# 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 =

**2. **If A’s 1 day's work =, 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 :

Hence,

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 =

A) 3 am | B) 12 pm |

C) 1 pm | D) 3 pm |

Explanation:

Work done by P and Q in the first two hours, working alternately

= First hour P + Second hour Q

work is completed in 2 hours

Then, the total time required to complete the work by P and Q working alternately=23=6hours

Thus, work will be completed at 3pm.

A) 24.5 days | B) 26.6 days |

C) 25 days | D) 20 days |

Explanation:

Arun has completed rd of the work in 8 days

Then he can complete the total work in

---- 8

1 ---- ?

= 24 days

But given Akhil is only 60% as efficient as Arun

Akhil =

Akhil can complete the total work in 40 days

Now, remaining 2/3rd of work can be completed in

1 ------ 40

------ ? 26.66 days.

A) 50 days | B) 48 days |

C) 47.5 days | D) 49 days |

Explanation:

50 men can build a tank in 40 days

Assume 1 man does 1 unit of work in 1 day

Then the total work is 50×40 = 2000 units

50 men work in the first 10 days and completes 50×10 = 500 units of work

45 men work in the next 10 days and completes 45×10 = 450 units of work

40 men work in the next 10 days and completes 40×10 = 400 units of work

35 men work in the next 10 days and completes 35×10 = 350 units of work

So far 500 + 450 + 400 + 350 = 1700 units of work is completed and

remaining work is 2000 - 1700 = 300 units

30 men work in the next 10 days. In each day, they does 30 units of work.

Therefore, additional days required =

Thus, total 10+10+10+10+10 = 50 days required.

**Sol : **In this kind of questions we find the work force required to complete the work in 1 day (or given unit of time) then we equate the work force to find the relationship between the efficiencies (or work rate) between the different workers.

Therefore 6B+8G = 6 days

6(6B+8G)= 1 day (inversely proportional)

36B+48G =1 ( by unitary method)

Again 14B+10G = 4days

56B+40G =1

so, here it is clear that either we employ 36B and 48G to finish the work in 1 day or 56B and 40G to finish the same job in 1 day. thus , we can say

36B+48G = 56B+40G

20B = 8G

G= 2.5B

Thus a Girl is 2.5 times as efficient as a boy.

Now, since 36B+48G = 1

36B+48 (2.5 B)=1

156B=1

i.e., to finish the job in 1 day 156 boys are required or the amount of work is 156 boys-days

Again 1G+1B=2.5B+1B=3.5B

Now, since 156 boys can finish the job in 1 day

so 1 boy can finish the job in 1 156 days

3.5 boys can finish the job in = days.

A) 120 days | B) 119 days |

C) 118 days | D) 117 days |

Explanation:

K's work in a day(1st day) = 1/30

L's work in a day(2nd day)= -1/60(demolishing)

hence in 2 days, combined work= 1/30 - 1/60

=1/60

since both works alternatively, K will work in odd days and L will work in even days.

1/60 unit work is done in 2 days

58/60 unit work will be done in 2 x 58 days = 116 days

Remaining work = 1-58/60

= 2/60

= 1/30

Next day, it will be K's turn and K will finish the remaining 1/30 work.

hence total days = 116 + 1 = 117.