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. |

- Rakesh, Rohan and Mohan can finish a task in \(10\) days, \(20\) days and \(25\) days respectively. If they start working together, then in how many days they will finish the task?
- \(4.2 \ days\)
- \(5.2 \ days\)
- \(6.5 \ days\)
- \(8.5 \ days\)

Answer: (b) \(5.2 \ days\)

Solution: Given, \(x = 10\), \(y = 20\), \(z = 25\), then $$ = \frac{xyz}{xy + yz + zx} $$ $$ = \frac{10 \times 20 \times 25}{10 \times 20 + 20 \times 25 + 25 \times 10} $$ $$ = \frac{5000}{200 + 500 + 250} $$ $$ = \frac{5000}{950} $$ $$ = 5.2 \ days $$

Solution: Given, \(x = 10\), \(y = 20\), \(z = 25\), then $$ = \frac{xyz}{xy + yz + zx} $$ $$ = \frac{10 \times 20 \times 25}{10 \times 20 + 20 \times 25 + 25 \times 10} $$ $$ = \frac{5000}{200 + 500 + 250} $$ $$ = \frac{5000}{950} $$ $$ = 5.2 \ days $$

- Three women K, L and M can finish a work in two days, four days and five days respectively. If they start working together, then in how many days they will finish the work?
- \(1.05 \ days\)
- \(2.20 \ days\)
- \(3.02 \ days\)
- \(4.06 \ days\)

Answer: (a) \(1.05 \ days\)

Solution: Given, \(x = 2\), \(y = 4\), \(z = 5\), then $$ = \frac{xyz}{xy + yz + zx} $$ $$ = \frac{2 \times 4 \times 5}{2 \times 4 + 4 \times 5 + 5 \times 2} $$ $$ = \frac{40}{38} $$ $$ = 1.05 \ days $$

Solution: Given, \(x = 2\), \(y = 4\), \(z = 5\), then $$ = \frac{xyz}{xy + yz + zx} $$ $$ = \frac{2 \times 4 \times 5}{2 \times 4 + 4 \times 5 + 5 \times 2} $$ $$ = \frac{40}{38} $$ $$ = 1.05 \ days $$

- P and Q can finish a work in \(5\) days, Q and R can finish the same work in \(10\) days and R & P can finish the same work in \(15\) days. If they start working together, then in how many days they will finish the work?
- \(6.05 \ days\)
- \(5.45 \ days\)
- \(7.58 \ days\)
- \(8.35 \ days\)

Answer: (b) \(5.45 \ days\)

Solution: Given, \(x = 5\), \(y = 10\) and \(z = 15\) then $$ = \frac{2xyz}{xy + yz + zx} $$ $$ = \frac{2 \times 5 \times 10 \times 15}{50 + 150 + 75} $$ $$ = \frac{1500}{275} $$ $$ = 5.45 \ days $$

Solution: Given, \(x = 5\), \(y = 10\) and \(z = 15\) then $$ = \frac{2xyz}{xy + yz + zx} $$ $$ = \frac{2 \times 5 \times 10 \times 15}{50 + 150 + 75} $$ $$ = \frac{1500}{275} $$ $$ = 5.45 \ days $$

- A and B can finish a work in \(8\) days, B and C can finish the same work in \(10\) days and C & A can finish the same work in \(12\) days. If they start working together, then in how many days they will finish the work?
- \(6.4 \ days\)
- \(5.4 \ days\)
- \(7.5 \ days\)
- \(8.3 \ days\)

Answer: (a) \(6.4 \ days\)

Solution: Given, \(x = 8\), \(y = 10\) and \(z = 12\) then $$ = \frac{2xyz}{xy + yz + zx} $$ $$ = \frac{2 \times 8 \times 10 \times 12}{80 + 120 + 96} $$ $$ = \frac{1920}{296} $$ $$ = 6.4 \ days $$

Solution: Given, \(x = 8\), \(y = 10\) and \(z = 12\) then $$ = \frac{2xyz}{xy + yz + zx} $$ $$ = \frac{2 \times 8 \times 10 \times 12}{80 + 120 + 96} $$ $$ = \frac{1920}{296} $$ $$ = 6.4 \ days $$

- A can finish a work in \(5\) days and B can finish the same work in \(10\) days. If they work together for \(3\) days and then A goes away, then in how many more days will B take to finish the work?
- \(4 \ days\)
- \(3 \ days\)
- \(2 \ days\)
- \(1 \ day\)

Answer: (d) \(1 \ day\)

Solution: Both can finish the part of work in \(3\) days $$ = \left(\frac{1}{5} + \frac{1}{10}\right) \times 3 $$ $$ = \frac{9}{10} $$ remaining work $$ = 1 - \frac{9}{10} $$ $$ = \frac{1}{10} $$ then B will finish the remaining work in days $$ = \frac{1/10}{1/10} $$ $$ = 1 \ day $$

Solution: Both can finish the part of work in \(3\) days $$ = \left(\frac{1}{5} + \frac{1}{10}\right) \times 3 $$ $$ = \frac{9}{10} $$ remaining work $$ = 1 - \frac{9}{10} $$ $$ = \frac{1}{10} $$ then B will finish the remaining work in days $$ = \frac{1/10}{1/10} $$ $$ = 1 \ day $$

- P can finish a task in \(10\) days and Q can finish the same work in \(15\) days. If they work together for \(5\) days and then P goes away, then in how many more days will Q take to finish the work?
- \(4.5 \ days\)
- \(3.5 \ days\)
- \(2.5 \ days\)
- \(1.5 \ days\)

Answer: (c) \(2.5 \ days\)

Solution: Both can finish the part of work in \(5\) days $$ = \left(\frac{1}{10} + \frac{1}{15}\right) \times 5 $$ $$ = \frac{5}{6} $$ remaining work $$ = 1 - \frac{5}{6} $$ $$ = \frac{1}{6} $$ then Q will finish the remaining work in days $$ = \frac{1/6}{1/15} $$ $$ = 2.5 \ days $$

Solution: Both can finish the part of work in \(5\) days $$ = \left(\frac{1}{10} + \frac{1}{15}\right) \times 5 $$ $$ = \frac{5}{6} $$ remaining work $$ = 1 - \frac{5}{6} $$ $$ = \frac{1}{6} $$ then Q will finish the remaining work in days $$ = \frac{1/6}{1/15} $$ $$ = 2.5 \ days $$

- \(10\) Men can finish a work in \(3\) days, \(8\) women can finish the same work in \(4\) days and \(6\) girls can finish the same work in \(5\) days. Find in how many days \(1\) man, \(1\) women and \(1\) girl working together can finish the work?
- \(25.5 \ days\)
- \(15.3 \ days\)
- \(12.2 \ days\)
- \(10.2 \ days\)

Answer: (d) \(10.2 \ days\)

Solution: one man can finish the work in days = \(10 \times 3\) = \(30\) days

one women can finish the work in days = \(8 \times 4\) = \(32\) days

one girl can finish the work in days = \(6 \times 5\) = \(30\) days

then one man, one women and one girl together can finish the part of work in one day $$ = \frac{1}{30} + \frac{1}{32} + \frac{1}{30} $$ $$ = \frac{94}{960} $$ Hence, one man, one women and one girl together can finish the work $$ = \frac{960}{94} $$ $$ = 10.2 \ days $$

Solution: one man can finish the work in days = \(10 \times 3\) = \(30\) days

one women can finish the work in days = \(8 \times 4\) = \(32\) days

one girl can finish the work in days = \(6 \times 5\) = \(30\) days

then one man, one women and one girl together can finish the part of work in one day $$ = \frac{1}{30} + \frac{1}{32} + \frac{1}{30} $$ $$ = \frac{94}{960} $$ Hence, one man, one women and one girl together can finish the work $$ = \frac{960}{94} $$ $$ = 10.2 \ days $$

- P is three times as efficient as Q and together they can finish a task in \(10\) days. Find number of days required by Q to finish the same task indivisually?
- \(39.9 \ days\)
- \(34.5 \ days\)
- \(32.6 \ days\)
- \(30.8 \ days\)

Answer: (a) \(39.9 \ days\)

Solution: Let P requires \(k\) days to finish the work

and Q requires \(3k\) days to finish the work, then $$ \frac{1}{k} + \frac{1}{3k} = \frac{1}{10} $$ $$ k = 13.3 \ days $$ Hence, Q requires \(3k\) = \(39.9\) days to finish the work.

Solution: Let P requires \(k\) days to finish the work

and Q requires \(3k\) days to finish the work, then $$ \frac{1}{k} + \frac{1}{3k} = \frac{1}{10} $$ $$ k = 13.3 \ days $$ Hence, Q requires \(3k\) = \(39.9\) days to finish the work.

- P can finish a work in \(6\) days and Q can finish the same work in \(8\) days, then find how many days are required to finish the work, if both are working together?
- \(3.4 \ days\)
- \(5.4 \ days\)
- \(6.5 \ days\)
- \(8.6 \ days\)

Answer: (a) \(3.4 \ days\)

Solution: Given, \(x = 6 \ days\), \(y = 8 \ days\)

If both P and Q start working together then $$ = \frac{xy}{x + y} $$ $$ = \frac{6 \times 8}{6 + 8} $$ $$ = \frac{48}{14} $$ $$ = 3.4 \ days $$ Hence, together they can finish the work in \(3.4 \ days\).

Solution: Given, \(x = 6 \ days\), \(y = 8 \ days\)

If both P and Q start working together then $$ = \frac{xy}{x + y} $$ $$ = \frac{6 \times 8}{6 + 8} $$ $$ = \frac{48}{14} $$ $$ = 3.4 \ days $$ Hence, together they can finish the work in \(3.4 \ days\).

- A women can finish a work in \(20\) days, but with the help of her son, she can finish the work in \(10\) days. Find in how many days her son can finish the work, alone?
- \(10 \ days\)
- \(15 \ days\)
- \(20 \ days\)
- \(22 \ days\)

Answer: (c) \(20 \ days\)

Solution: together, they can finish the part of work in one day = \(\frac{1}{10}\)

the women alone can finish the part of work in one day = \(\frac{1}{20}\)

then, her son alone can finish the part of work in one day $$ = \frac{1}{10} - \frac{1}{20} $$ $$ = \frac{1}{20} $$ Hence, her son alone can finish the work in \(20\) days.

Solution: together, they can finish the part of work in one day = \(\frac{1}{10}\)

the women alone can finish the part of work in one day = \(\frac{1}{20}\)

then, her son alone can finish the part of work in one day $$ = \frac{1}{10} - \frac{1}{20} $$ $$ = \frac{1}{20} $$ Hence, her son alone can finish 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