business mathematics

Chain Rule problems

Chain Rule problems

Problems

1)If 15 toys cost Rs.234, what do 35 toys cost ?

Sol: Let the required cost be Rs. x then
more toys more cost(direct proportion)
15:35:: 234:x
(15*x)=(234*35)
x=(234*35) /(15)= 546 Rs

2)If 36 men can do a piece of work in 25hours, in how many hours
will 15men do it?

Sol: Let the required number of hours be x.
less men more hours(Indirect proportion).
15:36::25:x
(15*x)=(36*25)
x=(36*25) /15
x=60
For 15 men it takes 60 hours.

3)If 9 engines consume 24metric tonnes of coal, when each is working
8 hours a day, how much coal will be required for 8 engines, each
running 13 hours a day, it being given that 3 engines of former type
consume as much as 4 engines of latter type?

Sol: Let 3 engines of former type consume 1 unit in 1 hour.
4 engines of latter type consume 1 unit in 1 hour.
1 engine of former type consumes 1/3 unit in 1 hour.
1 engine of latter type consumes ¼ unit in 1 hour.
Let required consumption of coal be x units.
Less engines, less coal consumed.(direct)
More working hours, more coal consumed(direct)
Less rate of consumption, less coal consumed (direct)

9:8
8:13 :: 24:x
1/3:1/4

(9*8*(1/3)*x)=(8*13*(1/4)*24)

24x=624

x=26 metric tonnes.

Complex Problems

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
=3/7
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

(4/7)*13*9*x=(3/7)*33*8*117
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-10)=(4x/5)*y
=> (x*y) -50x=(4(x*y)/5)
5(x*y)-4(x*y)=50x
x*y=50x
y=50

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