# Time and Work Questions

**FACTS AND FORMULAE FOR TIME AND WORK QUESTIONS**

**1. **If A can do a piece of work in n days, then A's 1 day's work =$\frac{1}{n}$

**2. **If A’s 1 day's work =$\frac{1}{n}$, then A can finish the work in n days.

**3. **A is thrice as good a workman as B, then:

Ratio of work done by A and B = 3 : 1.

Ratio of times taken by A and B to finish a work = 1 : 3.

NOTE :

$Efficiency\propto \frac{1}{Nooftimeunits}$

$\therefore Efficiency\times Time=Cons\mathrm{tan}tWork$

Hence, $Requiredtime=\frac{Work}{Efficiency}$

Whole work is always considered as 1, in terms of fraction and 100% , in terms of percentage.

In general, number of day's or hours = $\frac{100}{Efficiency}$

A) 1 day | B) 2 days |

C) 5 days | D) None of these |

Explanation:

Total work = 100+50 = 150man-days

In 8 days 100 man-days work has been completed. Now on 9th and 10th day there will be 25 workers. So in 2 days they wll complete additional 50 man- days work. Thus the work requires 2 more days.

A) 15 | B) 30 |

C) 25 | D) 10 |

Explanation:

Combined efficiency of all the three boats = 60 passenger/trip

Now, consider option(a)

15 trips and 150 passengers means efficiency of B1 = 10 passenger/trip

which means in carrying 50 passengers B1 must has taken 5 trips. So the rest trips equal to 5 (10-5 = 5) in which B2 and B3 together carried remaining 250 (300 - 50 = 250) Passengers.

Therefore the efficiency of B2 and B3 = 250/5 = 50 passenger/trip

Since, the combined efficiency of B1, B2 and B3 is 60. Which is same as given in the first statement hence option(a) is correct.

A) 64000 liters | B) 56000 liters |

C) 78000 liters | D) 60000 liters |

Explanation:

Let x liter be the per day filling and v litr be the capacity of the reservoir, then

90x + v = 40000 * 90 -----(1)

60x + v= 32000 * 60 ------(2)

solving eq.(1) and (2) , we get

x = 56000

Hence , 56000 liters per day can be used without the failure of supply.

A) 2 days | B) 2.5 days |

C) 2.25 days | D) 3 days |

Explanation:

1 man's 1 day work = 1/108

12 men's 6 day's work = 1/9 x 6 = 2/3

Remaining work = 1 - 2/3 = 1/3

16 men's 1 day work = 1/108 x 16 = 4/27

4/27 work is done by them in 1 day.

1/3 work is done by them in 27/4 x 1/3 = 9/4 days.

A) Rs.130 | B) Rs.185 |

C) Rs.70 | D) can't be determined |

Explanation:

Efficiency of kaushalya = 5%

Efficiency of kaikeyi = 4%

Thus, in 10 days working together they will complete only 90% of the work.

[(5+4)*10] =90

Hence, the remaining work will surely done by sumitra, which is 10%.

Thus, sumitra will get 10% of Rs. 700, which is Rs.70

A) 2 hours | B) 8 hours |

C) 6 hours | D) 4 hours |

Explanation:

Efficiency of Inlet pipe A = 4.16% $\left(\frac{100}{24}\right)$

Efficiency of Inlet pipe B = 5.83% $\left(\frac{100}{{\displaystyle 17\frac{1}{7}}}\right)$

Therefore, Efficiency of A and B together = 100 %

Now, if the efficiency of outlet pipe be x% then in 10 hours the capacity of tank which will be filled = 10 * (10 - x)

Now, since this amount of water is being emptied by 'C' at x% per hour, then

$\frac{10x\left(10-x\right)}{x}=2.5hrs$

=> x = 8

Therefore, in 10 hours 20% tank is filled only. Hence, the remaining 80% of the capacity will be filled by pipes A and B in 80/10 = 8 hours

A) 4:5 | B) 3:4 |

C) 4:3 | D) 2:3 |

Explanation:

(20 x 18) men can complete the work in in one day.

one man's one day work = 1/360

(18 x 15) women can complete the work in 1 day

1 woman's one day work = 1/270

So, required ratio = $\frac{1}{270}:\frac{1}{360}$= 4:3

A) 25 | B) 24 |

C) 23 | D) 21 |

Explanation:

(A+B+C) do 1 work in 10 days.

So (A+B+C)'s 1 day work=1/10 and as they work together for 4 days so workdone by them in 4 days=4/10=2/5

Remaining work=1-2/5=3/5

(B+C) take 10 more days to complete 3/5 work. So( B+C)'s 1 day work=3/50

Now A'S 1 day work=(A+B+C)'s 1 day work - (B+C)'s 1 day work=1/10-3/50=1/25

A does 1/25 work in in 1 day

Therefore 1 work in 25 days.