Saturday, 21 May 2011



1)A contract is to be completed in 46 days and 117 men were set to work,
each working 8 hours a day. After 33 days, 4/7 of the work is completed.
How many additional men may be employed so that the work may be completed
in time, each man now working 9 hours a day?
Sol: 4/7 of work is completed .
Remaining work=1- 4/7
Remaining period= 46-33
=13 days
Less work, less men(direct proportion)
less days, more men(Indirect proportion)
More hours/day, less men(Indirect proportion)

work 4/7:3/7
Days 13:33 :: 117:x
hrs/day 9:8

x=(3*33*8*117) / (4*13*9)
x=198 men
So, additional men to be employed=198 -117=81

2)A garrison had provisions for a certain number of days. After 10 days,
1/5 of the men desert and it is found that the provisions will now last
just as long as before. How long was that?

Sol: Let initially there be x men having food for y days.
After, 10 days x men had food for ( y-10)days
Also, (x -x/5) men had food for y days.
=> (x*y) -50x=(4(x*y)/5)

3)A contractor undertook to do a certain piece of work in 40 days. He
engages 100 men at the beginning and 100 more after 35 days and completes
the work in stipulated time. If he had not engaged the additional men,
how many days behind schedule would it be finished?

Sol: 40 days- 35 days=5 days
=>(100*35)+(200*5) men can finish the work in 1 day.
4500 men can finish it in 4500/100= 45 days
This s 5 days behind the schedule.

4)12 men and 18 boys,working 7 ½ hors a day, can do a piece if work in
60 days. If a man works equal to 2 boys, then how many boys will be
required to help 21 men to do twice the work in 50 days, working
9 hours a day?

Sol: 1man =2 boys
12men+18boys=>(12*2+18)boys=42 boys
let the required number of boys=x
21 men+x boys
=>((21*2)+x) boys
=>(42+x) boys
less days, more boys(Indirect proportion)
more hours per day, less boys(Indirect proportion)

days 50:60
hrs/day 9:15/2 :: 42:(42+x)
work 1:2
(42+x)= (60*15*42)/(50*9)= 84
x=84-42= 42
42 days behind the schedule it will be finished.

No comments:

Post a Comment