# 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 = $\frac{100}{Efficiency}$

150 men in 25 days do = $\frac{1}{4}$ work

Let 1 man in 1 day does =** x** work

Total work done by 150 men in 25 days = 150x * 25 = $\frac{1}{4}$ work => x = $\frac{1}{15000}$

Therefore, 100 men in 60 days do = 100 * 60x = **6000x** work = 6/15 = 2/5

Total work done =$\frac{1}{4}+\frac{2}{5}=\frac{13}{20}$

Therefore, Remaining work = 1 - $\frac{13}{20}$ = $\frac{7}{20}$

Remaining time = 130 - (25+60+10) = 35 days

Therefore, work is done in 25 days by 150 men.

Therefore, Work is done in 35 days by 150 men.

Hence, he should employ 50 more men.

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+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) 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) 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

$\left.\begin{array}{c}\mathrm{Till}10.00\mathrm{am}\mathrm{Pipe}\mathrm{P}\mathrm{filled}20\%\\ \mathrm{Till}10.00\mathrm{am}\mathrm{Pipe}\mathrm{Q}\mathrm{filled}8\%\\ \mathrm{Till}10.00\mathrm{am}\mathrm{Pipe}\mathrm{R}\mathrm{filled}2.5\%\end{array}\right\}30.5\%$

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) 17 + 4/7 days | B) 13 + 1/3 days |

C) 15 + 3/2 days | D) 16 days |

Explanation:

C alone can finish the work in 40 days.

As given C does half as much work as A and B together

=> (A + B) can do it in 20 days

(A + B)s 1 days wok = 1/20.

A's 1 days work : B's 1 days Work = 1/2 : 1 = 1:2(given)

A's 1 day’s work = (1/20) x (1/3) = (1/60) [Divide 1/20 in the raio 1:2]

B's 1 days work = (1/20) x (2/3) = 1/30

(A+B+C)'s 1 day's work = (1/60) + (1/30) + (1/40) = 9/120 = 3/40

All the three together will finish it in 40/3 = 13 and 1/3 days.

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) 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.