1. ## 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

2. Hello Ron
Originally Posted by rn5a
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 $\displaystyle 5$ days, the $\displaystyle 120$ men have $\displaystyle 25$ days of provisions left. A further $\displaystyle 5$ join them (making $\displaystyle 125$ men in all), which will reduce the number of days left in the ratio $\displaystyle 120:125$.

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 $\displaystyle \frac{120}{125}$.

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

3. 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

4. ## Ratios and multiplying factors

Hello Ron
Originally Posted by rn5a
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 $\displaystyle 10$% in a sale, and is now priced at £$\displaystyle 7.11$. What price was it offered at originally?
Using multiplying factors, the method is this:
The original price is $\displaystyle 100$%; the sale price is $\displaystyle 100 - 10 = 90$%; a ratio of $\displaystyle 100:90$. 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 $\displaystyle \frac{100}{90}$.

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

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# a garrison of 120 men has provisions for 30 days.At the end of 5days 5 more men join them.How many days can they sustein on the remaining provision

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