Thread: 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

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 days, the men have days of provisions left. A further join them (making men in all), which will reduce the number of days left in the ratio .

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 .

So the number of days left is days.

Grandad

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

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 % in a sale, and is now priced at £ . What price was it offered at originally?

Using multiplying factors, the method is this:

The original price is %; the sale price is %; a ratio of . 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 .

So the original price was £ £ .

See how easy that is?

Grandad