While there have already been some efficient approaches to this question posted in this thread, I thought that I would also show how this question can be solved quickly using the

weighted average mapping strategy (the link contains a list of questions that can be used to practice using the weighted average mapping strategy, also knows as the tug of war).

In general, the weighted average mapping strategy can be used to solve problems like this, where 1 more item is being added to a group. One weight will be the number in the original group (4 in this question), and the other weight will be 1, the size of the "group" of one being added; these are shown above the line in the diagram below. We know the average of the original group (78) and the final weighted average (80), so we can fill these in below the line in the diagram. We need to solve for the "average" of the group of 1 (which will be the score on the 5th test) by making the ratio of the weights equal to the ratio of the distances. The ratio of the weights is 4:1, and the smaller distance is 2, so the larger distance must be 8. Thus, the score of the 5th test is 80+8 = 88.

When I drew the diagram above by hand, I timed myself and found that it took me just 12 seconds. Please let me know if you have any questions, or if you want me to post a video solution!

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