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Anton [14]
3 years ago
12

provisions for 630 men to last for 25 days. How many men must be transferred to another camp so that the food lasts for 30 days?

​
Mathematics
1 answer:
algol [13]3 years ago
3 0

Answer:

105

Step-by-step explanation:

25 days = Food for 630 men

30 days = x           (Inverse variation)

30 * x = 630 * 25

x = 630 * 25/30

  = 21 * 25

  = 525 men.

630 - 525 = 105 men.

Therefore, 105 men must be transferred to another camp so that the food lasts for 30 days.

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Answer with the explanation step by step
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Answer:

The answer is A. \frac{3(x-21)}{(x+7)(x-7)}.

Step-by-step explanation:

To find the difference of this problem, start by simplifying the denominator, which will look like \frac{3}{x+7}-\frac{42}{(x+7)(x-7)}. Next, multiply \frac{3}{x+7} by \frac{x-7}{x-7}  to create a fraction with a common denominator in order to subtract from \frac{42}{(x+7)(x-7)}. The problem will now look like \frac{3}{x+7}*\frac{x-7}{x-7}-\frac{42}{(x+7)(x-7)}.  

Then, simplify the terms in the problem by first multiplying \frac{3}{x+7} and \frac{x-7}{x-7}, which will look like \frac{3(x-7)}{(x+7)(x-7)}-\frac{42}{(x+7)(x-7)}. The next step is to combine the numerators over the common denominator, which will look like \frac{3(x-7)-42}{(x+7)(x-7)}.

Next, simplify the numerator, and to simplify the numerator start by factoring 3 out of 3(x-7)-42, which will look like \frac{3(x-7-14)}{(x+7)(x-7)}. Then, subtract 14 from -7, which will look like \frac{3(x-21)}{(x+7)(x-7)}. The final answer will be \frac{3(x-21)}{(x+7)(x-7)}.    

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