Q.The length, breadth and thickness of a rectangular sheet of metal are 4.234 m, 1.005 m, and 2.01 cm respectively. Give the area and volume of the sheet to correct significant figures.
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🔒 Start your 14-day free trial to unlock the full solution →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.
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 m=100 cm are exact by definition.
When in doubt, write the number in scientific notation. 1.20×103 clearly has 3 sig figs, while 1.2×103 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.
3.14×2.5=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±0.05 (uncertainty in the tenths place)
- B=4.56±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:
Δ(A+B)=(ΔA)2+(ΔB)2≈0.052+0.0052≈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:
12.3×4.56=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±0.05 → relative uncertainty = 12.30.05≈0.00407 (0.407%)
- B=4.56±0.005 → relative uncertainty = 4.560.005≈0.00110 (0.110%)
For multiplication, relative uncertainties add (approximately):
A×BΔ(A×B)≈(AΔA)2+(BΔB)2
Plugging in:
≈0.004072+0.001102≈0.00422 (0.422%)
Now, the absolute uncertainty in the product:
Δ(A×B)≈0.00422×(12.3×4.56)≈0.00422×56.088≈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 …
Concept: Significant Figures Calculation — Because thickness is given along with length and breadth, this is a thin rectangular slab, so its "area" means the total surface area of all six faces, A=2(lb+bt+tl), and its volume is V=lbt. Both must be rounded to the least number of significant figures among the three measurements.
Step 1: Convert to the same unit.
Thickness =2.01 cm=0.0201 m (3 s.f.). Length =4.234 m (4 s.f.), breadth =1.005 m (4 s.f.). The least is 3 s.f. (from thickness), so both final answers are limited to 3 significant figures.
Step 2: Compute total surface area. …
The sheet is a thin rectangular slab, so its "area" means the total surface area of all six faces — length, breadth, and thickness all contribute. Using A=2(lb+bt+tl) and V=lbt, and rounding each to the significant figures set by the least precise measurement (thickness, with 3 significant figures), the total surface area is 8.72 m2 and the volume is 0.0855 m3.
Setting up
The sheet has three given dimensions:
- Length l=4.234 m
- Breadth b=1.005 m
- Thickness t=2.01 cm=0.0201 m
Because a thickness is given, this is not a flat two-dimensional rectangle — it is a thin rectangular slab (a cuboid) with six faces: two of size l×b, two of size b×t, and two of size t×l. "The area of the sheet" therefore means the total surface area of the slab, not just the area of its largest face. If only l×b were wanted, the thickness would never have been given at all.
Counting significant figures
- l=4.234 m → 4 significant figures
- b=1.005 m → 4 significant figures
- t=0.0201 m → 3 significant figures (leading zeros don't count; 2, 0, 1 do)
The least precise measurement is the thickness, with 3 significant figures. Since thickness enters both the area and volume calculations, both final answers are limited to 3 significant figures.
Total surface area
A=2(lb+bt+tl)
- lb=4.234×1.005=4.25517 m2
- bt=1.005×0.0201=0.0202005 m2
- tl=0.0201×4.234=0.0851034 m2
- Sum: 4.25517+0.0202005+0.0851034=4.3604839 m2
- A=2×4.3604839=8.7209678 m2
- Round to 3 significant figures: A=8.72 m2
Volume …
Method: Total Surface Area of a Thin Slab + Significant Figures
Method Name: Because length, breadth, and thickness are all given, the sheet is treated as a thin rectangular slab (a cuboid), not a flat 2-D rectangle. Its "area" means the total surface area of all six faces, A=2(lb+bt+tl), and its volume is V=lbt. Both results are then rounded using the Rule of Least Precise Measurement: a product carries only as many significant figures as its least precise factor.
Step 1: Identify significant figures in each given value
- Length l=4.234 m → 4 significant figures
- Breadth b=1.005 m → 4 significant figures
- Thickness t=2.01 cm → 3 significant figures
⚠️ Important: Thickness is in cm, while length and breadth are in m. Convert to the same unit before calculating.
Step 2: Convert thickness to metres
t=2.01 cm=2.01×10−2 m=0.0201 m
t still has 3 significant figures.
Step 3: Calculate the total surface area
Because thickness is given, the sheet is a slab with six faces — two of each pair (l,b), (b,t), (t,l):
A=2(lb+bt+tl)
lb=4.234×1.005=4.25517 m2
bt=1.005×0.0201=0.0202005 m2
tl=0.0201×4.234=0.0851034 m2
A=2(4.25517+0.0202005+0.0851034)=2×4.3604839=8.7209678 m2
Apply the significant figure rule: the least number of significant figures among l, b, t is 3 (from t), so round A to 3 significant figures:
A=8.72 m2
Step 4: Calculate the volume …
Here are the most common mistakes students make on this classic significant figures problem, along with the reasoning to avoid each.
Mistake 1: Forgetting to convert units before adding
The Error:
Students directly multiply 4.234×1.005×2.01 without noticing that thickness is in cm while length and breadth are in m. This gives a wildly wrong volume.
How to Avoid:
- Always check units first. Write them down beside each value.
- Convert everything to the same unit before any calculation.
- Here: 2.01 cm=0.0201 m.
Mistake 2: Using the wrong rule for multiplication/division
The Error:
Students apply the addition/subtraction rule (look at decimal places) to multiplication.
How to Avoid:
- Multiplication/Division: Round to the least number of significant figures, not decimal places.
Mistake 3: Counting significant figures incorrectly in the thickness
The Error:
Thinking 2.01 cm has 2 significant figures (because of the leading digit '2') or 4 significant figures (because of the trailing '01').
How to Avoid:
- Captive zeros between non-zero digits are significant. So 2.01 has 3 significant figures (2, 0, 1).
Mistake 4: Rounding intermediate results too early
The Error:
Rounding an intermediate product before finishing the calculation, which accumulates rounding error.
How to Avoid:
- Do the full calculation first with all digits, and round only the final answer.
Mistake 5: Computing only length × breadth and calling it "the area"
The Error:
Students multiply just the length and breadth, 4.234×1.005=4.255 m2, and report that as "the area of the sheet."
Why it's wrong:
A thickness is explicitly given — 2.01 cm — which means this is not a flat rectangle but a thin rectangular slab (a cuboid) with six faces. If "area" only meant l×b, the thickness would be completely irrelevant to the area calculation, and the question would never have given it. "The area of the sheet" here means the total surface area of the slab:
A=2(lb+bt+tl)
How to Avoid:
- Whenever a thickness (or any third dimension) is given alongside length and breadth for a physical "sheet" or "slab," compute the total surface area, not just one face.
- Compute all three face-pair products (lb, bt, tl), sum them, and double the sum:
A=2(4.234×1.005+1.005×0.0201+0.0201×4.234)=2(4.25517+0.0202005+0.0851034)≈8.72 m2 (3 s.f.)
Mistake 6: Reporting area or volume with too many or too few significant figures
The Error:
Giving area as 8.7209678 m2 (all raw digits) or keeping 4 significant figures because length and breadth each have 4.
Why it happens:
Not identifying which measurement has the least significant figures.
How to avoid:
- Identify the limiting factor: …
- TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQQ.The number of significant figures in 0.020260 is (A) 6 (B) 4 (C) 3 (D) 5
›Reveal solutionSolution
The key idea is that leading zeros are never significant, but trailing zeros after a decimal point are. For 0.020260, the significant figures are 2, 0, 2, 6, 0 — that’s 5 figures, so the answer is (D).
The concept here is significant figures (or significant digits), which tell us how precise a measurement is. The rules are simple but often misapplied:
- All non-zero digits are significant.
- Zeros between non-zero digits are significant.
- Leading zeros (to the left of the first non-zero digit) are not significant — they’re just placeholders.
- Trailing zeros after a decimal point are significant because they indicate the precision of the measurement.
In 0.020260, the zeros at the very beginning are just placeholders, but the zeros inside and at the end matter.
Let’s count step by step:
-
Identify the first non-zero digit.
The number is 0.020260. The first non-zero digit is the 2 after the decimal (the hundredths place). All zeros before it (the leading 0 before the decimal and the first 0 after the decimal) are not significant.
-
Count from that first non-zero digit onward.
Starting from the 2:
- 2 (significant)
- 0 (between 2 and 2, so significant)
- 2 (significant)
- 6 (significant)
- 0 (trailing zero after the decimal, so significant) …
- TG EAPCET 2026Set ap-2026-05-05-FN1 markMCQQ.The value of 3.00×10−20.004560×1200 in correct significant figures is (A) 1.8×102 (B) 1.824×102 (C) 182.40 (D) 182
›Reveal solutionSolution
The key idea is to apply the rule for significant figures in multiplication and division — the result must have the same number of significant figures as the term with the fewest. The final answer is 1.8×102.
The problem is about significant figures in a calculation. When you multiply or divide numbers, the result cannot be more precise than the least precise measurement involved. That means the number of significant figures in the final answer is determined by the term with the fewest significant figures.
Let’s identify the significant figures in each number:
- 0.004560 has 4 significant figures (the leading zeros don’t count, but the trailing zero after the decimal does).
- 1200 has 4 significant figures (no decimal point, so trailing zeros are ambiguous — but here it’s written without a decimal, so conventionally it has 4 significant figures; however, in many exam contexts, 1200 is taken as having 2 significant figures unless specified. Let’s check carefully: the problem gives 1200 without a decimal, so the safest interpretation is that it has 2 significant figures, because the trailing zeros are not significant. But wait — in the given expression, 1200 appears alongside 0.004560 (4 sig figs) and 3.00×10−2 (3 sig figs). The most common convention in such problems is that 1200 has 2 significant figures. We’ll verify by looking at the options — they all have 2, 3, or 4 significant figures, so we must decide.)
- 3.00×10−2 has 3 significant figures (the zeros after the decimal are significant).
Now, the term with the fewest significant figures is 1200 with 2 significant figures (if we take the standard convention). So the final answer must be rounded to 2 significant figures. …
- TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQQ.A piece of length 3.532 m is cut from a rod of length 43.4 m. The length of the remaining rod in metre is (up to correct significant figures) (A) 39.9 (B) 39.8 (C) 39.868 (D) 39.87
›Reveal solutionSolution
The key idea is that subtraction must respect the least precise decimal place in the original measurements. Since 43.4 m has one decimal place and 3.532 m has three, the result must be rounded to one decimal place, giving 39.9 m. The correct option is (A).
When we subtract two measured quantities, the result cannot be more precise than the least precise measurement. This is a fundamental rule of significant figures: the number of decimal places in the answer is limited by the measurement with the fewest decimal places. Here, 43.4 m has one decimal place (the tenths place is certain), while 3.532 m has three decimal places. So the difference must be reported to one decimal place only.
Let’s work through it step by step:
- Perform the subtraction exactly
43.4−3.532=39.868
This is the raw arithmetic result, but it implies a precision we don’t actually have.
-
Identify the limiting precision
The rod’s original length, 43.4 m, is given to the nearest tenth of a metre. That means the true length lies between 43.35 m and 43.45 m. The piece cut, 3.532 m, is given to the nearest thousandth, so its true length lies between 3.5315 m and 3.5325 m.
-
Consider the range of possible remainders
The smallest possible remainder occurs when the original rod is at its shortest and the cut piece is at its longest:
43.35−3.5325=39.8175
The largest possible remainder occurs when the original rod is at its longest and the cut piece is at its shortest:
43.45−3.5315=39.9185
So the true remainder lies somewhere between 39.8175 m and 39.9185 m.
- Round to the correct number of decimal places …
- TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQQ.A piece of length 3.532 m is cut from a rod of length 43.4 m. The length of the remaining rod in metre is (up to correct significant figures) (A) 39.8 (B) 39.868 (C) 39.9 (D) 39.87
›Reveal solutionSolution
The key idea is to apply the rules of significant figures in subtraction: the result must be rounded to the same decimal place as the least precise measurement. The remaining length is 39.9 m, so the correct option is (C).
We are given a rod of length 43.4 m and a piece of length 3.532 m is cut from it. The question asks for the length of the remaining rod, reported with the correct number of significant figures. This is not just a simple subtraction — we must respect the precision of the measurements.
Concept and Intuition:
When adding or subtracting measurements, the result cannot be more precise than the least precise measurement. Here, 43.4 m has one decimal place (tenths), while 3.532 m has three decimal places (thousandths). The subtraction should therefore be rounded to the tenths place. Many students mistakenly keep all digits from the calculation, but that would imply a false precision.
Step-by-step solution:
- Perform the subtraction exactly:
43.4−3.532=39.868
This is the raw arithmetic result.
-
Identify the least precise measurement:
- 43.4 m is measured to the nearest tenth (one decimal place).
- 3.532 m is measured to the nearest thousandth (three decimal places). The least precise is 43.4 m, with uncertainty in the tenths place.
-
Apply the rule for subtraction:
The result must be rounded to the same decimal place as the least precise measurement — the tenths place. Look at the digit in the hundredths place of 39.868, which is 6. Since 6 ≥ 5, we round up the tenths digit from 8 to 9.
-
Round the result: …
- TG EAPCET 2024Set ap-2024-05-07-FN1 markMCQQ.The density of a substance of mass 5.318 g which occupies a volume of 2.43 cm3 is (up to correct significant figures) (A) 2.188 g cm−3 (B) 2.180 g cm−3 (C) 2.19 g cm−3 (D) 2.18 g cm−3
›Reveal solutionSolution
Density is mass divided by volume; the result must be reported with the least number of significant figures from the given data. Here, 5.318 g has 4 significant figures and 2.43 cm³ has 3, so the density should have 3 significant figures: 2.19 g cm⁻³.
Concept & Intuition
The key idea is significant figures in division. When you multiply or divide measurements, the result cannot be more precise than the least precise measurement. The volume (2.43 cm³) has only 3 significant figures, so the density must be rounded to 3 significant figures — even if the raw calculation gives more digits. Many students forget this and just report the full calculator display, which leads to a wrong choice.
Step-by-step solution
- Write the formula for density Density ρ is mass m divided by volume V:
ρ=Vm
- Plug in the given values m=5.318g (4 significant figures) V=2.43cm3 (3 significant figures)
ρ=2.435.318
- Perform the division Using a calculator:
2.435.318≈2.188477...
- Apply the significant figure rule The volume has only 3 significant figures, so the density must be rounded to 3 significant figures. Look at the fourth digit: 2.188… — the third digit is 8, and the next digit is 8 ≥ 5, so we round up: 2.188→2.19 …
- TG EAPCET 2023Set ap-2023-05-10-FN1 markMCQQ.If two rods of lengths 13.244 cm and 11.2 cm are joined end to end, then their combined length to appropriate significant figures is (A) 24.444 cm (B) 24.44 cm (C) 24.4 cm (D) 2.444 cm
›Reveal solutionSolution
When adding measurements, the result's precision is limited by the least precise measurement. The combined length is found by adding the values and then rounding the sum to the same number of decimal places as the measurement with the fewest decimal places, which gives 24.4 cm.
When we perform calculations with measured quantities, it's crucial to express the result with an appropriate level of precision. This is where the concept of significant figures comes in. Significant figures indicate the reliability of a measurement. If a measurement is given as 11.2 cm, it implies that the measurement is reliable up to the first decimal place, and there's some uncertainty in the second decimal place and beyond.
When adding or subtracting measurements, the rule for significant figures is based on the number of decimal places. The idea is that the sum or difference cannot be more precise than the least precise measurement involved in the calculation. If one measurement is known only to the tenths place, the sum cannot be reliably known to the hundredths or thousandths place.
The specific rule is:
The result of an addition or subtraction should be rounded to the same number of decimal places as the measurement with the fewest decimal places.
Let's apply this to the given problem.
-
Identify the given lengths:
We are given two rod lengths:
- Length of the first rod, L1=13.244 cm
- Length of the second rod, L2=11.2 cm
-
Perform the addition:
To find the combined length, we add the two lengths:
Lcombined=L1+L2=13.244 cm+11.2 cm=24.444 cm
-
Determine the number of decimal places in each original measurement:
- L1=13.244 cm has three digits after the decimal point (2, 4, 4). So, it has 3 decimal places.
- L2=11.2 cm has one digit after the decimal point (2). So, it has 1 decimal place.
-
Apply the rule for significant figures in addition:
According to the rule, the result must be rounded to the same number of decimal places as the measurement with the fewest decimal places. Comparing 3 decimal places and 1 decimal place, the fewest is 1 decimal place.
Therefore, our calculated sum 24.444 cm must be rounded to 1 decimal place. …
-
- TG EAPCET 2023Set eng-2023-05-12-AN1 markMCQQ.The number of significant figures in 3.78×1022 kg is (A) 19 (B) 25 (C) 3 (D) 22
›Reveal solutionSolution
The number of significant figures in a number written in scientific notation is determined solely by the digits in the coefficient, not by the exponent. Here the coefficient is 3.78, which has three significant figures, so the answer is 3.
The key concept is significant figures (also called significant digits). These are the digits in a number that carry meaningful information about its precision. When a number is written in scientific notation — like 3.78×1022 — the exponent (1022) only tells us the order of magnitude (how large or small the number is). It does not affect how many digits are considered reliable or measured. All the significant figures are in the coefficient (the number before the ×10).
A common mistake is to think the exponent contributes to the count of significant figures. For example, someone might see 1022 and think “22” is part of the precision, but that’s wrong — the exponent is just a placeholder for the decimal point.
Let’s work through it:
-
Identify the coefficient.
The number is 3.78×1022. The coefficient is 3.78.
-
Count the digits in the coefficient.
3.78 has three digits: 3, 7, and 8. All are non-zero, so each is significant.
-
Ignore the exponent.
The 1022 part tells us the number is 378000000000000000000000 (378 followed by 20 zeros), but those zeros are not measured — they are just placeholders. The precision is still only three digits.
-
Select the matching option. …
-
- TG EAPCET 2023Set eng-2023-05-13-AN1 markMCQQ.The sum of three values 12.0, 19.034 and 2.0143 is equal to X. The number of significant figures in X is (A) 2 (B) 5 (C) 4 (D) 3
›Reveal solutionSolution
When adding numbers, the result should be reported with the same number of decimal places as the least precise measurement. The sum is 33.0483, but the least precise value (12.0) has one decimal place, so the sum rounds to 33.0, which has 3 significant figures. The correct option is (D).
The key idea here is the rule for significant figures in addition and subtraction: the result cannot have more decimal places than the measurement with the fewest decimal places. This is different from multiplication/division, where you count total significant figures. Many students mistakenly apply the multiplication rule to addition, which is the classic pitfall.
Let’s work through it step by step.
-
Identify the decimal places of each term.
- 12.0 has 1 decimal place (the tenths place).
- 19.034 has 3 decimal places (thousandths).
- 2.0143 has 4 decimal places (ten-thousandths). The least precise measurement is 12.0, with only 1 decimal place.
-
Perform the addition exactly.
12.0+19.034+2.0143=33.0483
This is the raw sum.
- Apply the addition rule. Since the least precise number has 1 decimal place, the sum must be rounded to 1 decimal place.
33.0483→33.0
(The digit after the tenths place is 4, so we round down.)
- Count the significant figures in the rounded result. …
-
- TG EAPCET 2022Set eng-2022-07-19-FN1 markMCQQ.The number of significant figures in 2.0400 is (A) 3 (B) 2 (C) 4 (D) 5
›Reveal solutionSolution
The key idea is that trailing zeros after a decimal point are significant. The number 2.0400 has 5 significant figures.
The concept here is significant figures — the digits in a number that carry meaningful information about its precision. The rules for counting them are straightforward, but the trickiest part is handling zeros. Zeros can be significant or not, depending on where they appear.
For 2.0400, let’s apply the rules step by step.
-
All non-zero digits are significant.
The digit
2is non-zero, so it counts. That’s 1 significant figure so far. -
Zeros between non-zero digits are significant.
Here, the zero between
2and4(the first zero after the decimal) is between two non-zero digits? Actually, check: the digits are2,.,0,4,0,0. The zero immediately after the decimal is between2and4? No —2is before the decimal,0is after. But the rule applies to zeros between non-zero digits within the number. Since2and4are separated only by a decimal point and a zero, that zero is indeed between them, so it is significant. That’s 2 significant figures now. -
Trailing zeros after a decimal point are significant.
Look at the end:
...400. The last two zeros come after the decimal and after the non-zero digit4. These are trailing zeros in a decimal number, so they are significant. That adds 2 more significant figures (the two zeros at the end). -
Count them all:
2→ 10(between 2 and 4) → 14→ 10(trailing) → 10(trailing) → 1 …
-
- TG EAPCET 2022Set eng-2022-07-20-AN1 markMCQQ.Round the number 234555359 to 3 significant figures. (A) 234000000 (B) 234000 (C) 235000000 (D) 234500000
›Reveal solutionSolution
Rounding to 3 significant figures means keeping the first three non-zero digits and adjusting the third based on the fourth digit. The result is 235000000.
The key idea behind significant figures is that they count only the digits that carry meaningful information about precision. For a large number like 234555359, the first three significant figures are the digits 2, 3, and 4 — because the number starts with a non-zero digit. The rest of the digits are placeholders that get replaced by zeros after rounding.
When rounding to a certain number of significant figures, you look at the digit immediately after the last one you want to keep. If that digit is 5 or greater, you round up the last kept digit; otherwise, you leave it as is.
-
Identify the first three significant figures.
The number is 234555359. The first digit (2) is the most significant, followed by 3, then 4. So the first three significant figures are 2, 3, and 4.
-
Locate the fourth digit (the one that decides rounding).
The fourth digit is the next one after the third significant figure. Here, after 2-3-4 comes the digit 5 (the fourth digit). So we look at this 5.
-
Apply the rounding rule.
Since the fourth digit is 5 (which is ≥ 5), we round up the third significant figure (which is 4) by 1. That makes it 5. All digits after the third significant figure become zeros.
-
Write the rounded number. …
-
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