Average (Arithmetic Mean)

The answer is C. Calling numbers “consecutive” means they must be next to each other in order. So we can imagine this sequence starting $5$, $6$, $7$ … and averaging $9.5$. Common sense tells us we’ll need to have some numbers above $9.5$, but how many? We could plug in the answers and work backward to check, but a smart approach here emphasizes the fact that, in an arithmetic (evenly spaced) sequence, the median always equals the mean. So the middle number will be the same as the average, and therefore the average will have an equal number of integers below it as above it. If we start counting at $5$, we’ll find that there are five integers below $9.5$, so there must also be five integers above it: $10$, $11$, $12$, $13$, and $14$.

You can also solve this problem by recognizing that in an arithmetic sequence, the average is equal to the smallest number plus the largest number divided by two. So in this case we could set up an equation, using $f$ for “first number” and $l$ for “last number”, as follows:

$$\begin{array}{rl}\frac{f+l}{2}& =\text{average}\\ \frac{5+l}{2}& =9.5\\ 5+l& =19\\ l& =14\end{array}$$

The answer is B. The CLT doesn’t allow a calculator because the makers of the test want you to exercise your mathematical reasoning and logic. We should feel free to estimate on this question, especially because most of the choices are relatively far apart from each other. Taking a moment to observe the three values we’re averaging, we notice that the lower two are quite close to each other, while the highest is much higher. We can rule out choice A is much too low, given that the much higher value of $36.5$ is involved.
Let’s round the highest answer up to $37$ so we can add whole numbers. Adding the three numbers we have now results in a sum of $72$. Happily, that number is divisible by $3$, and we need to divide by $3$ since there are three total values. This gives us $24$, and answer B is far closer to that value than are either choices C or D.

One other note: you might realize that the two smaller values are going to pull down the answers quite a bit in this scenario. After all, there are too smaller numbers and only one larger one. Think of this like a weighted average, a “tug of war” between the smaller numbers and the larger one. The result should certainly be closer to $17$ and $18$ than it is to $36.5$. In this light, the two largest answer choices appear considerably too high.

The answer is C. Like our approach question to this lesson, this question shows the importance of calculating the sum in an average question. Using the sum formula given in this module, we multiply $8\times 16$ to get the current sum of $128$. This puts in a position to accept new entries into the system and simply add them to our existing total. The 15-year-old and the 18-year-old bring a total of $33$ years, so our new sum is $161$. Fortunately, our new number of students is $10$, and dividing by $10$ only involves moving the decimal place left by one. $161÷10=16.1$.

The answer is A. To answer this challenging question we need to keep our sights on the word “must”; any of the statistical values that do not have to change can be eliminated. Starting with the mean, or average, we find perhaps the most straightforward conclusion: doubling one of the numbers must change the mean because it changes the sum. As long as the number of elements in the set doesn’t change, the mean will be larger. Before we continue, note that we can already eliminate choices C and D because statement I is true. Interestingly, we don’t even have to test the median, because the only thing we need to distinguish between choices A and B is a conclusion about the mode!

To address the mode, we recall that the mode is the value that appears the most in a list. In this question, each number is different, so each number is evenly the mode. If doubling the fourth-lowest number makes the result the same as another number in the list, then the mode would change: the number that appears twice would be the new mode. The problem is that we don’t know whether doubling the fourth-smallest number makes it equal to another number or not. Since we can’t know for sure, we eliminate statement III and therefore answer B. If you’re curious about why statement II doesn’t have to be true (the inevitably conclusion if A is the answer!), consider this: we know that the fifth number in this list would be the median since it’s the middle of nine numbers. The number we’re doubling is just one before the median in order. So wouldn’t the median change because the fourth-smallest number would hop over the fifth number in value when doubled? We don’t know that for sure because there could be a large gap between the fourth and fifth numbers. For example, if the fourth-smallest number was $10$ and the median was $30$, doubling the fourth-lowest number would still leave it smaller than the median, so the median wouldn’t change.

The answer is B. By now you may be getting used to the first step on challenging average problems: find the sum. Since $20\times 75=1500$, we know that the total score of all 20 tests was $1,500$. We can now subtract $94$ and arrive at our new total of $1,406$. We know this total needs to be divided by the number of tests, but remember that this number is now $19$, not $20$.

Dividing such large numbers might look challenging, but keep in mind that the question doesn’t say anything about rounding, so the answer must be exact. Furthermore, we can use our math reasoning to conclude that, if we remove a number higher than the average, the average must go down. Focusing on choices A and B, we see that choice A is an odd number. Is that number multiplied by $19$ going to give us an even number result? No, it never could. So it must be the case that $1,406÷19=74$. Feel free to check our logic by working out the math!