Maths Week Scotland 2024 - Challenge 4 - Test Scores

Part of MathsMaths Week Scotland 2024

Challenge 4 - Test Scores

Challenge 4 is all about working out averages.

Maths teacher Chris Smith and pupils from Grange Academy are here to explain.

The Maths Week Scotland Daily Challenges have been set by the Scottish Mathematical Council.

So here's the challenge:

Mairi is about to take the last of a series of tests.

If she gets 23 in this test, it will improve her average by 1.

If she scores 39, it will improve her average by 3.

What was Mairi’s average score before the last test?

A pile of test papers with 'Average = ?' and two papers next to them, one which shows a score of 23 and 'average +1' and another below which shows a score of 39 and 'average + 3'

Need a hint?

  • Think about the differences between the scores, what that can tell us?
  • You don’t need to know the number of tests but it might help
  • Could you use algebra to make two equations that you can compare?

Solution

Worked out the answer? Here's how you can do it.

Two equations stacked: 39 average +3 subtract 23 average +1

Step 1

We know that a score of 23 will improve Mairi's average by 1.

And a score of 39 will improve her average by 3.

We can subtract and compare the difference.

Two equations stacked: 39 average +3 subtract 23 average +1
39 average +3 subtract 23 average +1 = 16 average +2

Step 2

39 - 23 = 16

3 - 1 = 2

So 16 marks increases the average by 2.

39 average +3 subtract 23 average +1 = 16 average +2
8 average +1 appears below the last sum

Step 3

This means 8 marks increases the average by 1.

Remember that a score of 23 will also increase the average by 1.

8 average +1 appears below the last sum
Pile of test papers with average = 15 alongside.

Step 4

So if we subtract 8 from 23, we will get back to the average.

23 - 8 = 15

So Mari's previous average score was 15.

Pile of test papers with average = 15 alongside.

Solution using algebra

Another way to solve this is by using algebra.

If Mairi has to sit \(n\) tests overall, then she has currently completed \(n - 1\) tests.

The symbol for a common average is \(\bar{x}\), so Mairi's current total score is \((n - 1)\bar{x}\).

If Mairi sits the final test and scores \(23\), we can write this by adding \(23\) to the expression:

\((n-1)\bar{x}+23\)

We also know that this increases her average by one. We can write this as:

\(n(\bar{x}+1)\)

Turn this into an equation:

\(n(\bar{x}+1)=(n-1)\bar{x}+23\)

Simplify:

\(n\bar{x}+n=n\bar{x}-\bar{x}+23\)

\(n=-\bar{x}+23\)

Rearrange:

\(n+\bar{x}=23\)

If Mairi sits the final test and scores \(39\), we can write this as:

\((n-1)\bar{x}+39\)

We also know that this increases her average by three. We can write this as:

\(n(\bar{x}+3)\)

Turn this into an equation:

\(n(\bar{x}+3)=(n-1)\bar{x}+23\)

Simplify:

\(n\bar{x}+3n=n\bar{x}-\bar{x}+39\)

\(3n=-\bar{x}+39\)

Rearrange:

\(3n+\bar{x}=39\)

We can subtract one of these equations from the other to find the value of \(n\):

\(3n+\bar{x}-(n+\bar{x})=39-23\)

\(2n=16\)

\(n=8\)

We can now put the value of \(n\) into one of the equations to find the value of \(\bar{x}\).

\(n+\bar{x}=23\)

\(15+\bar{x}=23\)

\(\bar{x}=8\)

So Mairi sits eight tests overall. She has already sat seven tests and her current average score is 15.

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