# Time and Work Questions

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 From the second day onwards, one worker was withdrawn each day. The job was finished when the last worker was withdrawn. Had no worker been withdrawn at any stage, the group would have finished the job in 55% of the time.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.

10 2881
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 2839
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 2824
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.

7 2704
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