Why do I disagree with my calculator?

Question

I followed the order of opperations but my answer disagrees with my calulator's.

Problem: $331.91 - 1.03 - 19.90 + 150.00$

Calculator answer: $460.98$
My answer: $162.98$

Why the discrepancy?

Answer 1

It is clear that you did the operation $$331.91 - 1.03 - (19.90 + 150.00) $$ while the calculator instead simply did $$(331.91 - 1.03 - 19.90) + 150.00$$ A strict order of operations (PEMDAS) is in some cases not really the actual convention used in actual mathematics in some sense. When you see something like

$$3.2345-9000+2345$$ the intent of the author is $$(3.2345-9000)+2345$$ as subtraction is really just addition by a negative (as division is just multiplacation by an inverse) and it is quite convenient to just work left to right in these cases.

Answer 2

The calculator is correct of course (at least in magnitude of the answer, I haven't actually calculate it, and I won't). I bet you are having problems with those negative terms.

For this particular exercise and infinite more like this one, you can take another approach, more intuitive than just following rules like a robot. Look at what you are being asked: it's a sum. A sum with negative and positive numbers. We can already see that the positive numbers exceed in magnitude the negative ones, so there is no way in hell that your answer is correct.

Structurally you could add all positives, add all negatives and then add those two groups, just like you can identify the 'subject' and 'predicate' part in a grammatical sentence. In my experience, that has always minimize sign-related errors.

Answer 3

"I follow the order of operation."

Addition (+) and subtraction (-) have the same priority; and should be evaluated from left to right.

Another responder noted you did the addition first, then the subtractions.

Answer 4

Obviously calculator is correct.Steps to do it manually: since + and - have same priority we can solve them in any order so add $331.91$ and $150.00$ (since they have same sign) and add $1.03$ and $19.90$ (same sign) $$331.91+150.00=481.91$$ and$$-1.03-19.90=-20.93$$ then $$481.91-20.93=460.98$$ Another way: $$331.91-1.03-19.90+150.00\implies 330.88-19.90+150.00\implies 310.98+150.00 \implies 460.98$$

Answer 5

As PVAL has pointed out, the problem seems to be a misinterpretation of the PEMDAS mnemonic for operator precedence. You seem to be interpreting it to mean:

1. Parentheses
2. Exponentiation
3. Multiplication
4. Division
5. Addition
6. Subtraction

And thus interpreting the expression $331.91−1.03−19.90+150.00$ as $331.91−1.03−(19.90+150.00) = 160.98$.

However, the actual convention used in mathematics (and in most computer and calculator programming languages) is:

1. Parentheses
2. Exponentiation
3. Multiplication and Division
4. Addition and Subtraction

That is, the + and - operators have equal precedence, and are evaluated in strict left-to-right order. So, $331.91−1.03−19.90+150.00$ means $((331.91−1.03)−19.90)+150.00 = 460.98$.

Also, note that the Muliplication/Division precedence level works slightly differently in mathematical notation than in programming languages, due to implicit multiplication and two-dimensional fraction notation. In a spreadsheet or a computer program, the fraction $\frac{ab}{cd}$ must be written in a "linear" form with explicit multiplication as a * b / (c * d), where the parentheses are required because * and / have equal precedence (just like + and -), and so the unparenthesized expression a * b / c * d would be interpreted as $\frac{a \times b}{c} \times {d}$. In mathematical typesetting, however, the expression $ab/cd$ is ambiguous and prone to get interpreted as $\frac{ab}{cd}$ instead of $\frac{ab}{c} d$.

Answer 6

To add a bit to the other answers, especially in applied disciplines it is often common to first do an approximation. In your terms given by $$331.91 −1.03−19.90+150.00$$ rounding to the next $50$ gives $$300 + 150$$ so your results should be in the ballpark of $450$. If this is not the case something went completely wrong. This simple example can of course be generalized to other operations but the approach stays the same.