**Answer –** Option C

**Explanation –**

\(\frac{5 * 7}{5 + 7} = \frac{35}{12} = 2 \frac{11}{12}\) days

**2. Rajan can do a piece of work in 10 days, Rakesh can do in 12 days and Mukesh do the same work in 15 days. In how many days they can finish the work, working together**

**Answer –** Option C

**Explanation –**

\(\frac{1}{10} + \frac{1}{12} + \frac{1}{15} = \frac {6 + 5 + 1}{60} = \frac {1}{4}\)

**3. Sohan and Rohan can do a piece of work in 10 days, Rohan and Mohan can do in 12 days, Sohan and Mohan in 15 days. The time taken by them to do this work together is**

**Answer –** Option D

**Explanation –**

(S + R)’s 1 days work = \(\frac{1}{10}\)

(R + M)’s 1 days work = \(\frac{1}{12}\)

(M + S)’s 1 days work = \(\frac{1}{15}\)

2(S + R + M)’s 1 day work = \(\frac{1}{4}\)

i.e, (S + R +M)’s 1 days work = \(\frac{1}{4}\)

Hence time taken by them to do this work together = 8 days

**4. Sunil and Santosh together can finish a work in 30 days. They worked for it for 20 days and then Santosh left. The remaining work was done by Sunil alone in 20 more days, Santosh alone can finish the work in**

**Answer –** Option D

**Explanation –**

After 20 days,

Balance work = \(\frac{1}{3}\)

20x = \(\frac{1}{3}\)

x = \(\frac{1}{60}\)

Again, x + y = x = \(\frac{1}{30}\)

x = \(\frac{1}{60}\) + y = x = \(\frac{1}{30}\)

i.e, y = x = \(\frac{1}{60}\)

i.e, Santosh can finish in 60 days.

**5. Rambabu, Santosh and Sunil can finish a piece of work in 10, 12 and 15 days respectively. If Santosh stops doing the work after 2 days, Rambabu and Sunil will finish the work in**

**Answer –** Option B

**Explanation –**

2(\(\frac{{R}_{b}}{\frac{{S}_{A}}{{S}_{L}}}\)) = 2 (\(\frac{{1}{10}}{\frac{\frac{1}{12}}{\frac{1}{15}}}\))

Balance work to be done by \({R}_{b}\))and SL\({S}_{L}\))

= 1 – \(\frac {1}{2}\) = \(\frac {1}{2}\)

(Rb + SL)’s 1 day work

\(\frac {1}{10} + \frac {1}{15} = \frac {3 + 2}{30} = \frac {1}{6}\)

i.e, (Rb + SL) will take 3 days to do \(\frac {1}{2}\) of the work

**6. A man undertakes to do a certain work in 150 days. He employs 200 workers. He discovers that only a quarter of the work is done is 50 days. In order to complete the work on schedule, he must additionally employ**

**Answer –** Option B

**Explanation –**

200 persons do \(\frac {1}{4}\) of the work in 50 days

i.e, 1 person does \(\frac {1}{4}\) of work in 50 * 200 days

1 person does \(\frac {3}{4}\) of work in 3 * 50 * 200 days

n persons do \(\frac {3}{4}\) of the work in \(\frac {3 * 50 * 200}{n}\) days

i.e, \(\frac {3 * 50 * 200}{n}\) = 100

n = 300

Already 200 workers are there, hence 100 more persons are to be employed.

**7. Rajan can do a piece of work in 30 days and Mukesh in 20 days. Mukesh alone at it for 10 days and then leaves. Rajan alone can finish the remaining work in**

**Answer –** Option A

**Explanation –**

Mukesh’s 10 day’s work = \(\frac {10}{20}\) = \(\frac {1}{2}\)

Balance work = 1 – \(\frac {1}{2}\) = \(\frac {1}{2}\)

Hence \(\frac {1}{2}\) work Rajan can do in \(\frac {30}{2}\)

= 15 days

**8. A group of 10 students working one hour per day complete a piece of work in 12 days. If there are 12 students in the group and they work one hour per day, then will be able to complete the work in**

**Answer –** Option C

**Explanation –**

Students | Days |
---|---|

10 | 12 |

12 | x |

i.e, 12 : x :: 12 : 10,

\(\frac {12}{x}\) = \(\frac {12}{10}\)

x = 10 days

**9. Ram and Sunil can finish a work in 8 and 16 hours repectively. If they work at it alternatively for an hour Ram beginning first, the work will be finish in**

**Answer –** Option C

**Explanation –**

\(\frac{1}{8} + \frac{1}{16} + \frac{1}{8} + \frac{1}{16}…… = 1 = \frac{2 + 1 + 2+ 1 + …}{16} = 1\)

i.e, \(10 \frac{1}{2}\)

**10. 12 men complete a work in 18 days. 6 days after they had started working, 4 men joined them. How many days will all of them take to complete the remaining work ?**

**Answer –** Option D

**Explanation –**

12 men’s 6 days work = \(\frac{6}{18}\)\( \frac{1}{3}\)

i.e, \( \frac{1}{\frac{2}{3}}\) = \(\frac{12 * 18}{16 * ?}\)

? = 9 days

**11. 72 men can build a wall 280 m long in 42 days. The number of persons who would take 36 days to build a similar will 100 m in length will be**

**Answer –** Option B

**Explanation –**

\(\frac{280}{100}\) = \(\frac{72 * 42}{? * 36}\)

? = 30 men.

**12. Sunil can do a work in 36 days, Ram in 54 days and Balu in 72 days. They star t ed wor king together but before the work was to be over, Sunil left 8 days before and 12 days before Ram. Balu will complete the work alone in**

**Answer –** Option B

**13. Rajan can do a piece of job in 6 days, Sunil in 8 days and Mukesh in 12 days. Sunil and Mukesh worked together for 2 days and t hen Rajan replaces Mukesh. Now how long the new partner will have to work to complete the job ?**

**Answer –** Option D

**14. 25 men were employed to do a piece of work which they could finish in 20 days. But the men dropped off by. 5 at the end of every 10 days. In what time will the work be completed ?**

**Answer –** Option B

**Explanation –**

\(\frac{1}{1}\) = \(\frac{25 * 20}{25 * 10 * 20 * 10 *15 * D}\)

i.e, D = \( 3\frac{1}{3}\) days

Hence time to complete the work = \( 23\frac{1}{3}\) days

**15. If 10 men can do a work in 6 days and 15 women can do the same in 5 days, then 8 men and
5 women can together do the work in**

**Answer –** Option C

**Explanation –**

10 men can do the work in 6 days.

If the work is to be finished in 5 days, then number of women required = 15

If the work is to be finished in 6 days, then number of women required

= \( \frac{15 * 5}{6}\) = \( \frac{25}{2}\)

i.e, 10 men = \(\frac{25}{2}\)women

**Answer –** Option B

**Explanation –**

Required number of men will be

\(\frac {4 * 5 * 3}{4}\) = 15

**2. When Ram and Mohan work together, they complete a work in 4 days. If Ram alone can complete this work in 12 days then in how many days Mohan alone can complete this work ?**

**Answer –** Option B

**Explanation –**

Let the total work be 12 units such that work done by Ram in one day be 1 unit.

i.e, Work done by Mohan in one day

= \(\frac {12}{4}\) – 1 = 2 units.

**3. Pipe ‘F can fill a tank in 36 hours and pipe ‘Q’ can fill this tank in 45 hours. If both the pipes are opened simultaneously, then how much time will be taken to fill this tank ?**

**Answer –** Option A

**Explanation –**

Let the capacity of the tank be 180 units i.e, LCM of 36 and 45 such that efficiencies of the two pipes = 5 units and 4 units.

i.e, required time = \(\frac {180}{5 + 4}\) = 20hrs

**4. To complete a work P takes 50% more time than Q. If together they take 18 days to complete the work, how much time shall Q take to do it?**

**Answer –** Option A

**Explanation –**

Let the time taken by P and Q to complete the work alone be 3x and 2x such that total work be 6x units and their respective efficiencies be 2 units and 3 units.

\(\frac {6x}{\frac{3}{2}} = 18\)

x = 15

Hence, Q will complete the work in 2x

= 2 × 15 = 30 days

**5. Two pipes can fill a tank in 20 minutes and 30 minutes respectively. If both the pipes are opened simultaneously, then the tank will be filled in**

**Answer –** Option B

**Explanation –**

Let the capacity of the tank be 60 units i.e.

LCM of 20 and 30 such that their respective efficiencies be 3 units and 2 units.

Hence, required time = \(\frac {60}{3 + 2}\) = 12 minutes

**6. A and B can do a piece of work in 8 days. A alone can do the same work in 12 days. The number of days in which B alone can do the same work is**

**Answer –** Option B

**Explanation –**

Let the total work be 24 units i.e. LCM of 8 and 12.

work done by A and B in one day = \(\frac {24}{8}\) = 3units

and that by A alone in one day = \(\frac {24}{12}\) = 2 units

Hence, work done by B in one day = 1 unit

and required time = \(\frac {24}{1}\) = 24 days.

**7. P and Q can do a piece of work in 12 days, Q and R in 15 days and R and P in 20 days. In how
many days P alone can do the same work?**

**Answer –** Option

**Explanation –**

Let the total work be 60 units i.e. LCM of 12, 15 and 20.

work done by P and Q in one day = \(\frac {60}{12}\) = 5 units

and that by Q and R in one day = \(\frac {60}{15}\) = 4 units

and that by P and R in one day = \(\frac {60}{20}\) = 3 units

Thus, work done by P, Q and R in 1 day = \(\frac {5 + 4 + 3 }{2}\) = 6

and that by P in one day = 6 4(work done by Q and R in 1 day) = 2 units

Hence, required time = \(\frac {60}{2} = 30\) days

**8. A and B can do a piece of work in 24 days. If efficiency of A is double than B, then in how
many days, A alone can do the same work?**

**Answer –** Option B

**Explanation –**

Set the work done by A and B value in one day be 2 units and 1 units respectively

Total work = 24 × (2 + 1) = 72 units and

Required time = \(\frac {72}{2}\) = 36 units

**9. P and Q can do a piece of work in 12 days, Q and R in 15 days and R and P in 20 days. In how
many days Q alone can do the same work?**

**Answer –** Option A

**Explanation –**

Set the total work be 60 unit i.e. LCM of 12, 15 and 20

Work done in one day

by P and Q = 5 units ….(1)

by Q and R = 4 units ….(2)

and by R and P = 3 units ….(3)

Adding (1), (2) and (3) we get

2(P + Q + R) = 5 + 4 + 3

P + Q + R = \(\frac {12}{2}\) = 6 units

Thus, work done y Q in one day

= (P + Q + R) – (P + R)

= 6 – 3 = 3 units

Hence, required time = \(\frac {60}{3}\) = 20 units

**10. A can do a piece of work in 15 days and B ran do the same work in 10 days. If they work together, number of days required to complete the same work is**

**Answer –** Option B

**Explanation –**

Work done by A in 1day = \(\frac {1}{15}\)

Work done by in B in 1day = \(\frac {1}{10}\)

Let number of days required = x

\((\frac {1}{15} + \frac {1}{10})\)x = 1

i.e, x = 6

**11. P and O can do a piece of work in 10 days, Q and R in 12 days and R and P in 15 days. In how many days P alone can do the same work?**

**Answer –** Option A

**Explanation –**

Let P do work in P days

\(\frac{A}{Q}\)[ \(\frac{\frac{1}{P}}{\frac{1}{Q}}\)] = \(\frac{1}{10}\)……(1)

\(\frac{\frac{1}{Q}}{\frac{1}{R}}\) = \(\frac{1}{12}\)…….(2)

\(\frac{\frac{1}{P}}{\frac{1}{R}}\) = \(\frac{1}{12}\)……..(3)

Solving (1), (2), (3)

P = 24

**12. A and B can do a piece of work in 24 days. If efficiency of A is double than B, then in how many days B alone can do the same work?**

**Answer –** Option A

**Explanation –**

Efficiency ratio A : B = 2 : 1

Time ratio A : B = 1 : 2

i.e x & 2x

\(\frac{A}{Q}\)\(\frac{\frac{1}{x}}{\frac{1}{2x}}\)= \(\frac{1}{24}\)

x = 36

i.e, B can do work in 2x days = 72

**13. P and Q can do a piece of work in 12 days, Q and in R in 15 lays and R and P in 20 days. In how many days R alone can do the same work?**

**Answer –** Option B

**Explanation –**

Let work done in 1 day by P, Q, R

= \(\frac{1}{p}\), \(\frac{1}{q}\), \(\frac{1}{r}\)

i.e, \(\frac{\frac{1}{p}}{\frac{1}{q}}\) = \(\frac{1}{12}\)…(1)

\(\frac{\frac{1}{q}}{\frac{1}{r}}\) = \(\frac{1}{15}\)…..(2)

\(\frac{\frac{1}{r}}{\frac{1}{p}}\) = \(\frac{1}{20}\)…..(3)

By solving (4) & (2),

\(\frac{1}{r} – \frac{1}{p}\) = \(\frac{1}{15} – \frac{1}{12}\)

solving (3) & (4) :

\(\frac{2}{r} = \frac{1}{15} – \frac{1}{12} + \frac{1}{20}\) = r = 60

i.e, Number of days in which r can do work = 60

**14. A can do a piece of work in 20 day and B can do the same work in 30 days. If they work together the number of days required to do the same work**

**Answer –** Option D

**Explanation –**

Work done by A in 1 day = \(\frac{1}{20}\)

___________ B _______ = \(\frac{1}{30}\)

Let no. of days required = D

i.e, (\(\frac{\frac{1}{20}}{\frac{1}{30}}\))D = 1

D = 12

**15. P and Q can do a piece of work in 10 days. Q and R in 12 days and R and P in 15 days. In how many days R alone can do the same work?,**

**Answer –** Option C

**Explanation –**

Let R do the work in R days

\(\frac{A}{Q}\)\(\frac{\frac{1}{P}}{\frac{1}{Q}}\)= \(\frac{1}{10}\) ….(1)

\(\frac{\frac{1}{Q}}{\frac{1}{R}}\)= \(\frac{1}{12}\)…(2)

\(\frac{\frac{1}{P}}{\frac{1}{R}}\)= \(\frac{1}{15}\)…(3)

solving (1), (2), (3):-

R = 40

finish the work together?

**Answer –** Option B

**Explanation –**

Let time taken by A, B, C combined is D days.

i.e, (\(\frac{1}{12}\) ÷ \(\frac{1}{15}\) ÷ \(\frac{1}{20}\))D = 1

(\(\frac{1}{5}\)) * D = 1

D = 5 days

**2. Pipe ‘P’ can fill a tank in 10 hours and Pipe ‘Q’ can fill this tank in 12 hours. Pipe ‘R’ can empty the full tank in 20 hours. I f all the three pipes are operated simultaneously, then in how much time this tank will be filled ?**

**Answer –** Option B

**Explanation –**

Let capacity of tank be 120 l

i.e, filling speed of

P = \(\frac {120}{10}\)

i.e, filling speed of

Q = \(\frac {120}{12}\)

emptying speed of

R = \(\frac {120}{20}\) = 6 l/hr

i.e, \(\frac {120}{1 2+ 10 – 6}\) = \(\frac {15}{2}\)hours

**3. 36 men can complete a work in 18 days. In how many days will 27 men complete the same work**

**Answer –** Option D

**Explanation –**

M1D1 = M2 D2

36 × 18 = 27 × D2

D2= 24 days

**4. A man, a woman and a boy can together complete a piece of work in 3 days. I f a man alone can do it in 6 days and a boy alone in 18 days, how long will a woman take to complete the work?**

**Answer –** Option A

**Explanation –**

Let the work be completed by a woman in x days.

According to the question

(\(\frac{1}{6}\) ÷ \(\frac{1}{2}\) ÷ \(\frac{1}{18}\)) * 3 = 1

x = 9 days

**5. A tap can fill a cistern in 8 hours and another tap can empty it in 16 hours. If both the taps are open, the time taken to fill the tank will be**

**Answer –** Option C

**Explanation –**

Let capacity of tank be 16 litres

filling speed = \(\frac{16}{8}\) = 2 l/hr

emptying speed = \(\frac{16}{16}\) = 1 l/hr

i.e, Time = \(\frac{16}{2 – 1}\) = 16 hours

**6. A team of 100 workers is supposed to do a work in 40 days. After 35 days, 100 more workers were employed and the work was finished on. Time. How many days would have it been delayed if additional workers were not employed?**

**Answer –** Option D

**Explanation –**

35 × 100 + 5 × (100 + 100) = 100 × N

N = 45 days

No. of days object =(45 – 40) days = 5 days

**7. A, B and C can do a work 6,12 and 24 days respectively. They all begin together. A continues to work till it is finished, C leaves off 2 days and B one day before completion. In what time (in days) is the work finished?**

**Answer –** Option B

**Explanation –**

one days work of A = \(\frac{1}{6}\)

one days work of B = \(\frac{1}{12}\)

one days work of C = \(\frac{1}{24}\)

(\(\frac{1}{6}\) ÷ \(\frac{1}{12}\) ÷ \(\frac{1}{24}\))x÷\(\frac{1}{6}\) ÷ \(\frac{1}{12}\) * 1 + \(\frac{1}{6}\) = 1

x = 2 days

Ans = 2 + 1 + 1 = 4 days

**8. Two persons working 2 hours a day assemble 3 machines in 3 days. The number of machines assembled by 5 persons working 4 hours a day in 4 days is**

**Answer –** Option D

**Explanation –**

\(\frac{2 * 3 * 2}{3}\) = \(\frac{5 * 4* 4}{x}\) x = 20 machines

**9. Two pipes A and B can fill a tank with water in 2 hours and 32 hours respectively. Both the pipes are open together. I f the tank is filled up in 14 hours, then the first pipe must be turned off after**

**Answer –** Option C

**Explanation –**

Portion of tank filled by (A +B) + position filled by B =1

= (\(\frac{1}{24}\) ÷ \(\frac{1}{32}\)) ÷ (14 – T)\(\frac{1}{32}\) = 1

T = \(\frac{27}{2}\) = 13 \(\frac{1}{2}\)

**10. Two pipes A and B running together can fill acistern in 6 \(\frac{2}{3}\) minutes. Pipe B takes 3 minutes more than A to fill it. The time, in minutes, taken by B alone to fill the cistern is**

**Answer –** Option C

**Explanation –**

\(\frac{1}{x}\) ÷ \(\frac{1}{x + 3}\) ÷ \(\frac{3}{20}\)

\(\frac{x + X + 3}{x(X + 3)}\) = \(\frac{3}{20}\)

\(\frac{2x + 3}{x(X + 3)}\) = \(\frac{3}{20}\)

40x + 60 = \({3x}^{2}\) + 9x

\({3x}^{2}\)– 31x – 60 = 0

x = 12 minutes

Pipe A takes 12 minutes

Pipe B takes 15 minutes