# 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) 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) 2:6:3 | B) 2:3:6 |

C) 1:2:3 | D) 3:1:2 |

Explanation:

A : B : C

Ratio of efficiency 3 : 1 : 2

Ratio of No.of days 1/3 : 1/1 : 1/2

or 2 : 6 : 3

Hence A is correct.

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) 1/24 days | B) 7/24 days |

C) 24/7 days | D) 4 days |

Explanation:

(A+B+C)'s 1 day's work = (1/24 + 1/6 + 1/12) = 7/24

so, A,B and C together will complete the work in 24/7 days.

A) 50 | B) 40 |

C) 45 | D) 10 |

Explanation:

Let the number of workers be x.

Now, Using work equivalence method,

X + (X-1) + (X-2)+ . . . . + 1 = X *55% of X

=> [X * (X+1)] / 2 = X * (55X/100) [because, Series is in AP. Sum of AP = {No. of terms (first term+ last term)/2} ]

Therefore, X = 10

A) 4 days | B) 5 days |

C) 6 days | D) 7 days |

Explanation:

Work donee by A and B in the first two hours, working alternatively = First hour A + Second hour B = (1/4) + (1/12) = 1/3.

Thus, the total time required to complete the work = 2 (3) = 6 days

As the pipes are operating alternatively, thus their 2 minutes job is = $\frac{1}{4}+\frac{1}{6}=\frac{5}{12}$

In the next 2 minutes the pipes can fill another $\frac{5}{12}$ part of cistern.

Therefore, In 4 minutes the two pipes which are operating alternatively will fill $\frac{5}{12}+\frac{5}{12}=\frac{5}{6}$Remaining part = $1-\frac{5}{6}=\frac{1}{6}$

Pipe A can fill $\frac{1}{4}$ of the cistern in 1 minute

Pipe A can fill $\frac{1}{6}$ of the cistern in = $\frac{2}{3}$ min

Therefore, Total time taken to fill the Cistern

**4 + $\frac{2}{3}$ minutes.**