Class 8 Mathematics Federal Board of Intermediate and Secondary Education (FBISE) (FBISE) 📚 Model Textbook of Mathematics Grade 8 (FBISE / NBF)

Class 8 Mathematics - Ch 2: Estimation, Rounding Off, Significant Figures & Approximation Error (FBISE)

📖 Chapter 2: Estimation and Approximation 📅 Updated: Sep 19, 2026

Teacher & Student Roadmap Grade 8 Mathematics • FBISE / National Curriculum (NBF)

Instructional Guide: Unit 02 Estimation and Approximation (تخمینہ اور تقریب)

Target Student Learning Outcomes (SLOs)

Understand mathematical vocabulary related to estimation and approximation.
Round whole numbers to the nearest 10, 100, 1000, and to greatest place values.
Round rational numbers and decimals to specified decimal places (d.p.) and nearest whole numbers.
Master all rules of Significant Figures (s.f.) for non-zeros, captive zeros, trailing zeros, and leading zeros.
Calculate and analyze approximation errors and upper/lower bounds in perimeter, area, and volume.
Apply estimation techniques to verify real-world computations, budgets, and mental arithmetic.

Common Student Misconceptions & Traps

Leading Zeros vs Significant Figures: Thinking $0.0034$ has 4 s.f. Leading zeros in numbers $< 1$ are placeholders only ($0.0034$ has only 2 s.f.).
Trailing Zeros with Decimals: Dropping trailing zeros like writing $6.8$ instead of $6.80$ when asked for 2 d.p. or 3 s.f. Trailing zeros after decimal indicate precision!
Decimal Places vs Significant Figures: Confusing 2 d.p. with 2 s.f. $0.00429$ to 2 d.p. is $0.00$, but to 2 s.f. is $0.0043$.
Error Bounds in Calculations: Assuming max area error is simply calculating with rounded numbers, rather than computing $(l_{\max} \times w_{\max}) - \text{nominal area}$.

3-Step Concept Mastery Strategy

Step 1: Identify Target Place: Underline the target place value (e.g. tenths place, 3rd significant digit).
Step 2: Inspect the Deciding Digit (Immediate Right): If digit is $\ge 5$, round UP $(+1)$. If $< 5$, round DOWN (keep digit unchanged).
Step 3: Fill / Truncate: For whole numbers, replace following digits with zeros. For decimals, drop following digits.

Curriculum Note: This unit aligns with FBISE Grade 8 Mathematics (NBF - National Curriculum 2022). All decimal rounding, significant figures taxonomy, error analysis, and 100% solved exercises are presented with step-by-step calculations.

Unit 2: Estimation & Approximation

A complete conceptual guide covering rounding whole numbers and decimals, significant figures taxonomy, calculation estimation, and error bound analysis in practical geometry.

1. Estimation and Rounding Off Whole Numbers

In daily life, exact figures are often unnecessary or impossible to determine immediately. When planning a wedding feast for 180 invited guests, the caterer prepares food for approximately 160 to 200 people. This rough calculation is called estimation (تخمینہ) or approximation (تقریب).

Three Ways to Round Off Numbers:

To the nearest 10, 100, 1000, or greatest place.
To a specified number of decimal places (d.p.).
To a specified number of significant figures (s.f.).

A. Rounding to Nearest Tens (10)

Rule: Look at the digit in the ones (units) place.
• If ones digit is $0, 1, 2, 3, \text{ or } 4$ ($< 5$), replace ones digit with $0$.
• If ones digit is $5, 6, 7, 8, \text{ or } 9$ ($\ge 5$), increase tens digit by $1$ and replace ones digit with $0$.

Example 1: Round $43$ to nearest tens.
Ones digit is $3 < 5 \implies \mathbf{40}$.

Example 2: Round $70006$ to nearest tens.
Ones digit is $6 \ge 5 \implies \mathbf{70010}$.

B. Rounding to Nearest Hundreds (100)

Rule: Look at the digit in the tens place.
• If tens digit $< 5$, replace tens and ones digits with zeros.
• If tens digit $\ge 5$, increase hundreds digit by $1$ and replace tens and ones digits with zeros.
Key Fact: Numbers ending in $01\text{ to }49$ round downwards; numbers ending in $50\text{ to }99$ round upwards.

Example 1: Round $5637$ to nearest hundreds.
Tens digit is $3 < 5 \implies \mathbf{5600}$.

Example 2: Round $9673$ to nearest hundreds.
Tens digit is $7 \ge 5 \implies 9600 + 100 = \mathbf{9700}$.

C. Rounding to Nearest Thousands (1000)

Rule: Look at the digit in the hundreds place.
• If hundreds digit $< 5$, replace hundreds, tens, and ones digits with zeros.
• If hundreds digit $\ge 5$, increase thousands digit by $1$ and replace lower digits with zeros.

Example 1: Round $17349$ to nearest 1000.
Hundreds digit is $3 < 5 \implies \mathbf{17000}$.

Example 2: Round $15836$ to nearest 1000.
Hundreds digit is $8 \ge 5 \implies \mathbf{16000}$.

Solved Textbook Check Points (Page 21)

Check Point 1: Check which of the following numbers rounded off upwards or downwards to nearest hundreds:

(i) $7342$: Tens digit is $4 < 5 \implies$ Rounded downwards to $7300$.
(ii) $5789$: Tens digit is $8 \ge 5 \implies$ Rounded upwards to $5800$.
(iii) $3565$: Tens digit is $6 \ge 5 \implies$ Rounded upwards to $3600$.
(iv) $4772$: Tens digit is $7 \ge 5 \implies$ Rounded upwards to $4800$.
(v) $8676$: Tens digit is $7 \ge 5 \implies$ Rounded upwards to $8700$.

Check Point 2: A hockey match is watched by 47361 people. Write this number correct to the nearest:

(a) $10,000$: Thousands digit is $7 \ge 5 \implies \mathbf{50,000}$.
(b) $1,000$: Hundreds digit is $3 < 5 \implies \mathbf{47,000}$.
(c) $10$: Ones digit is $1 < 5 \implies \mathbf{47,360}$.

2. Rounding Off Decimals to Required Degree of Accuracy (Decimal Places)

In science and financial transactions, we round off long strings of decimal digits to a manageable degree of accuracy.

A. Nearest Whole Number (0 d.p.)

Look at the tenths digit (1st digit after decimal point).
• $8.74 \to$ tenths digit is $7 \ge 5 \implies \mathbf{9}$.
• $97.547 \to$ tenths digit is $5 \ge 5 \implies \mathbf{98}$.
• $14.47 \to$ tenths digit is $4 < 5 \implies \mathbf{14}$.

B. Nearest Tenths (1 d.p.)

Look at the hundredths digit (2nd digit after decimal).
• $19.354 \to$ hundredths is $5 \ge 5 \implies \mathbf{19.4}$.
• $13.34 \to$ hundredths is $4 < 5 \implies \mathbf{13.3}$.

C. Nearest Hundredths (2 d.p.)

Look at the thousandths digit (3rd digit after decimal).
• $403.389 \to$ thousandths is $9 \ge 5 \implies \mathbf{403.39}$.
• $67.024 \to$ thousandths is $4 < 5 \implies \mathbf{67.02}$.

3. Significant Figures (s.f.) — Concepts, Rules & Visual Guide

Definition: In physical measurement, significant figures (نمایاں اعداد) are all the accurately known (certain) digits plus one estimated (uncertain) digit.
Example: If a speedometer reads between $120.4\text{ km/h}$ and $120.5\text{ km/h}$, an estimated speed of $120.46\text{ km/h}$ contains 5 significant figures (4 certain digits: $1, 2, 0, 4$ and 1 estimated digit: $6$).

Summary of Significant Figures Rules

Rule Category	Condition	Status	Example & s.f. Count
Non-Zero Digits	All digits from $1$ to $9$	Always Significant	$8762 \implies \mathbf{4\text{ s.f.}}$
Captive (Middle) Zeros	Zeros trapped between non-zero digits	Always Significant	$602.005 \implies \mathbf{6\text{ s.f.}}$
Trailing Zeros (with Decimal)	Zeros at the end of decimal number	Always Significant	$70.00 \implies \mathbf{4\text{ s.f.}}$, $72.04010 \implies \mathbf{7\text{ s.f.}}$
Leading Zeros (Numbers $< 1$)	Zeros before first non-zero digit	Never Significant	$0.00171 \implies \mathbf{3\text{ s.f.}}$ (2 non-sig zeros after decimal)
Trailing Zeros (Whole Numbers)	Zeros at end without a decimal point	Not Significant (Placeholders)	$83,00 \implies \mathbf{2\text{ s.f.}}$

Solved Textbook Check Point (Page 26): Number of Significant Figures

(i) $53.214$: All digits are non-zero $\implies \mathbf{5\text{ s.f.}}$
(ii) $7.2051$: Zero is trapped between 2 and 5 $\implies \mathbf{5\text{ s.f.}}$
(iii) $60.003$: Trapped zeros between 6 and 3 $\implies \mathbf{5\text{ s.f.}}$
(iv) $0.0001269$: Four leading zeros ($0.000$) are placeholders $\implies \mathbf{4\text{ s.f.}}$ ($1, 2, 6, 9$).

4. Approximation Error & Error Bounds in Practical Geometry

When a measurement is rounded to the nearest unit, the actual true value lies within a range called the error bounds.
• Lower Bound (LB) $= \text{Nominal Value} - 0.5 \times \text{unit of precision}$
• Upper Bound (UB) $= \text{Nominal Value} + 0.5 \times \text{unit of precision}$
• Maximum Possible Error $= |\text{Calculated Extreme} - \text{Nominal Value}|$

Textbook Example: Area Error for Circle ($r = 21\text{ cm}$ to nearest cm)

Radius interval: $20.5\text{ cm} \le r < 21.5\text{ cm}$ ($\pi \approx \frac{22}{7}$).
• Case 1 (Lower Bound): $A_{\min} = \frac{22}{7} \times (20.5)^2 = 1320.7857\text{ cm}^2 \approx \mathbf{1320\text{ cm}^2}$ (to 3 s.f.).
• Case 2 (Nominal): $A_{\text{nominal}} = \frac{22}{7} \times (21)^2 = 1386\text{ cm}^2 \approx \mathbf{1390\text{ cm}^2}$ (to 3 s.f.).
• Case 3 (Upper Bound): $A_{\max} = \frac{22}{7} \times (21.5)^2 = 1452.7857\text{ cm}^2 \approx \mathbf{1450\text{ cm}^2}$ (to 3 s.f.).
• Error at LB: $1390 - 1320.7857 = 69.21\text{ cm}^2$.
• Error at UB: $1452.7857 - 1390 = 62.79\text{ cm}^2$.
• Maximum Possible Error: $\mathbf{69.21\text{ cm}^2}$ (occurs at $r = 20.5\text{ cm}$).

Exercise 2.1 — Step-by-Step Complete Solutions

Question 1: Round off the numbers to the indicated decimal place.

(i) $3.386$ (1 d.p.): Look at 2nd decimal digit ($8 \ge 5$). Add 1 to tenths digit $\implies \mathbf{3.4}$

(ii) $3.247$ (1 d.p.): Look at 2nd decimal digit ($4 < 5$). Tenths digit remains 2 $\implies \mathbf{3.2}$

(iii) $4.266$ (2 d.p.): Look at 3rd decimal digit ($6 \ge 5$). Add 1 to hundredths $\implies \mathbf{4.27}$

(iv) $0.375$ (2 d.p.): Look at 3rd decimal digit ($5 \ge 5$). Add 1 to hundredths $\implies \mathbf{0.38}$

(v) $4.68$ (1 d.p.): Look at 2nd decimal digit ($8 \ge 5$). Add 1 to tenths digit $\implies \mathbf{4.7}$

(vi) $7.5328$ (2 d.p.): Look at 3rd decimal digit ($2 < 5$). Hundredths digit remains 3 $\implies \mathbf{7.53}$

(vii) $2.34$ (1 d.p.): Look at 2nd decimal digit ($4 < 5$). Tenths digit remains 3 $\implies \mathbf{2.3}$

(viii) $5.735$ (2 d.p.): Look at 3rd decimal digit ($5 \ge 5$). Add 1 to hundredths $\implies \mathbf{5.74}$

(ix) $0.18375$ (3 d.p.): Look at 4th decimal digit ($7 \ge 5$). Add 1 to thousandths $\implies \mathbf{0.184}$

(x) $0.27286$ (2 d.p.): Look at 3rd decimal digit ($2 < 5$). Hundredths digit remains 7 $\implies \mathbf{0.27}$

(xi) $0.36265$ (4 d.p.): Look at 5th decimal digit ($5 \ge 5$). Add 1 to 4th digit $\implies \mathbf{0.3627}$

(xii) $0.17561$ (4 d.p.): Look at 5th decimal digit ($1 < 5$). 4th digit remains 6 $\implies \mathbf{0.1756}$

Question 2: Estimate the answer to each of the following calculations.

(i) $3.7 \times 12.2$:
$3.7 \approx 4$, $12.2 \approx 12 \implies 4 \times 12 = \mathbf{48}$ (or to 1 s.f.: $4 \times 10 = 40$)

(ii) $56 \times 183$:
$56 \approx 60$, $183 \approx 200 \implies 60 \times 200 = \mathbf{12000}$

(iii) $32.7 \times 502$:
$32.7 \approx 30$, $502 \approx 500 \implies 30 \times 500 = \mathbf{15000}$

(iv) $12.26 \times 75.4$:
$12.26 \approx 10$, $75.4 \approx 80 \implies 10 \times 80 = \mathbf{800}$ (or $12 \times 75 = 900$)

(v) $13.82 \times 3.82$:
$13.82 \approx 14$, $3.82 \approx 4 \implies 14 \times 4 = \mathbf{56}$

(vi) $104.7 \div 23.81$:
$104.7 \approx 100$, $23.81 \approx 20 \implies 100 \div 20 = \mathbf{5}$

(vii) $44.31 \div 1.876$:
$44.31 \approx 44$ (or $40$), $1.876 \approx 2 \implies 44 \div 2 = \mathbf{22}$

(viii) $69.37 \div 7.49$:
$69.37 \approx 70$, $7.49 \approx 7 \implies 70 \div 7 = \mathbf{10}$

(ix) $14.023 \div 6.816$:
$14.023 \approx 14$, $6.816 \approx 7 \implies 14 \div 7 = \mathbf{2}$

(x) $105.732 \div 9.652$:
$105.732 \approx 100$ (or $110$), $9.652 \approx 10 \implies 100 \div 10 = \mathbf{10}$ (or $11$)

Question 3: Estimate the value of:

(i) $15.1 + 36.02 - 8.9$:
Round each number to nearest integer:
$15.1 \approx 15$, $36.02 \approx 36$, $8.9 \approx 9$
$\text{Estimated Value} = 15 + 36 - 9 = 51 - 9 = \mathbf{42}$ (Actual: $42.22$)

(ii) $\sqrt{16.3 \times 24.8}$:
Round numbers inside radical:
$16.3 \approx 16$, $24.8 \approx 25$
$\text{Estimated Value} = \sqrt{16 \times 25} = \sqrt{16} \times \sqrt{25} = 4 \times 5 = \mathbf{20}$ (Actual: $\sqrt{404.24} \approx 20.105$)

Question 4: Sidra wrote this calculation: $14.62 \times 401 = 586.262$.

(a) Use estimation to check why Sidra was wrong:
Estimate each number: $14.62 \approx 15$ (or $10$), $401 \approx 400$.
$\text{Estimated Product} = 15 \times 400 = \mathbf{6000}$.
Since $586.262$ is roughly $600$ (which is $10$ times smaller than $6000$), Sidra's answer is completely wrong.

(b) Determine correct answer and describe Sidra's mistake:
• Calculator Answer: $14.62 \times 401 = \mathbf{5862.62}$.
• Mistake Description: Sidra misplaced the decimal point. She put $3$ decimal places ($586.262$) instead of $2$ decimal places ($5862.62$), effectively dividing her answer by $10$.

Question 5: The correct answer of $16.3 \times 25.7$ is given below along with 3 wrong answers. Use estimation to decide which is the correct answer:
(i) $41.891$ (ii) $418.91$ (iii) $4189.1$ (iv) $41891$

Solution:
Estimate: $16.3 \approx 16$ (or $20$), $25.7 \approx 25$ (or $25$).
$\text{Estimated product} = 16 \times 25 = 400$ (or $20 \times 25 = 500$).
Among the options, only (ii) $418.91$ is close to $400$.
Correct Option: $\mathbf{(ii)\ 418.91}$

Question 6: Use estimation to decide how much the following calculations differ from their actual values.

(i) $16.4 \times 7321 = 120,000$ (given estimate):
• Estimate: $16 \times 7000 = 112,000$ (or $20 \times 7000 = 140,000$ or rounded to $120,000$).
• Actual value: $16.4 \times 7321 = 120,064.4$.
• Difference: $|120,064.4 - 120,000| = \mathbf{64.4}$ (or from $112,000$: $8,064.4$).

(ii) $65.332 \times 10.3 = 650$ (given estimate):
• Estimate: $65 \times 10 = 650$.
• Actual value: $65.332 \times 10.3 = 672.9196$.
• Difference: $|672.9196 - 650| = \mathbf{22.9196}$.

(iii) $197 \times 3576 = 700,000$ (given estimate):
• Estimate: $200 \times 3500 = 700,000$ (or $200 \times 3600 = 720,000$).
• Actual value: $197 \times 3576 = 704,472$.
• Difference: $|704,472 - 700,000| = \mathbf{4,472}$.

(iv) $437.81 \div 2.27 = 190$ (given estimate):
• Estimate: $440 \div 2.2 = 200$ (or $400 \div 2 = 200$ or $190$).
• Actual value: $437.81 \div 2.27 = 192.8678\dots$
• Difference: $|192.8678 - 190| \approx \mathbf{2.87}$.

Question 7: At a school the average pocket money spent by each student during break time is Rs. 20. There are 1500 students in the school. Estimate the total amount spent by students each day.

Given Data:
Average pocket money per student $= \text{Rs. } 20$
Number of students $= 1500$
Calculation:
$$\text{Estimated Total Spent} = 1500 \times 20 = \mathbf{\text{Rs. } 30,000}$$ Conclusion: The total estimated pocket money spent by all students each day is Rs. 30,000.

Exercise 2.2 — Step-by-Step Complete Solutions

Question 1: Find the exact number of significant figures of the following numbers.

(i) $7271$: All digits non-zero $\implies \mathbf{4\text{ s.f.}}$

(ii) $936.12$: All digits non-zero $\implies \mathbf{5\text{ s.f.}}$

(iii) $701.009$: Captive zeros between non-zeros $\implies \mathbf{6\text{ s.f.}}$

(iv) $80.00$: Trailing zeros after decimal $\implies \mathbf{4\text{ s.f.}}$

(v) $75.03020$: Trapped & trailing zeros after decimal $\implies \mathbf{7\text{ s.f.}}$

(vi) $0.00225$: Leading zeros are placeholders $\implies \mathbf{3\text{ s.f.}}$ ($2, 2, 5$)

(vii) $0.000370$: Leading zeros not sig, trailing zero is sig $\implies \mathbf{3\text{ s.f.}}$ ($3, 7, 0$)

(viii) $0.002370$: Leading zeros not sig, trailing zero is sig $\implies \mathbf{4\text{ s.f.}}$ ($2, 3, 7, 0$)

(ix) $0.0003670$: Leading zeros not sig, trailing zero is sig $\implies \mathbf{4\text{ s.f.}}$ ($3, 6, 7, 0$)

(x) $0.026400$: Leading zeros not sig, trailing zeros are sig $\implies \mathbf{5\text{ s.f.}}$ ($2, 6, 4, 0, 0$)

Question 2: Find the exact number of significant and non-significant figures of the following numbers.

Part	Number	Significant Figures	Non-Significant Figures	Explanation
(i)	$8.986$	4	Nil (0)	All non-zeros
(ii)	$93.8463$	6	Nil (0)	All non-zeros
(iii)	$1009.001$	7	Nil (0)	Captive zeros between non-zeros
(iv)	$10.90$	4	Nil (0)	Captive and trailing decimal zero
(v)	$30.30210$	7	Nil (0)	Captive & trailing zeros after decimal
(vi)	$0.003450$	4 ($3,4,5,0$)	3 ($0.00$)	Leading zeros are placeholders
(vii)	$0.03710$	4 ($3,7,1,0$)	2 ($0.0$)	Leading zeros are placeholders
(viii)	$0.029700$	5 ($2,9,7,0,0$)	2 ($0.0$)	Leading zeros are placeholders
(ix)	$0.000370$	3 ($3,7,0$)	4 ($0.000$)	Leading zeros are placeholders
(x)	$0.02400$	4 ($2,4,0,0$)	2 ($0.0$)	Leading zeros are placeholders

Question 3: Write each of the following numbers correct to 3 significant figures.

(i) $49315$: 4th digit is $1 < 5 \implies \mathbf{49300}$

(ii) $505861$: 4th digit is $8 \ge 5 \implies \mathbf{506000}$

(iii) $20.357$: 4th digit is $5 \ge 5 \implies \mathbf{20.4}$

(iv) $0.004295$: 4th sig digit is $5 \ge 5 \implies \mathbf{0.00430}$

Question 4: Round off each of the following to: (a) 1 s.f., (b) 2 s.f., (c) 3 s.f.

Number	(a) 1 Sig Fig	(b) 2 Sig Figs	(c) 3 Sig Figs
(i) $0.003284$	$0.003$	$0.0033$	$0.00328$
(ii) $3.0829$	$3$	$3.1$	$3.08$
(iii) $302.104$	$300$	$300$	$302$
(iv) $1382.955$	$1000$	$1400$	$1380$
(v) $9.302$	$9$	$9.3$	$9.30$
(vi) $3.9991$	$4$	$4.0$	$4.00$
(vii) $40.001$	$40$	$40$	$40.0$
(viii) $0.0001256$	$0.0001$	$0.00013$	$0.000126$
(ix) $3.4072$	$3$	$3.4$	$3.41$
(x) $64.321$	$60$	$64$	$64.3$

Question 5: Round off the following measurements to indicated significant figures.

(i) $1273.866$ to 6 s.f.: 7th digit is $6 \ge 5 \implies \mathbf{1273.87}$

(ii) $203.102$ to 4 s.f.: 5th digit is $0 < 5 \implies \mathbf{203.1}$

(iii) $1.0718$ to 2 s.f.: 3rd digit is $7 \ge 5 \implies \mathbf{1.1}$

(iv) $0.003674$ to 1 s.f.: 2nd sig digit is $6 \ge 5 \implies \mathbf{0.004}$

(v) $0.0001266$ to 3 s.f.: 4th sig digit is $6 \ge 5 \implies \mathbf{0.000127}$

Question 6: A rectangular window of a room has sides with lengths of $40\text{ m}$ and $50\text{ m}$ correct to the nearest meter. Calculate the maximum and minimum possible values of: (i) the perimeter, (ii) the area, (iii) the maximum and minimum error while calculating perimeter and area.

Given Intervals (correct to nearest meter $\implies \text{tolerance} = \pm 0.5\text{ m}$):
• Width: $w = 40\text{ m} \implies 39.5\text{ m} \le w < 40.5\text{ m}$
• Length: $l = 50\text{ m} \implies 49.5\text{ m} \le l < 50.5\text{ m}$
• Nominal Perimeter: $P_{\text{nominal}} = 2(50 + 40) = 2(90) = \mathbf{180\text{ m}}$
• Nominal Area: $A_{\text{nominal}} = 50 \times 40 = \mathbf{2000\text{ m}^2}$

(i) Maximum and Minimum Possible Perimeter:
• $\text{Minimum Perimeter } (P_{\min}) = 2(l_{\min} + w_{\min}) = 2(49.5 + 39.5) = 2(89) = \mathbf{178\text{ m}}$
• $\text{Maximum Perimeter } (P_{\max}) = 2(l_{\max} + w_{\max}) = 2(50.5 + 40.5) = 2(91) = \mathbf{182\text{ m}}$

(ii) Maximum and Minimum Possible Area:
• $\text{Minimum Area } (A_{\min}) = l_{\min} \times w_{\min} = 49.5 \times 39.5 = \mathbf{1960.25\text{ m}^2}$
• $\text{Maximum Area } (A_{\max}) = l_{\max} \times w_{\max} = 50.5 \times 40.5 = \mathbf{2045.25\text{ m}^2}$

(iii) Maximum and Minimum Error in Perimeter and Area:
• Perimeter Errors:
• $\text{Error at lower bound} = |178 - 180| = 2\text{ m}$
• $\text{Error at upper bound} = |182 - 180| = 2\text{ m}$
• $\text{Minimum error} = \mathbf{0\text{ m}}$ (at exact measurement), $\text{Maximum error} = \mathbf{2\text{ m}}$.
• Area Errors:
• $\text{Error at lower bound} = |1960.25 - 2000| = 39.75\text{ m}^2$
• $\text{Error at upper bound} = |2045.25 - 2000| = 45.25\text{ m}^2$
• $\text{Minimum error} = \mathbf{0\text{ m}^2}$ (at exact measurement), $\text{Maximum error} = \mathbf{45.25\text{ m}^2}$ (at upper bound).

Review Exercise 2 — Step-by-Step Complete Solutions

Question 1: Encircle the correct answer in the following questions (MCQs).

(i) The round off value of $108.76$ to nearest whole number is:
Tenths digit is $7 \ge 5 \implies \mathbf{(b)\ 109}$

(ii) The round off value of $0.7868$ to nearest whole number is:
Tenths digit is $7 \ge 5 \implies \mathbf{(c)\ 1}$

(iii) The value of $43.006$ to nearest 1-decimal place is:
Hundredths digit is $0 < 5 \implies \mathbf{43.0}$

(iv) The value of $18.253$ to nearest 1 decimal place is:
Hundredths digit is $5 \ge 5 \implies \mathbf{(a)\ 18.3}$

(v) If estimated value of $\sqrt{103.4}$ is $10$, then its actual value is:
$\sqrt{103.4} \approx 10.1685\dots \approx \mathbf{(b)\ 10.2}$

(vi) The value $4.3062$ has ..... significant figures:
All non-zeros and trapped zero count $\implies \mathbf{(c)\ 5}$

(vii) The value $0.0001366$ has ..... non-significant figures:
Four leading placeholder zeros ($0.000$) $\implies \mathbf{(c)\ 4}$

(viii) If we round off $6.785$ to 2-decimal places, we get $6.80$ which has ..... significant figures:
$6.80$ contains 3 digits ($6, 8, 0$) $\implies \mathbf{(b)\ 3}$

(ix) All zeros between non-zero numbers are always:
Captive zeros are always $\implies \mathbf{(c)\ \text{Significant}}$

(x) The value $1273.866$ correct to 6 significant figures is:
7th digit is $6 \ge 5 \implies \mathbf{(a)\ 1273.87}$

Question 2: Find decimal approximation.

(i) $9.6$ to nearest whole number: Tenths is $6 \ge 5 \implies \mathbf{10}$

(ii) $16.74$ to nearest tenth (1 d.p.): Hundredths is $4 < 5 \implies \mathbf{16.7}$

(iii) $0.64$ to 2-decimal places: Already has 2 d.p. $\implies \mathbf{0.64}$

(iv) $0.3757$ to 3-decimal places: 4th digit is $7 \ge 5 \implies \mathbf{0.376}$

(v) $0.81642$ to 4-decimal places: 5th digit is $2 < 5 \implies \mathbf{0.8164}$

(vi) $0.98765$ to 5-decimal places: Already has 5 d.p. $\implies \mathbf{0.98765}$

Question 3: Inzamam-Ul-Haq Batting Average

Average is calculated as $57.5752$. Round this to 1-decimal place.
• Look at hundredths digit: $7 \ge 5$.
• Add 1 to tenths digit ($5+1=6$).
Answer: $\mathbf{57.6}$

Question 4: Insaf Jeweler Profit

Annual profit is $\$147.837\text{ million}$. Round this to:
• (i) 1-decimal place: Hundredths is $3 < 5 \implies \mathbf{\$147.8\text{ million}}$
• (ii) 2-decimal place: Thousandths is $7 \ge 5 \implies \mathbf{\$147.84\text{ million}}$
• (iii) 3-decimal place: Stated value $\implies \mathbf{\$147.837\text{ million}}$

Question 5: Write the number $194.8693$ correct to:

(i) 3 d.p.: $\mathbf{194.869}$

(ii) 3 s.f.: $\mathbf{195}$

(iii) 2 s.f.: $\mathbf{190}$

(iv) 1 d.p.: $\mathbf{194.9}$

(v) 4 d.p.: $\mathbf{194.8693}$

(vi) 5 s.f.: $\mathbf{194.87}$

(vii) 2 d.p.: $\mathbf{194.87}$

(viii) 1 s.f.: $\mathbf{200}$

Question 6: Write each of the following numbers correct to 3-significant figures.

(i) $87215$: $\mathbf{87200}$

(ii) $707862$: $\mathbf{708000}$

(iii) $26.352$: $\mathbf{26.4}$

(iv) $0.004157$: $\mathbf{0.00416}$

(v) $0.2563$: $\mathbf{0.256}$

(vi) $0.003735$: $\mathbf{0.00374}$

(vii) $8.5324$: $\mathbf{8.53}$

(viii) $6.0163$: $\mathbf{6.02}$

(ix) $4.7324$: $\mathbf{4.73}$

(x) $19.2062$: $\mathbf{19.2}$

(xi) $0.025960$: $\mathbf{0.0260}$

(xii) $134578$: $\mathbf{135000}$

Question 7: Round off the following measurements to indicated significant figures.

(i) $7.385$ to 2 s.f.: $\mathbf{7.4}$

(ii) $1382$ to 1 s.f.: $\mathbf{1000}$

(iii) $2793$ to 2 s.f.: $\mathbf{2800}$

(iv) $18003$ to 3 s.f.: $\mathbf{18000}$

(v) $19.888$ to 3 s.f.: $\mathbf{19.9}$

(vi) $180.28$ to 4 s.f.: $\mathbf{180.3}$

(vii) $36.8613$ to 2 s.f.: $\mathbf{37}$

(viii) $190.28$ to 4 s.f.: $\mathbf{190.3}$

(ix) $149.26$ to 3 s.f.: $\mathbf{149}$

(x) $681.42$ to 4 s.f.: $\mathbf{681.4}$

Question 8: Use estimation to decide which of the following calculations are definitely wrong / right.

(i) $15.2 \times 6120 = 930240$:
Estimate: $15 \times 6000 = 90,000$.
Given value is $930,240$ (about 10 times too large).
Verdict: Definitely Wrong! (Actual: $93,024$)

(ii) $65.224 \times 12.4 = 5.26$:
Estimate: $65 \times 10 = 650$.
Given value is $5.26$ (it was divided instead of multiplied!).
Verdict: Definitely Wrong! (Actual: $808.7776$)

(iii) $192 \times 4587 = 880704$:
Estimate: $200 \times 4500 = 900,000$.
Given value $880,704$ is close to $900,000$.
Verdict: Correct!

(iv) $346.92 \times 2.36 = 14.7$:
Estimate: $350 \times 2 = 700$.
Given value is $14.7$ (decimal error/division).
Verdict: Definitely Wrong! (Actual: $818.7312$)

Question 9: Estimating Words in a Book

A book has $328$ pages with an average of $270.3$ words on each page. Estimate the number of words in the book.
• Round pages: $328 \approx 300$ (or $330$ or $300$ to 1 s.f.)
• Round words per page: $270.3 \approx 300$ (or $270$)
• $\text{Estimated Words} = 300 \times 300 = \mathbf{90,000\text{ words}}$
(Using 2 s.f.: $330 \times 270 = \mathbf{89,100\text{ words}}$, Actual: $88,658.4$)

Question 10: Circumference of a Circle

Estimate the circumference of a circle to 2-decimal places with a radius of $23.7\text{ cm}$.
• Formula: $C = 2\pi r$
• Using $\pi \approx 3.14159$ (or $\frac{22}{7}$):
$$C = 2 \times 3.14159 \times 23.7 = 148.911\dots \approx \mathbf{148.91\text{ cm}}$$ (Using $\pi = \frac{22}{7}$: $C = \frac{44 \times 23.7}{7} = \frac{1042.8}{7} = 148.9714\dots \approx \mathbf{148.97\text{ cm}}$)

5. Study Cues, Rhymes & Frequently Asked Questions

🎵 Rhyme: The Rounding Roller-Coaster

Find your place and look to the right,
Four or less, sleep tight (keep it tight)!
Five or more, raise the score (add one more)!
All numbers behind, zeros or out of sight!

🔍 The 3 Zero Rules in a Nutshell

Trapped Zeros: $5005 \implies$ Count them (Significant).
Leading Zeros: $0.005 \implies$ Ignore them (Placeholders).
Trailing Zeros with Dot: $5.00 \implies$ Count them (Precision).

Frequently Asked Conceptual Questions (FAQs)

1. What is the difference between decimal places (d.p.) and significant figures (s.f.)?

Decimal places count the number of digits specifically located to the right of the decimal point. Significant figures count all meaningful digits starting from the first non-zero digit regardless of where the decimal point is. For example, $0.0052$ has 4 decimal places, but only 2 significant figures.

2. Why are trailing zeros in $5.00$ significant but trailing zeros in $500$ not significant?

In $5.00$, the zeros indicate that measurement was carefully conducted to the hundredths place. In $500$ without a decimal point, the zeros are merely holding place value and could represent a rounded figure between $450$ and $550$.

3. Why does error increase when calculating Area compared to Perimeter?

Perimeter is calculated through addition $2(l+w)$, where errors simply add linearly. Area is calculated through multiplication $(l \times w)$, where errors multiply with dimensions, creating a much larger (propagated) discrepancy.

More Chapter Notes for Class 8 (FBISE)

Mathematics

Mathematics • Chapter 1 FBISE

Class 8 Mathematics - Ch 1: Real Numbers, Decimal Classification, Properties, Inequalities & Absolute Values (FBISE)

Real Numbers

Mathematics • Chapter 3 FBISE

Class 8 Mathematics - Ch 3: Square & Square Roots, Cubes & Cube Roots Mastery Guide (FBISE)

Square and Square Roots, Cubes and Cube Roots

Mathematics • Chapter 4 FBISE

Class 8 Mathematics - Ch 4: Financial Arithmetic Mastery Guide (FBISE)

Financial Arithmetic

Mathematics • Chapter 5 FBISE

Class 8 Mathematics - Ch 5: Sets Mastery Guide (FBISE)

Sets

Mathematics • Chapter 6 FBISE

Class 8 Mathematics - Ch 6: Algebra & Sequence Mastery Guide (FBISE)

Algebra and Sequence

Mathematics • Chapter 7 FBISE

Class 8 Mathematics - Ch 7: Linear Equations & Inequalities Mastery Guide (FBISE)

Linear Equations and Inequalities

Mathematics • Chapter 8 FBISE

Class 8 Mathematics - Ch 8: Geometric Transformations Mastery Guide (FBISE)

Transformations

Mathematics • Chapter 9 FBISE

Class 8 Mathematics Ch 9: Geometry Mastery Guide (FBISE)

Geometry

Mathematics • Chapter 10 FBISE

Class 8 Mathematics Ch 10: Practical Geometry Mastery Guide (FBISE)

Practical Geometry

Mathematics • Chapter 11 FBISE

Chapter 11: Mensuration (پیمائش)

Mensuration

Mathematics • Chapter 12 FBISE

Chapter 12: Data Handling — Comprehensive Solved Notes & Solution Manual

Data Handling

Mathematics • Chapter 13 FBISE

Chapter 13: Probability — Comprehensive Solved Notes & Solution Manual

Probability

Self-Assessment Practice

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⏱️ Class 8 Model Examination Federal Board of Intermediate and Secondary Education (FBISE)

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Class 8 Mathematics - Ch 2: Estimation and Approximation Chapter Mock Test

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