### APTITUDE- Problems on CHAIN RULE

**CHAIN RULE**

**1.DIRECT PROPORTION:** Two Quantities are said to be directly proportional,if on the increase (or decrease) of th one, the other increases(or decreases) to the same extent.

Ex:(i) Cost is directly proportional to the number of articles.(More articles,More cost).

(ii)Work done is directly proportional to the number of men working on it. (More men, more work).

**2.INDIRECT PROPORTION:** Two Quantities are said to be indirectly proportional,if on the increase of the one , the other decreases to the same extent and vice-versa.

Ex:(i) The time taken by a car covering a certain distance is inversely

proportional to th speed of the car.(More speed, less is the time taken to cover the distance).

(ii)Time taken to finish a work is inversely proportional to the number

of persons working at it.(More persons, less is the time taken to finish a job).

NOTE: In solving Questions by chain rule, we compare every item with the term to be found out.

**SIMPLE 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 Â¼ 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 Â½ 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.

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