# 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) 17 men | B) 14 men |

C) 13 men | D) 16 men |

Explanation:

M x T / W = Constant

where, M= Men (no. of men)

T= Time taken

W= Work load

So, here we apply

M1 x T1/ W1 = M2 x T2 / W2

Given that, M1 = 4 men, T1 = 7 hours ; T2 = 2 hours, we have to find M2 =?

Note that here, W1 = W2 = 1 road, ie. equal work load.

Clearly, substituting in the above equation we get, M2 = 14 men.

A) 4 days | B) 6 days |

C) 8 days | D) 10 days |

Explanation:

Ratio of rates of working of A and B =2:1. So, ratio of times taken =1:2

Therefore, A's 1 day's work=1/9

B's 1 day's work=1/18

(A+B)'s 1 day's work= 1/9 + 1/18 = 1/6

so, A and B together can finish the work in 6 days

A) 18 days | B) 24 days |

C) 30 days | D) 36 days |

Explanation:

2(A+B+C)'s 1 day work = 1/30 + 1/24 + 1/20 = 1/8

=>(A+B+C)'s 1 day's work= 1/16

work done by A,B and C in 10 days=10/16 = 5/8

Remaining work= 3/8

A's 1 day's work= $\left(\frac{1}{16}-\frac{1}{24}\right)=\frac{1}{48}$

Now, 1/48 work is done by A in 1 day.

So, 3/8 work wil be done by A in =48 x (3/8) = 18 days

A) 215 days | B) 225 days |

C) 235 days | D) 240 days |

Explanation:

Given that

**(10M + 15W) x 6 days = 1M x 100 days**

=> 60M + 90W = 100M

=> 40M = 90W

=> **4M = 9W.**

From the given data,

1M can do the work in 100 days

=> 4M can do the same work in 100/4= 25 days.

=> 9W can do the same work in 25 days.

=> 1W can do the same work in **25 x 9 = 225 days.**

Hence, 1 woman can do the same work in** 225 days.**

A) 27 days | B) 12 days |

C) 25 days | D) 18 days |

Explanation:

$\frac{3}{4}\times (x-2)x=(x+7)(x-10)$

$\Rightarrow {x}^{2}-6x-280=0$

=> x= 20 and x=-14

so, the acceptable values is x=20

Therefore, Total work =(x-2)x = 18 x 20 =360 unit

Now 360 = 30 x k

=> k=12 days

A) 4.5 | B) 5 |

C) 6 | D) 9 1/3 |

Explanation:

A : C

Efficiency 5 : 3

No of days 3x : 5x

Given that, 5x-6 =3x => x = 3

Number of days taken by A = 9

Number of days taken by C = 15

B : C

Days 2 : 3

Therefore, Number of days taken by B = 10

Work done by B and C in initial 2 days = $2\left[\frac{1}{10}+\frac{1}{15}\right]$= 1/3

Thus, Rest work =2/3

Number of days required by A to finish 2/3 work = (2/3) x 9 = 6 days

A) 10 % | B) 14 ( 2/7 )% |

C) 20 % | D) Can't be determined |

Explanation:

Let he initially employed x workers which works for D days and he estimated 100 days for the whole work and then he doubled the worker for (100-D) days.

D * x +(100- D) * 2x= 175x

=> D= 25 days

Now , the work done in 25 days = 25x

Total work = 175x

Therefore, workdone before increasing the no of workers = $\frac{25x}{175x}\times 100$ % = $14\frac{2}{7}\%$

(A+B)'s two days work = $\frac{1}{40}+\frac{1}{50}=\frac{9}{200}$

Evidently, the work done by A and B duing 22 pairs of days

i.e in 44 days = $22\times \frac{9}{200}=\frac{198}{200}$

Remaining work = $1-\frac{198}{200}$= 1/100

Now on 45th day A will have the turn to do 1/100 of the work and this work A will do in $40\times \frac{1}{100}=\frac{2}{5}$

Therefore, Total time taken = 44$\frac{2}{5}$daya