The Power of Estimation

There's an old saying that says, “Almost only counts in horseshoes and hand grenades.” In grade school and high school, you were probably taught that math had to be precise; maybe you even had that unforgivingly drilled into you. Well, now you're preparing for the Quant, and the rules are different. On Quant Math, “almost” can be good enough to count.

It's a fact that you cannot use a calculator on the Quant. Therefore, it's also a fact that the writers of the QAUNT can't expect you do long calculator-type calculations on the Quant. They can't expect you to multiply & divide ugly four-digit numbers and get an exact answer – but they can, and will, expect you to estimate.

When should I estimate on QAUNT math?

The short answer is: whenever you would have to be a Will-Hunting-type savant to figure out the exact result in your head, that's a clue that you should ditch the exact answer altogether and try estimating. The QAUNT may give you the green light by using the words “the estimated value of” or “approximately.” Another clue is the spread of the answer choices. If the answer choices are all very close together, well, then it's going to take some precision to distinguish among them. But, if the answer choices are widely spaced, estimating will get you close enough to the right answer.

What the QAUNT will and won't ask

Here is an example of a question that will not appear on the QAUNT Math: “Jill invests \${10000} in an account that pays an annual rate of 3.96\%, compounding semi-annually. Figure out the exact amount she has after two years.” True, that might have been a question in high school math, but definitely not on the QAUNT. First of all, it's not Problem Solving or Data Sufficiency, so it's not the right question type. Moreover, nobody short of a savant-sadist is going to expect you come up to the exact answer that question without a calculator. You will absolutely not have to do a problem like that.

Here, though, is a suspiciously similar question, and one that the QAUNT could pose:

1. Jill invests \${10000} in an account that pays an annual rate of 3.96\%, compounding semi-annually. Approximately how much does she have in her account after two years?

(A) \${10079.44}

(B) \${10815.83}

(C) \${12652.61}

(D) \${14232.14}

(E) \${20598.11}

Solution: first of all, notice the magic word “approximately” — the test-writer is letting us know estimation is perfectly fine. Furthermore, the answer choices are nicely spread out, which will facilitate estimating.

OK, get ready for some fast & furious estimation. The interest rate 3.96\% is an ugly number, so I'm going to approximate that as 4\%. It compounds semiannually, so that means that there's 2\% every six months, and that happens four times in two years. Well, 2\% of \${10000} is \${200}. If you get \${200}, or a little more, on four occasions, that's a little more than \${800} in interest. We expect an answer slightly higher than \${10800}, so of course (B) is just right.

Notice, I estimated so that everything up until the last sum was single-digit math. Single-digit calculations are a good standard for which to strive when you are practicing estimation.

By the way, if you find the bank that will do answer (E), double your money in only two years, that's terrific, but it probably is something wildly illegal, a Ponzi scheme or worse! In the real world, that just doesn't happen. On word problems, especially in financial situations, you should always have your antenna up for what's realistic or unrealistic.

Practice Question

2. ACME's manufacturing costs for sets of horseshoes include a \${11,450} initial outlay, and \${19.75} per set. They can sell the sets for \${52.50}. If profit is revenue from sales minus manufacturing costs, and the company produces and sells 987 sets of horseshoes, what was their profit?

(A) \${20,874.25}

(B) \${30,943.25}

(C) \${41,308.50}

(D) \${51,817.50}

(E) \${53,624.25}

Answer and Explanation

So here's some more fast and furious estimation. Initial manufacturing outlay: round that from \${11450} to \${10000}. Cost per set: round to \${20}. Sales revenue per set: \${50}. Number produced & sold: 1000. OK, now we're in business.

cost = 10000 + 20*1000 = 10000 + 20000 = \${30000}.

sales \space revenue = 50*1000 = \${50000}.

profit = (sales \space revenue) - (cost) = \${50000} - \${30000} = \${20000}

Answer choice (A) is the only answer even close to that. Single digit calculations all the way, and it was enough to get the answer!

If you would like to share your thoughts on this or ask a question, please let us know in the Comments section below!