The trap that costs marks on significant figures is deceptively simple: aspirants apply the wrong rounding rule to the wrong arithmetic operation. Multiplication and division require rounding to the fewest significant figures among the inputs. Addition and subtraction require rounding to the fewest decimal places. Swapping these two rules is a high-frequency mistake on NEET (NCERT Class 11 Physics Chapter 1, pages 4–5).
What are significant figures? Every measured quantity carries uncertainty. Significant figures are the digits in a measurement that are known reliably plus the first uncertain digit (NCERT Class 11 Physics Chapter 1, page 3). They encode the precision of the instrument that produced the number.
Counting rules (NCERT conventions):
The two rounding rules:
Watch out: When a subtraction problem like 9.99 m − 0.0099 m appears, the fewest decimal places is 2 (from 9.99). The answer is 9.98 m — not 9.9801 m (keeping all digits) and not 1.0 × 10¹ m (wrongly applying the sig-fig rule). This is the exact trap NEET has tested.
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
How many significant figures are in the measurement 0.00470 kg?
Answer: D. The leading zeros (0.00) are not significant — they only indicate the position of the decimal. The digits 4, 7, and the trailing 0 after 7 are all significant, giving 3 significant figures (NCERT Class 11 Physics Chapter 1, page 4).
Why A is wrong: A (2) counts only the non-zero digits 4 and 7, ignoring the trailing zero after the decimal point. A trailing zero after a decimal point IS significant — it indicates measured precision.
Why B is wrong: B (5) counts all five digits including the leading zeros. Leading zeros before the first non-zero digit are never significant.
Why C is wrong: C (4) incorrectly counts one of the leading zeros as significant. Leading zeros in 0.00470 serve only as placeholders and are never significant.
The number 8.0 × 10⁴ has how many significant figures?
Answer: A. In scientific notation, all digits in the coefficient are significant. The coefficient 8.0 contains two digits (8 and 0), so the number has 2 significant figures (NCERT Class 11 Physics Chapter 1, page 5).
Why B is wrong: B (1) ignores the trailing zero in 8.0. In scientific notation, the zero after the decimal point in the coefficient is significant — it was deliberately written to show precision.
Why C is wrong: C (4) incorrectly counts the exponent's digits or the implied zeros in 80000. In scientific notation, only the coefficient determines significant figures.
Why D is wrong: D (5) treats the expanded form 80000 and counts all five digits. The purpose of scientific notation is to make sig-fig count unambiguous — only the coefficient 8.0 matters.
Which of the following numbers has exactly 4 significant figures?
Answer: C. In 5.000, all four digits are significant: the non-zero digit 5 and the three trailing zeros after the decimal point each indicate measured precision (NCERT Class 11 Physics Chapter 1, page 4).
Why A is wrong: A (0.0050) has only 2 significant figures. The leading zeros (0.00) are not significant, leaving only 5 and the trailing 0.
Why B is wrong: B (5000) is ambiguous without a decimal point — it could have 1, 2, 3, or 4 significant figures depending on the measurement context. Without additional notation (e.g. 5.000 × 10³), the sig-fig count is undefined.
Why D is wrong: D (0.005) has only 1 significant figure. All four zeros are leading zeros serving as decimal placeholders.
A rectangular plate has length 4.234 m and breadth 1.05 m. The area of the plate, expressed with the correct number of significant figures, is:
Answer: B. This is a multiplication, so the result takes the fewest significant figures of the inputs. The length 4.234 has 4 sig figs; the breadth 1.05 has 3 sig figs. The calculator gives 4.234 × 1.05 = 4.4457. Rounded to 3 significant figures: 4.45 m² (NCERT Class 11 Physics Chapter 1, pages 4–5). This pattern matches NEET 2022 (sig-fig product rule).
Why A is wrong: A (4.44570) keeps all digits from the calculator without rounding. In multiplication, the result must be rounded to the fewest significant figures among the inputs (3, from 1.05). (Trap: keeping the raw calculator answer.)
Why C is wrong: C (4.446) rounds to 4 significant figures, matching the length's sig-fig count instead of the breadth's. The rule requires the FEWEST sig figs, which is 3 from 1.05.
Why D is wrong: D (4.4) rounds to only 2 significant figures. Neither input has as few as 2 sig figs — the minimum is 3 (from 1.05). Over-rounding loses precision that the measurement supports.
The result of 28.4 m + 2.175 m, expressed with the correct number of significant figures, is:
Answer: D. This is an addition, so the result takes the fewest decimal places of the inputs. 28.4 has 1 decimal place; 2.175 has 3 decimal places. The sum is 30.575, rounded to 1 decimal place: 30.6 m (NCERT Class 11 Physics Chapter 1, pages 4–5).
Why A is wrong: A (30.575) keeps all decimal places from the raw addition without rounding. Addition requires rounding to the fewest decimal places among the inputs (1, from 28.4). (Trap: treating the raw sum as final.)
Why B is wrong: B (30.58) rounds to 2 decimal places. This does not match either input's decimal-place count as the limiting factor — 28.4 has only 1 decimal place.
Why C is wrong: C (31) rounds to 0 decimal places, which is more aggressive than the rule requires. The input 28.4 has 1 decimal place, so the answer should retain 1 decimal place. (Trap: applying the multiplication sig-fig rule — 28.4 has 3 sig figs, so rounding to 2 sig figs gives 31.)
A student measures two lengths as 12.50 m and 2.1 m and subtracts them: 12.50 − 2.1 = 10.40. The correctly rounded result is:
Answer: A. Subtraction uses the decimal-place rule. 12.50 has 2 decimal places; 2.1 has 1 decimal place. The fewest is 1 decimal place. The raw difference 10.40 rounded to 1 decimal place is 10.4 m (NCERT Class 11 Physics Chapter 1, pages 4–5).
Why B is wrong: B (10) rounds to 0 decimal places, which is more aggressive than the rule requires. The limiting input 2.1 has 1 decimal place, not 0. (Trap: confusing sig-fig count with decimal-place count — 2.1 has 2 sig figs, but the subtraction rule uses decimal places, not sig figs.)
Why C is wrong: C (10.40) retains 2 decimal places, matching 12.50's decimal-place count. The rule requires the FEWEST decimal places, which is 1 from 2.1, not 2 from 12.50.
Why D is wrong: D (1.0 × 10¹) has 2 significant figures, but significant figures are not the governing criterion for subtraction. The decimal-place rule applies, giving 10.4 m.
Two rods are measured as 5.2 cm and 3.34 cm. Their difference is calculated as 5.2 − 3.34 = 1.86 cm. A student reports the answer as 1.9 cm. Another student reports it as 2 cm. Which student is correct and why?
Answer: B. Subtraction uses the fewest-decimal-places rule. 5.2 has 1 decimal place; 3.34 has 2 decimal places. The minimum is 1 decimal place. 1.86 rounded to 1 decimal place is 1.9 cm, so the first student is correct (NCERT Class 11 Physics Chapter 1, pages 4–5).
Why A is wrong: A incorrectly applies the multiplication/division rule (fewest significant figures) to a subtraction. The second student rounded 1.86 to 2 sig figs, getting 1.9, then further rounded to 1 sig fig, getting 2 — this double error stems from using the wrong rule entirely. (Trap: sig-fig rule swap — applying the multiplication rule to subtraction.)
Why C is wrong: C claims no rounding is needed. Every arithmetic result from measured quantities must be rounded according to the appropriate rule. Keeping 1.86 implies 2 decimal places, which exceeds the precision of the less precise input (5.2, which has only 1 decimal place).
Why D is wrong: D (1.860) adds a trailing zero, implying 3 decimal places — even more precision than the raw result. This contradicts the fundamental purpose of sig-fig rounding.
A student computes the area of a square sheet with side 2.1 cm as 2.1 × 2.1 = 4.41 cm². She then subtracts the area of a small hole, 0.065 cm², from the sheet area. What is the correct final answer?
Answer: C. Step 1 (multiplication): 2.1 × 2.1 = 4.41; 2.1 has 2 sig figs, so the intermediate product is 4.4 cm² (2 sig figs). Step 2 (subtraction): 4.4 − 0.065; 4.4 has 1 decimal place, 0.065 has 3 decimal places. Fewest decimal places is 1. Raw difference = 4.335, rounded to 1 decimal place = 4.3 cm² (NCERT Class 11 Physics Chapter 1, pages 4–5).
Why A is wrong: A (4.345) uses the unrounded product 4.41 and subtracts 0.065 to get 4.345, then keeps all digits. This skips rounding at both steps — multiplication was not rounded to 2 sig figs, and subtraction was not rounded to fewest decimal places.
Why B is wrong: B (4.35) partially rounds the multiplication step (keeping 4.41 at 3 sig figs instead of 2) and then rounds the subtraction to 2 decimal places instead of 1. Both the intermediate and final rounding are wrong.
Why D is wrong: D (4.4) correctly rounds the multiplication (4.4 cm²) but then ignores the subtraction entirely. The hole area 0.065 cm² is small but must still be subtracted before reporting the final answer.
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Given
Two measured lengths: L₁ = 9.99 m (2 decimal places, 3 sig figs) and L₂ = 0.0099 m (4 decimal places, 2 sig figs).
Required
Find L₁ − L₂ expressed with the correct number of significant figures.
Concept
When measured quantities are subtracted, the result must be rounded to the fewest decimal places among the inputs — not the fewest significant figures. This is the addition/subtraction rounding convention (NCERT Class 11 Physics Chapter 1, page 5).
Formula
For Z = A ± B: round Z to the fewest decimal places among A and B.
Substitution
Raw subtraction: 9.99 − 0.0099 = 9.9801 m.
Decimal places: L₁ has 2 decimal places; L₂ has 4 decimal places. The fewest is 2.
Calculation
9.9801 rounded to 2 decimal places → 9.98 m.
No exact constants are involved in this problem — both numbers are measured quantities with inherent uncertainty.
Final answer
9.98 m (3 significant figures, 2 decimal places).
Note: if we had wrongly applied the multiplication/division rule (fewest sig figs = 2, from 0.0099), we would get 1.0 × 10¹ m — a drastically different and incorrect answer.
Common trap
The most common error is applying the "fewest significant figures" rule to this subtraction. L₂ = 0.0099 has only 2 significant figures, which tempts students to round the answer to 2 sig figs (giving 10 m or 1.0 × 10¹ m). But subtraction demands the decimal-place rule, not the sig-fig rule.
Similar NEET-style question
Two masses are measured: m₁ = 25.0 g and m₂ = 0.034 g. Express (m₁ − m₂) with the correct significant figures.
Answer: 25.0 − 0.034 = 24.966 → fewest decimal places is 1 (from 25.0) → 25.0 g.
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All the reliably-known digits in a measured quantity, plus the first uncertain (estimated) digit, are called significant figures. Significant figures indicate the precision of measurement.
-- NCERT Class 11 Physics, Ch. 1, p. 3Rules: (i) all non-zero digits are significant; (ii) zeros between non-zero digits are significant; (iii) leading zeros (before the first non-zero digit) in a decimal number are NOT significant; (iv) trailing zeros in a number without a decimal point are NOT significant; (v) trailing zeros after a decimal point ARE significant. Exact numbers (counted, definitions) have unlimited significant figures.
-- NCERT Class 11 Physics, Ch. 1, p. 4Multiplication/division: result has the same number of significant figures as the input with the fewest significant figures. Addition/subtraction: result has the same number of decimal places as the input with the fewest decimal places.
-- NCERT Class 11 Physics, Ch. 1, p. 5A cube has side measured as 7.203 m (4 sig figs). Surface area 6 × 7.203² = 311.299254 m² rounded to 311.3 m² (4 sig figs). Volume 7.203³ = 373.714754 m³ rounded to 373.7 m³ (4 sig figs).
-- NCERT Class 11 Physics, Ch. 1, p. 6These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
Category: Similar Terms
Student applies the 'fewest significant figures' rule (which governs multiplication and division) to a sum or difference. Subtraction of two measured numbers must instead reflect the FEWEST decimal places.
Question involves addition/subtraction of measured numbers with very different magnitudes or decimal-place counts (e.g. 9.99 - 0.0099). Distractors offer answers rounded by sig-fig rule rather than decimal-place rule.
Memorise: multiplication/division → fewest SIGNIFICANT FIGURES; addition/subtraction → fewest DECIMAL PLACES. Always identify which arithmetic operation is being performed before applying any rule.
Root cause: concept gap
For multiplication/division, the result has the fewest significant figures of any input. For addition/subtraction, the result has the fewest decimal places of any input. Decimal places ≠ significant figures.
Distractor truncates a sum to too few significant figures by applying the multiplication rule.
More in Units and Measurements: 7 exam traps and mistakes · 3 formulas · 7 question patterns from its other lessons.
Taking into account of the significant figures, what is the value of 9.99 m−0.0099 m ?
no rounding
Calculator answer kept verbatim
applies mult rule to subtraction
Default to 'fewest sig figs' without distinguishing subtraction's decimal-places rule
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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