Aptitude - Chain Rule - Discussion

No equations nothing required, simple understanding
For better connection with the question, lets take boys and rice bags instead of cows and husk bags.

40 boys eat 40 bags of rice in 40 days

the gang of 40 boys eat 1 bag in 1 day
" " 40 boys eat 2 bags in 2 days
" " " " " 3 bags in 3 days

Similarly, group of 40 boys eats 40 bags in 40 days.

Think if 39 people are not there 1 boy can't eat 1 bag in 1 day it takes 40days for him to complete,

As good as saying 1 boy will take 40 days to eat 1 bag.

40 cows/1 cow = 40 days/1 day (direct proportionate).

Let 40 cows/1 cow = c.

40 days/1 day = d.

c/40 bags = d/x bags.

Putting value of c = 40 & d = 40, we get x = 40;

Question will get rather tough when values differ (say 40 cows, 30 bags, 20 days and we have to find 1 cow to eat 1 bag in how many days.

Solution will be:

c = 40, d = 20 putting in direct relation.

= 40/30 = 20/x.

=> x = 15.

Direct hit method will not be applicable in such type of calculation based method.

@Navin.

I think this is the most simple and procedural method. Just think and got down on increasing one parameter how it affects the other parameter as simple as that,

But yet another initiative way to this problem:

40 cows eat 40 bags in days = 40.

So,1 cow will eat 40 bags in days = 40 * 40 (how? One cow will obviously take more time to eat as compared to 40 of them!).

1 cow will eat just 1 bag in days= (40 *40) /40 (how? 1 bag will require less time to be finished!).

= 40.

EASIEST METHOD:

First, we fix the data 40 days:
40 cows ----> 40 bags
1 cow -----> ? bags
? = 40 ÷ 40
=1 bag.

40 days ----> 1 bag
1 day -----> ? bag
? = 1 ÷ 40.
= 1/40 bag.

1 day ------> 1/40 bag.
? day -------> 1 bag.
? = 1 ÷ 1/40.
= 40 days.

40 cows eat 40 bags in 1 day
Therefore work done by 40 cows = 40/40 = 1
work done by 1 cow = 1/40
Therefore 1 cow can do 1/40 of whole work
Hence 1/40 + 1/40 + 1/40.....40 times = 1 that is the whole work
Therefore 1 cow will take 40 days to eat 1 bag.

Simple thing is one 40cows eat 40bags in 40days.
So, that mean 1cow eating 1bag 40days..

I want to solve this problem from reverse...lets think if 1cow eat 1 bag in 1day..for 40days 40cows*40bags=1600.

Now you got it?

40 cows 40 bags 40 days.

Considering cows and days.
40 cows 40 day.
=> 1 cow 1600 day(indirect proportion).

Above result is for 40 days.

So for 1 day ans = (1600)/40.
= 40 days.

40 cows eat 40 bags in 40 days
40 cows eat 40/40 bags in 1 days (i.e.40 cows eat 1 bags in 1 days)
therefore,
1 cow eat 1/40 bags in 1day
and
1 cow eat 1 bags in 40 day

Holy cowwa cowwa!!!

40C-40D----40B
1C-1D----(40B/(40C*40D))
i.e. 1C-1D----1/40B

Now since 1 cow in 1 day consume 1/40 bags,that means

1/40B consumed in- 1Dday by cow is- 1
So 1B in 1D is 1/(1/40)==40 days

40 cows eat 40 bags in 40 days, i.e 40 cows eat 1 bag in 1day.

In that 1 bag, each cow eats 1/40th in 1 day.

So, how many 1/40th make 1 bag.

1/40 * x = 1,

x = 40.