# Continuing from last post, the tutor gives an example of a question involving the mean.

You will often find a question such as this on a qualifying exam:

Bill writes six math tests. They all have equal weighting towards his final mark. Suppose, after five tests, his average (mean) mark is 71, yet he needs a final average of at least 73. What mark must he manage on his sixth test?

Solution:

Let x be the mark he needs on his sixth test. If his average after five tests is 71, the equivalent situation is that he got 71 on each of those five. Our general formula for average (mean) is as follows:

mean=(sum of all values)/(number of values)

In Bill’s case,

mean=(71+71+71+71+71+x)/6=73

which simplifies to

(355+x)/6=73

Multiplying both sides by 6 gives

355+x=438

Subtract 355 from both sides, giving

x=83

Apparently, Bill needs 83 on his sixth test to bring his average from 71 to 73. Here’s one way to see it: he needs 73 on his last, plus 2 points for each of the five he got 71 on. 73+5(2)=83.

I’ll be further discussing the mean with applications in future posts:)

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

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