# 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 =$\inline \frac{1}{n}$

2. If A’s 1 day's work =$\inline \frac{1}{n}$, 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 :

$\inline \dpi{100} \fn_jvn Efficiency \propto \frac{1}{number\; of\; time\; units}$

$\inline \dpi{100} \fn_jvn \therefore Efficiency \times time=constant\; work$

Hence, $\inline \dpi{100} \fn_jvn Required \; time = \frac{work}{efficiency}$

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 = $\inline \fn_jvn \frac{100}{efficiency}$

Q:

A is thrice efficient as B and C is twice as efficient as B. what is the ratio of number of days taken by A,B and C, when they work individually?

 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               $\inline&space;\fn_jvn&space;\left&space;\{$ $\inline&space;\fn_jvn&space;\frac{1}{3}$     :    $\inline&space;\fn_jvn&space;\frac{1}{1}$    :   $\inline&space;\fn_jvn&space;\frac{1}{2}$  $\inline&space;\fn_jvn&space;\left&space;\right&space;\}$

or                                              2    :    6    :    3

Hence A is correct.     $\inline&space;\fn_jvn&space;\left&space;[&space;\because&space;Time&space;\prec&space;\frac{1}{Eficiency}&space;\right&space;]$

10 3482
Q:

A, B and C can complete a piece of work in 24,6 and 12 days respectively.Working together, they will complete the same work in:

 A) 1/24 days B) 7/24 days C) 24/7 days D) 4 days

Explanation:

(A+B+C)'s 1 day's work =$\inline&space;{\color{Black}\left&space;(&space;\frac{1}{24}+\frac{1}{6}+\frac{1}{12}&space;\right&space;)=\frac{7}{24}}$

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

8 3366
Q:

(x-2) men can do a piece of work in x days and (x+7) men can do 75% of the same work in (x-10)days. Then in how many days can (x+10) men finish the work?

 A) 27 days B) 12 days C) 25 days D) 18 days

Explanation:

$\inline&space;\fn_jvn&space;\frac{3}{4}\times&space;(x-2)x=(x+7)(x-10)$

$\inline&space;\fn_jvn&space;\Rightarrow$    $\inline&space;\fn_jvn&space;x^{2}-6x-280=0$

$\inline&space;\fn_jvn&space;\Rightarrow$    x= 20   and   x=-14

so, the acceptable values is x=20

$\inline&space;\fn_jvn&space;\therefore$ Total work = $\inline&space;\fn_jvn&space;(x-2)\times&space;x$ = 18 x 20 =360 unit

Now   360 = 30 x k          $\inline&space;\fn_jvn&space;\because&space;(30=20+10)$

$\inline&space;\fn_jvn&space;\Rightarrow$  k=12 days

8 3356
Q:

A group of workers was put on a job. From the second day onwards, one worker was withdrawn each day. The job was finished when the last worker was withdrawn. Had no worker been withdrawn at any stage, the group would have finished the job in 55% of the time. How many workers were there in the group?

 A) 50 B) 40 C) 45 D) 10

Explanation:

It can be solved easily through option.

$\inline&space;\left&space;(&space;10+9+8+....+1&space;\right&space;)=10\left&space;(&space;10\times&space;\frac{55}{100}&space;\right&space;)$

55 = 55     Hence correct

Alternatively:

$\inline&space;\frac{n(n+1)}{2}=n\times&space;\frac{55n}{100}$

=> n= 10

In Both cases total work is 55man-days.

11 3260
Q:

The ratio of efficiency of A is to C is 5:3. The ratio of number of days taken by B is to C is 2:3. A takes 6 days less than C, when A and C completes the work individually. B and C started the work and left after 2 days. The number of days taken by A to finish the remaining work is:

 A) 4.5 B) 5 C) 6 D) 9 1/3

Explanation:

A     :    B    :    C

Efficiency     $\inline&space;\fn_jvn&space;\rightarrow$         10    :    9    :     6

No of days   $\inline&space;\fn_jvn&space;\rightarrow$         9x    :  10x   :    15x

$\inline&space;\fn_jvn&space;\Rightarrow$       15x-9x = 6

$\inline&space;\fn_jvn&space;\therefore$             x = 1

Number of days taken b A = 9

Number of days taken by B= 10

Number of days taken by C = 15

work done by B and C in initial 2 days = $\inline&space;\fn_jvn&space;\frac{2\times&space;1}{6}$ = $\inline&space;\fn_jvn&space;\frac{1}{3}$

rest work =$\inline&space;\fn_jvn&space;\frac{2}{3}$

Number of days required by A to finish $\inline&space;\fn_jvn&space;\frac{2}{3}$ work = $\inline&space;\fn_jvn&space;\frac{2/3}{1/9}$ = 6 days