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Math Help - Inverse Variation

  1. #1
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    Question Inverse Variation

    Here's a problem on inverse variation which I am not able to solve. Can someone please explain me how to do it?

    A garrison of 120 men has provision for 30 days. At the end of 5 days, 5 more men joined them. How many days can they sustain on the remaining provison?

    Thanks,

    Ron
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  2. #2
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    Hello Ron
    Quote Originally Posted by rn5a View Post
    Here's a problem on inverse variation which I am not able to solve. Can someone please explain me how to do it?

    A garrison of 120 men has provision for 30 days. At the end of 5 days, 5 more men joined them. How many days can they sustain on the remaining provison?

    Thanks,

    Ron
    At the end of the first 5 days, the 120 men have 25 days of provisions left. A further 5 join them (making 125 men in all), which will reduce the number of days left in the ratio 120:125.

    Since the number of days must be made smaller, we create a multiplying factor (a fraction) out of the two numbers in the ratio by putting the smaller number on top; i.e. we multiply the number of days by \frac{120}{125}.

    So the number of days left is 25\times\frac{120}{125}= 24 days.

    Grandad
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  3. #3
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    Thanks mate for your help. I have some doubts lingering in my mind.....

    Suppose I re-phrase the question:

    A garrison of 500 men has provision for 30 days. If 100 more men join the garrison, for how many days will the provison last now?

    This is how I would go about it:

    Let the no. of days be d. Thus,

    500 X 30 = (500 + 100) X d
    15000 = 600d
    d = 25

    This is how I usually do it. Now coming back to my first question & your solution:

    At the end of the first 5 days, 120 men have 25 days of provision left. Thus the solution will be the following:

    Let the no. of days be d. Thus

    120 X (30-5) = (120 + 5) X d
    120 X 25 = 125d
    d = 24

    Have I got it right?

    Thanks,

    Ron
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  4. #4
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    Ratios and multiplying factors

    Hello Ron
    Quote Originally Posted by rn5a View Post
    Thanks mate for your help. I have some doubts lingering in my mind.....

    Suppose I re-phrase the question:

    A garrison of 500 men has provision for 30 days. If 100 more men join the garrison, for how many days will the provison last now?

    This is how I would go about it:

    Let the no. of days be d. Thus,

    500 X 30 = (500 + 100) X d
    15000 = 600d
    d = 25

    This is how I usually do it. Now coming back to my first question & your solution:

    At the end of the first 5 days, 120 men have 25 days of provision left. Thus the solution will be the following:

    Let the no. of days be d. Thus

    120 X (30-5) = (120 + 5) X d
    120 X 25 = 125d
    d = 24

    Have I got it right?

    Thanks,

    Ron
    Yes, your algebraic methods are absolutely fine, and if you're happy with that method, stick to it.

    I like to use what is a now sometimes regarded as the old-fashioned method of multiplying factors (as I explained in my first post) because they often avoid algebra, and lead very quickly to the answer. Here's another example:
    A shirt is reduced by 10% in a sale, and is now priced at 7.11. What price was it offered at originally?
    Using multiplying factors, the method is this:
    The original price is 100%; the sale price is 100 - 10 = 90%; a ratio of 100:90. We need to create a multiplying factor using these numbers that will make the sale price bigger - to get back to the original price. So the bigger number goes on top to make the multiplying factor \frac{100}{90}.

    So the original price was 7.11\times\frac{100}{90}= 7.90.
    See how easy that is?

    Grandad
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