Ratios and multiplying factors

Hello Ron Quote:

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?

Grandad