part of the tank filled in one hour $$ = \frac{1}{4} + \frac{1}{6} - \frac{1}{10} $$ $$ = \frac{19}{60} $$ Hence, the tank will be filled in \(\frac{60}{19}\) hours or \(3.1\) hours.

- One tap can fill a tank in \(6\) hours and another tap can empty the tank in \(8\) hours. How long will they take to fill the tank, if both the taps are opened?
- \(20 \ hours\)
- \(21 \ hours\)
- \(23 \ hours\)
- \(24 \ hours\)

Answer: (d) \(24 \ hours\)

Solution: net part of the tank filled in one hour $$ = \frac{1}{6} - \frac{1}{8} $$ $$ = \frac{1}{24} $$ Hence, tank will be filled in \(24\) hours.

Solution: net part of the tank filled in one hour $$ = \frac{1}{6} - \frac{1}{8} $$ $$ = \frac{1}{24} $$ Hence, tank will be filled in \(24\) hours.

- Two taps A and B can fill a tank in \(2\) hours and \(4\) hours, respectively. If both taps are opened together, then find the time required to fill the tank?
- \(\frac{3}{4} \ hours\)
- \(\frac{4}{3} \ hours\)
- \(\frac{5}{2} \ hours\)
- \(\frac{2}{5} \ hours\)

Answer: (b) \(\frac{4}{3}\)

Solution: net part of the tank filled in one hour $$ = \frac{1}{2} + \frac{1}{4} $$ $$ = \frac{3}{4} $$ Hence, tank will be filled in \(\frac{4}{3}\) hours.

Solution: net part of the tank filled in one hour $$ = \frac{1}{2} + \frac{1}{4} $$ $$ = \frac{3}{4} $$ Hence, tank will be filled in \(\frac{4}{3}\) hours.

Lec 1: Time and Work Case (1) and Case (2)
Exercise-1
Lec 2: Time and Work Case (3) and Case (4)
Exercise-2
Lec 3: Concept of Positive and Negative work
Exercise-3
Lec 4: Concept of Pipes and Cisterns
Exercise-4
Lec 5: Concept of Efficiency
Exercise-5