# Percentage problem

• Dec 4th 2009, 06:24 PM
nvwxgn
Percentage problem
I need help with the following percentage question:

In a 400-people population, it has been done examinations to diagnose anemia and intestinal parasitism. The results made it clear that:

- 80% of the people with anemia have also intestinal parasitism;

- 50% of the people with intestinal parasitism have also anemia;

- 220 people have neither anemia nor intestinal parasitism.

Out of the 400 people, what is the percentage of people with anemia?

I know that the answer is 25%, but how do I get at it?
• Dec 4th 2009, 09:44 PM
TKHunny
Quote:

Originally Posted by nvwxgn
I need help with the following percentage question:

In a 400-people population, it has been done examinations to diagnose anemia and intestinal parasitism. The results made it clear that:

- 80% of the people with anemia have also intestinal parasitism;

- 50% of the people with intestinal parasitism have also anemia;

- 220 people have neither anemia nor intestinal parasitism.

Out of the 400 people, what is the percentage of people with anemia?

I know that the answer is 25%, but how do I get at it?

You must start defining.

A = # of folks with anemia
P = # of folks with parasites
B = # of folks with both
220 = # of folks with neither

We have immediately:

1) A + P - B = 400-220 = 180
2) A(0.80) = B
3) P(0.50) = B

Solving - Substitute 2) and 3) into 1)

(b/(0.8) + B/(0.5) - B) = 180 ==> B = 80

Back-Substitute into 2)

A = 80/(0.8) = 100

Note: I did not see how to solve this problem until AFTER I wrote clear and sufficient definitions. Only after those four rows did I see where this was going and that the information given was sufficient.

Note: Notice how I did NOT solve for P. P was not the most convenient first solution (B) and P was not the item asked for (A). Again, the defintions, clear and sufficient, made it possible to know what to go for first.
• Dec 4th 2009, 10:58 PM
bigwave
you can also do this with just one varible

let A = people with anemia

since half the people with intestinal parasitism have A and those with A .8 have intestinal parasitism

then
$
A + .8A = 180$

so $A = 100$

and $\frac{400}{100} = .25$

I should mention that I also drew a diagram of what was going on... then it all came down to something
obviously simple. word problems really need to be transformed into a way that is clear what is happening
• Dec 5th 2009, 12:19 PM
nvwxgn
Good. Thanks a lot