100 Percent of Daily Allowance of Iron

Date: 09/30/2001 at 18:30:48
From: Amy
Subject: Percentage Word Problem

A common foodstuff is found to contain .00125% iron. The serving size 
is 87.0 grams. If the recommended daily allowance is 18mg of iron, how 
many servings would a person have to eat to get 100% of the daily 
allowance of iron?

About all I know at this point is .018 grams = 100% iron. I thought 
maybe if you found out how many times .00125 goes into 100, for 100%, 
you could figure something out, but I'm not sure. Please help.

Amy

Date: 10/04/2001 at 10:49:43
From: Doctor Ian
Subject: Re: Percentage Word Problem

Hi Amy, 

If the foodstuff is 0.00125% iron, then there are 0.00125 grams of 
iron for every 100 grams of the foodstuff. This follows from the 
definition of 'percent', so make sure you see why it's true. (Write 
back if you don't see why it's true!)

0.00125 grams is the same as 1.25 milligrams:

  0.00125   0.0125   0.125   1.25
  ------- = ------ = ----- = ----
     1        10      100    1000

So there are 1.25 milligrams in 100 grams of foodstuff.  

If a person needs 18 milligrams of iron, how many times does he need 
to eat 1.25 milligrams of iron? Well, if he eats 1.25 milligrams 10 
times, that's 12.5 milligrams, which means he has to eat 5.5 more 
milligrams.  

If he eats 1.25 milligrams 4 more times, that's 5 milligrams, which 
gets him up to 17.5 milligrams.  

So he has to eat 1.25 milligrams of iron about 14 times to get up to 
18 milligrams of iron. 

Now, each time he does that, he has to eat 100 grams of the foodstuff.  
That adds up to about 1400 grams of foodstuff. 

Now the question is: How many 87-gram servings does he have to eat in 
order to eat 1400 grams? Again, if he eats 10 servings, that's 870 
grams, which leaves him about 530 short. If he eats 3 more, that's 261 
grams, which still leaves him 270 short. If you keep going, you'll end 
up with the approximate number of servings. 

It's important to note that I'm not just looking for the magic, 
one-shot equation that will solve the problem. Instead, I'm trying to 
think through each step, so that I understand what's going on, and I 
can have some confidence that I'm on the right track.  

At the end of all this, I might come up with an equation, or I might 
not. But if I do, I'll know exactly where it came from.  

There is also another way I could go about this, which would be to 
play with dimensions. For example, I start out knowing that a daily 
requirement is 18 milligrams:

  18 mg = (mg per serving) * (number of servings)

So if I can figure out how many mg are in a serving, I'm in good 
shape. How can I do that?  Well, I have to start with what the problem 
gives me, which is mg per 100 grams:

               1.25 mg iron      ? g foodstuff
  18 mg iron = --------------- * -------------- * (number of servings)
               100 g foodstuff   ? servings

I don't know what numbers go in the middle yet, but I know what the 
dimensions have to be in order for the dimensions I don't want to 
cancel out properly. 

Now, in fact, having written it out this way, I can see that I _do_ 
know what numbers have to replace the question marks.  The problem 
tells me that a serving is 87 grams of foodstuff.  So I know that

               1.25 mg iron      87 g foodstuff
  18 mg iron = --------------- * -------------- * (number of servings)
               100 g foodstuff   1 serving
      
Now I can see that my dimensions will cancel out the way I need them 
to:

                                    xxxxxxxxxxx
               1.25 mg iron      87 g foodstuff
  18 mg iron = --------------- * -------------- * (number of servings)
               100 g foodstuff   1 serving
                   xxxxxxxxxxx

               (1.25 * 87) mg iron
             = ------------------- * (number of servings)
               (100 * 1) servings

Now, just looking at the numbers, I can see that 87 is a little 
smaller than 100, and 1.25 is a little larger than 1, so this whole 
thing should come out to be about equal to 1:

            ~ 
            =  1 * (number of servings)

which means that I'd need about 18 servings. Does that seem 
reasonable? Well, working the other way, we got around 16 servings, so 
this is in the ball park. If you use a calculator to work out the 
exact answer, you'll get something slightly different, but not _too_ 
different.  

As I said, there are lots of different ways to go about solving any 
problem, and any solution that you _know_ must be correct is 
preferable to a solution that you _hope_ is correct. For example, here 
is a very simple, if not very elegant, way to solve the problem:

  A common foodstuff is found to contain .00125% iron. 
  A serving size for the foodstuff is 87.0 grams. 

Each serving must then contain
 
                      0.00125
    0.00125% * 87 g = ------- * 87 g          (definition of percent)
                        100

                    = 0.0010875 grams of iron


  If the recommended daily allowance is 18mg of iron, 

18 milligrams is 18 thousandths of a gram, or 18/1000 grams, or 0.018 
grams. 

  how many servings of this foodstuff would a person have to 
  eat to get 100% of the daily allowance of iron?

This is the same as asking: How many times do I need to eat 0.0010875 
grams of iron at a time, in order to eat a total of 0.018 grams of 
iron? Which is another way of asking, how many times does 0.0010875 go 
into 0.018?  

So in this solution, I haven't done anything tricky at all. I just had 
to understand what a percentage is, and how to compute the number of 
times you have to eat some small amount of stuff in order to eat some 
total amount of it. 

Although teachers often insist that you solve a particular kind of 
problem using a particular set of steps, in the real world only two 
things matter: (1) that you come up with the right answer, and (2) 
that you _know_ it's the right answer, without having to ask anyone 
else to check that for you. If that means that you have to draw 
pictures, or even count on your fingers, then that's perfectly okay.  

I hope this helps. Write back if you'd like to talk about this some 
more, or if you have any other questions. 

- Doctor Ian, The Math Forum
  http://mathforum.org/dr.math/