SOLUTION: The 250 soldiers in a camp had enough food for forty days, After 10 days, 50 soldiers joined them. For how long will the remaining food last them? Solve using variation.

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Question 1012045: The 250 soldiers in a camp had enough food for forty days, After 10 days, 50 soldiers joined them. For how long will the remaining food last them? Solve using variation.
Found 2 solutions by ValorousDawn, MathTherapy:
Answer by ValorousDawn(53) About Me  (Show Source):
You can put this solution on YOUR website!
I don't even think variation is a legitimate way to solve this problem but here's something.
Lets say each soldier eats 1 food unit every day. Forty days for 250 soldiers is 10,000 food units. After 10 days, they are down to 10,000-250*10=8,750. If there are now 250%2B50=300+soldiers, they then subsequently will eat 300 food units a day. They will have food for 8750%2F300=29.16 days. If you must round, you'd round down, for 29 days.

Answer by MathTherapy(10551) About Me  (Show Source):
You can put this solution on YOUR website!

The 250 soldiers in a camp had enough food for forty days, After 10 days, 50 soldiers joined them. For how long will the remaining food last them? Solve using variation.
Let S be number of soldiers, k = constant of variation, j = amount of food, and T = time 
S+=+k%28j%29%2FT
250+=+k%281%29%2F40
250+=+k%2F40
k = 250(40), or 10,000
Let fraction consumed after 10 days, be j
S+=+k%28j%29%2FT
250+=+10000%28j%29%2F10
250+=+10000j%2F10
250+=+1000j
j, or fraction consumed after 10 days = 250%2F1000, or 1%2F4
With 1%2F4 of food consumed 3%2F4 remains, and 300 (250 + 50) soldiers are in camp
S+=+k%28j%29%2FT
300+=+10000%283%2F4%29%2FT
300+=+7500%2FT
300T = 7,500 ------- Cross-multiplying
T, or time the remaining 3%2F4 of food will last = 7500%2F300, or highlight_green%2825%29 days