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. |

- M can finish a work in five days and N can finish the same work in \(10\) days, then find how many days are required to finish the work, if both are working together?
- \(3.3 \ days\)
- \(4.4 \ days\)
- \(5.5 \ days\)
- \(6.6 \ days\)

Answer: (a) \(3.3 \ days\)

Solution: Given, \(x = 5 \ days\), \(y = 10 \ days\)

If both M and N start working together then $$ = \frac{xy}{x + y} $$ $$ = \frac{5 \times 10}{5 + 10} $$ $$ = \frac{50}{15} $$ $$ = 3.3 \ days $$ Hence, together they can finish the work in \(3.3 \ days\).

Solution: Given, \(x = 5 \ days\), \(y = 10 \ days\)

If both M and N start working together then $$ = \frac{xy}{x + y} $$ $$ = \frac{5 \times 10}{5 + 10} $$ $$ = \frac{50}{15} $$ $$ = 3.3 \ days $$ Hence, together they can finish the work in \(3.3 \ days\).

- Ram can finish a task in two days and Shyam can finish the same task in five days. If they start working together, then how many days are required to finish the task?
- \(5.2 \ days\)
- \(4.5 \ days\)
- \(2.5 \ days\)
- \(1.4 \ days\)

Answer: (d) \(1.4 \ days\)

Solution: Given, \(x = 2 \ days\), \(y = 5 \ days\)

If both Ram and Shyam start working together then $$ = \frac{xy}{x + y} $$ $$ = \frac{2 \times 5}{2 + 5} $$ $$ = \frac{10}{7} $$ $$ = 1.4 \ days $$ Hence, together they can finish the task in \(1.4 \ days\).

Solution: Given, \(x = 2 \ days\), \(y = 5 \ days\)

If both Ram and Shyam start working together then $$ = \frac{xy}{x + y} $$ $$ = \frac{2 \times 5}{2 + 5} $$ $$ = \frac{10}{7} $$ $$ = 1.4 \ days $$ Hence, together they can finish the task in \(1.4 \ days\).

- A man can finish \(50 \ \%\) of a work in \(10\) days. Find how many days will he take to complete the work five times?
- \(100 \ days\)
- \(110 \ days\)
- \(120 \ days\)
- \(150 \ days\)

Answer: (a) \(100 \ days\)

Solution: The man can finish \(\frac{1}{2}\) part of the work in \(10\) days

then one part of work he can finish $$ = \frac{10}{1/2} $$ $$ = 20 \ days $$ Hence, he will complete five times of work in days $$ = 20 \times 5 $$ $$ = 100 \ days $$

Solution: The man can finish \(\frac{1}{2}\) part of the work in \(10\) days

then one part of work he can finish $$ = \frac{10}{1/2} $$ $$ = 20 \ days $$ Hence, he will complete five times of work in days $$ = 20 \times 5 $$ $$ = 100 \ days $$

- John and Jack together can finish a task in \(5 \ days\). If John alone can finish the same task in \(10 \ days\), then how many days will Jack take to finish the same task alone?
- \(25 \ days\)
- \(15 \ days\)
- \(10 \ days\)
- \(5 \ days\)

Answer: (c) \(10 \ days\)

Solution: Both can finish the task in one day = \(\frac{1}{5}\)

John alone can finish the task in one day = \(\frac{1}{10}\)

then Jack alone can finish the task in one day $$ = \frac{1}{5} - \frac{1}{10} $$ $$ = \frac{1}{10} $$ Hence, Jack alone can finish the task in \(10\) days.

Solution: Both can finish the task in one day = \(\frac{1}{5}\)

John alone can finish the task in one day = \(\frac{1}{10}\)

then Jack alone can finish the task in one day $$ = \frac{1}{5} - \frac{1}{10} $$ $$ = \frac{1}{10} $$ Hence, Jack alone can finish the task in \(10\) days.

- Vishal is twice as efficient as Rohit. Together, they can finish a work in \(10 \ days\). Find in how many days can Rohit alone finish the same task?
- \(50 \ days\)
- \(30 \ days\)
- \(20 \ days\)
- \(10 \ days\)

Answer: (b) \(30 \ days\)

Solution: Let Vishal can finish the task in \(x\) days and

Rohit can finish the task in \(2x\) days

Together, they can finish the task in days $$ \frac{xy}{x + y} = 10 $$ $$ \frac{x \times 2x}{x + 2x} = 10 $$ $$ \frac{2x^2}{3x} = 10 $$ $$ \frac{2x}{3} = 10 $$ $$ x = 15 $$ Hence, Rohit alone can finish the task in \(2x\) days = \(30\) days

Solution: Let Vishal can finish the task in \(x\) days and

Rohit can finish the task in \(2x\) days

Together, they can finish the task in days $$ \frac{xy}{x + y} = 10 $$ $$ \frac{x \times 2x}{x + 2x} = 10 $$ $$ \frac{2x^2}{3x} = 10 $$ $$ \frac{2x}{3} = 10 $$ $$ x = 15 $$ Hence, Rohit alone can finish the task in \(2x\) days = \(30\) days

- If three friends A, B and C can complete a work in \(10\), \(8\) and \(2\) days respectively. Find in how many days they can together finish the work together?
- \(\frac{40}{29} \ days\)
- \(\frac{29}{40} \ days\)
- \(\frac{27}{38} \ days\)
- \(\frac{38}{27} \ days\)

Answer: (a) \(\frac{40}{29} \ days\)

Solution: All three together can finish the work in one day $$ = \frac{1}{10} + \frac{1}{8} + \frac{1}{2} $$ $$ = \frac{29}{40} $$ Hence, together they can finish the work in \(\frac{40}{29}\) days.

Solution: All three together can finish the work in one day $$ = \frac{1}{10} + \frac{1}{8} + \frac{1}{2} $$ $$ = \frac{29}{40} $$ Hence, together they can finish the work in \(\frac{40}{29}\) days.

- A man can finish a work in \(10\) days, but with the help of his friend, he can finish the work in \(5\) days. Find in how many days, his friend can finish the work, alone?
- \(20 \ days\)
- \(15 \ days\)
- \(10 \ days\)
- \(8 \ days\)

Answer: (c) \(10 \ days\)

Solution: together, they can finish the part of work in one day = \(\frac{1}{5}\)

the man alone can finish the part of work in one day = \(\frac{1}{10}\)

then, his friend alone can finish the part of work in one day $$ = \frac{1}{5} - \frac{1}{10} $$ $$ = \frac{1}{10} $$ Hence, his friend alone can finish the work in \(10\) days.

Solution: together, they can finish the part of work in one day = \(\frac{1}{5}\)

the man alone can finish the part of work in one day = \(\frac{1}{10}\)

then, his friend alone can finish the part of work in one day $$ = \frac{1}{5} - \frac{1}{10} $$ $$ = \frac{1}{10} $$ Hence, his friend alone can finish the work in \(10\) days.

- A man can finish a work in two days and a women can finish the same work in four days. If they work together, then in how many days they can finish the work?
- \(\frac{3}{4} \ days\)
- \(\frac{4}{3} \ days\)
- \(\frac{2}{3} \ days\)
- \(\frac{3}{2} \ days\)

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

Solution: the man and women together can finish the part of work in one day $$ = \frac{1}{2} + \frac{1}{4} $$ $$ = \frac{3}{4} $$ Hence, together, they can finish the work in \(\frac{4}{3}\) days.

Solution: the man and women together can finish the part of work in one day $$ = \frac{1}{2} + \frac{1}{4} $$ $$ = \frac{3}{4} $$ Hence, together, they can finish the work in \(\frac{4}{3}\) days.

- Two friends M and N together can finish a task in \(20\) days. If M alone can finish the task in \(5\) days, then in how many days N alone can finish the same work?
- \(6.6 \ days\)
- \(5.6 \ days\)
- \(4.5 \ days\)
- \(3.2 \ days\)

Answer: (a) \(6.6 \ days\)

Solution: Given, \(x = 20\), \(y = 5\), then $$ = \frac{xy}{x - y} $$ $$ = \frac{20 \times 5}{20 - 5} $$ $$ = \frac{100}{15} $$ $$ = 6.6 \ days $$

Solution: Given, \(x = 20\), \(y = 5\), then $$ = \frac{xy}{x - y} $$ $$ = \frac{20 \times 5}{20 - 5} $$ $$ = \frac{100}{15} $$ $$ = 6.6 \ days $$

- A women and a girl can finish a work in \(12\) days. If a women alone can finish the work in \(4\) days, then in how many days the girl alone can finish the work?
- \(5 \ days\)
- \(6 \ days\)
- \(7 \ days\)
- \(8 \ days\)

Answer: (b) \(6 \ days\)

Solution: Given, \(x = 12\), \(y = 4\), then $$ = \frac{xy}{x - y} $$ $$ = \frac{12 \times 4}{12 - 4} $$ $$ = \frac{48}{8} $$ $$ = 6 \ days $$

Solution: Given, \(x = 12\), \(y = 4\), then $$ = \frac{xy}{x - y} $$ $$ = \frac{12 \times 4}{12 - 4} $$ $$ = \frac{48}{8} $$ $$ = 6 \ days $$

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