Estimation

Round each number to 1 significant figure: 345 ≈ 300 and 7 is already 1 significant figure. So the estimate is \(300 \times 7 = 2100\).

The actual answer is \(345 \times 7 = 2415\). The estimate of 2100 is in the right ballpark — it tells us the answer is in the low thousands, which is enough to spot errors like 24150 or 241.5 from misplaced decimal points or extra zeros.

Round to 1 significant figure: 0.48 ≈ 0.5 and 62 ≈ 60. The estimate is \(0.5 \times 60 = 30\).

The actual answer is \(0.48 \times 62 = 29.76\), so the estimate of 30 is remarkably close. This question is useful because students often struggle to round decimals to 1 significant figure — they need to recognise that 0.48 has two significant figures (4 and 8), and rounding to one gives 0.5.

Round to 1 significant figure: 8.7 ≈ 9 and 0.032 ≈ 0.03. The estimate is \(9 \div 0.03 = 300\). To see why: \(9 \div 0.03\) is the same as asking “how many lots of 0.03 fit into 9?” Since \(0.03 \times 300 = 9\), the answer is 300.

The actual answer is \(8.7 \div 0.032 = 271.875\). The key skill here is recognising that 0.032 has only two significant figures (3 and 2) — the leading zeros are not significant. Students who think 0.032 rounds to 0.0 will get stuck.

Round each value to 1 significant figure: 5.2 ≈ 5, 38 ≈ 40, and 0.49 ≈ 0.5. The estimate becomes \(\dfrac{5 \times 40}{0.5} = \dfrac{200}{0.5} = 400\).

The actual answer is \(\dfrac{5.2 \times 38}{0.49} \approx 403.3\), so the estimate is very accurate here. This shows how estimation works with multi-step calculations — round every value first, then calculate with the simpler numbers.

Round the base to 1 significant figure: 5.1 ≈ 5. The power stays the same. The estimate is \(5^3 = 5 \times 5 \times 5 = 125\).

The actual answer is \(5.1^3 = 132.651\). A common misconception is trying to round the power as well, but in estimation, we only round the numbers being operated on, not the structural mathematical operations themselves.

Example: 0.00823

Another: 0.00751

Creative: 0.00849 — right at the upper boundary, or 0.007500001 — barely rounds up to 0.008 rather than staying at 0.007.

Trap: 0.08 — a student might confuse 0.08 with 0.008, but 0.08 is ten times larger and is already written to 1 significant figure. The leading zeros in 0.008 are not significant — a common source of confusion.

Example: \(3.5 \times 6.5 \to 4 \times 7 = 28\). Actual: 22.75. Both round up, so the estimate is larger.

Another: \(85 \times 25 \to 90 \times 30 = 2700\). Actual: 2125. Again, both round up.

Creative: \(0.65 \times 0.45 \to 0.7 \times 0.5 = 0.35\). Actual: 0.2925. Both numbers are less than 1, and both round up — so the overestimate effect compounds through multiplication.

Trap: \(6.5 \times 1.4 \to 7 \times 1 = 7\). Actual: 9.1. A student might think “6.5 rounds up so the estimate must be bigger.” But 1.4 rounds down from 1.4 to 1 — losing nearly a third of its value. This rounding down dominates, making the estimate an underestimate, not an overestimate.

Example: 34 and 28 — both round to 30.

Another: 0.072 and 0.068 — both round to 0.07.

Creative: 450 and 549 — both round to 500, yet they are 99 apart! This shows how much information rounding to 1 significant figure can discard.

Trap: 35 and 45 — a student might think “both end in 5 so they round the same way.” But 35 rounds to 40 (the 5 tells you to round the 3 up to 4) and 45 rounds to 50 (the 5 tells you to round the 4 up to 5). They round to different values.

Example: \(5.4 – 4.8 \to 5 – 5 = 0\). Actual: 0.6.

Another: \(7.2 – 6.8 \to 7 – 7 = 0\). Actual: 0.4.

Creative: \(0.54 – 0.51 \to 0.5 – 0.5 = 0\). Actual: 0.03. Both decimals round to the same value, wiping out a real difference. This shows a major limitation of estimation for subtracting close numbers.

Trap: \(5.1 – 4.1 \to 5 – 4 = 1\). Actual: 1. A student might think “the numbers are close, so the estimate must be zero” — but here the first significant figures are different (5 vs 4), so the estimate gives 1, not 0. It doesn’t satisfy the condition.

Example: \(\sqrt{15}\). Rounding 15 to 1 s.f. gives \(\sqrt{20}\), which is approximately 4.47 — not an easy mental calculation! Rounding to the nearest square number (\(\sqrt{16}\)) gives a clean estimate of 4.

Another: \(\sqrt{82}\). Rounding to 1 s.f. gives \(\sqrt{80}\), which isn’t an integer. Rounding to \(\sqrt{81}\) cleanly estimates to 9.

Creative: \(\sqrt{0.8}\). Rounding to 1 s.f. keeps it as \(\sqrt{0.8}\) (unhelpful), whereas changing it to a nearby manageable number like \(\sqrt{0.81}\) yields 0.9.

Trap: \(\sqrt{38}\) — a student might mistakenly round this to \(\sqrt{40}\) following the 1 s.f. rule and stop. This perfectly illustrates that while 1 s.f. rounding is standard for most operations, estimating roots requires rounding to nearby square numbers instead.

It depends on whether the number rounds up or down. For example, 74 rounds to 70 (smaller), but 86 rounds to 90 (larger). And 50 rounds to 50 (unchanged).

This tackles the common belief that “rounding always makes things smaller.” In fact, rounding can increase, decrease, or preserve a value. It just depends on the digit after the first significant figure.

By definition, 1 significant figure means keeping only one non-zero digit. For example: 345 → 300 (one non-zero digit: 3), 0.067 → 0.07 (one non-zero digit: 7), 99 → 100 (one non-zero digit: 1).

Students might say “sometimes” because 300 looks like it has three digits — but it has only one non-zero digit. The trailing zeros are placeholders, not significant. Similarly, 0.07 has only one non-zero digit; the leading zeros are there to show the place value.

For two-digit numbers, they match. For example, 73 rounds to 70 using either method. But for three-digit numbers they diverge: 734 to the nearest 10 is 730, while 734 to 1 significant figure is 700.

This targets the key confusion between significant figures and place-value rounding. Rounding to 1 s.f. is not the same as rounding to the nearest 10, 100, or any fixed column — the column it rounds to depends on the size of the number.

Rounding to 1 significant figure keeps the first non-zero digit, so the result is always non-zero (for any positive number). For example, 0.004 rounds to 0.004, 0.047 rounds to 0.05, and 3 stays as 3. No matter how small the number, it always has a first significant figure.

Students who confuse significant figures with place-value rounding might think 0.3 “rounds to 0” (rounding to the nearest whole number), or that 0.047 “rounds to 0.0” (rounding to 1 decimal place). But rounding to 1 significant figure is fundamentally different — it finds the first non-zero digit and rounds there.

(Note: In physics or engineering, an exceptionally small number might be considered negligible, but purely mathematically, rounding to 1 s.f. guarantees we hunt down that first non-zero digit, no matter how microscopic it is).

It works fairly well if the numbers are of similar size and round in opposite directions. For example, \(41 + 52 \approx 40 + 50 = 90\) (actual is 93).

But it fails disastrously when adding numbers of vastly different magnitudes. For example, \(45,000 + 12\). Rounding both to 1 significant figure gives \(50,000 + 10 = 50,010\), which is further from the true answer (45,012) than if we hadn’t rounded the 12 at all! Addition estimation requires preserving the magnitude of the largest number.

The student has misidentified the first significant figure. They treated the 0 in the tenths column as the first significant figure and rounded there. But leading zeros are never significant — the first significant figure in 0.067 is the 6 (in the hundredths column). The 7 then tells us to round up.

The correct rounding is \(0.067 \approx 0.07\), giving an estimate of \(0.07 \times 40 = 2.8\). The actual answer is \(0.067 \times 42 = 2.814\). The student’s answer of 4 is nearly double the true value — a significant error caused by misunderstanding what “significant” means.

The answer happens to be correct — 480 is a good estimate — but the reasoning reveals a flawed “keep the first digit” shortcut. The student isn’t rounding; they’re truncating. For these numbers, truncating coincidentally gives the correct result because 6.3 rounds down to 6 and 78 rounds up to 80.

This shortcut breaks badly for numbers like 0.067. “Keeping the first digit” would give 0 (or 0.0), not 0.07. It also fails for 8.7 — keeping the first digit gives 8, but rounding to 1 significant figure gives 9. The correct method is rounding to 1 significant figure: identify the first non-zero digit, look at the next digit, and round up or down accordingly.

The student has rounded 0.52 to the nearest whole number instead of to 1 significant figure. They treated all the values as needing to become whole numbers. But 0.52 to 1 significant figure is 0.5, not 0. The first significant figure is 5 (in the tenths column), and the 2 tells us to round down.

The correct estimate is \(\dfrac{200 \times 4}{0.5} = \dfrac{800}{0.5} = 1600\). The actual answer is approximately 1571. This is a critical misconception: rounding to 1 significant figure does not mean rounding to the nearest whole number. For numbers less than 1, the rounded value will also be less than 1.

The student has confused significant figures with decimal places and place-value rounding. They rounded 8.35 to 1 decimal place (getting 8.4) and 29 to the nearest 10 (getting 30). Neither of these is rounding to 1 significant figure.

8.35 rounded to 1 significant figure is 8 (the first significant figure is 8; the 3 after it tells us to round down). 29 rounded to 1 significant figure is 30 (this one happens to be correct). The correct estimate is \(8 \times 30 = 240\). The actual answer is \(8.35 \times 29 = 242.15\). The student’s answer of 252 isn’t wildly off, but their method shows a fundamental misunderstanding of what “1 significant figure” means.