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. |

- P can complete a work in \(5\) days and Q can complete the same work in \(10\) days, while R can completely destroy the work in \(20\) days. If they start working at the same time, then in how many days will the work be completed?
- \(1 \ day\)
- \(2 \ days\)
- \(3 \ days\)
- \(4 \ days\)

Answer: (d) \(4 \ days\)

Solution: The part of work completed by P, Q and R in one day $$ = \frac{1}{5} + \frac{1}{10} - \frac{1}{20} $$ $$ = \frac{1}{4} $$ Hence, the work will be completed in \(4\) days

Solution: The part of work completed by P, Q and R in one day $$ = \frac{1}{5} + \frac{1}{10} - \frac{1}{20} $$ $$ = \frac{1}{4} $$ Hence, the work will be completed in \(4\) days

- A child is trying to reach at the top of a tree which is \(53\) meters of height. Child can climb upto \(5\) meters in first hour but climb down to \(1\) meter in next hour. Find the time required by the child to reach the top of the tree?
- \(22 \ hours\)
- \(25 \ hours\)
- \(28 \ hours\)
- \(30 \ hours\)

Answer: (b) \(25 \ hours\)

Solution: The child climbing up \(5\) meters in first hour and climbing down \(1\) meter in next hour, then

work done by the child will be \(4\) meters in \(2\) hours

so the child can be climbed \(48\) meters in \(24\) hours

Hence, the child climbed last five meters in one hour then, total time required by the child to climb the tree = \(24 + 1\) = \(25\) hours

Solution: The child climbing up \(5\) meters in first hour and climbing down \(1\) meter in next hour, then

work done by the child will be \(4\) meters in \(2\) hours

so the child can be climbed \(48\) meters in \(24\) hours

Hence, the child climbed last five meters in one hour then, total time required by the child to climb the tree = \(24 + 1\) = \(25\) hours

- A boy is trying to reach at the top of a plateform which is \(25\) meters of height. boy can climb upto \(5\) meters in first hour but climb down to \(3\) meter in next hour. Find the time required by the boy to reach the top of the plateform?
- \(21 \ hours\)
- \(26 \ hours\)
- \(27 \ hours\)
- \(29 \ hours\)

Answer: (a) \(21 \ hours\)

Solution: The boy climbing up \(5\) meters in first hour and climbing down \(3\) meter in next hour, then

work done by the boy will be \(2\) meters in \(2\) hours

so the boy can be climbed \(20\) meters in \(20\) hours

Hence, he climbed last five meters in one hour then, total time required by the child to climb the tree = \(20 + 1\) = \(21\) hours

Solution: The boy climbing up \(5\) meters in first hour and climbing down \(3\) meter in next hour, then

work done by the boy will be \(2\) meters in \(2\) hours

so the boy can be climbed \(20\) meters in \(20\) hours

Hence, he climbed last five meters in one hour then, total time required by the child to climb the tree = \(20 + 1\) = \(21\) hours

- A pipe can fill the tank in \(6\) hours and pipe B can empty the tank in \(12\) hours. Find the time required to fill the tank, if both pipes are running simultaneously?
- \(10 \ hours\)
- \(12 \ hours\)
- \(15 \ hours\)
- \(16 \ hours\)

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

Solution: Part of the tank filled in one hour $$ = \frac{1}{6} - \frac{1}{12} $$ $$ = \frac{1}{12} $$ Hence, the tank will be filled in \(12\) hours

Solution: Part of the tank filled in one hour $$ = \frac{1}{6} - \frac{1}{12} $$ $$ = \frac{1}{12} $$ Hence, the tank will be filled in \(12\) hours

- Pipe P can fill the tank in \(8\) hours and pipe Q can empty the tank in \(10\) hours. Find the time required to fill the tank, if both pipes are running simultaneously?
- \(20 \ hours\)
- \(30 \ hours\)
- \(40 \ hours\)
- \(50 \ hours\)

Answer: (c) \(40 \ hours\)

Solution: Part of the tank filled in one hour $$ = \frac{1}{8} - \frac{1}{10} $$ $$ = \frac{1}{40} $$ Hence, the tank will be filled in \(40\) hours

Solution: Part of the tank filled in one hour $$ = \frac{1}{8} - \frac{1}{10} $$ $$ = \frac{1}{40} $$ Hence, the tank will be filled in \(40\) hours

- Vishal can build a wall in \(15\) days and Rohan can build the same wall in \(20\) days. If they start working together, then in how many days they will build the wall?
- \(\frac{60}{7} \ days\)
- \(\frac{7}{60} \ days\)
- \(\frac{60}{9} \ days\)
- \(\frac{9}{60} \ days\)

Answer: (a) \(\frac{60}{7} \ days\)

Solution: Both can build the part of wall in one day $$ = \frac{1}{15} + \frac{1}{20} $$ $$ = \frac{7}{60} $$ Hence, they can build the wall in \(\frac{60}{7}\) days

Solution: Both can build the part of wall in one day $$ = \frac{1}{15} + \frac{1}{20} $$ $$ = \frac{7}{60} $$ Hence, they can build the wall in \(\frac{60}{7}\) days

- A is twice as efficient as B, and finish the work \(10\) days earlier than B. Find number of days required to finish the work by B?
- \(20 \ days\)
- \(26 \ days\)
- \(25 \ days\)
- \(28 \ days\)

Answer: (a) \(20 \ days\)

Solution: Let A requires \(x\) days and B requires \(2x\) days to finish the work, then $$ 2x - x = 10 $$ $$ x = 10 \ days $$ $$ 2x = 20 \ days $$ Hence, B requires \(20\) days to finish the work.

Solution: Let A requires \(x\) days and B requires \(2x\) days to finish the work, then $$ 2x - x = 10 $$ $$ x = 10 \ days $$ $$ 2x = 20 \ days $$ Hence, B requires \(20\) days to finish the work.

- A and B together can do a piece of work in \(20\) days and A alone can do it in \(30\) days. In how many days can B do it alone?
- \(40 \ days\)
- \(50 \ days\)
- \(60 \ days\)
- \(70 \ days\)

Answer: (c) \(60 \ days\)

Solution: both can do part of work in one day = \(\frac{1}{20}\)

A alone can do the part of work in one day = \(\frac{1}{30}\)

then B alone can do the part of work in one day $$ = \frac{1}{20} - \frac{1}{30} $$ $$ = \frac{1}{60} $$ Hence, b alone can complete the work in \(60\) days.

Solution: both can do part of work in one day = \(\frac{1}{20}\)

A alone can do the part of work in one day = \(\frac{1}{30}\)

then B alone can do the part of work in one day $$ = \frac{1}{20} - \frac{1}{30} $$ $$ = \frac{1}{60} $$ Hence, b alone can complete the work in \(60\) days.

- \(3\) men can do a work in \(4\) days. How many men are needed to do the work in \(6\) days?
- \(2 \ men\)
- \(3 \ men\)
- \(4 \ men\)
- \(5 \ men\)

Answer: (a) \(2 \ men\)

Solution: Given, \(3\) men can do the work in \(4\) days, so $$ 3 \times 4 = 12 $$ then $$ Man \times day = 12 $$ $$ M \times 6 = 12 $$ $$ M = 2 $$ Hence, \(2\) men are needed to do the work in \(6\) days.

Solution: Given, \(3\) men can do the work in \(4\) days, so $$ 3 \times 4 = 12 $$ then $$ Man \times day = 12 $$ $$ M \times 6 = 12 $$ $$ M = 2 $$ Hence, \(2\) men are needed to do the work in \(6\) days.

- \(5\) men can do a work in \(10\) days. How many men are needed to do the work in \(20\) days?
- \(10 \ days\)
- \(20 \ days\)
- \(30 \ days\)
- \(40 \ days\)

Answer: (b) \(20 \ days\)

Solution: Given, \(5\) men can do the work in \(10\) days, so $$ 5 \times 10 = 50 $$ then $$ Man \times day = 50 $$ $$ M \times 20 = 50 $$ $$ M = 2.5 $$ Hence, \(2.5\) men are needed to do the work in \(20\) days.

Solution: Given, \(5\) men can do the work in \(10\) days, so $$ 5 \times 10 = 50 $$ then $$ Man \times day = 50 $$ $$ M \times 20 = 50 $$ $$ M = 2.5 $$ Hence, \(2.5\) men are needed to do the work in \(20\) days.

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