# A Test Dilemma: Do As You’re Told, or Do What’s Right?

#### (A new question of the week)

Some questions we get, while small, raise interesting issues. In a question we got last month, there are several little issues pertaining to how the final answer should be chosen; as is often the case, it seems that a diligent student who cares about accuracy might be penalized. This has important implications for the design of tests. Let’s take a look.

## The question

A question says,

What is the smallest amount of money that a man needs to deposit annually for there to be an amount balance of over $15,000 at 6% interest after 12 years? (Round your answer to the nearest dollar.) According to the formula provided, the answer in decimal is$889.155. When 889 is plugged in the formula, the balance is less than $15,000. When 890 is plugged in the formula, the balance is over$15,000. So, in this case, should the answer be 889 according to the instruction stated inside the parentheses or 890 according to the context?

To do a quick check before answering, I brought up Excel and used its FV (future value) formula to find the amount. With deposits of $889, the final amount comes to$14,997.38, which is indeed too little; it doesn’t answer the question! With deposits of $890, we end up with$15,014.25, which is enough. Jing Zhu is right (assuming payments are at the end of each year, which we’ll be looking at below).

We haven’t been told what formula was provided; that will be one issue (and is a reason we prefer that you show us the entire problem you are working on, include such context as formulas you use). The main  issue is how to round.

## Rounding issues

I replied:

That’s a good question! I think it is a matter of interpretation.

What is the smallest whole number of dollars that a man needs to deposit annually for there to be an amount balance of over $15,000 at 6% interest after 12 years? My view of the question as stated is that the answer to the question is$889.18, and then we follow the directions for reporting our answer, by rounding it to $889. There is another issue here, having to do with details that are missing from the problem. Is the interest compounded annually? When are the deposits made — at the beginning or end of each year? You say a formula was provided, so perhaps you didn’t have to know these details in order to choose a formula. Based on your answer, I suppose the formula provided was some form of the first formula in the “Regular Deposits” section of this Ask Dr. Math FAQ: Interest Calculations This agrees with my thoughts, but says it more plainly: As a teacher, we would rather change the wording of the question so that the correct answer is clearly correct; as a student, we would take the wording literally and round after finding the truly correct answer. And why does his wording result in a better question? This way, the rounding is an inherent part of the problem, not an afterthought tacked on for the sake of uniform answers. It takes rounding seriously, requiring thought about which way to round in order to fulfill the needs of the problem, which can be important in real life. And in so doing, it prevents the conflict we’ve seen in trying to answer it. As for the formula, the FAQ says this: Suppose you opened an account at a bank which was paying an annual interest rate of i (a fraction, equivalent to 100i%). You make a deposit of M at the end of each of q equal time periods each year (including the end of the last period). The interest is compounded once per period. Then the value P of the account at the end of n years is given by P = M([1+(i/q)]nq-1)(q/i) This formula, expressed more clearly, is $$P=M\left(\left[1+\left(\frac{i}{q}\right)\right]^{nq}-1\right)\frac{q}{i}$$ which can be rearranged slightly to $$P=\frac{Mq}{i}\left(\left[1+\left(\frac{i}{q}\right)\right]^{nq}-1\right)$$ For annual payments, $$q = 1$$ (because payments are made once a year), and the formula becomes $$P=\frac{M}{i}\left(\left[1+i\right]^{n}-1\right)$$ This formula is derived from the formula for the sum of a geometric series, which is particularly easy to see in this case of annual payments. If payments are made at the end of each year, for n years, then the last payment earns no interest, adding only M to the final amount; the next-to-last payment grows for 1 year, adding $$M(1+i)$$ to the total; and so on, until the first payment grows over $$n – 1$$ years, contributing $$M(1+i)^{n-1}$$. The total is therefore $$M+M(1+i)+…+M(1+i)^{n-1} = M\frac{\left(1+i\right)^{n}-1}{(1+i)-1}$$ which leads to our formula. ## The wrong formula? Jing Zhu answered us, providing the entire problem: Thank you for your reply. I attached the original question for your reference. This is an SAT prep question, and because of the word “over” (“over$15,000), I think we should use inequality to solve this problem. As you can see, the final answer turns out to be greater than 889.155. Therefore, I’d argue that the only correct answer should be 890.

The formula is the same as ours, where $$M$$ is called $$P_0$$, and $$i$$ is called $$r$$. But …

I replied,

Thanks. Interestingly, though the formula you were given is the same one we give in our FAQ that Dr. Rick referred to, their description is wrong. This formula (for an “ordinary annuity”) applies when payments are made at the end of each period (year), not at the beginning as they say (which is called an “annuity due”, and is mentioned after the main discussion in our FAQ). But we have to go by their formula, so we can ignore that discrepancy.

When payments are at the start of each period, each earns an extra period of interest, so the total is multiplied by $$\left(1+\frac{i}{q}\right)$$, leading to the FAQ’s formula, $$P=M\left(\left[1+\left(\frac{i}{q}\right)\right]^{nq}-1\right)\left(1+\frac{q}{i}\right)$$ as shown in the FAQ.

The problem is valid if we change their word “beginning” to “end”. So the work will be the same whether we see this error or not. But if a student knew the correct formula, the problem would be still more confusing!

Here is the work to solve problem 38, as I would write it:

$$\frac{P_0}{0.06}\left(\left[1+0.06\right]^{12}-1\right)>15,000$$

$$P_0>\frac{(0.06)15,000}{\left[1.06\right]^{12}-1}$$

$$P_0>\frac{(0.06)15,000}{1.012196}=889.15544$$

So the smallest amount (in pennies) that satisfies this is found by rounding up to $889.16; and the smallest amount (in whole dollars) is found by rounding up to$890.

## Wisdom for test-takers

I continued:

Your use of an inequality, and your work itself, are good. We are left with the same dilemma we discussed: to take it literally and round the correct answer of $889.16 to the (technically incorrect) answer,$889, just because they told you to; or to answer Dr. Rick’s modified question, which makes more practical sense, and say $890. If I were you, I would answer$889, which seems to be what they are asking for, knowing that you are wiser than that. They said to find the correct answer, and then to round that to the nearest dollar, so that is what you do.

If this were given on an actual test, and it was a question on which you show work, you could explain all this in writing and certainly get credit. If you could only give the numerical answer, then I would enter \$890.

And it’s quite likely that the actual test would not have such a dilemma built in. (Sometimes there have been bad problems, and they have been discovered and accounted for.)

A similar point was made by teacher CTB in a comment to a post from last year, while I was writing this post:

Having myself also worked as an examiner (for A-level Maths – Stats module), it is unlikely that in an exam you would be penalised for any sensible decision you make to get to an answer. If you write down your assumption somewhere in your solution, so that the examiner can read it, you should get credit for your calculations.

When you can show work, you can be much more confident of being judged fairly, and therefore less anxious about how to answer.