how to solve chain rule problems

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?

a)      50 days                b) 30 days          c) 40 days           d) 60 days

Answer:

Let “x” be the number of days the provisions would last for the garrison of “y” men

We can re phrase the question now.

“ there is provision for y men for ( x – 10) days, and there is provision for ( y – y/5 = 4y/5) men for x number of days”

So, x = (x – 10) * {y/(4y/5)}

Or, x = (5/4)( x – 10)

Or, 4x = 5x – 50

Or, x = 50 days. answer is a.

If 40 men can make 30 boxes in 20 days, How many more men are needed to make 60 boxes in 25 days?

a)      28           b) 24                      c) 22                       d) 26

Answer:

40 x (60/30) x (20/25) =  64 men

So, extra men = 64 – 40 = 24. Answer is b.

Note: for making more number of boxes you will need more number of men. So fraction must be more than 1............... (60/ 30)

In more number of days you will need less number of men.

So, the fraction should be less than 1............................(20/25)

In a camp, food was sufficient for 2000 people for 54 days. After 15 days, more people came and the food last only for 20 more days. How many people came?

a)      1900       b) 1800                 c) 1940                  d) 2000

Answer:

We will re-phrase the question

There is food for 2000 people for (54 – 15 = 39) days. Food lasted for 20 days. So, how many people where there.

= 2000 x (39/20) = 3900

So, extra people who joined was = 3900 – 2000 = 1900

answer is a