At Midnight the Temperature Was 6°F: The Temperature Dropped 2.5°F Each Hour for 6 Hours, Then Rose 4¾°F — The Final Temperature Was −4.25°F (Option C)

The final temperature was −4.25°F, which is choice C. Here's the arithmetic behind it, in the order it happened. The midnight reading was 6°F. A drop of 2.5°F per hour across 6 hours removes 15°F total, landing the thermometer at −9°F at 6 a.m. Then a rise of 4¾°F pushes the reading up to −4.25°F. The condition that trips most people is the mixed fraction: 4¾°F means 4.75°F, and the two movements are sequential, not simultaneous. Any answer other than −4.25°F comes from a slip in one of those steps.

Why the Answer Is −4.25°F and Not One of the Other Three Choices

Each wrong option maps to a specific mistake, so working backward from them is a useful in this sort of problem. The distractor set here is well designed. Choice A (8.25°F) and choice D (25.75°F) come from adding the movements to 6°F instead of applying the drop first and then the rise in sequence, or from misreading 4¾ as a larger number. Choice B (−13.75°F) comes from subtracting the rise from the post-drop temperature in the wrong direction.

The 20.25°F spread between 25.75°F and −4.25°F is the size of the error you make if you ignore the word "dropped." That's not a rounding issue. It's a sign issue, and sign errors are what most standardized-test temperature items are built to catch.

Negative temperatures follow the same order of operations as positive ones, which is why the problem doesn't actually require knowing anything about weather. You subtract 15 from 6 and get −9. You add 4.75 and get −4.25. The number line does the work.

The Step-by-Step Calculation, With Each Intermediate Value

Run the problem in the order the words give it, and record every intermediate reading. That habit catches sign errors before they compound. It also mirrors how a meteorologist reconstructs an hourly temperature log from a start value and a rate.

1: Write the midnight reading. Start value: 6°F at 12:00 a.m. 2: Compute the total drop. −2.5°F per hour × 6 hours = −15°F. 3: Apply the drop. 6°F − 15°F = −9°F. This is the reading after the sixth hour, at 6:00 a.m. 4: Convert the mixed fraction. 4¾°F = 4.75°F. 5: Apply the rise. −9°F + 4.75°F = −4.25°F. 6: Check the magnitude. The drop was 15°F, the rise was 4.75°F, so the net movement is −10.25°F. Add that to 6°F and you get −4.25°F. Both routes agree.

If you want a single expression for a spreadsheet, use =6-(2.5*6)+4.75. Excel and Google Sheets both return −4.25. Enter 4 3/4 as 4.75, not as three separate entries.

Failure point worth naming: if you round 4¾ to 5°F, you get −4°F and pick choice C anyway by luck, but the reasoning is wrong. If you round to 4.7, you get −4.3 and may talk yourself out of C. Keep the exact 4.75 until the final line.

Modeling the Six Hours as a Colapsed Temperature Log

The problem describes a steady rate over six hours, which lets you build an hour-by-hour table. That table is the check that catches arithmetic slips and the clearest way to show a student why the order matters.

Two things in that table are worth pointing at. First, the reading crosses zero between 2:00 and 3:00 a.m., which is exactly where most students lose the sign. Second, the final rise of 4.75°F is nearly twice as large as any single hourly drop, so it wipes out almost two hours' worth of cooling in one movement. That's a good sanity check on the realism of a textbook problem: a 4.75°F jump in one step is plausible with a warm front or a change in wind direction, but it isn't a steady process, and the problem treats it as a single event rather than a rate.

Why "Dropped 2.5°F Each Hour" Means Multiplication, Not Six Subtractions

The phrase "each hour" signals a rate, and a rate over a stated interval means multiply. Six separate subtractions of 2.5 give the same answer as one multiplication, but the multiplication is faster and less error-prone. The table above shows the six subtractions explicitly so you can see they match.

There's a subtlety in language here that trips up English-language learners and native speakers alike. "Dropped 2.5°F each hour for 6 hours" could be read as "dropped to 2.5°F," which would be a completely different problem. The word "each" rescues it: each hour, the drop was 2.5°F. Read it as "the rate was 2.5°F per hour, downward."

A second subtlety: the 6 hours is a duration, not a count of readings. If you count seven readings (midnight through 6 a.m. inclusive), you might be tempted to apply seven drops. Don't. The interval is 6 hours, so there are six hourly changes. The midnight reading is the starting point, not the result of a change.

Converting 4¾°F Correctly, and the Fraction Errors That Cost Points

Mixed fractions are the single most common source of error in this problem. 4¾ is not 43, not 4.3, and not 4.34. It is 4 + 3/4, which is 4.75. Three quarters is 0.75 because 3 ÷ 4 = 0.75. Nothing more complicated than that.

If you're working on paper and want to avoid decimals entirely, convert everything to quarters. The start is 24 quarters (6 × 4). The drop is 15°F, which is 60 quarters. The rise is 19 quarters (4¾ × 4). So: 24 − 60 + 19 = −17 quarters, and −17 ÷ 4 = −4.25. That's the same answer through a different route, and it avoids the decimal conversion altogether. Some students find quarter-units easier because every value in the problem is already a multiple of a quarter or a half.

What the −4.25°F Result Means in Real Weather Terms

A final reading of −4.25°F is cold but not extreme by continental standards. For context, the National Weather Service treats wind chill as the more safety-relevant number once temperatures drop below freezing, because moving air strips heat from exposed skin faster than still air does. At −4°F with a 10 mph wind, the wind chill lands near −18°F, which is the range where frostbite on exposed skin can begin in roughly 30 minutes.

That's the practical reason temperature-rate problems show up in cold-weather training. A forecast that says "temperature falling 2.5°F per hour overnight" tells a road crew or a livestock operation how fast conditions will deteriorate, and the final value tells them whether they'll be working in single digits or below zero. The arithmetic in this problem is the same arithmetic a dispatcher runs to decide when to send crews out.

One edge case worth flagging: if the "rise" had come before the drop instead of after, the answer would be different only if the rise changed the starting point. Order matters when you're applying changes to a running total. Here, the problem states the sequence clearly, so there's no ambiguity. But in a real forecast, a warm front arriving at 4 a.m. would interrupt the steady cooling, and the log would no longer be a straight line.

Checking Your Work: The Net-Change Method and Sign Audit

The fastest independent check is to compute the net change once and apply it to the start value. The drop is −15°F and the rise is +4.75°F, so the net is −10.25°F. Start at 6°F, subtract 10.25, and you're at −4.25°F. If your step-by-step route and your net-change route disagree, one of them has a sign error.

Sign audits matter more than most people think. Write every movement with its sign in front of it: −15 and +4.75. Then the sum is just arithmetic on signed numbers, and there's no way to accidentally subtract a rise or add a drop.

A second check: estimate first. The drop is 15°F from a start of 6°F, so you know the intermediate is negative, somewhere around −9°F. The rise is less than 5°F, so the final must be between −9°F and −4°F. Only choice C falls in that window. Estimation alone eliminates A, B, and D before you do any exact arithmetic. Most test-taking guides recommend this for exactly this reason: it's faster and it catches gross errors.

Where This Problem Sits in the Standard Temperature-Word-Problem Family

This item belongs to a well-known family: a start value, a rate applied over an interval, then a one-time change. The family shows up in middle school math, in GED and HiSET practice sets, in ASVAB arithmetic reasoning sections, and in nursing dosage calculations where the "temperature" is replaced by a drug concentration. Once you recognize the shape, the method is fixed: apply the rate, then apply the one-time change, then check the net.

The family has variants worth knowing. Some versions give the rate in degrees per hour and ask for the time when the temperature reaches a target, which requires division instead of multiplication. Others give the start and end values and ask for the rate, which requires subtraction then division. The version here is the easiest configuration, which is why it appears early in most curricula.

A version-specific note: some textbooks write negative temperatures with the degree symbol before the minus sign, as in −4.25°F, and some write it as -4.25 °F with a space. Both are acceptable in most style guides, but the number is identical. Don't let formatting distract from the math.

Common Mistakes Ranked by How Often They Appear

Ranking the errors is more useful than listing them, because it tells you where to spend your checking time. In classroom use, the sign error on the drop is the most frequent, followed by the fraction conversion, followed by misreading the duration.

The seventh-drop error deserves a note. If you count the readings from midnight to 6 a.m. inclusively and apply a drop to each, you get 7 × 2.5 = 17.5°F of cooling, landing at −11.5°F before the rise. The problem says "for 6 hours," so six drops is correct. Count intervals, not clock strikes.

FAQ

What if the temperature had risen 4¾°F before the 6-hour drop instead?

The answer would be different only in sequence, not in arithmetic, and the final value depends on which start value the rise applies to. If the rise came first, midnight would effectively read 10.75°F before cooling begins, and six hours of −2.5°F per hour would land at −4.25°F anyway. That symmetry surprises people: adding then subtracting the same magnitudes in either order gives the same endpoint because addition commutes. The order only changes the intermediate readings, not the final one.

Does −4.25°F mean 4.25 degrees below zero on the Fahrenheit scale?

Yes. The minus sign means below zero on the Fahrenheit scale, which is 32 degrees below the freezing point of water. In Celsius terms, −4.25°F converts to about −20.14°C using the formula (F − 32) × 5/9. That's a hard freeze by any measure, and it's the kind of reading that shows up in upper-Midwest January forecasts.

Why isn't the answer −13.75°F?

−13.75°F comes from subtracting the 4.75°F rise from the −9°F intermediate instead of adding it. The problem says the temperature "rose 4¾°F," and a rise always moves the reading upward, toward zero and beyond. Subtracting a rise inverts its direction and produces an answer that's 9.5°F too cold. If the problem had said "fell 4¾°F" after the drop, then −13.75°F would be correct.

Is this a typical standardized-test question, and which tests use it?

It's a standard arithmetic reasoning item of the kind used in GED Mathematical Reasoning, HiSET, and ASVAB practice sets, and it appears in middle-school ratio and rate units. The exact wording varies, but the structure — start value, rate over an interval, one-time change — is one of the most reused templates in test prep. The mixed-fraction detail is the main difficulty knob; easier versions use whole numbers.

What's the total temperature change from midnight to the end?

The net change is −10.25°F, computed as 4.75 − 15. That's a cooling of just over 10 degrees across the whole period, even though the drop alone was 15 degrees. The rise recovers nearly a third of the cooling. If you're tracking a cold snap, the net change is the number that matters for pipe-freeze risk, not the peak drop.

Can I solve this without converting 4¾ to a decimal?

Yes, and some people find it cleaner. Work in quarter-degrees: start at 24 quarters, subtract 60 quarters for the drop, add 19 quarters for the rise, ending at −17 quarters, which is −4.25°F. Every value in the problem is a multiple of a quarter or a half, so the quarter-unit method avoids decimal conversion entirely and reduces the chance of a rounding slip.

What if the 2.5°F drop applied only for 5 hours?

Then the drop would be 12.5°F, the intermediate would be −6.5°F, and the final after the 4.75°F rise would be −1.75°F, which isn't among the choices. That tells you the problem intends the full 6 hours, and it's a useful check: if your arithmetic lands on a value not in the option set, re-read the duration before re-doing the subtraction. The options are designed to bracket the correct answer, not to include every possible misreading.

How would I explain this to someone who's never seen a negative temperature?

Use money as the analogy. Starting with $6, losing $2.50 six times puts you at −$9, meaning you owe $9. Then someone gives you $4.75, and you owe $4.25. The debt is the negative temperature. The analogy holds because signed numbers behave identically whether they represent dollars or degrees, and it gives a concrete mental image for why adding a positive to a negative moves you toward zero without necessarily crossing it.