Probing Questions: Rounding to Significant Figures
Probing Questions

Rounding to Significant Figures

Questions designed to stretch thinking, reveal misconceptions, and spark mathematical reasoning.

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Convince Me That…

Students must construct a mathematical argument for why each statement is true.

1
Convince me that 0.00372 rounded to 2 significant figures is 0.0037
๐Ÿ’ก Possible Argument

The leading zeros (0.00) are not significant — they are just placeholders showing how small the number is. The first significant figure is 3 and the second is 7. To round to 2 s.f., look at the third significant figure: 2. Since 2 < 5, we round down, giving 0.0037.

A common mistake is to count the zeros and say the answer is 0.00 (confusing significant figures with decimal places) or to include the leading zeros in the count and think the first two significant figures are 0 and 0. The key principle: start counting significant figures from the first non-zero digit.

2
Convince me that 4372 rounded to 2 significant figures is 4400, not 44
๐Ÿ’ก Possible Argument

The first two significant figures of 4372 are 4 and 3. The next digit is 7 (≥ 5), so we round up: 43 becomes 44. But we must keep the number in the thousands — we need placeholder zeros to preserve its size. The answer is 4400, not 44.

Think of it on a number line: 4372 sits between 4300 and 4400. Since 4372 is closer to 4400, that’s where it rounds. Writing “44” would place us between 40 and 50 — a completely different part of the number line.

4300 4400 4372 44? 44 is way over here near 0!
3
Convince me that 3.50 has 3 significant figures, not 2
๐Ÿ’ก Possible Argument

In 3.50, the digits 3, 5, and 0 are all significant. The trailing zero after the decimal point counts because it was deliberately included to show precision — it tells us the measurement is accurate to the nearest hundredth, not just the nearest tenth.

Compare 3.5 and 3.50: they have the same value, but 3.5 has 2 s.f. while 3.50 has 3 s.f. The difference matters in science and measurement: 3.50 says “we measured carefully enough to know the hundredths digit, and it was 0.” That trailing zero carries information about precision.

4
Convince me that 0.09961 rounded to 2 significant figures is 0.10
๐Ÿ’ก Possible Argument

The first significant figure is 9 and the second is 9. The third significant figure is 6 (≥ 5), so we round up. But rounding 99 up gives 100 — the 9 in the second position becomes 10, which carries over. So 0.099… becomes 0.10.

The trailing zero in 0.10 is important — it shows the answer has been rounded to 2 significant figures. Writing just 0.1 would suggest only 1 significant figure. This is a case where rounding pushes a number across a “boundary” — from the thousandths into the tenths.

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Give an Example Of…

Think carefully — the fourth box is a trap! Give a non-example that looks right but isn’t.

1
Give an example of a number that has exactly 3 significant figures
An example
Another example
One no-one else will think of
A sneaky non-example
๐Ÿ’ก Possible Answers

Example: 4.72

Another: 0.00305

Creative: 10.0 (trailing zero counts) or Standard Form: \(1.20 \times 10^4\) (the coefficient 1.20 has 3 s.f.).

Trap: 0.52 — a student might count “0, 5, 2” and think there are 3 significant figures. But the leading zero is not significant, so 0.52 has only 2 s.f.

2
Give an example of two different numbers that both round to 4.7 to 2 significant figures
An example
Another example
One no-one else will think of
A sneaky non-example
๐Ÿ’ก Possible Answers

Example: 4.72 and 4.68

Another: 4.74 and 4.66

Creative: 4.7049 and 4.6501 — pushing to opposite ends of the range that rounds to 4.7.

Trap: 4.75 and 4.65 — a student might pair these up symmetrically, but 4.75 to 2 s.f. gives 4.8 (the third s.f. is 5, which rounds up), not 4.7. Meanwhile 4.65 does correctly round to 4.7.

3
Give an example of a number that changes by more than 100 when rounded to 1 significant figure
An example
Another example
One no-one else will think of
A sneaky non-example
๐Ÿ’ก Possible Answers

Example: 1490 → 1000 (change of 490)

Another: 5600 → 6000 (change of 400)

Creative: 14 999 → 10 000 (change of 4999) — the bigger the number and the closer the leading digits are to a boundary, the larger the possible change.

Trap: 5040 → 5000 (change of only 40). A student might think “it’s a big number so rounding must change it by a lot” but the change here is only 40, well under 100.

4 โœฆ
Give an example of a number where rounding to 1 significant figure moves it into a different power of ten
An example
Another example
One no-one else will think of
A sneaky non-example
๐Ÿ’ก Possible Answers

Example: 960 → 1000 (jumps from hundreds to thousands)

Another: 0.0097 → 0.01 (jumps from thousandths to hundredths)

Creative: 0.9996 → 1 — crosses from below 1 to exactly 1, jumping from the tenths into the units.

Trap: 470 → 500. A student might think “it jumped from 470 to 500 — that’s a big change!” but both numbers are still in the hundreds. No change in power of ten.

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Always, Sometimes, Never

Is the statement always true, sometimes true, or never true? Students should justify their decision with examples.

1
Rounding a number to fewer significant figures gives a smaller result
SOMETIMES

It’s true that 3.42 rounded to 2 s.f. gives 3.4, which is smaller. But 3.67 rounded to 2 s.f. gives 3.7, which is larger than the original.

Rounding can make a number larger (when rounding up), smaller (when rounding down), or leave it unchanged (when the dropped digits are all zeros, as in 3.40 → 3.4). The direction depends on the digit being examined, not on the act of rounding itself.

2
Leading zeros in a decimal are significant figures
NEVER

Leading zeros in decimals like 0.003 or 0.0742 are never significant. They are placeholders that show position — how far right of the decimal point the significant digits are. The number 0.003 has just 1 significant figure (the 3), not 4.

Students sometimes think more digits means more significant figures, but leading zeros exist only to indicate scale, not precision. You can verify this using standard form: 0.003 = 3 × 10−3. Converting to standard form strips away leading zeros entirely, confirming they carry no information about measurement accuracy.

3
Rounding to 1 significant figure always gives a number with just one non-zero digit
ALWAYS

By definition, 1 significant figure means keeping just one significant digit. Since significant figures start from the first non-zero digit, the result always has exactly one non-zero digit, plus any placeholder zeros needed to preserve magnitude.

Examples: 47 → 50 (one non-zero digit: 5), 0.0083 → 0.008 (one non-zero digit: 8), 6.7 → 7 (one non-zero digit: 7), 3200 → 3000 (one non-zero digit: 3). The placeholder zeros in results like 50 or 3000 are not non-zero digits — they simply hold the number’s position on the number line.

4
If two numbers round to the same value to 2 significant figures, they must also round to the same value to 1 significant figure
SOMETIMES

It works in many cases: 4.32 and 4.37 both round to 4.3 (2 s.f.) and both round to 4 (1 s.f.).

But consider 0.0348 and 0.0351. Both round to 0.035 to 2 s.f., but to 1 s.f., 0.0348 rounds to 0.03 while 0.0351 rounds to 0.04 — they go in different directions! This happens whenever two numbers straddle a 1 s.f. rounding boundary while agreeing at 2 s.f.

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Odd One Out

Which is the odd one out? Can you make a case for each one? There’s no single right answer!

1
Which is the odd one out?
0.030
2.0
500
๐Ÿ’ก A Case for Each
500 is the odd one out — it’s the only one where the number of significant figures is ambiguous. It could be 1, 2, or 3 s.f. depending on context. To fix this, mathematicians use Standard Form: \(5 \times 10^2\) (1 s.f.), \(5.0 \times 10^2\) (2 s.f.), or \(5.00 \times 10^2\) (3 s.f.).
0.030 is the odd one out — it’s the only one less than 1, and the only one containing leading placeholder zeros. Both 2.0 and 500 are greater than 1 with no leading zeros.
2.0 is the odd one out — it’s the only one with no placeholder zeros at all. Every zero in 2.0 is significant. By contrast, 0.030 has leading placeholder zeros (0.0) and 500 has trailing placeholder zeros.
2
Which is the odd one out?
0.065
0.650
0.605
๐Ÿ’ก A Case for Each
0.065 is the odd one out — it’s the only one with exactly 2 significant figures. Both 0.650 and 0.605 have 3 significant figures.
0.650 is the odd one out — it’s the only one with a trailing significant zero (its last significant digit is 0). In 0.065 and 0.605, the final digit is a non-zero number (5 in both cases).
0.605 is the odd one out — it’s the only one that rounds down when rounded to 1 significant figure (0.605 → 0.6, since the second digit is 0). The other two both round up: 0.065 → 0.07 and 0.650 → 0.7.
3
Which is the odd one out?
3.45
3.55
3.050
๐Ÿ’ก A Case for Each
3.55 is the odd one out — it’s the only one that rounds up when rounded to 1 significant figure (3.55 → 4). Both 3.45 and 3.050 round down to 3.
3.050 is the odd one out — it’s the only one with 4 significant figures (3, 0, 5, 0). Both 3.45 and 3.55 have exactly 3 significant figures.
3.45 is the odd one out — it’s the only one where rounding to 2 s.f. first and then to 1 s.f. gives a different answer than rounding directly to 1 s.f. Cascade: 3.45 → 3.5 → 4. Direct: 3.45 → 3. The other two give consistent results either way (3.55 → 3.6 → 4 and directly → 4; 3.050 → 3.1 → 3 and directly → 3). This highlights the danger of rounding in stages.
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Explain the Mistake

Each example contains a deliberate error targeting a common misconception. Can you find where and why the reasoning goes wrong?

1
Round 0.04736 to 3 significant figures.
A student writes:

Answer: 0.047

Reasoning: “I counted three digits after the decimal point: 0, 4, 7. So the answer is 0.047.”

๐Ÿ” The Mistake

The student has confused significant figures with decimal places. They counted 3 digits after the decimal point (giving 3 d.p., not 3 s.f.). These are different things: “0.047” is 0.04736 rounded to 3 decimal places, and it only has 2 significant figures.

The first 3 significant figures of 0.04736 are 4, 7, 3 (leading zeros don’t count). The fourth significant figure is 6 (≥ 5), so we round up: the correct answer is 0.0474.

2
How many significant figures does 0.050 have?
A student writes:

Answer: 1 significant figure

Reasoning: “Trailing zeros don’t count in decimals, they only count in whole numbers like 500. So in 0.050, the last zero doesn’t count. Just the 5 is significant.”

๐Ÿ” The Mistake

The student is confusing integer rules with decimal rules. In integers (like 500), trailing zeros are indeed often ambiguous or not significant. But in decimals (like 0.050), trailing zeros are always significant.

Why write the zero if it doesn’t add value? It’s there to show precision. The ‘5’ and the ‘0’ are both significant, so the answer is 2 significant figures.

3
Round 2.449 to 1 significant figure.
A student writes:

Answer: 3

Reasoning: “I’ll round step by step. The 9 rounds the 4 up to 5, giving 2.45. Then the 5 rounds the 4 up to 5, giving 2.5. Then the 5 rounds the 2 up to 3. So the answer is 3.”

๐Ÿ” The Mistake

The student has used cascade rounding (also called “double rounding” or “creeping rounding”) — rounding digit by digit from right to left. Each intermediate rounding changes the next digit, creating a chain reaction that inflates the final answer. This is one of the most dangerous rounding misconceptions because it looks systematic and careful.

The correct method: look only at the digit immediately after the rounding position. To round 2.449 to 1 s.f., the first significant figure is 2 and the next digit is 4. Since 4 < 5, round down. The correct answer is 2, not 3. Never round in stages — always go straight from the original number.

4
Write 0.05291 correct to 2 significant figures.
A student writes:

Answer: 0.05

Reasoning: “I counted 2 significant figures from the left: the 0 and the 5. The next digit is 2, which is less than 5, so I rounded down to 0.05.”

๐Ÿ” The Mistake

The student counted a leading zero as a significant figure. They treated the “0” in the hundredths column as the first significant figure and the “5” as the second. But leading zeros are never significant — they are just placeholders.

The first significant figure is 5 and the second is 2. Looking at the next digit (9 ≥ 5), we round up: the correct answer is 0.053.

0.05291 Ignore START HERE 1st 2nd