- 24 men working 5 hours a day can finish a work in 12 days. Find the number of men required to finish the same work in 8 days working 6 hours a day?
(24*5*12)/1 = (x*8*6)/1
- The work done by a man, a woman and a child is in the ratio of 3 : 2 : 1. There are 20 men, 30 women and 48 children in a factory. Their weekly wages amount to Rs 840, which is divided in the ratio of work done by the men, women and children. What will be the wages of 10 men, 20 women and 30 children for 1 week?
Ratio of wages of 20Men, 30Women and 48Children
=3*20 : 2*30 : 1*48
=5: 5: 4.
Wages of 20 Men per week =5/14 *840=Rs300.
Then wages of a man per week =300/20=Rs15.
Similarly, wages of a woman per week=Rs10
wages of a children per week=Rs5.
Total wages per week =10*15 + 10*20 + 5*30
- A and B working separately can do a piece of work in 6 and 8 days, respectively. If they work for a day alternately with A beginning, the work would be complete in
7days6 3/4days10days7 2/3daysNoneOption B
B—–8——–3 (LCM 24)
In 2 days both work 7unit.
Then in 6days they completed 21unit work.
Remaining 3unit work completed by A in 3/4days.
Total days=6 3/4days.
- 12 children and 15 men complete a certain piece of work in 7 days. If each child takes twice the time taken by a man to finish the work, in how many days will 14 men finish the same work?
12days11 1/3days10 1/2days11daysNoneOption C
14M—–?==> (21*7)/14=10 1/2days.
- A can do a piece of work in 15 days, B in 20 days. A and B work at it together for 6 days and then C finishes it in 3 days, in how many days could C have done it alone?
B—–20—–3 LCM 60.
A and B work together 7 unit in 1day.
Then in 6 days both completed 7*6=42unit.
Remaining 18unit completed by C in 3 days.
Then 60unit—–? ==>10days.
- 4 men can do a piece of work in 2 hours, which 5 women could do in 3 hours, or 10 children in 4 hours. How long would 1 man, 1 woman and 1 child together take to do the work?
5 7/12days5days4days4 8/13daysNoneOption D
1man can do a work in 2*4 =8hrs
1woman can do a work in 3*5=15hrs.
1children can do a work in 4*10=40hrs.
1 man, 1 woman and 1 child together take to do the work
=(8*15*40) / [(8*15) +(15*40) + (40*8)] =(8*15*40) /(120+600+320)
- Pipe A and Pipe B can fill a cistern in 12 hours and 15 hours respectively. When a third pipe C which works as an outlet pipe is also open then the cistern can be filled in 20 hours. The outlet pipe can empty a full cistern in
Let the pipe C empty the tank in x hrs.
1/12 +1/15 -1/x=1/20
- C is twice efficient as A, B takes thrice as many days as C. A takes 12 days to finish the work alone. If they work in pairs (i.e., AB, BC, CA) starting with AB on the first day then BC on the second day and AC on the third and so on, then how many days are required to finish the work?
6days5 1/9days7 2/3days8daysNoneOption B
A takes 12 days.
(A+B)’s 1 day work =1/12+1/18 =5/36
(B+C)’s 1 day work =1/18 +1/6 =8/36
(C+A)’s 1 day work =1/12+1/6 =9/36
In 5 days they worked =(5+8+9+5+8)/36
Remaining 1/36 work done by CA 1/36*36/9=1/9days.
Total days=5 1/9.
- If 12Men and 16Boys can do a piece of work in 5days. 13 Men and 24 Boys can do it in 4 days, Then the ratio of the work done by a man to that of a boy is?
(12M+16B) *5 = (13M+24B) *4
60M+80B = 52M+96B
- A and B can do a piece of work in 45 days and 40 days respectively. They began to do the work together but A leaves after some days and then B completed the remaining work in 23 days. The number of days after which A left the work was
B—–40—–9unit LCM 360
B work for 23days. B complete =23*9=207unit.
Remaining work =153.
Required time =153/17=9days.