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Question from garth, a parent:

A maths teacher promised a learner that she would pay him R8 for each problem solved correctly,
but she would fine him R5 every incorrect solution.
At the end of 26 problems neither owed any money to the other.
How many problems did the learner solve correctly?

We have two responses for you

Hi Garth,

I would use algebra to solve this problem.

Suppose the learner solved p problems correctly. He got R8 for each problem he solved correctly and hence he received R8 × p.

For how many did he give an incorrect solution? He received a fine of R5 for each incorrect solution so what was his total fine?

Since neither the learner nor the teacher owed money at the end the amount the learner received was equal to his total fine. Solve for p.

Penny

 

Garth,

I'm going to suggest an old-fashioned method that some 19th century school textbooks used to teach. It can be used for a lot of problems like this one.

Can you find _some_ M and N such that M wrong answers balance N right answers? (Hint: play around with the numbers in the original problem.) For what total number of problems would this be a correct answer?

Now scale this solution to get a solution to the real problem.

Good Hunting!
RD

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