The numbers 2.745 and 2.735 on rounding off to 3 significant figures will give (a) 2.75 and 2.74 (b) 2.74 and

NCERT Exemplar · Q4

Q. The numbers 2.745 and 2.735 on rounding off to 3 significant figures will give

(a) 2.75 and 2.74

(b) 2.74 and 2.73

(d) 2.74 and 2.74

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Concept understanding — Significant Figures Calculation

Significant Figures: The Art of Honest Measurement

Imagine you're measuring the length of a table with a ruler that has marks every millimeter. You see the table edge falls somewhere between 152.3 cm and 152.4 cm. You estimate it as 152.35 cm. But here's the truth: you're certain about 152.3, pretty sure about the 0.05, and guessing about anything beyond that. Significant figures are simply a way to communicate how much of that number you actually know.

The Core Idea

Every measurement has uncertainty. Significant figures (or "sig figs") are the digits in a number that carry meaningful information about its precision. They include all the digits you're sure of, plus one more that you estimate.

Note

A digit is "significant" if removing it would change the precision of the measurement. Zeros can be tricky — they might just be placeholders.

The Rules (Memorize These)

1. Non-zero digits are always significant

$123.45$ has 5 sig figs. Simple.

2. Zeros between non-zero digits are significant

$1002$ has 4 sig figs. The zeros are "sandwiched" — they're part of the measurement.

3. Leading zeros are never significant

$0.00123$ has 3 sig figs. Those zeros just tell you where the decimal point is.

4. Trailing zeros are significant only if there's a decimal point

$1200$ has 2 sig figs (no decimal — zeros are placeholders)
$1200.$ has 4 sig figs (decimal tells us those zeros were measured)
$1200.0$ has 5 sig figs

5. Exact numbers have infinite sig figs

If you count 5 apples, that's exactly 5 — no uncertainty. Conversion factors like $1 \text{ m} = 100 \text{ cm}$ are exact by definition.

Tip

When in doubt, write the number in scientific notation. $1.20 \times 10^3$ clearly has 3 sig figs, while $1.2 \times 10^3$ has 2.

Why This Matters: Calculations

When you multiply or add measurements, the uncertainty propagates. You can't claim more precision than your least precise measurement.

Multiplication and Division

The result should have the same number of sig figs as the measurement with the fewest sig figs.

$3.14 \times 2.5 = 7.85$ but you report $7.9$ (2 sig figs, because 2.5 has only 2)

Addition and Subtraction

The result should have the same decimal places as the measurement with the fewest decimal places.

$12.11 + 18.0 = 30.11$ but you report $30.1$ (one decimal place, because 18.0 has one) …

Why this formula?

Significant Figures: Why the Rules Work

Let’s start with the core idea: significant figures (sig figs) are a way to honestly report how precise a measurement is. The rules for addition/subtraction and multiplication/division aren’t arbitrary — they come directly from how uncertainty propagates through calculations.

1. The Fundamental Idea: Uncertainty is the Key

Every measurement has an uncertainty (error). When we say a length is 12.3 cm, we mean:

The true value lies somewhere between 12.25 cm and 12.35 cm (assuming ±0.05 cm uncertainty).
The last digit (3) is uncertain; the digits before it (1 and 2) are certain.

Why this matters: When we combine measurements, the uncertainty in the result depends on the uncertainties of the inputs. Sig fig rules are a shortcut for this uncertainty propagation.

2. Rule for Addition and Subtraction

Statement: The result should have the same number of decimal places as the measurement with the fewest decimal places.

Example:

$12.3 + 4.56 = 16.86$ → round to 16.9 (one decimal place, like 12.3)

Why this holds

Consider two measurements:

$A = 12.3 \pm 0.05$ (uncertainty in the tenths place)
$B = 4.56 \pm 0.005$ (uncertainty in the hundredths place)

When we add:

Certain digits: 12.3 has certainty up to the tenths place. 4.56 has certainty up to the hundredths place.
The weaker link: The tenths place of A is uncertain. So in the sum, the hundredths place (from B) is meaningless — because we don’t even know the tenths place of A exactly.

Mathematically, the absolute uncertainty in the sum is:

\Delta(A+B) = \sqrt{(\Delta A)^2 + (\Delta B)^2} \approx \sqrt{0.05^2 + 0.005^2} \approx 0.0502

This uncertainty is ~0.05, which affects the tenths place. So reporting the hundredths place is false precision.

Key takeaway: The result’s last significant digit is in the same decimal place as the least precise measurement’s last digit.

3. Rule for Multiplication and Division

Statement: The result should have the same number of significant figures as the measurement with the fewest significant figures.

Example:

$12.3 \times 4.56 = 56.088$ → round to 56.1 (three sig figs, like both inputs)

Why this holds

Let’s use relative uncertainty (percentage error):

$A = 12.3 \pm 0.05$ → relative uncertainty = $\frac{0.05}{12.3} \approx 0.00407$ (0.407%)
$B = 4.56 \pm 0.005$ → relative uncertainty = $\frac{0.005}{4.56} \approx 0.00110$ (0.110%)

For multiplication, relative uncertainties add (approximately):

\frac{\Delta(A \times B)}{A \times B} \approx \sqrt{\left(\frac{\Delta A}{A}\right)^2 + \left(\frac{\Delta B}{B}\right)^2}

Plugging in:

\approx \sqrt{0.00407^2 + 0.00110^2} \approx 0.00422 \text{ (0.422\%)}

Now, the absolute uncertainty in the product:

\Delta(A \times B) \approx 0.00422 \times (12.3 \times 4.56) \approx 0.00422 \times 56.088 \approx 0.237

This uncertainty (~0.2) affects the tenths place of the result. So the result 56.088 has uncertainty in the first decimal — meaning only three digits (5, 6, and the uncertain 1) are meaningful. That’s three sig figs, matching the input with fewer sig figs (both have three here).

Key takeaway: The number of sig figs in the result is limited by the least precise measurement’s number of sig figs, because relative uncertainty is dominated by the measurement with the largest relative error.

4. Why These Rules Are Different …

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Both 2.745 and 2.735 round to 2.74 when the standard round-half-to-even rule (used by NCERT for a dropped digit of exactly 5) is applied — the answer is option (D).

The rounding rule for a dropped digit of exactly 5

When the digit right after the last figure you keep is exactly 5 with nothing beyond it, always rounding up introduces a small upward bias over many numbers. NCERT's Class-11 Physics textbook prescribes the round-half-to-even ("banker's rounding") convention for this exact situation: look at the last retained digit — if it is already even, leave it as is; if it is odd, round it up to make it even.

(This applies only when the dropped digit is exactly 5 with no further non-zero digits after it — an ordinary digit $\geq 5$ followed by more non-zero digits still always rounds up.)

Rounding 2.745 to 3 significant figures

Keeping 3 significant figures means keeping $2.74$ and dropping the trailing $5$ . The last retained digit is $4$ , which is already even, so it stays unchanged:

2.745 \to 2.74

Rounding 2.735 to 3 significant figures …

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Here are the most common mistakes students make when rounding 2.745 and 2.735 to 3 significant figures, along with how to avoid each.

Mistake 1: Forgetting the "Rule for 5" (Rounding to the Nearest Even)

The Error:

Many students think that when the digit to be dropped is exactly 5, you always round up.

They round 2.745 → 2.75 (correct)
But they also round 2.735 → 2.74 (incorrect here — see below)

Why it’s wrong:

The standard convention (especially in Indian exams like JEE, NEET, and CBSE) is:

If the digit to be dropped is exactly 5, round to the nearest even digit.

2.745: The digit before 5 is 4 (even). Dropping the 5 keeps it even → 2.74 (not 2.75).
2.735: The digit before 5 is 3 (odd). Dropping the 5 rounds up to make it even → 2.74.

How to Avoid:

Memorise: "Round to even when it's exactly 5."
Practice with a few examples:
- 3.45 → 3.4 (4 is even)
- 3.35 → 3.4 (3 is odd, so round up to 4)

Mistake 2: Counting Significant Figures Incorrectly

The Error:

Students sometimes think 2.745 has 4 significant figures and try to round to 3 by looking at the wrong digit.

Why it’s wrong:

2.745 has 4 significant figures (2, 7, 4, 5).
To round to 3 significant figures, you keep the first three digits (2, 7, 4) and look at the fourth digit (5) to decide.

How to Avoid:

Always count from the first non-zero digit from the left.
For 2.745: digits are 2, 7, 4, 5 → 4 sig figs.
For 2.735: digits are 2, 7, 3, 5 → 4 sig figs.
The last digit you keep is the 3rd sig fig. The next digit tells you what to do.

Mistake 3: Applying the "Round Up for 5" Rule Blindly

The Error:

Students apply a blanket rule: "If the next digit is 5 or more, round up."

This gives: 2.745 → 2.75 and 2.735 → 2.74.

Why it’s wrong:

This rule works for most cases, but not when the digit to be dropped is exactly 5 with no non-zero digits after it. The "round to even" rule overrides it.

How to Avoid:

When the digit to be dropped is exactly 5 (and nothing after it), use the even-digit rule.
If there are non-zero digits after the 5 (e.g., 2.7451), then you always round up because it's more than halfway.

Mistake 4: Confusing "3 Significant Figures" with "3 Decimal Places"

The Error:

Students think rounding to 3 significant figures means keeping 3 digits after the decimal point.

Why it’s wrong: …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Council of Higher Secondary Education, Manipur (Higher Secondary 1st Year) 2026Set ANNUAL1 markMCQ
$Q. A piece of paper is found to be 5.32 cm long and 2.4 cm broad. The area of the paper expressed with proper significant figures is -$
(a) 12.768 cm^2
(b) 12.76 cm^2
(c) 12.8 cm^2
(d) 13 cm^2
›Reveal solutionSolution

$5.32 \times 2.4 = 12.768$ , but rounded to the fewer significant figures of 2.4 (2 s.f.), the area is 13 cm².

Area $= \text{length} \times \text{breadth} = 5.32\ cm \times 2.4\ cm = 12.768\ cm^2$ .

Rule of significant figures in multiplication/division: the result must be reported with the same number of significant figures as the measurement with the fewest significant figures.

…

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Council of Higher Secondary Education, Manipur (Higher Secondary 1st Year) 2025Set ANNUAL1 markMCQ
$Q. The significant numbers of 4200 kg is$
(a) 2
(b) 4
(c) 3
(d) zero
›Reveal solutionSolution
Written as 4200 kg (no decimal point), only 4 and 2 count as significant digits — 2 significant figures.

The rule for significant figures on a number with trailing zeros and no decimal point is that those trailing zeros are NOT counted as significant, because we cannot tell from the way the number is written whether they were actually measured or are simply place-holding zeros needed to show the magnitude of the quantity.

For 4200 kg:

The non-zero digits 4 and 2 are always significant.

The two zeros after them, with no decimal point present, are ambiguous — by convention they are treated as not significant. …
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Council of Higher Secondary Education, Manipur (Higher Secondary 1st Year) 2025Set ANNUAL1 mark
$Q. What is meant by significant figures?$
›Reveal solutionSolution
Significant figures are the meaningful digits in a number that convey how precisely a quantity was measured.

When a quantity is measured using any instrument, the digits which are known reliably plus the one digit that is uncertain (estimated) are together called significant figures. For example, if a length is reported as 12.35 cm, the digits 1, 2, 3 are certain and the last digit 5 is uncertain (estimated), giving four significant figures in total.

Rules for counting significant figures:

All non-zero digits are significant (285 cm has 3 sig figs).

Zeros between two non-zero digits are significant (2.005 has 4 sig figs).

Leading zeros (before the first non-zero digit) are NOT significant (0.03 has 1 sig fig).

Trailing zeros after a decimal point ARE significant (0.200 has 3 sig figs).

Trailing zeros in a number without a decimal point may or may not be significant, and are best expressed in scientific notation to avoid ambiguity. …
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Council of Higher Secondary Education, Manipur (Higher Secondary 1st Year) 2024Set ANNUAL1 markMCQ
$Q. The number of significant figures in the scientific notation 1.23 \times 10^5 is -$
(a) One
(b) Three
(c) Five
(d) Eight
›Reveal solutionSolution

Counting the digits in the coefficient 1.23 (the power of 10 is not counted) gives three significant figures.

In scientific notation $N \times 10^n$ , the exponent 10^n merely fixes the decimal place/order of magnitude and carries no information about precision. Only the digits in the coefficient N are significant.

…

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Council of Higher Secondary Education, Manipur (Higher Secondary 1st Year) 2021Set ANNUAL1 mark
$Q. What is the precision of measurement?$
›Reveal solutionSolution
Precision measures how close repeated readings of the same quantity are to each other, regardless of whether they are close to the true value.

In any experimental measurement, two related but distinct ideas describe reliability:

Accuracy: how close a measured value is to the true/accepted value.

Precision: how close repeated measurements of the same quantity are to one another (i.e., how reproducible the readings are). …
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Q.The numbers 2.745 and 2.735 on rounding off to 3 significant figures will give

Concept understanding — Significant Figures Calculation

Significant Figures: The Art of Honest Measurement

The Core Idea

The Rules (Memorize These)

1. Non-zero digits are always significant

2. Zeros between non-zero digits are significant

3. Leading zeros are never significant

4. Trailing zeros are significant only if there's a decimal point

5. Exact numbers have infinite sig figs

Why This Matters: Calculations

Multiplication and Division

Addition and Subtraction

Why this formula?

Significant Figures: Why the Rules Work

1. The Fundamental Idea: Uncertainty is the Key

2. Rule for Addition and Subtraction

Why this holds

3. Rule for Multiplication and Division

Why this holds

4. Why These Rules Are Different …

The rounding rule for a dropped digit of exactly 5

Rounding 2.745 to 3 significant figures

Rounding 2.735 to 3 significant figures …

Method: The "Even-Digit Rule" (also called Banker's Rounding or Round-Half-to-Even)

Steps

Step 1: Identify the digit to keep

Step 2: Apply the Even-Digit Rule

Mistake 1: Forgetting the "Rule for 5" (Rounding to the Nearest Even)

Mistake 2: Counting Significant Figures Incorrectly

Mistake 3: Applying the "Round Up for 5" Rule Blindly

Mistake 4: Confusing "3 Significant Figures" with "3 Decimal Places"

Q.The numbers 2.745 and 2.735 on rounding off to 3 significant figures will give

The rounding rule for a dropped digit of exactly 5

Rounding 2.745 to 3 significant figures

Rounding 2.735 to 3 significant figures …

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