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Advanced Math/% Improvement (Corrected)


QUESTION: Hi Randy - I think my previous email was scrambled so I'm sending this newer version. Thanks.

Hi Randy - I have two questions.  I'm teaching a class and I would like to take a baseline and measure the percent improvement.  So let's say the baseline is a set of homework assignments and the student only scores 5% correctly.  Three months later the student takes the same homework assignment and scores 100% correctly. So would I score the improvement as follows:

(100-5)/5 = (95/5) = 19X  so the improvement is 19 Fold

But could I also express it as a percentage improvement as well? If so, how?

My second question is as follows:

What is the student scores 0% initially? Then we'd have:
(100-0/0) = (100/0) = But we can't divide by 0.  How can we handle the situation here?

ANSWER: For the first question a factor of 19 translates into 1800%. To arrive at this result, consider the case where the student gets a score of 30 on the first test and a score of 60 on the second. This is a 2-fold increase and is usually represented as a 100% improvement. That is, the improvement is the sum of 100% of the old score plus the old score.

Similarly, a factor of 3 means the improvement is 2-times the original plus the original. This is represented as a 200% improvement. Continuing this way, a factor of 19 would be represented by adding 18 times the original to the original. The factor of 18 is 1800% of the original.

For the second question, you're right that you can't divide by 0. In this situation, it doesn't really make sense to calculate an improvement score. In fact, it seems to me, that even if the first score is non-zero but really small (like 1 out of 100), the improvement score would not be very meaningful.

Hope this helps.

---------- FOLLOW-UP ----------

QUESTION: Hi Randy - I follow most of your explanation except for: " That is, the improvement is the sum of 100% of the old score plus the old score." and " Continuing this way, a factor of 19 would be represented by adding 18 times the original to the original."

Would the way I calculated 19X be perfectly fine to use?

In my classroom, it is very possible that a student may score a 0.  In such a case, I would offer one-to-one coaching to the student.  In such a situation as we've seen, we can't divide by zero, can you make a suggestion to handle such a situation? (For example, should we artificially make the initially score 5%?)  Thank you kindly.

Roberta, I'm afraid my previous answer was not quite right (not to mention confusing). Your method of calculating the improvement is correct. Converting it to a percentage just means multiplying the factor by 100. In my example of getting a 30 and then a 60 would be

(60-30)/30 = 1 --> 100%.

Sorry for the confusion.

Regarding students getting a zero on the first test, setting a baseline of 5% sounds reasonable albeit arbitrary.

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randy patton


college mathematics, applied math, advanced calculus, complex analysis, linear and abstract algebra, probability theory, signal processing, undergraduate physics, physical oceanography


26 years as a professional scientist conducting academic quality research on mostly classified projects involving math/physics modeling and simulation, data analysis and signal processing, instrument development; often ocean related

J. Physical Oceanography, 1984 "A Numerical Model for Low-Frequency Equatorial Dynamics", with M. Cane

M.S. MIT Physical Oceanography, B.S. UC Berkeley Applied Math

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