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There is food for 800 men in a fortress for 30 days. After 10 days 400 more men join in. How long the food will last while food per man is 2/3 from before.

I listed ratios as following 

Days    Men            Food
30      800            1   (all food)
x       1200 (+400)    2/3 (since 1/3 of food has been consumed already by 800 men)

30 : x ::   1200 : 800   inversely proportional
            1    : 2/3   directly proportional
 x * 1200 * 1 = 30 * 800 * (2/3)
 x = 13.33

I get the answer while the book tells me it is 20.

I can figure out what I am doing wrong.

Sayam Qazi
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  • Your proportion is incomplete: you have two-thirds the amount of food you started with after the first ten days, but it is going to be consumed two-thirds as rapidly each day ("2/3 of the food per man") from then on. So you need another factor of 2/3 in your equation. – colormegone Mar 18 '16 at 17:59
  • @RecklessReckonner I have found out the solution, What should I do now? – Sayam Qazi Mar 18 '16 at 18:02
  • If you're asking what to do with your equation, you have the number of days the remaining food will last ( x ) times 1200 men equal to 30 days times 800 men times 2/3 of the original amount of food. But the rate at which it will be consumed per man in those x days is 2/3 of the original rate. So there should be another factor of 2/3 on the left hand side, given the way you set this up. – colormegone Mar 18 '16 at 18:10

1 Answers1

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Suppose the daily ration starts at R per man. So there must be 24000R initially. After 10 days there is 16000R left. The number of men increases to 1200 or 40R/3 per man. The ration is reduced to 2R/3, so the food will last for 20 days.

almagest
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