Q.The number of significant figures in 0.06900 is

Showing the 12 most recent of 14 on this concept.

1. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.The number of significant figures present in $0.010200 \times 10^3$ is (A) 7 (B) 5 (C) 6 (D) 10

   ›Reveal solutionSolution

   Significant-figure rules: leading zeros never count, but trailing zeros after a decimal point always count. $0.010200$ has 5 significant figures, and multiplying by $10^3$ doesn't change that count.

   Concept and Intuition

   Significant figures reflect the precision of a measurement. Zeros purely used to fix the decimal point's position (leading zeros in a number $<1$) carry no information about precision and are excluded. But once you're past the first nonzero digit, EVERY digit — including trailing zeros written after a decimal point — is considered a deliberately recorded (significant) digit.

   Step-by-Step Solution

   1. Write out $0.010200$: digits are $0,0,1,0,2,0,0$ after the decimal point.
   2. The two leading zeros (before the '1') are NOT significant — they only locate the decimal point.
   3. Starting from the first nonzero digit '1' through to the end: $1,0,2,0,0$ — these ARE all significant (trailing zeros after a decimal point count) — that's 5 digits.
   4. Multiplying by $10^3$ just rescales the number (equivalent to moving the decimal point); it introduces no new measured digits and removes none, so the significant-figure count stays 5.
   5. Answer: 5 significant figures.

   Common Mistakes

   • Miscounting the leading zeros as significant.
   • Thinking multiplying by a power of 10 changes the number of significant figures.

   ✓Final answer

   The correct option is (B) — 5.

   ANSWER: B

2. AP EAPCET 2026Set ap-2026-05-19-FN1 markMCQ
   Q.From the point of view of significant figures, which of the following is/are correct?

   (i) $11.3\ \text{cm} + 4\ \text{cm} = 15.3\ \text{cm}$

   (ii) $4.53\ \text{m} - 1.2\ \text{m} = 3.3\ \text{m}$

   (iii) $5.45\ \text{kg} - 3.2\ \text{kg} = 2.25\ \text{kg}$

   (iv) $84.8\ \text{cm} + 48.6\ \text{cm} = 133\ \text{cm}$ (A)

   (ii) only (B)

   (iv) only (C)

   (i) &

   (iii) only (D)

   (ii) &

   (iv) only

   ›Reveal solutionSolution

   The addition/subtraction rule keeps only as many decimal places as the least

   precise term; checking all four, only statement (ii) obeys it. Answer: (A).

   Concept and Intuition

   When adding or subtracting measured quantities, the result cannot be more

   precise than the least precise input. Concretely: round the final sum/ difference to the same number of decimal places as the term with the fewest decimal places (not the fewest significant figures — decimal places are what

   matter here).

   Step-by-Step Solution

   1. (i) $11.3\ \text{cm} + 4\ \text{cm}$: raw sum $=15.3$. "4 cm" has 0 decimal places, so the answer must be rounded to 0 decimals: 15 cm. The given value 15.3 keeps a spurious decimal — statement (i) is incorrect.
   2. (ii) $4.53 - 1.2$: raw difference $=3.33$. "1.2" has 1 decimal place (fewer than 4.53's 2), so round to 1 decimal: 3.3. This matches the given value exactly — statement (ii) is correct.
   3. (iii) $5.45 - 3.2$: raw difference $=2.25$. "3.2" has 1 decimal place, so round to 1 decimal: 2.3. The given value 2.25 keeps an extra decimal — statement (iii) is incorrect.
   4. (iv) $84.8 + 48.6$: raw sum $=133.4$. Both terms have 1 decimal place, so the sum should also keep 1 decimal: 133.4. The given value 133 drops the decimal — statement (iv) is incorrect.
   5. Only (ii) obeys the significant-figures addition rule → option (A).

   Common Mistakes

   • Applying the multiplication rule (fewest significant figures) instead of the correct addition/subtraction rule (fewest decimal places).
   • Not noticing that "4 cm" is effectively a 0-decimal-place measurement.

   ✓Final answer

   The correct option is (A) — (ii) only.

   ANSWER: A

3. AP EAPCET 2026Set ap-2026-05-20-AN1 markMCQ
   Q.Which of the following have same number of significant figures? (A) 0.0025 (B) 0.0430 (C) 5005 (D) 500.0 (E) 2.003. The correct answer is (A) A & B only (B) B, C & D only (C) C, D & E only (D) A, C & D only

   ›Reveal solutionSolution

   Counting significant figures carefully (leading zeros never count; trailing zeros count only with a decimal point or between nonzero digits) shows C, D, E all have 4 significant figures.

   Concept and Intuition

   The rules for significant figures: (i) all non-zero digits are significant;

   (ii) zeros between non-zero digits are significant;

   (iii) leading zeros (before the first non-zero digit) are never significant;

   (iv) trailing zeros are significant only if the number has a decimal point.

   Step-by-Step Solution

   1. (A) 0.0025: leading zeros (0.00) don't count; significant digits are "2","5" $\to$ 2 s.f.
   2. (B) 0.0430: leading zeros don't count; "4","3","0" — the trailing zero counts because there's a decimal point $\to$ 3 s.f.
   3. (C) 5005: all digits, including the internal zeros, are significant $\to$ 4 s.f.
   4. (D) 500.0: has a decimal point, so all four digits (5,0,0,0) are significant $\to$ 4 s.f.
   5. (E) 2.003: all four digits significant $\to$ 4 s.f.
   6. Matching count of 4: C, D, E.

   Common Mistakes

   • Treating trailing zeros in 500.0 as ambiguous the way they would be in a bare integer "5000" (without a decimal point) — the explicit decimal point here removes the ambiguity and makes them significant.
   • Undercounting (B) by ignoring that its trailing zero (after the decimal point) is significant.

   ✓Final answer

   The correct option is (C) — C, D & E only.

   ANSWER: C

4. AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
   Q.Observe the following I) 0.0063 II) 132.00 III) 1004 The number of significant figures in I, II and III is respectively (A) 4, 3, 5 (B) 4, 5, 4 (C) 4, 3, 4 (D) 2, 5, 4

   ›Reveal solutionSolution

   Applying the standard significant-figure rules to 0.0063, 132.00, and 1004 gives 2, 5, and 4 significant figures respectively.

   Concept and Intuition

   Significant figures rules: (1) all non-zero digits are significant; (2) zeros between two non-zero digits ("captive" zeros) are significant; (3) leading zeros (before the first non-zero digit) are never significant — they only locate the decimal point; (4) trailing zeros are significant only if the number contains a decimal point.

   Step-by-Step Solution

   1. 0.0063: the leading zeros (0.00) merely position the decimal point and are not significant; only 6 and 3 count → 2 significant figures.
   2. 132.00: this number has a decimal point, so the trailing zeros after "132" are significant; digits are 1,3,2,0,0 → 5 significant figures.
   3. 1004: the zeros are sandwiched between the non-zero digits 1 and 4 (captive zeros), so they are significant; digits are 1,0,0,4 → 4 significant figures.
   4. So the counts, in order I, II, III, are 2, 5, 4.

   Common Mistakes

   • Counting leading zeros in 0.0063 as significant (giving a wrong count of 4 or 5).
   • Forgetting that trailing zeros after a decimal point (as in 132.00) are always significant.

   ✓Final answer

   The correct option is (D) — 2, 5, 4.

   ANSWER: D

5. AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
   Q.The number of significant figures in the simplification of $\dfrac{0.501}{0.05}(0.312 - 0.03)$ is (A) 1 (B) 3 (C) 2 (D) 5

   ›Reveal solutionSolution

   The least-precise data ($0.05$, $0.03$) carry one significant figure, so the answer has 1 significant figure.

   Concept and Intuition

   For $\times$ and $\div$, the final result keeps the same number of significant figures as the quantity with the fewest. Leading zeros only fix the decimal point and are never significant, so $0.05$ and $0.03$ each have exactly one significant figure.

   Step-by-Step Solution

   1. Evaluate the value: $0.312-0.03 = 0.282$ and $\dfrac{0.501}{0.05} = 10.02$, giving $10.02 \times 0.282 \approx 2.83$.
   2. Count significant figures in the data: $0.501$ has 3, $0.05$ has 1, $0.312$ has 3, $0.03$ has 1.
   3. The operation is dominated by division and multiplication, so the result is limited by the smallest count, which is $1$ (from $0.05$ and $0.03$).
   4. Hence the simplification is reported to $1$ significant figure ($\approx 3$).

   Common Mistakes

   • Counting the leading zeros in $0.05$ or $0.03$ as significant.
   • Reporting the raw calculator value ($2.83$, i.e. 3 figures) without applying the least-significant-figure rule.

   ✓Final answer

   The correct option is (A) — 1 significant figure.

   ANSWER: A

   Note

   This solution was worked out by our team and cross-checked by a second independent solve. The official answer key for this question could not be confirmed, so please cross-verify with the official paper where possible.

6. AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
   Q.Consider the following A) 0.0025 B) 500.0 C) 2.0034 Number of significant figures in A, B and C respectively, are (A) 5, 4, 4 (B) 2, 4, 2 (C) 4, 3, 2 (D) 2, 4, 5

   ›Reveal solutionSolution

   This tests the significant-figure rules for leading zeros, trailing zeros after a decimal point, and captive zeros; the counts are 2, 4, 5.

   Concept and Intuition

   Significant figures reflect the precision of a measurement: leading zeros (before the first non-zero digit) are never significant since they only locate the decimal point; zeros between non-zero digits (captive zeros) are always significant; trailing zeros are significant only if the number has an explicit decimal point.

   Step-by-Step Solution

   1. $A = 0.0025$: the leading zeros (0.00) merely fix the decimal position and are not significant; only "25" counts → 2 significant figures.
   2. $B = 500.0$: the number is written with an explicit decimal point, so ALL digits including the trailing zeros are significant → "5","0","0","0" → 4 significant figures.
   3. $C = 2.0034$: every digit from the first non-zero digit onward is significant, including the captive zeros between 2 and 34 → "2","0","0","3","4" → 5 significant figures.
   4. So (A, B, C) = (2, 4, 5), matching option (D).

   Common Mistakes

   • Treating 500.0 as having only 1 or 3 significant figures by ignoring the decimal point rule.
   • Miscounting captive zeros in 2.0034 as insignificant.

   ✓Final answer

   The correct option is (D) — 2, 4, 5.

   ANSWER: D

7. AP EAPCET 2025Set eng-2025-05-26-FN1 markMCQ
   Q.The number of significant figures in 0.03240 is (A) 5 (B) 4 (C) 6 (D) 3

   ›Reveal solutionSolution

   This tests the significant-figures rules for leading vs. trailing zeros: leading zeros are placeholders (never significant), while a zero written after the decimal point past the first nonzero digit is a measured, significant digit.

   Concept and Intuition

   Significant figures represent the precision of a measurement. Zeros purely used to fix the decimal point's position (leading zeros, e.g. in $0.032$) carry no information about precision and are excluded. But once you've reached the first nonzero digit, every digit after it — including trailing zeros written explicitly after a decimal point — is understood to have been measured and counted.

   Step-by-Step Solution

   1. Write out $0.03240$ digit by digit: $0,0,3,2,4,0$.
   2. The first two zeros (before the $3$) are leading zeros — not significant, since they only locate the decimal point.
   3. Starting from the first nonzero digit $3$, all subsequent digits count: $3,2,4,0$.
   4. The final $0$ is written explicitly after the decimal expansion, so per convention it IS significant (it signals precision to that digit).
   5. Total significant figures $=4$.

   Common Mistakes

   • Miscounting the leading zeros as significant, which would wrongly give 6.
   • Dropping the trailing zero as "just a zero," which would wrongly give 3.

   ✓Final answer

   The correct option is (B) — 4.

   ANSWER: B

8. AP EAPCET 2024Set eng-2024-05-21-AN1 markMCQ
   Q.The length of the side of a cube is $1.2 \times 10^{-2}$ m. Its volume up to correct significant figures is (A) $1.732 \times 10^{-6}\ \mathrm{m^3}$ (B) $1.73 \times 10^{-6}\ \mathrm{m^3}$ (C) $1.70 \times 10^{-6}\ \mathrm{m^3}$ (D) $1.7 \times 10^{-6}\ \mathrm{m^3}$

   ›Reveal solutionSolution

   $(1.2\times10^{-2})^3 = 1.728\times10^{-6}$, rounded to the 2 significant figures of the given data $\Rightarrow 1.7\times10^{-6}\ \mathrm{m^3}$.

   Concept and Intuition

   A calculated quantity can be no more precise than the least precise measurement used to compute it. Since the side length $1.2\times10^{-2}$ m has only 2 significant figures, the volume must also be reported to 2 significant figures, regardless of how many digits the raw calculation produces.

   Step-by-Step Solution

   1. $V=a^3=(1.2\times10^{-2})^3 = 1.728\times10^{-6}\ \mathrm{m^3}$ (raw calculation).
   2. The given side length has 2 significant figures ($1.2$).
   3. Rounding $1.728$ to 2 significant figures gives $1.7$.
   4. So $V=1.7\times10^{-6}\ \mathrm{m^3}$.

   Common Mistakes

   • Reporting all the calculator digits ($1.732$ or $1.73$) without rounding to match the input precision.
   • Rounding to 3 significant figures instead of 2.

   ✓Final answer

   The correct option is (D) — $1.7 \times 10^{-6}\ \mathrm{m^3}$.

   ANSWER: D

9. AP EAPCET 2023Set ap-2023-05-22-AN1 markMCQ
   Q.The most accurate measurement among the following is (A) $20 \times 10^{-3}$ m (B) $200 \times 10^{-4}$ m (C) $2 \times 10^{-2}$ m (D) $0.02$ m

   ›Reveal solutionSolution

   The number of significant figures determines measurement precision/accuracy; $200\times10^{-4}$ m carries 3 significant figures, the most among the four choices, making it the most accurate.

   Concept and Intuition

   When a quantity is expressed in standard scientific notation $a \times 10^n$, every digit in the coefficient $a$ is treated as significant (this is precisely why scientific notation removes the ambiguity of trailing zeros). More significant figures in a measurement imply a smaller least count / finer precision of the measuring instrument, and hence greater accuracy.

   Step-by-Step Solution

   1. $20\times10^{-3}$ m — coefficient "20" has 2 significant figures.
   2. $200\times10^{-4}$ m — coefficient "200" has 3 significant figures (since all digits in a scientific-notation mantissa are significant).
   3. $2\times10^{-2}$ m — coefficient "2" has 1 significant figure.
   4. $0.02$ m — has 1 significant figure (the leading zeros are not significant).
   5. Comparing sig figs: 2, 3, 1, 1 respectively — option (B) has the most (3), making it the most precisely/accurately expressed measurement.

   Common Mistakes

   • Assuming all four options represent numerically equal values and are therefore "equally accurate" — accuracy here is about precision of expression (sig figs), not numeric equivalence.
   • Miscounting trailing zeros in a scientific-notation mantissa as insignificant.

   ✓Final answer

   The correct option is (B) — $200 \times 10^{-4}$ m.

   ANSWER: B

10. AP EAPCET 2023Set eng-2023-05-17-FN1 markMCQ
    Q.The number of significant figures in 4.870 m is (A) 3 (B) 4 (C) 2 (D) 1

    ›Reveal solutionSolution

    A trailing zero after a decimal point is always significant, so $4.870$ has 4 significant figures.

    Concept and Intuition

    Significant-figure rules: all non-zero digits are significant; zeros between non-zero digits are significant; trailing zeros are significant only if there is a decimal point present, because such a zero represents a deliberately measured/reported digit of precision (not just a placeholder).

    Step-by-Step Solution

    1. Write out the digits of $4.870$: $4$, $8$, $7$, $0$.
    2. $4$ is non-zero — significant.
    3. $8$ is non-zero — significant.
    4. $7$ is non-zero — significant.
    5. The trailing $0$ comes after a decimal point, so it is a measured digit of precision, not a placeholder — significant.
    6. Total significant figures $=4$.

    Common Mistakes

    • Discarding the trailing zero as "just a zero" — this rule only applies to trailing zeros in a number without a decimal point (e.g., $4870$ m is ambiguous/could be 3 or 4 sig figs), not to $4.870$.

    ✓Final answer

    The correct option is (B) — 4.

    ANSWER: B

11. AP EAPCET 2022Set ap-2022-07-11-AN1 markMCQ
    Q.The number of significant figures in the quantity 5.6200 J is (A) 3 (B) 5 (C) 2 (D) 4

    ›Reveal solutionSolution

    Tests the rules of significant figures; trailing zeros after a decimal point count as significant, so $5.6200$ has 5 significant figures.

    Concept and Intuition

    Significant figures represent the precision of a measured quantity. The rules state: (i) all non-zero digits are significant;

    (ii) zeros between non-zero digits are significant;

    (iii) trailing zeros in a number with a decimal point are significant (they were deliberately recorded to show precision);

    (iv) leading zeros are never significant.

    Step-by-Step Solution

    1. Write out the digits of $5.6200$: $5, 6, 2, 0, 0$.
    2. $5$ and $6$ are non-zero digits — significant.
    3. $2$ is a non-zero digit between other digits — significant.
    4. The two trailing zeros come after the decimal point, so by rule they are also significant (they indicate the measurement was precise to that last digit).
    5. Total significant digits = $5$.

    Common Mistakes

    • Wrongly dropping trailing zeros after a decimal point, thinking they are "just placeholders" — they are not; they carry precision information.
    • Confusing this with trailing zeros in a number without a decimal point (e.g., $5600$), which would be ambiguous/not necessarily significant.

    ✓Final answer

    The correct option is (B) — 5.

    ANSWER: B

12. AP EAPCET 2022Set eng-2022-07-04-FN1 markMCQ
    Q.If $N_A$, $N_B$ and $N_C$ are the number of significant figures in $A = 0.001204$ m, $B = 43120000$ m and $c = 1.200$ m respectively then (A) $N_A = N_B = N_C$ (B) $N_A > N_B > N_C$ (C) $N_A < N_B < N_C$ (D) $N_A > N_B < N_C$

    ›Reveal solutionSolution

    Applying the significant-figure rules (leading zeros never count, zeros between non-zero digits always count, trailing zeros count only with a decimal point) gives $N_A=N_B=N_C=4$.

    Concept and Intuition

    Significant figures reflect measurement precision. The rules: (i) all non-zero digits are significant;

    (ii) zeros sandwiched between non-zero digits are significant;

    (iii) leading zeros (before the first non-zero digit) are never significant;

    (iv) trailing zeros are significant only if the number has an explicit decimal point.

    Step-by-Step Solution

    1. $A=0.001204$ m: The zeros before "1" are leading zeros — not significant. The digits $1,2,0,4$ remain; the "0" between "2" and "4" is between non-zero digits, so it counts. $N_A=4$.
    2. $B=43120000$ m: No decimal point is shown, so trailing zeros are ambiguous and conventionally not counted. Significant digits are $4,3,1,2$. $N_B=4$.
    3. $C=1.200$ m: A decimal point is present, so the trailing zeros ARE significant. Digits $1,2,0,0$ all count. $N_C=4$.
    4. Comparing: $N_A=N_B=N_C=4$.

    Common Mistakes

    • Assuming trailing zeros in $B$ (no decimal point) are automatically significant — they are not, by the standard convention.
    • Missing that the internal zero in $A$ (between 2 and 4) is significant, undercounting $N_A$.

    ✓Final answer

    The correct option is (A) — $N_A=N_B=N_C$.

    ANSWER: A