# 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 =

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.**

A) 12 days | B) 18 days |

C) 24 days | D) 30 days |

Explanation:

A + B = C + D

| | | |

Ratio of efficiency 10x + 5x 9x + 6x

|________| |_________|

15x 15x

Therefore , ratio of efficiency of A:C =10:9

Therefore, ratio of days taken by A:C = 9:10

Therefore, number of days taken by A = 18 days

A) 4:18 pm | B) 3:09 pm |

C) 12:15 pm | D) 11:09 am |

Explanation:

Efficiency of P= 100/20= 5% per hour

Efficiency of Q= 100/25= 4% per hour

Efficiency of R= 100/40= 2.5% per hour

Efficiency of S=100/50= 2% per hour

Cistern filled till 10 am by P, Q and R

$\left.\begin{array}{c}Till10.00amPipePfilled20\%\\ Till10.00amPipeQfilled8\%\\ Till10.00amPipeRfilled2.5\%\end{array}\right\}30.5\%$

Thus, at 10 am pipe P,Q and R filled 30.5% of the cistern.

Rest of cistern to be filled = 100 - 30.5 = 69.5%

Now, the time taken by P,Q,R and S together to fill the remaining capacity of the cistern

=69.5 / (5+4+2.5+2) = 5 Hours and 9 minutes(approx)

Therefore, total time =4 hrs + 5hrs 9 mins

= 9 hrs and 9 mins

It means cistern will be filled up at 3:09 pm

A) A | B) B |

C) C | D) can't be determined |

Explanation:

A + B= 70%

B + C =50%

$\left[\because (A+B)+(B+C)-(A+B+C)=B\right]$

=> B= 20% A= 50% and C=30%

Hence A is most efficient

(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

A) 3 hours | B) 4 hours |

C) 5 hours | D) None of these |

Explanation:

Rate of leakage = 8.33% per hour

Net efficiency = 50 - (16.66 + 8.33)= 25%

Time required = 100/25 = 4 hours

A) 6 | B) 8 |

C) 9 | D) 12 |

Explanation:

$\begin{array}{ccccc}& & Amit& & Bharat\\ No.ofDays& & 45& & 40\\ Efficiency& & 2.22Percent(=\frac{1}{45})& & 2.5Percent(=\frac{1}{40})\end{array}$

Amit did the work in 56 days = $56\times \frac{1}{45\times 2}=\frac{28}{45}$

Therefore, Rest work 17/45 was done by Amit and Bharath = $\frac{{\displaystyle \raisebox{1ex}{$17$}\!\left/ \!\raisebox{-1ex}{$45$}\right.}}{{\displaystyle \raisebox{1ex}{$17$}\!\left/ \!\raisebox{-1ex}{$360$}\right.}}$ = 8 days

( since Amit and Bharath do the work in one day = $\frac{1}{45}+\frac{1}{40}=\frac{17}{360}$)

A) 9 days | B) 11 days |

C) 13 days | D) 15 days |

Explanation:

Ratio of times taken by A and B = 100 : 130 = 10 : 13.

Suppose B takes x days to do the work.

Then, 10 : 13 :: 23 : x => x = ( 23 x 13/10 ) => x = 299 /10.

A's 1 day's work = 1/23 ;

B's 1 day's work = 10/299 .

(A + B)'s 1 day's work = ( 1/23 + 10/299 ) = 23/299 = 113 .

Therefore, A and B together can complete the work in 13 days.