From Concepts to Mastery

Ontario Grade 7–8 Scientific Notation — Teacher Read-Aloud Master Course

A complete, highly detailed, read-aloud teaching course with static diagrams, interactive decimal-jump animations, dynamic wording, 120 graduated practice questions, detailed solutions, and a 30-question mastery test.

Ontario Grades 7–8English only120 practice questions30-question final testOffline interactive HTMLColour-coded exponent counting

Version note: Solutions are written for explicit classroom narration. Use the purple read-aloud buttons for browser speech.

Course Map

1. Ontario Grade 7–8 Learning Progression

Grade 7 bridge: represent and compare whole numbers up to one billion, including expanded form using powers of ten. Grade 8 mastery: represent and compare very large and very small numbers using scientific notation and explain real-world uses.

Enrichment: multiplication, division, addition, subtraction, multi-step operations, significant digits, error analysis, bounds, and modelling extend beyond the minimum expectation and prepare students for Grade 9 mathematics and science.

APlace value
BPowers of ten
CDecimal jumps
DScientific notation
EOperations
FModelling
Read aloud: Scientific notation is a place-value language. It does not change a number. It separates the number into two jobs: the coefficient tells us the precise digits, and the power of ten tells us the size or scale. Every time the decimal point crosses one place, the power of ten changes by one.
Core Meaning

2. What Scientific Notation Means

A non-zero number in normalized scientific notation has the form

a × 10n

Coefficient rule: 1 ≤ |a| < 10.   Exponent rule: n is an integer.

The coefficient, a

It carries the meaningful digits. In 6.34 × 107, the coefficient is 6.34.

The exponent, n

It records the scale. It tells how many powers-of-ten jumps connect the coefficient to the original number.

Teacher Read-Aloud Script

The exponent is not guessed from the number of visible zeros. It is counted from decimal-point movement. First create a valid coefficient. Then count every one-place movement. Finally decide the sign from the direction and the size of the original number.

Place-Value Foundation

3. Powers of Ten: The Complete Ladder

ExponentPowerDecimal valueMeaning
61061,000,000one million
5105100,000one hundred thousand
410410,000ten thousand
31031,000one thousand
2102100one hundred
110110ten
01001the centre of the ladder
−110−10.1one tenth
−210−20.01one hundredth
−310−30.001one thousandth
−410−40.0001one ten-thousandth
−510−50.00001one hundred-thousandth
−610−60.000001one millionth
Move one step up: multiply by 10.
Move one step down: divide by 10.
Zero power: 100 = 1.
Negative power: 10−n = 1 ÷ 10n.
10⁻³10⁻²10⁻¹10⁰10¹10²10³ 0.0010.010.11101001,000 Exponent increases; value × 10 each step Exponent decreases; value ÷ 10 each step
Most Important Visual Skill

4. Count Decimal Jumps Without Confusion

Large number → coefficient

Move decimal LEFT

Exponent is POSITIVE

+n

Small number → coefficient

Move decimal RIGHT

Exponent is NEGATIVE

−n
Do not count zeros. Count the one-place gaps crossed by the decimal point. A zero can be crossed, but the count is about movement, not about how many zeros happen to appear.

Dynamic Decimal-Jump Counter

Each jump receives its own colour and number.

Original number
Target: one non-zero digit before the decimal
Dynamic wording

1Make a legal coefficient

Exactly one non-zero digit must be to the left of the decimal point.

2Count coloured jumps

One gap crossed equals one count.

3Choose the sign

Large original: positive. Small original: negative.

4Reverse-check

Use the exponent to move back and reconstruct the original number.

Dynamic Graph

5. Explore 10n on an Exponent Number Line

Dynamic wording

Conversion Methods

6. Standard Form ↔ Scientific Notation

Standard → Scientific

  1. Find the first non-zero digit.
  2. Place the new decimal immediately after it.
  3. Count one-place decimal jumps.
  4. Use +n for a large original number and −n for a small original number.
  5. Check 1 ≤ |a| < 10.
Example: 508,000 = 5.08 × 105
Example: 0.000508 = 5.08 × 10−4

Scientific → Standard

  1. Read the sign of the exponent.
  2. Positive exponent: move the coefficient decimal right.
  3. Negative exponent: move it left.
  4. Move exactly |n| places.
  5. Insert placeholder zeros where needed.
Example: 3.17 × 104 = 31,700
Example: 3.17 × 10−4 = 0.000317

Why the signs can feel “backwards”

When converting a large standard number to scientific notation, you move the decimal left to make the coefficient smaller, but the exponent is positive because the power of ten must enlarge the coefficient back to the original number.

When converting a small decimal, you move the decimal right to make the coefficient larger, but the exponent is negative because the power of ten must shrink the coefficient back to the original small number.

Comparison and Ordering

7. Compare by Exponent First, Then Coefficient

Are both numbers positive and normalized?
↓
Different exponents?
Yes → greater exponent means greater numberNo → compare coefficients
Example: 2.1 × 107 > 9.9 × 106 because exponent 7 > 6.
Example: 4.81 × 10−5 > 4.79 × 10−5 because the exponents match and 4.81 > 4.79.
Negative exponent warning: −4 is greater than −5. Therefore 10−4 is ten times greater than 10−5.
Operations

8. Add, Subtract, Multiply, and Divide

Multiplication

Multiply coefficients. Add exponents.

(a × 10m)(b × 10n) = ab × 10m+n

Division

Divide coefficients. Subtract exponents.

(a × 10m) ÷ (b × 10n) = (a ÷ b) × 10m−n

Addition

First make exponents equal. Then add coefficients.

3.2 × 105 + 4.1 × 104 = 3.61 × 105

Subtraction

First make exponents equal. Then subtract coefficients.

7.0 × 10−3 − 8 × 10−4 = 6.2 × 10−3
Most common operation error: exponents are not added during ordinary addition. The “add exponents” rule belongs to multiplication of powers with the same base.

Normalize after every operation

18 × 104 is equivalent to 1.8 × 105. Moving the coefficient decimal left one place requires increasing the exponent by one.

0.42 × 10−3 is equivalent to 4.2 × 10−4. Moving the coefficient decimal right one place requires decreasing the exponent by one.

Precision and Technology

9. Significant Digits, Estimation, and Calculator E-Notation

Significant digits

Begin counting at the first non-zero digit. Leading zeros only locate the decimal point; they are not significant.

Order of magnitude

The exponent gives the basic scale, but a coefficient near 10 may make the nearest power of ten one level higher.

E-notation

6.2E−5 means 6.2 × 10−5. The E does not mean “times ten”; it means “times ten raised to.”

Rounding example: 9.996 × 104 to three significant digits is 1.00 × 105, not 10.0 × 104 as a final normalized answer.
Error Clinic

10. Common Misconceptions and Repairs

Error: Counting only zeros.
Repair: Mark the original and target decimal points; count gaps crossed.
Error: Writing 0.63 × 105 as final form.
Repair: Coefficient must be at least 1 and less than 10.
Error: Positive exponent for a small decimal.
Repair: Reverse-check: positive powers enlarge the coefficient.
Error: Comparing coefficients before exponents.
Repair: Compare scale first.
Error: Adding exponents during addition.
Repair: Match exponents, then add coefficients.
Error: Forgetting parentheses in m − (−n).
Repair: Write the subtraction line before simplifying.
30 Questions

Easy Practice

30 questions. Question numbers are identical in all three versions.

Question 1Grade 7 bridgeConvert standard form to scientific notation1 mark
Write 7,200 in scientific notation.
Detailed solution and read-aloud reasoning
  1. Locate the original decimal point. In a whole number it is understood to be after the final digit; in a small decimal it is already visible.
  2. Move the decimal point left until exactly one non-zero digit remains to its left. The coefficient becomes 7.2.
  3. Count every one-place jump: 3 jumps.
  4. Because the decimal moved left, the exponent is positive: 3.
  5. Write the result: 7.2 × 103.
  6. Check by moving the decimal in 7.2 3 places in the opposite direction. This reconstructs 7,200.
Answer: 7.2 × 103
Question 2Grade 7 bridgeConvert standard form to scientific notation1 mark
Write 45,000 in scientific notation.
Detailed solution and read-aloud reasoning
  1. Locate the original decimal point. In a whole number it is understood to be after the final digit; in a small decimal it is already visible.
  2. Move the decimal point left until exactly one non-zero digit remains to its left. The coefficient becomes 4.5.
  3. Count every one-place jump: 4 jumps.
  4. Because the decimal moved left, the exponent is positive: 4.
  5. Write the result: 4.5 × 104.
  6. Check by moving the decimal in 4.5 4 places in the opposite direction. This reconstructs 45,000.
Answer: 4.5 × 104
Question 3Grade 7 bridgeConvert standard form to scientific notation1 mark
Write 630,000 in scientific notation.
Detailed solution and read-aloud reasoning
  1. Locate the original decimal point. In a whole number it is understood to be after the final digit; in a small decimal it is already visible.
  2. Move the decimal point left until exactly one non-zero digit remains to its left. The coefficient becomes 6.3.
  3. Count every one-place jump: 5 jumps.
  4. Because the decimal moved left, the exponent is positive: 5.
  5. Write the result: 6.3 × 105.
  6. Check by moving the decimal in 6.3 5 places in the opposite direction. This reconstructs 630,000.
Answer: 6.3 × 105
Question 4Grade 7 bridgeConvert standard form to scientific notation1 mark
Write 9,810,000 in scientific notation.
Detailed solution and read-aloud reasoning
  1. Locate the original decimal point. In a whole number it is understood to be after the final digit; in a small decimal it is already visible.
  2. Move the decimal point left until exactly one non-zero digit remains to its left. The coefficient becomes 9.81.
  3. Count every one-place jump: 6 jumps.
  4. Because the decimal moved left, the exponent is positive: 6.
  5. Write the result: 9.81 × 106.
  6. Check by moving the decimal in 9.81 6 places in the opposite direction. This reconstructs 9,810,000.
Answer: 9.81 × 106
Question 5Grade 7 bridgeConvert standard form to scientific notation1 mark
Write 72,500,000 in scientific notation.
Detailed solution and read-aloud reasoning
  1. Locate the original decimal point. In a whole number it is understood to be after the final digit; in a small decimal it is already visible.
  2. Move the decimal point left until exactly one non-zero digit remains to its left. The coefficient becomes 7.25.
  3. Count every one-place jump: 7 jumps.
  4. Because the decimal moved left, the exponent is positive: 7.
  5. Write the result: 7.25 × 107.
  6. Check by moving the decimal in 7.25 7 places in the opposite direction. This reconstructs 72,500,000.
Answer: 7.25 × 107
Question 6Grade 7 bridgeConvert standard form to scientific notation1 mark
Write 408,000,000 in scientific notation.
Detailed solution and read-aloud reasoning
  1. Locate the original decimal point. In a whole number it is understood to be after the final digit; in a small decimal it is already visible.
  2. Move the decimal point left until exactly one non-zero digit remains to its left. The coefficient becomes 4.08.
  3. Count every one-place jump: 8 jumps.
  4. Because the decimal moved left, the exponent is positive: 8.
  5. Write the result: 4.08 × 108.
  6. Check by moving the decimal in 4.08 8 places in the opposite direction. This reconstructs 408,000,000.
Answer: 4.08 × 108
Question 7Grade 7 bridgeConvert standard form to scientific notation1 mark
Write 999,000,000 in scientific notation.
Detailed solution and read-aloud reasoning
  1. Locate the original decimal point. In a whole number it is understood to be after the final digit; in a small decimal it is already visible.
  2. Move the decimal point left until exactly one non-zero digit remains to its left. The coefficient becomes 9.99.
  3. Count every one-place jump: 8 jumps.
  4. Because the decimal moved left, the exponent is positive: 8.
  5. Write the result: 9.99 × 108.
  6. Check by moving the decimal in 9.99 8 places in the opposite direction. This reconstructs 999,000,000.
Answer: 9.99 × 108
Question 8Grade 7 bridgeConvert standard form to scientific notation1 mark
Write 1,000,000,000 in scientific notation.
Detailed solution and read-aloud reasoning
  1. Locate the original decimal point. In a whole number it is understood to be after the final digit; in a small decimal it is already visible.
  2. Move the decimal point left until exactly one non-zero digit remains to its left. The coefficient becomes 1.
  3. Count every one-place jump: 9 jumps.
  4. Because the decimal moved left, the exponent is positive: 9.
  5. Write the result: 1 × 109.
  6. Check by moving the decimal in 1 9 places in the opposite direction. This reconstructs 1,000,000,000.
Answer: 1 × 109
Question 9Grade 8Convert standard form to scientific notation1 mark
Write 0.6 in scientific notation.
Detailed solution and read-aloud reasoning
  1. Locate the original decimal point. In a whole number it is understood to be after the final digit; in a small decimal it is already visible.
  2. Move the decimal point right until exactly one non-zero digit remains to its left. The coefficient becomes 6.
  3. Count every one-place jump: 1 jump.
  4. Because the decimal moved right, the exponent is negative: -1.
  5. Write the result: 6 × 10-1.
  6. Check by moving the decimal in 6 1 place in the opposite direction. This reconstructs 0.6.
Answer: 6 × 10-1
Question 10Grade 8Convert standard form to scientific notation1 mark
Write 0.08 in scientific notation.
Detailed solution and read-aloud reasoning
  1. Locate the original decimal point. In a whole number it is understood to be after the final digit; in a small decimal it is already visible.
  2. Move the decimal point right until exactly one non-zero digit remains to its left. The coefficient becomes 8.
  3. Count every one-place jump: 2 jumps.
  4. Because the decimal moved right, the exponent is negative: -2.
  5. Write the result: 8 × 10-2.
  6. Check by moving the decimal in 8 2 places in the opposite direction. This reconstructs 0.08.
Answer: 8 × 10-2
Question 11Grade 8Convert standard form to scientific notation1 mark
Write 0.0042 in scientific notation.
Detailed solution and read-aloud reasoning
  1. Locate the original decimal point. In a whole number it is understood to be after the final digit; in a small decimal it is already visible.
  2. Move the decimal point right until exactly one non-zero digit remains to its left. The coefficient becomes 4.2.
  3. Count every one-place jump: 3 jumps.
  4. Because the decimal moved right, the exponent is negative: -3.
  5. Write the result: 4.2 × 10-3.
  6. Check by moving the decimal in 4.2 3 places in the opposite direction. This reconstructs 0.0042.
Answer: 4.2 × 10-3
Question 12Grade 8Convert standard form to scientific notation1 mark
Write 0.00091 in scientific notation.
Detailed solution and read-aloud reasoning
  1. Locate the original decimal point. In a whole number it is understood to be after the final digit; in a small decimal it is already visible.
  2. Move the decimal point right until exactly one non-zero digit remains to its left. The coefficient becomes 9.1.
  3. Count every one-place jump: 4 jumps.
  4. Because the decimal moved right, the exponent is negative: -4.
  5. Write the result: 9.1 × 10-4.
  6. Check by moving the decimal in 9.1 4 places in the opposite direction. This reconstructs 0.00091.
Answer: 9.1 × 10-4
Question 13Grade 8Convert standard form to scientific notation1 mark
Write 0.000073 in scientific notation.
Detailed solution and read-aloud reasoning
  1. Locate the original decimal point. In a whole number it is understood to be after the final digit; in a small decimal it is already visible.
  2. Move the decimal point right until exactly one non-zero digit remains to its left. The coefficient becomes 7.3.
  3. Count every one-place jump: 5 jumps.
  4. Because the decimal moved right, the exponent is negative: -5.
  5. Write the result: 7.3 × 10-5.
  6. Check by moving the decimal in 7.3 5 places in the opposite direction. This reconstructs 0.000073.
Answer: 7.3 × 10-5
Question 14Grade 8Convert standard form to scientific notation1 mark
Write 0.00000508 in scientific notation.
Detailed solution and read-aloud reasoning
  1. Locate the original decimal point. In a whole number it is understood to be after the final digit; in a small decimal it is already visible.
  2. Move the decimal point right until exactly one non-zero digit remains to its left. The coefficient becomes 5.08.
  3. Count every one-place jump: 6 jumps.
  4. Because the decimal moved right, the exponent is negative: -6.
  5. Write the result: 5.08 × 10-6.
  6. Check by moving the decimal in 5.08 6 places in the opposite direction. This reconstructs 0.00000508.
Answer: 5.08 × 10-6
Question 15Grade 8Convert standard form to scientific notation1 mark
Write 0.00000064 in scientific notation.
Detailed solution and read-aloud reasoning
  1. Locate the original decimal point. In a whole number it is understood to be after the final digit; in a small decimal it is already visible.
  2. Move the decimal point right until exactly one non-zero digit remains to its left. The coefficient becomes 6.4.
  3. Count every one-place jump: 7 jumps.
  4. Because the decimal moved right, the exponent is negative: -7.
  5. Write the result: 6.4 × 10-7.
  6. Check by moving the decimal in 6.4 7 places in the opposite direction. This reconstructs 0.00000064.
Answer: 6.4 × 10-7
Question 16Grade 8Convert standard form to scientific notation1 mark
Write 0.000000009 in scientific notation.
Detailed solution and read-aloud reasoning
  1. Locate the original decimal point. In a whole number it is understood to be after the final digit; in a small decimal it is already visible.
  2. Move the decimal point right until exactly one non-zero digit remains to its left. The coefficient becomes 9.
  3. Count every one-place jump: 9 jumps.
  4. Because the decimal moved right, the exponent is negative: -9.
  5. Write the result: 9 × 10-9.
  6. Check by moving the decimal in 9 9 places in the opposite direction. This reconstructs 0.000000009.
Answer: 9 × 10-9
Question 17Grade 7 bridgeConvert scientific notation to standard form1 mark
Write 3.5 × 104 in standard notation.
Detailed solution and read-aloud reasoning
  1. The coefficient is 3.5 and the exponent is 4.
  2. A positive exponent tells us to move the decimal point right.
  3. Move it exactly 4 places, inserting zeros where needed.
  4. The standard form is 35,000.
  5. Check: converting 35,000 back to scientific notation gives 3.5 × 104.
Answer: 35,000
Question 18Grade 7 bridgeConvert scientific notation to standard form1 mark
Write 8.02 × 106 in standard notation.
Detailed solution and read-aloud reasoning
  1. The coefficient is 8.02 and the exponent is 6.
  2. A positive exponent tells us to move the decimal point right.
  3. Move it exactly 6 places, inserting zeros where needed.
  4. The standard form is 8,020,000.
  5. Check: converting 8,020,000 back to scientific notation gives 8.02 × 106.
Answer: 8,020,000
Question 19Grade 7 bridgeConvert scientific notation to standard form1 mark
Write 1.009 × 108 in standard notation.
Detailed solution and read-aloud reasoning
  1. The coefficient is 1.009 and the exponent is 8.
  2. A positive exponent tells us to move the decimal point right.
  3. Move it exactly 8 places, inserting zeros where needed.
  4. The standard form is 100,900,000.
  5. Check: converting 100,900,000 back to scientific notation gives 1.009 × 108.
Answer: 100,900,000
Question 20Grade 7 bridgeConvert scientific notation to standard form1 mark
Write 9.9 × 109 in standard notation.
Detailed solution and read-aloud reasoning
  1. The coefficient is 9.9 and the exponent is 9.
  2. A positive exponent tells us to move the decimal point right.
  3. Move it exactly 9 places, inserting zeros where needed.
  4. The standard form is 9,900,000,000.
  5. Check: converting 9,900,000,000 back to scientific notation gives 9.9 × 109.
Answer: 9,900,000,000
Question 21Grade 8Convert scientific notation to standard form1 mark
Write 4.6 × 10-2 in standard notation.
Detailed solution and read-aloud reasoning
  1. The coefficient is 4.6 and the exponent is -2.
  2. A negative exponent tells us to move the decimal point left.
  3. Move it exactly 2 places, inserting zeros where needed.
  4. The standard form is 0.046.
  5. Check: converting 0.046 back to scientific notation gives 4.6 × 10-2.
Answer: 0.046
Question 22Grade 8Convert scientific notation to standard form1 mark
Write 7.05 × 10-4 in standard notation.
Detailed solution and read-aloud reasoning
  1. The coefficient is 7.05 and the exponent is -4.
  2. A negative exponent tells us to move the decimal point left.
  3. Move it exactly 4 places, inserting zeros where needed.
  4. The standard form is 0.000705.
  5. Check: converting 0.000705 back to scientific notation gives 7.05 × 10-4.
Answer: 0.000705
Question 23Grade 8Convert scientific notation to standard form1 mark
Write 2.3 × 10-6 in standard notation.
Detailed solution and read-aloud reasoning
  1. The coefficient is 2.3 and the exponent is -6.
  2. A negative exponent tells us to move the decimal point left.
  3. Move it exactly 6 places, inserting zeros where needed.
  4. The standard form is 0.0000023.
  5. Check: converting 0.0000023 back to scientific notation gives 2.3 × 10-6.
Answer: 0.0000023
Question 24Grade 8Convert scientific notation to standard form1 mark
Write 9.01 × 10-8 in standard notation.
Detailed solution and read-aloud reasoning
  1. The coefficient is 9.01 and the exponent is -8.
  2. A negative exponent tells us to move the decimal point left.
  3. Move it exactly 8 places, inserting zeros where needed.
  4. The standard form is 0.0000000901.
  5. Check: converting 0.0000000901 back to scientific notation gives 9.01 × 10-8.
Answer: 0.0000000901
Question 25Grade 8Recognize normalized scientific notation1 mark
Which expression is written in correct scientific notation: 0.45 × 106, 4.5 × 105, or 45 × 104?
Detailed solution and read-aloud reasoning

A scientific-notation coefficient must satisfy 1 ≤ |a| < 10. Only 4.5 meets this condition. Therefore the correct expression is 4.5 × 105.

Answer: 4.5 × 105
Question 26Grade 7 bridgeUnderstand zero exponent1 mark
Complete the statement: 100 = ___.
Detailed solution and read-aloud reasoning

Any non-zero base raised to the zero power equals 1. In the powers-of-ten pattern, 1000, 100, 10, 1 shows that dividing by 10 once more gives 100 = 1.

Answer: 1
Question 27Grade 8Understand negative powers of ten1 mark
Complete the statement: 10−3 = ___.
Detailed solution and read-aloud reasoning

A negative exponent means reciprocal: 10−3 = 1 ÷ 103 = 1 ÷ 1000 = 0.001.

Answer: 0.001
Question 28Grade 7 bridgeCount decimal-place jumps1 mark
How many one-place decimal jumps change 5.7 into 57,000?
Detailed solution and read-aloud reasoning

5.7 → 57 → 570 → 5,700 → 57,000. The decimal point moves four places to the right. This is multiplication by 104.

Answer: 4 jumps to the right
Question 29Grade 8Determine exponent sign1 mark
State the sign of the exponent when 0.00038 is written in scientific notation.
Detailed solution and read-aloud reasoning

The number is between 0 and 1. To form 3.8, the decimal point moves right four places. The original small number is recovered by multiplying by 10−4, so the exponent is negative.

Answer: Negative
Question 30Grade 7 bridgeDetermine exponent sign1 mark
State the sign of the exponent when 62,000,000 is written in scientific notation.
Detailed solution and read-aloud reasoning

The number is greater than 10. To form 6.2, the decimal point moves left seven places. The original large number is recovered by multiplying by 107, so the exponent is positive.

Answer: Positive
30 Questions

Medium Practice

30 questions. Question numbers are identical in all three versions.

Question 31Grade 8Compare numbers in scientific notation2 marks
Insert <, >, or =: 4.2 × 107 ___ 3.9 × 107.
Detailed solution and read-aloud reasoning
  1. Confirm both coefficients are between 1 and 10 in absolute value.
  2. The exponents are equal, so compare the coefficients: 4.2 is greater than 3.9.
  3. Therefore, 4.2 × 107 > 3.9 × 107.
  4. Magnitude check: the standard values are 42,000,000 and 39,000,000.
Answer: 4.2 × 107 > 3.9 × 107
Question 32Grade 8Compare numbers in scientific notation2 marks
Insert <, >, or =: 6.1 × 105 ___ 9.8 × 104.
Detailed solution and read-aloud reasoning
  1. Confirm both coefficients are between 1 and 10 in absolute value.
  2. The exponents are different. Since 5 is greater than 4, the first number is greater.
  3. Therefore, 6.1 × 105 > 9.8 × 104.
  4. Magnitude check: the standard values are 610,000 and 98,000.
Answer: 6.1 × 105 > 9.8 × 104
Question 33Grade 8Compare numbers in scientific notation2 marks
Insert <, >, or =: 2.75 × 10-3 ___ 8.1 × 10-4.
Detailed solution and read-aloud reasoning
  1. Confirm both coefficients are between 1 and 10 in absolute value.
  2. The exponents are different. Since -3 is greater than -4, the first number is greater.
  3. Therefore, 2.75 × 10-3 > 8.1 × 10-4.
  4. Magnitude check: the standard values are 0.00275 and 0.00081.
Answer: 2.75 × 10-3 > 8.1 × 10-4
Question 34Grade 8Compare numbers in scientific notation2 marks
Insert <, >, or =: 9.01 × 10-6 ___ 1.2 × 10-5.
Detailed solution and read-aloud reasoning
  1. Confirm both coefficients are between 1 and 10 in absolute value.
  2. The exponents are different. Since -6 is less than -5, the first number is less.
  3. Therefore, 9.01 × 10-6 < 1.2 × 10-5.
  4. Magnitude check: the standard values are 0.00000901 and 0.000012.
Answer: 9.01 × 10-6 < 1.2 × 10-5
Question 35Grade 8Compare numbers in scientific notation2 marks
Insert <, >, or =: 5 × 108 ___ 5 × 108.
Detailed solution and read-aloud reasoning
  1. Confirm both coefficients are between 1 and 10 in absolute value.
  2. The exponents are equal, so compare the coefficients: 5.00 equals 5.
  3. Therefore, 5 × 108 = 5 × 108.
  4. Magnitude check: the standard values are 500,000,000 and 500,000,000.
Answer: 5 × 108 = 5 × 108
Question 36Grade 8Compare numbers in scientific notation2 marks
Insert <, >, or =: 1.09 × 10-2 ___ 1.9 × 10-2.
Detailed solution and read-aloud reasoning
  1. Confirm both coefficients are between 1 and 10 in absolute value.
  2. The exponents are equal, so compare the coefficients: 1.09 is less than 1.9.
  3. Therefore, 1.09 × 10-2 < 1.9 × 10-2.
  4. Magnitude check: the standard values are 0.0109 and 0.019.
Answer: 1.09 × 10-2 < 1.9 × 10-2
Question 37Grade 8Compare numbers in scientific notation2 marks
Insert <, >, or =: 7.4 × 103 ___ 0.74 × 104.
Detailed solution and read-aloud reasoning
  1. Confirm both coefficients are between 1 and 10 in absolute value.
  2. The exponents are different. Since 3 is less than 4, the first number is less.
  3. Therefore, 7.4 × 103 = 0.74 × 104.
  4. Magnitude check: the standard values are 7,400 and 7,400.
Answer: 7.4 × 103 = 0.74 × 104
Question 38Grade 8Compare numbers in scientific notation2 marks
Insert <, >, or =: 3.6 × 10-7 ___ 36 × 10-8.
Detailed solution and read-aloud reasoning
  1. Confirm both coefficients are between 1 and 10 in absolute value.
  2. The exponents are different. Since -7 is greater than -8, the first number is greater.
  3. Therefore, 3.6 × 10-7 = 36 × 10-8.
  4. Magnitude check: the standard values are 0.00000036 and 0.00000036.
Answer: 3.6 × 10-7 = 36 × 10-8
Question 39Grade 8Order numbers in scientific notation2 marks
Order from least to greatest: 3.2 × 105, 9.1 × 104, 1.7 × 106.
Detailed solution and read-aloud reasoning
  1. Compare exponents first. A greater exponent gives a greater positive number.
  2. If exponents match, compare coefficients.
  3. Order from least to greatest: 9.1 × 104 < 3.2 × 105 < 1.7 × 106.
  4. Optional check in standard form: 320,000, 91,000, 1,700,000.
Answer: 9.1 × 104 < 3.2 × 105 < 1.7 × 106
Question 40Grade 8Order numbers in scientific notation2 marks
Order from least to greatest: 4.5 × 10-3, 8.2 × 10-4, 2.1 × 10-2.
Detailed solution and read-aloud reasoning
  1. Compare exponents first. A greater exponent gives a greater positive number.
  2. If exponents match, compare coefficients.
  3. Order from least to greatest: 8.2 × 10-4 < 4.5 × 10-3 < 2.1 × 10-2.
  4. Optional check in standard form: 0.0045, 0.00082, 0.021.
Answer: 8.2 × 10-4 < 4.5 × 10-3 < 2.1 × 10-2
Question 41Grade 8Order numbers in scientific notation2 marks
Order from least to greatest: 6.01 × 107, 6.1 × 107, 5.99 × 107.
Detailed solution and read-aloud reasoning
  1. Compare exponents first. A greater exponent gives a greater positive number.
  2. If exponents match, compare coefficients.
  3. Order from least to greatest: 5.99 × 107 < 6.01 × 107 < 6.1 × 107.
  4. Optional check in standard form: 60,100,000, 61,000,000, 59,900,000.
Answer: 5.99 × 107 < 6.01 × 107 < 6.1 × 107
Question 42Grade 8Order numbers in scientific notation2 marks
Order from least to greatest: 9.9 × 10-6, 1.01 × 10-4, 7.5 × 10-5.
Detailed solution and read-aloud reasoning
  1. Compare exponents first. A greater exponent gives a greater positive number.
  2. If exponents match, compare coefficients.
  3. Order from least to greatest: 9.9 × 10-6 < 7.5 × 10-5 < 1.01 × 10-4.
  4. Optional check in standard form: 0.0000099, 0.000101, 0.000075.
Answer: 9.9 × 10-6 < 7.5 × 10-5 < 1.01 × 10-4
Question 43Grade 8Order numbers in scientific notation2 marks
Order from least to greatest: 2.4 × 103, 0.24 × 104, 24 × 102.
Detailed solution and read-aloud reasoning
  1. Compare exponents first. A greater exponent gives a greater positive number.
  2. If exponents match, compare coefficients.
  3. Order from least to greatest: 2.4 × 103 < 0.24 × 104 < 24 × 102.
  4. Optional check in standard form: 2,400, 2,400, 2,400.
Answer: 2.4 × 103 < 0.24 × 104 < 24 × 102
Question 44Grade 8Order numbers in scientific notation2 marks
Order from least to greatest: 7.07 × 10-8, 7.7 × 10-8, 6.99 × 10-8.
Detailed solution and read-aloud reasoning
  1. Compare exponents first. A greater exponent gives a greater positive number.
  2. If exponents match, compare coefficients.
  3. Order from least to greatest: 6.99 × 10-8 < 7.07 × 10-8 < 7.7 × 10-8.
  4. Optional check in standard form: 0.0000000707, 0.000000077, 0.0000000699.
Answer: 6.99 × 10-8 < 7.07 × 10-8 < 7.7 × 10-8
Question 45Grade 8 extensionMultiply or divide in scientific notation3 marks
Calculate and express the answer in scientific notation: (2 × 103) × (4 × 105).
Detailed solution and read-aloud reasoning
  1. Use the scientific-notation operation rule: multiply the coefficients and add the exponents.
  2. (2 × 4) × 103+5 = 8 × 108.
  3. Normalize if necessary so the coefficient is at least 1 and less than 10.
  4. Final answer: 8 × 108.
  5. Reasonableness check: estimate the coefficients and powers separately; the order of magnitude agrees with 108.
Answer: 8 × 108
Question 46Grade 8 extensionMultiply or divide in scientific notation3 marks
Calculate and express the answer in scientific notation: (3.5 × 104) × (2 × 102).
Detailed solution and read-aloud reasoning
  1. Use the scientific-notation operation rule: multiply the coefficients and add the exponents.
  2. (3.5 × 2) × 104+2 = 7 × 106.
  3. Normalize if necessary so the coefficient is at least 1 and less than 10.
  4. Final answer: 7 × 106.
  5. Reasonableness check: estimate the coefficients and powers separately; the order of magnitude agrees with 106.
Answer: 7 × 106
Question 47Grade 8 extensionMultiply or divide in scientific notation3 marks
Calculate and express the answer in scientific notation: (8 × 107) ÷ (4 × 103).
Detailed solution and read-aloud reasoning
  1. Use the scientific-notation operation rule: divide the coefficients and subtract the exponents.
  2. (8 ÷ 4) × 107-(3) = 2 × 104.
  3. Normalize if necessary so the coefficient is at least 1 and less than 10.
  4. Final answer: 2 × 104.
  5. Reasonableness check: estimate the coefficients and powers separately; the order of magnitude agrees with 104.
Answer: 2 × 104
Question 48Grade 8 extensionMultiply or divide in scientific notation3 marks
Calculate and express the answer in scientific notation: (6.3 × 10-2) × (3 × 10-4).
Detailed solution and read-aloud reasoning
  1. Use the scientific-notation operation rule: multiply the coefficients and add the exponents.
  2. (6.3 × 3) × 10-2+-4 = 18.9 × 10-6.
  3. Normalize if necessary so the coefficient is at least 1 and less than 10.
  4. Final answer: 1.89 × 10-5.
  5. Reasonableness check: estimate the coefficients and powers separately; the order of magnitude agrees with 10-5.
Answer: 1.89 × 10-5
Question 49Grade 8 extensionMultiply or divide in scientific notation3 marks
Calculate and express the answer in scientific notation: (9 × 10-5) ÷ (3 × 10-2).
Detailed solution and read-aloud reasoning
  1. Use the scientific-notation operation rule: divide the coefficients and subtract the exponents.
  2. (9 ÷ 3) × 10-5-(-2) = 3 × 10-3.
  3. Normalize if necessary so the coefficient is at least 1 and less than 10.
  4. Final answer: 3 × 10-3.
  5. Reasonableness check: estimate the coefficients and powers separately; the order of magnitude agrees with 10-3.
Answer: 3 × 10-3
Question 50Grade 8 extensionMultiply or divide in scientific notation3 marks
Calculate and express the answer in scientific notation: (7.2 × 106) × (8 × 102).
Detailed solution and read-aloud reasoning
  1. Use the scientific-notation operation rule: multiply the coefficients and add the exponents.
  2. (7.2 × 8) × 106+2 = 57.6 × 108.
  3. Normalize if necessary so the coefficient is at least 1 and less than 10.
  4. Final answer: 5.76 × 109.
  5. Reasonableness check: estimate the coefficients and powers separately; the order of magnitude agrees with 109.
Answer: 5.76 × 109
Question 51Grade 8 extensionAdd or subtract in scientific notation4 marks
Calculate and express the answer in scientific notation: 3.4 × 106 + 2.1 × 106.
Detailed solution and read-aloud reasoning
  1. Addition and subtraction require matching powers of ten. Choose 106.
  2. Rewrite: 3.4 × 106 = 3.4 × 106 and 2.1 × 106 = 2.1 × 106.
  3. Combine coefficients: (3.4 + 2.1) × 106 = 5.5 × 106.
  4. Normalize the coefficient if needed.
  5. Final answer: 5.5 × 106.
  6. Check in standard form: 3,400,000 + 2,100,000 = 5,500,000.
Answer: 5.5 × 106
Question 52Grade 8 extensionAdd or subtract in scientific notation4 marks
Calculate and express the answer in scientific notation: 8.9 × 104 - 3.7 × 104.
Detailed solution and read-aloud reasoning
  1. Addition and subtraction require matching powers of ten. Choose 104.
  2. Rewrite: 8.9 × 104 = 8.9 × 104 and 3.7 × 104 = 3.7 × 104.
  3. Combine coefficients: (8.9 - 3.7) × 104 = 5.2 × 104.
  4. Normalize the coefficient if needed.
  5. Final answer: 5.2 × 104.
  6. Check in standard form: 89,000 - 37,000 = 52,000.
Answer: 5.2 × 104
Question 53Grade 8 extensionAdd or subtract in scientific notation4 marks
Calculate and express the answer in scientific notation: 6.25 × 10-3 + 1.8 × 10-3.
Detailed solution and read-aloud reasoning
  1. Addition and subtraction require matching powers of ten. Choose 10-3.
  2. Rewrite: 6.25 × 10-3 = 6.25 × 10-3 and 1.8 × 10-3 = 1.8 × 10-3.
  3. Combine coefficients: (6.25 + 1.8) × 10-3 = 8.05 × 10-3.
  4. Normalize the coefficient if needed.
  5. Final answer: 8.05 × 10-3.
  6. Check in standard form: 0.00625 + 0.0018 = 0.00805.
Answer: 8.05 × 10-3
Question 54Grade 8 extensionAdd or subtract in scientific notation4 marks
Calculate and express the answer in scientific notation: 9.02 × 10-5 - 4.6 × 10-5.
Detailed solution and read-aloud reasoning
  1. Addition and subtraction require matching powers of ten. Choose 10-5.
  2. Rewrite: 9.02 × 10-5 = 9.02 × 10-5 and 4.6 × 10-5 = 4.6 × 10-5.
  3. Combine coefficients: (9.02 - 4.6) × 10-5 = 4.42 × 10-5.
  4. Normalize the coefficient if needed.
  5. Final answer: 4.42 × 10-5.
  6. Check in standard form: 0.0000902 - 0.000046 = 0.0000442.
Answer: 4.42 × 10-5
Question 55Grade 8Apply scientific notation in context2 marks
The distance travelled by a signal is 2,450,000 m. Write this distance in scientific notation.
Detailed solution and read-aloud reasoning
  1. Place the decimal after the first non-zero digit: 2.45.
  2. The decimal moved left 6 places, so the exponent is +6.
  3. 2,450,000 m = 2.45 × 106 m.
Answer: 2.45 × 106
Question 56Grade 8Apply scientific notation in context2 marks
A cell has a width of 0.0000148 m. Write the width in scientific notation.
Detailed solution and read-aloud reasoning
  1. Move the decimal right until the coefficient is 1.48.
  2. Count 5 places.
  3. The original number is less than 1, so use −5.
  4. 0.0000148 m = 1.48 × 10−5 m.
Answer: 1.48 × 10−5
Question 57Grade 8Apply scientific notation in context2 marks
Round 7.846 × 105 to two significant digits.
Detailed solution and read-aloud reasoning
  1. The first two significant digits are 7 and 8.
  2. The next digit is 4, so do not round 8 upward.
  3. 7.846 × 105 ≈ 7.8 × 105.
Answer: 7.8 × 105
Question 58Grade 8Apply scientific notation in context2 marks
Round 3.976 × 10−7 to three significant digits.
Detailed solution and read-aloud reasoning
  1. Keep 3, 9, and 7.
  2. The next digit is 6, so round 7 up to 8.
  3. 3.976 × 10−7 ≈ 3.98 × 10−7.
Answer: 3.98 × 10−7
Question 59Grade 8Apply scientific notation in context2 marks
A computer stores 6.4 × 108 bytes. Another stores 9.5 × 107 bytes. Which stores more?
Detailed solution and read-aloud reasoning
  1. Compare exponents: 8 is greater than 7.
  2. Therefore 6.4 × 108 is greater even though 6.4 is less than 9.5.
  3. The first computer stores more.
Answer: 6.4 × 108 bytes
Question 60Grade 8Apply scientific notation in context2 marks
A dust particle has a mass of 4.2 × 10−9 kg. Write this in standard notation.
Detailed solution and read-aloud reasoning
  1. The exponent −9 means move the decimal left 9 places.
  2. Insert eight zeros between the decimal point and 4.
  3. 4.2 × 10−9 kg = 0.0000000042 kg.
Answer: 0.0000000042 kg
30 Questions

Difficult Practice

30 questions. Question numbers are identical in all three versions.

Question 61Grade 8 extensionAdd or subtract in scientific notation4 marks
Calculate and express the answer in scientific notation: 4.6 × 107 + 3.2 × 106.
Detailed solution and read-aloud reasoning
  1. Addition and subtraction require matching powers of ten. Choose 107.
  2. Rewrite: 4.6 × 107 = 4.6 × 107 and 3.2 × 106 = 0.32 × 107.
  3. Combine coefficients: (4.6 + 0.32) × 107 = 4.92 × 107.
  4. Normalize the coefficient if needed.
  5. Final answer: 4.92 × 107.
  6. Check in standard form: 46,000,000 + 3,200,000 = 49,200,000.
Answer: 4.92 × 107
Question 62Grade 8 extensionAdd or subtract in scientific notation4 marks
Calculate and express the answer in scientific notation: 8.1 × 105 - 2.75 × 104.
Detailed solution and read-aloud reasoning
  1. Addition and subtraction require matching powers of ten. Choose 105.
  2. Rewrite: 8.1 × 105 = 8.1 × 105 and 2.75 × 104 = 0.275 × 105.
  3. Combine coefficients: (8.1 - 0.275) × 105 = 7.825 × 105.
  4. Normalize the coefficient if needed.
  5. Final answer: 7.825 × 105.
  6. Check in standard form: 810,000 - 27,500 = 782,500.
Answer: 7.825 × 105
Question 63Grade 8 extensionAdd or subtract in scientific notation4 marks
Calculate and express the answer in scientific notation: 6.03 × 10-3 + 9.5 × 10-4.
Detailed solution and read-aloud reasoning
  1. Addition and subtraction require matching powers of ten. Choose 10-3.
  2. Rewrite: 6.03 × 10-3 = 6.03 × 10-3 and 9.5 × 10-4 = 0.95 × 10-3.
  3. Combine coefficients: (6.03 + 0.95) × 10-3 = 6.98 × 10-3.
  4. Normalize the coefficient if needed.
  5. Final answer: 6.98 × 10-3.
  6. Check in standard form: 0.00603 + 0.00095 = 0.00698.
Answer: 6.98 × 10-3
Question 64Grade 8 extensionAdd or subtract in scientific notation4 marks
Calculate and express the answer in scientific notation: 7.2 × 10-5 - 3.8 × 10-6.
Detailed solution and read-aloud reasoning
  1. Addition and subtraction require matching powers of ten. Choose 10-5.
  2. Rewrite: 7.2 × 10-5 = 7.2 × 10-5 and 3.8 × 10-6 = 0.38 × 10-5.
  3. Combine coefficients: (7.2 - 0.38) × 10-5 = 6.82 × 10-5.
  4. Normalize the coefficient if needed.
  5. Final answer: 6.82 × 10-5.
  6. Check in standard form: 0.000072 - 0.0000038 = 0.0000682.
Answer: 6.82 × 10-5
Question 65Grade 8 extensionAdd or subtract in scientific notation4 marks
Calculate and express the answer in scientific notation: 1.25 × 109 + 8.9 × 108.
Detailed solution and read-aloud reasoning
  1. Addition and subtraction require matching powers of ten. Choose 109.
  2. Rewrite: 1.25 × 109 = 1.25 × 109 and 8.9 × 108 = 0.89 × 109.
  3. Combine coefficients: (1.25 + 0.89) × 109 = 2.14 × 109.
  4. Normalize the coefficient if needed.
  5. Final answer: 2.14 × 109.
  6. Check in standard form: 1,250,000,000 + 890,000,000 = 2,140,000,000.
Answer: 2.14 × 109
Question 66Grade 8 extensionAdd or subtract in scientific notation4 marks
Calculate and express the answer in scientific notation: 5.04 × 106 - 7.8 × 105.
Detailed solution and read-aloud reasoning
  1. Addition and subtraction require matching powers of ten. Choose 106.
  2. Rewrite: 5.04 × 106 = 5.04 × 106 and 7.8 × 105 = 0.78 × 106.
  3. Combine coefficients: (5.04 - 0.78) × 106 = 4.26 × 106.
  4. Normalize the coefficient if needed.
  5. Final answer: 4.26 × 106.
  6. Check in standard form: 5,040,000 - 780,000 = 4,260,000.
Answer: 4.26 × 106
Question 67Grade 8 extensionAdd or subtract in scientific notation4 marks
Calculate and express the answer in scientific notation: 9.99 × 10-7 + 1.2 × 10-8.
Detailed solution and read-aloud reasoning
  1. Addition and subtraction require matching powers of ten. Choose 10-7.
  2. Rewrite: 9.99 × 10-7 = 9.99 × 10-7 and 1.2 × 10-8 = 0.12 × 10-7.
  3. Combine coefficients: (9.99 + 0.12) × 10-7 = 10.11 × 10-7.
  4. Normalize the coefficient if needed.
  5. Final answer: 1.011 × 10-6.
  6. Check in standard form: 0.000000999 + 0.000000012 = 0.000001011.
Answer: 1.011 × 10-6
Question 68Grade 8 extensionAdd or subtract in scientific notation4 marks
Calculate and express the answer in scientific notation: 4.005 × 10-4 - 7.5 × 10-6.
Detailed solution and read-aloud reasoning
  1. Addition and subtraction require matching powers of ten. Choose 10-4.
  2. Rewrite: 4.005 × 10-4 = 4.005 × 10-4 and 7.5 × 10-6 = 0.075 × 10-4.
  3. Combine coefficients: (4.005 - 0.075) × 10-4 = 3.93 × 10-4.
  4. Normalize the coefficient if needed.
  5. Final answer: 3.93 × 10-4.
  6. Check in standard form: 0.0004005 - 0.0000075 = 0.000393.
Answer: 3.93 × 10-4
Question 69Grade 8 extensionMulti-step operations5 marks
Evaluate and write the result in scientific notation: ((2.4 × 105)(3.5 × 10−2)) ÷ (7 × 101).
Detailed solution and read-aloud reasoning
  1. Follow the order of operations: brackets, multiplication/division, then addition/subtraction.
  2. Multiply first: 2.4 × 3.5 = 8.4 and 105 × 10−2 = 103. Then divide by 7 × 101: (8.4 ÷ 7) × 103−1 = 1.2 × 102.
  3. Final answer: 1.2 × 102.
  4. Check in standard form: 120.
Answer: 1.2 × 102
Question 70Grade 8 extensionMulti-step operations5 marks
Evaluate and write the result in scientific notation: (9 × 10−4) ÷ (3 × 102) + 2.5 × 10−6.
Detailed solution and read-aloud reasoning
  1. Follow the order of operations: brackets, multiplication/division, then addition/subtraction.
  2. Divide first: 3 × 10−6. The exponents now match, so add coefficients: (3 + 2.5) × 10−6 = 5.5 × 10−6.
  3. Final answer: 5.5 × 10-6.
  4. Check in standard form: 0.0000055.
Answer: 5.5 × 10-6
Question 71Grade 8 extensionMulti-step operations5 marks
Evaluate and write the result in scientific notation: (6.4 × 107 − 8 × 106) ÷ 7.
Detailed solution and read-aloud reasoning
  1. Follow the order of operations: brackets, multiplication/division, then addition/subtraction.
  2. Rewrite 8 × 106 as 0.8 × 107. Subtract: 5.6 × 107. Divide by 7: 0.8 × 107 = 8 × 106.
  3. Final answer: 8 × 106.
  4. Check in standard form: 8,000,000.
Answer: 8 × 106
Question 72Grade 8 extensionMulti-step operations5 marks
Evaluate and write the result in scientific notation: (1.2 × 10−3 + 4.8 × 10−4) × 5 × 102.
Detailed solution and read-aloud reasoning
  1. Follow the order of operations: brackets, multiplication/division, then addition/subtraction.
  2. Rewrite 4.8 × 10−4 as 0.48 × 10−3. Add to get 1.68 × 10−3. Multiply: 8.4 × 10−1.
  3. Final answer: 8.4 × 10-1.
  4. Check in standard form: 0.84.
Answer: 8.4 × 10-1
Question 73Grade 8 extensionMulti-step operations5 marks
Evaluate and write the result in scientific notation: (7.5 × 108)(4 × 10−5) − 2 × 104.
Detailed solution and read-aloud reasoning
  1. Follow the order of operations: brackets, multiplication/division, then addition/subtraction.
  2. Multiply: 30 × 103 = 3 × 104. Subtract 2 × 104 to obtain 1 × 104.
  3. Final answer: 1 × 104.
  4. Check in standard form: 10,000.
Answer: 1 × 104
Question 74Grade 8 extensionMulti-step operations5 marks
Evaluate and write the result in scientific notation: (8.4 × 10−7) ÷ (2.1 × 10−3) × 5 × 102.
Detailed solution and read-aloud reasoning
  1. Follow the order of operations: brackets, multiplication/division, then addition/subtraction.
  2. Divide: 4 × 10−4. Multiply by 5 × 102: 20 × 10−2 = 2 × 10−1.
  3. Final answer: 2 × 10-1.
  4. Check in standard form: 0.2.
Answer: 2 × 10-1
Question 75Grade 8 extensionMulti-step operations5 marks
Evaluate and write the result in scientific notation: (3 × 106 + 4.5 × 105) ÷ (1.5 × 102).
Detailed solution and read-aloud reasoning
  1. Follow the order of operations: brackets, multiplication/division, then addition/subtraction.
  2. Rewrite 4.5 × 105 as 0.45 × 106. Add: 3.45 × 106. Divide: (3.45 ÷ 1.5) × 104 = 2.3 × 104.
  3. Final answer: 2.3 × 104.
  4. Check in standard form: 23,000.
Answer: 2.3 × 104
Question 76Grade 8 extensionMulti-step operations5 marks
Evaluate and write the result in scientific notation: (5.6 × 10−2 − 7 × 10−3) ÷ (7 × 10−2).
Detailed solution and read-aloud reasoning
  1. Follow the order of operations: brackets, multiplication/division, then addition/subtraction.
  2. Rewrite 7 × 10−3 as 0.7 × 10−2. Subtract: 4.9 × 10−2. Divide by 7 × 10−2: 0.7 = 7 × 10−1.
  3. Final answer: 7 × 10-1.
  4. Check in standard form: 0.7.
Answer: 7 × 10-1
Question 77Grade 8Error analysis and missing values3 marks
A student writes 0.000052 = 5.2 × 105. Identify and correct the error.
Detailed solution and read-aloud reasoning
  1. The coefficient 5.2 is correct.
  2. The original number is less than 1, so the exponent must be negative.
  3. The decimal moves right 5 places to form 5.2.
  4. Correct statement: 0.000052 = 5.2 × 10−5.
Answer: 5.2 × 10−5
Question 78Grade 8Error analysis and missing values3 marks
A student writes 46.2 × 104 as a final answer. Normalize it.
Detailed solution and read-aloud reasoning
  1. 46.2 is not a valid coefficient because it is at least 10.
  2. Move its decimal one place left: 46.2 → 4.62.
  3. Compensate by increasing the exponent by 1: 104 → 105.
  4. 46.2 × 104 = 4.62 × 105.
Answer: 4.62 × 105
Question 79Grade 8Error analysis and missing values3 marks
Find n: 7.3 × 10n = 0.000073.
Detailed solution and read-aloud reasoning
  1. Starting at 7.3, move the decimal left until reaching 0.000073.
  2. There are 5 one-place moves.
  3. Moving left from the coefficient means a negative exponent.
  4. n = −5.
Answer: n = −5
Question 80Grade 8Error analysis and missing values3 marks
Find n: 0.084 × 10n = 8.4 × 106.
Detailed solution and read-aloud reasoning
  1. Rewrite 0.084 as 8.4 × 10−2.
  2. Then 0.084 × 10n = 8.4 × 10n−2.
  3. Match exponents: n − 2 = 6.
  4. n = 8.
Answer: n = 8
Question 81Grade 8Error analysis and missing values3 marks
Find the missing coefficient: ___ × 10−4 = 0.000768.
Detailed solution and read-aloud reasoning
  1. Convert 0.000768 to scientific notation.
  2. Move the decimal right 4 places: 7.68.
  3. Therefore 7.68 × 10−4 = 0.000768.
Answer: 7.68
Question 82Grade 8Error analysis and missing values3 marks
Explain why 3.2 × 10−7 is greater than 9.8 × 10−8.
Detailed solution and read-aloud reasoning
  1. For positive numbers, the greater exponent gives the greater magnitude.
  2. −7 is greater than −8.
  3. 3.2 × 10−7 = 0.00000032, while 9.8 × 10−8 = 0.000000098.
  4. Therefore the first number is greater.
Answer: Because −7 > −8
Question 83Grade 8 extensionReal-world multi-step application4 marks
A satellite sends 3.6 × 106 bytes each minute for 2.5 × 102 minutes. How many bytes are sent?
Detailed solution and read-aloud reasoning
  1. Identify the operation from the context.
  2. Multiply rate by time: (3.6 × 2.5) × 106+2 = 9 × 108 bytes.
  3. Final answer: 9 × 108.
  4. Check the unit and order of magnitude.
Answer: 9 × 108
Question 84Grade 8 extensionReal-world multi-step application4 marks
A sample contains 8.4 × 10−3 g divided equally among 2.1 × 103 containers. What mass is in each container?
Detailed solution and read-aloud reasoning
  1. Identify the operation from the context.
  2. Divide total mass by number of containers: (8.4 ÷ 2.1) × 10−3−3 = 4 × 10−6 g.
  3. Final answer: 4 × 10-6.
  4. Check the unit and order of magnitude.
Answer: 4 × 10-6
Question 85Grade 8 extensionReal-world multi-step application4 marks
A road is 7.5 × 104 m long. A crew completes 2.8 × 103 m. How much remains?
Detailed solution and read-aloud reasoning
  1. Identify the operation from the context.
  2. Match exponents: 2.8 × 103 = 0.28 × 104. Subtract: 7.22 × 104 m.
  3. Final answer: 7.22 × 104.
  4. Check the unit and order of magnitude.
Answer: 7.22 × 104
Question 86Grade 8 extensionReal-world multi-step application4 marks
A bacterium has mass 6 × 10−13 kg. Find the mass of 4.5 × 107 bacteria.
Detailed solution and read-aloud reasoning
  1. Identify the operation from the context.
  2. Multiply: 27 × 10−6 = 2.7 × 10−5 kg.
  3. Final answer: 2.7 × 10-5.
  4. Check the unit and order of magnitude.
Answer: 2.7 × 10-5
Question 87Grade 8 extensionReal-world multi-step application4 marks
A file shrinks from 9.6 × 108 bytes to one eighth of its size. Find the new size.
Detailed solution and read-aloud reasoning
  1. Identify the operation from the context.
  2. Divide the coefficient by 8: 9.6 ÷ 8 = 1.2. The exponent remains 8, so the new size is 1.2 × 108 bytes.
  3. Final answer: 1.2 × 108.
  4. Check the unit and order of magnitude.
Answer: 1.2 × 108
Question 88Grade 8 extensionReal-world multi-step application4 marks
A microscope enlarges an object of length 3.2 × 10−5 m by a factor of 2.5 × 104. Find the displayed length.
Detailed solution and read-aloud reasoning
  1. Identify the operation from the context.
  2. Multiply: 8 × 10−1 m = 0.8 m.
  3. Final answer: 8 × 10-1.
  4. Check the unit and order of magnitude.
Answer: 8 × 10-1
Question 89Grade 8 extensionReal-world multi-step application4 marks
Two measurements are 4.09 × 10−6 m and 7.5 × 10−7 m. Find their total.
Detailed solution and read-aloud reasoning
  1. Identify the operation from the context.
  2. Rewrite 7.5 × 10−7 as 0.75 × 10−6. Add: 4.84 × 10−6 m.
  3. Final answer: 4.84 × 10-6.
  4. Check the unit and order of magnitude.
Answer: 4.84 × 10-6
Question 90Grade 8 extensionReal-world multi-step application4 marks
A machine makes 1.25 × 105 parts per day for 3.2 × 101 days, but rejects 4 × 104 parts. How many acceptable parts remain?
Detailed solution and read-aloud reasoning
  1. Identify the operation from the context.
  2. Produced: 4 × 106. Rejected: 0.04 × 106. Acceptable: 3.96 × 106 parts.
  3. Final answer: 3.96 × 106.
  4. Check the unit and order of magnitude.
Answer: 3.96 × 106
30 Questions

Extremely Difficult Practice

30 questions. Question numbers are identical in all three versions.

Question 91Grade 8 enrichmentAdvanced multi-step operations6 marks
Evaluate: [(3.6 × 108)(2.5 × 10−3) + 7.5 × 105] ÷ (1.5 × 102). Give a normalized scientific-notation answer.
Detailed solution and read-aloud reasoning
  1. Plan the order of operations before calculating.
  2. First multiply to get 9 × 105. Add 7.5 × 105 to get 16.5 × 105 = 1.65 × 106. Divide by 1.5 × 102: 1.1 × 104.
  3. Normalize every intermediate result when it helps prevent exponent errors.
  4. Final answer: 1.1 × 104.
  5. Independent decimal check: 11,000.
Answer: 1.1 × 104
Question 92Grade 8 enrichmentAdvanced multi-step operations6 marks
Evaluate: [(8.4 × 10−6) ÷ (2.1 × 10−2)] − 3.5 × 10−4. Give a normalized scientific-notation answer.
Detailed solution and read-aloud reasoning
  1. Plan the order of operations before calculating.
  2. Divide to get 4 × 10−4. Subtract 3.5 × 10−4: 0.5 × 10−4 = 5 × 10−5.
  3. Normalize every intermediate result when it helps prevent exponent errors.
  4. Final answer: 5 × 10-5.
  5. Independent decimal check: 0.00005.
Answer: 5 × 10-5
Question 93Grade 8 enrichmentAdvanced multi-step operations6 marks
Evaluate: (9.6 × 107 − 1.2 × 107) ÷ [(3.5 × 102)(4 × 101)]. Give a normalized scientific-notation answer.
Detailed solution and read-aloud reasoning
  1. Plan the order of operations before calculating.
  2. Numerator: 8.4 × 107. Denominator: 14 × 103 = 1.4 × 104. Divide: 6 × 103.
  3. Normalize every intermediate result when it helps prevent exponent errors.
  4. Final answer: 6 × 103.
  5. Independent decimal check: 6,000.
Answer: 6 × 103
Question 94Grade 8 enrichmentAdvanced multi-step operations6 marks
Evaluate: [(5 × 10−4)2] ÷ (2.5 × 10−9). Give a normalized scientific-notation answer.
Detailed solution and read-aloud reasoning
  1. Plan the order of operations before calculating.
  2. Square first: 25 × 10−8 = 2.5 × 10−7. Divide by 2.5 × 10−9: 1 × 102.
  3. Normalize every intermediate result when it helps prevent exponent errors.
  4. Final answer: 1 × 102.
  5. Independent decimal check: 100.
Answer: 1 × 102
Question 95Grade 8 enrichmentAdvanced multi-step operations6 marks
Evaluate: [(4.8 × 105) ÷ (1.2 × 10−3)] × (2.5 × 10−4). Give a normalized scientific-notation answer.
Detailed solution and read-aloud reasoning
  1. Plan the order of operations before calculating.
  2. Divide first: 4 × 108. Multiply: 10 × 104 = 1 × 105.
  3. Normalize every intermediate result when it helps prevent exponent errors.
  4. Final answer: 1 × 105.
  5. Independent decimal check: 100,000.
Answer: 1 × 105
Question 96Grade 8 enrichmentAdvanced multi-step operations6 marks
Evaluate: (6.03 × 10−7 + 9.7 × 10−8 − 4.5 × 10−9). Give a normalized scientific-notation answer.
Detailed solution and read-aloud reasoning
  1. Plan the order of operations before calculating.
  2. Use 10−7: 6.03 + 0.97 − 0.045 = 6.955. Result: 6.955 × 10−7.
  3. Normalize every intermediate result when it helps prevent exponent errors.
  4. Final answer: 6.955 × 10-7.
  5. Independent decimal check: 0.0000006955.
Answer: 6.955 × 10-7
Question 97Grade 8 enrichmentAdvanced multi-step operations6 marks
Evaluate: [(7.2 × 109) ÷ (9 × 10−2)] ÷ (4 × 105). Give a normalized scientific-notation answer.
Detailed solution and read-aloud reasoning
  1. Plan the order of operations before calculating.
  2. First division: 0.8 × 1011 = 8 × 1010. Second division: 2 × 105.
  3. Normalize every intermediate result when it helps prevent exponent errors.
  4. Final answer: 2 × 105.
  5. Independent decimal check: 200,000.
Answer: 2 × 105
Question 98Grade 8 enrichmentAdvanced multi-step operations6 marks
Evaluate: [(2.4 × 10−3)(5 × 106)] + [(1.8 × 104) ÷ (3 × 10−2)]. Give a normalized scientific-notation answer.
Detailed solution and read-aloud reasoning
  1. Plan the order of operations before calculating.
  2. First term: 12 × 103 = 1.2 × 104. Second term: 0.6 × 106 = 6 × 105. Add: 0.12 × 105 + 6 × 105 = 6.12 × 105.
  3. Normalize every intermediate result when it helps prevent exponent errors.
  4. Final answer: 6.12 × 105.
  5. Independent decimal check: 612,000.
Answer: 6.12 × 105
Question 99Grade 8 enrichmentAdvanced multi-step operations6 marks
Evaluate: (1.005 × 108 − 9.95 × 107) ÷ (2.5 × 102). Give a normalized scientific-notation answer.
Detailed solution and read-aloud reasoning
  1. Plan the order of operations before calculating.
  2. Rewrite 9.95 × 107 as 0.995 × 108. Difference: 0.010 × 108 = 1 × 106. Divide by 2.5 × 102: 4 × 103.
  3. Normalize every intermediate result when it helps prevent exponent errors.
  4. Final answer: 4 × 103.
  5. Independent decimal check: 4,000.
Answer: 4 × 103
Question 100Grade 8 enrichmentAdvanced multi-step operations6 marks
Evaluate: [(3 × 10−2)3] ÷ [(9 × 10−5)(1 × 10−2)]. Give a normalized scientific-notation answer.
Detailed solution and read-aloud reasoning
  1. Plan the order of operations before calculating.
  2. Numerator: 27 × 10−6 = 2.7 × 10−5. Denominator: 9 × 10−7. Divide: 0.3 × 102 = 3 × 101.
  3. Normalize every intermediate result when it helps prevent exponent errors.
  4. Final answer: 3 × 101.
  5. Independent decimal check: 30.
Answer: 3 × 101
Question 101Grade 8 enrichmentComplex modelling6 marks
A sensor records 2.4 × 105 readings per hour for 3.5 × 102 hours. It discards 7.5% of all readings. How many readings remain?
Detailed solution and read-aloud reasoning
  1. Translate each sentence into an operation and preserve units.
  2. Total = 8.4 × 107. Remaining fraction = 1 − 0.075 = 0.925. Remaining = 8.4 × 107 × 0.925 = 7.77 × 107.
  3. Final answer: 7.77 × 107.
  4. Check whether the size and unit are realistic for the context.
Answer: 7.77 × 107
Question 102Grade 8 enrichmentComplex modelling6 marks
A tank contains 6.3 × 104 L. It leaks 2.5 × 101 L/min for 1.8 × 103 min, then receives 7.5 × 103 L. Find the final volume.
Detailed solution and read-aloud reasoning
  1. Translate each sentence into an operation and preserve units.
  2. Leak loss = 4.5 × 104 L. After leaking: 1.8 × 104 L. Add 0.75 × 104 L to get 2.55 × 104 L.
  3. Final answer: 2.55 × 104.
  4. Check whether the size and unit are realistic for the context.
Answer: 2.55 × 104
Question 103Grade 8 enrichmentComplex modelling6 marks
A microorganism doubles every cycle. It begins with 3.2 × 103 organisms. How many are present after 12 doublings?
Detailed solution and read-aloud reasoning
  1. Translate each sentence into an operation and preserve units.
  2. Twelve doublings multiply by 212 = 4096. Then 3.2 × 103 × 4096 = 13,107.2 × 103 = 1.31072 × 107.
  3. Final answer: 1.31072 × 107.
  4. Check whether the size and unit are realistic for the context.
Answer: 1.31072 × 107
Question 104Grade 8 enrichmentComplex modelling6 marks
A spacecraft travels 7.2 × 106 km at 3.0 × 104 km/h, rests 18 h, then travels 1.8 × 106 km at the same speed. Find the total elapsed time.
Detailed solution and read-aloud reasoning
  1. Translate each sentence into an operation and preserve units.
  2. First travel time = 2.4 × 102 h. Second travel time = 6 × 101 h. Add 18 h: 240 + 60 + 18 = 318 h = 3.18 × 102 h.
  3. Final answer: 3.18 × 102.
  4. Check whether the size and unit are realistic for the context.
Answer: 3.18 × 102
Question 105Grade 8 enrichmentComplex modelling6 marks
A square chip has side length 4.5 × 10−3 m. Find its area in scientific notation.
Detailed solution and read-aloud reasoning
  1. Translate each sentence into an operation and preserve units.
  2. Area = side² = (4.5²) × 10−6 = 20.25 × 10−6 = 2.025 × 10−5 m².
  3. Final answer: 2.025 × 10-5.
  4. Check whether the size and unit are realistic for the context.
Answer: 2.025 × 10-5
Question 106Grade 8 enrichmentComplex modelling6 marks
A cube has edge length 2 × 10−4 m. Find its volume.
Detailed solution and read-aloud reasoning
  1. Translate each sentence into an operation and preserve units.
  2. Volume = edge³ = 2³ × 10−12 = 8 × 10−12 m³.
  3. Final answer: 8 × 10-12.
  4. Check whether the size and unit are realistic for the context.
Answer: 8 × 10-12
Question 107Grade 8 enrichmentComplex modelling6 marks
A data centre has 4.8 × 103 servers. Each processes 6.25 × 107 operations per second. Efficiency is 96%. Find useful operations per second.
Detailed solution and read-aloud reasoning
  1. Translate each sentence into an operation and preserve units.
  2. Raw rate = 30 × 1010 = 3 × 1011. Useful rate = 0.96 × 3 × 1011 = 2.88 × 1011.
  3. Final answer: 2.88 × 1011.
  4. Check whether the size and unit are realistic for the context.
Answer: 2.88 × 1011
Question 108Grade 8 enrichmentComplex modelling6 marks
A medicine dose is 3.5 × 10−6 g/kg. Find the total dose for 240 patients averaging 62 kg each.
Detailed solution and read-aloud reasoning
  1. Translate each sentence into an operation and preserve units.
  2. Total mass represented = 240 × 62 = 14,880 kg. Dose = 3.5 × 10−6 × 14,880 = 52,080 × 10−6 g = 5.208 × 10−2 g.
  3. Final answer: 5.208 × 10-2.
  4. Check whether the size and unit are realistic for the context.
Answer: 5.208 × 10-2
Question 109Grade 8 enrichmentEstimation, bounds, and precision5 marks
Without exact multiplication, estimate the order of magnitude of (8.9 × 107)(2.2 × 10−3).
Detailed solution and read-aloud reasoning
  1. Estimate 8.9 ≈ 9 and 2.2 ≈ 2.
  2. Coefficient product ≈ 18, which is about 1.8 × 10.
  3. Powers give 104; the extra factor 10 raises the order to 105.
  4. The exact product is 1.958 × 105, confirming the estimate.
Answer: 105
Question 110Grade 8 enrichmentEstimation, bounds, and precision5 marks
A value is rounded to 4.6 × 105 to two significant digits. Give the lower and upper bounds.
Detailed solution and read-aloud reasoning
  1. The rounding unit is 0.1 × 105 = 10,000.
  2. Half a rounding unit is 5,000.
  3. Lower bound: 460,000 − 5,000 = 455,000.
  4. Upper bound: 460,000 + 5,000 = 465,000, not included.
  5. 455,000 ≤ value < 465,000.
Answer: 455,000 ≤ value < 465,000
Question 111Grade 8 enrichmentEstimation, bounds, and precision5 marks
A value is rounded to 7.20 × 10−4 to three significant digits. Give its bounds.
Detailed solution and read-aloud reasoning
  1. The last kept place is 0.01 × 10−4.
  2. Half that unit is 0.005 × 10−4.
  3. Subtract and add: 7.195 × 10−4 and 7.205 × 10−4.
  4. The lower bound is included; the upper bound is excluded.
Answer: 7.195 × 10−4 ≤ value < 7.205 × 10−4
Question 112Grade 8 enrichmentEstimation, bounds, and precision5 marks
Which is the better estimate for 6.02 × 108 + 9.7 × 107: 7 × 108 or 7 × 109? Explain.
Detailed solution and read-aloud reasoning
  1. 9.7 × 107 is about 1 × 108.
  2. 6.02 × 108 is about 6 × 108.
  3. The sum is about 7 × 108, not 10 times larger.
  4. Exact sum: 6.99 × 108.
Answer: 7 × 108
Question 113Grade 8 enrichmentEstimation, bounds, and precision5 marks
A calculator shows 9.9996E7. Round to three significant digits and write in scientific notation.
Detailed solution and read-aloud reasoning
  1. E7 means × 107.
  2. 9.9996 rounded to three significant digits becomes 10.0.
  3. Normalize 10.0 × 107 as 1.00 × 108.
  4. Keep the two trailing zeros to show three significant digits.
Answer: 1.00 × 108
Question 114Grade 8 enrichmentEstimation, bounds, and precision5 marks
Find the percentage difference between 5.0 × 106 and 4.8 × 106, using 5.0 × 106 as the reference.
Detailed solution and read-aloud reasoning
  1. Difference = 0.2 × 106.
  2. Divide by the reference: (0.2 × 106) ÷ (5.0 × 106) = 0.04.
  3. Convert to percent: 0.04 × 100% = 4%.
Answer: 4%
Question 115Grade 8 enrichmentReasoning and proof5 marks
Construct a number in scientific notation that is greater than 7.9 × 10−6 but less than 8.0 × 10−6.
Detailed solution and read-aloud reasoning
  1. The exponents already match.
  2. Choose any coefficient strictly between 7.9 and 8.0.
  3. For example, 7.95 works.
  4. 7.9 × 10−6 < 7.95 × 10−6 < 8.0 × 10−6.
Answer: Example: 7.95 × 10−6
Question 116Grade 8 enrichmentReasoning and proof5 marks
Find all integer exponents n for which 3.2 × 10n lies between 100,000 and 100,000,000.
Detailed solution and read-aloud reasoning
  1. 100,000 = 1 × 105; 100,000,000 = 1 × 108.
  2. 3.2 × 105, 3.2 × 106, and 3.2 × 107 lie inside the interval.
  3. n = 4 is too small and n = 8 is too large.
  4. n = 5, 6, or 7.
Answer: n = 5, 6, or 7
Question 117Grade 8 enrichmentReasoning and proof5 marks
A positive number has scientific notation a × 10−4. What range of standard values is possible?
Detailed solution and read-aloud reasoning
  1. Normalized scientific notation requires 1 ≤ a < 10.
  2. Multiply the entire inequality by 10−4.
  3. 1 × 10−4 ≤ value < 10 × 10−4.
  4. 0.0001 ≤ value < 0.001.
Answer: 0.0001 ≤ value < 0.001
Question 118Grade 8 enrichmentReasoning and proof5 marks
Explain why multiplying a positive number by 10−3 moves its decimal point left three places rather than right.
Detailed solution and read-aloud reasoning
  1. 10−3 is not negative ten cubed; it is the reciprocal of 103.
  2. 10−3 = 1/1000 = 0.001.
  3. Multiplying by 0.001 makes a number one thousand times smaller.
  4. On a place-value chart, becoming smaller shifts every digit three places to the right, which makes the decimal point appear three places to the left relative to the digits.
Answer: Because 10−3 = 0.001 = 1/1000
Question 119Grade 8 enrichmentReasoning and proof5 marks
Two students claim that 0.42 × 108 and 42 × 106 are different because their coefficients and exponents differ. Decide.
Detailed solution and read-aloud reasoning
  1. Normalize 0.42 × 108: move coefficient decimal right one place and lower exponent by 1 → 4.2 × 107.
  2. Normalize 42 × 106: move coefficient decimal left one place and raise exponent by 1 → 4.2 × 107.
  3. Therefore the expressions are equal.
Answer: They are equal; both are 4.2 × 107
Question 120Grade 8 enrichmentReasoning and proof5 marks
Create two different non-normalized scientific-notation expressions equal to 6.3 × 105.
Detailed solution and read-aloud reasoning
  1. Move the coefficient decimal one place right and decrease the exponent by 1: 63 × 104.
  2. Move the coefficient decimal one place left and increase the exponent by 1: 0.63 × 106.
  3. Both preserve the value because coefficient and power changes compensate.
Answer: Examples: 63 × 104 and 0.63 × 106
Cumulative Assessment

Final Mastery Test

30 questionsSuggested time: 90–120 minutesTotal: 129 marksNo calculator first; calculator verification second

Show all decimal-jump counts, exponent changes, normalization, units, and final answers.

Part A — Representation and Comparison

Test 12 marks
Write 805,000,000 in scientific notation.
Marking solution
  1. Move the decimal left 8 places to form 8.05.
  2. The exponent is positive because the original is large.
  3. 8.05 × 108.
Answer: 8.05 × 108
Test 22 marks
Write 0.000000349 in scientific notation.
Marking solution
  1. Move the decimal right 7 places to form 3.49.
  2. The exponent is negative because the original is between 0 and 1.
  3. 3.49 × 10-7.
Answer: 3.49 × 10-7
Test 32 marks
Write 7.006 × 108 in standard notation.
Marking solution
  1. Move the decimal right 8 places.
  2. Insert zeros where needed.
  3. 700,600,000.
Answer: 700,600,000
Test 42 marks
Write 2.81 × 10-7 in standard notation.
Marking solution
  1. Move the decimal left 7 places.
  2. Insert zeros where needed.
  3. 0.000000281.
Answer: 0.000000281
Test 54 marks
Order from least to greatest: 9.4 × 10-6, 1.02 × 10-4, 7.8 × 10-5, 8.99 × 10-6.
Marking solution
  1. Compare exponents first: −6 values are smallest, then −5, then −4.
  2. Among the −6 values, compare 8.99 and 9.4.
  3. 8.99 × 10-6 < 9.4 × 10-6 < 7.8 × 10-5 < 1.02 × 10-4.
Answer: 8.99 × 10-6 < 9.4 × 10-6 < 7.8 × 10-5 < 1.02 × 10-4
Test 63 marks
A student claims 6.7 × 107 < 9.1 × 106 because 6.7 < 9.1. Correct the reasoning.
Marking solution
  1. Compare exponents before coefficients.
  2. 7 is greater than 6, so the first number has the greater order of magnitude.
  3. 6.7 × 107 > 9.1 × 106.
Answer: 6.7 × 107 > 9.1 × 106

Part B — Operations

Test 74 marks
Calculate (4.8 × 106)(2.5 × 10−4).
Marking solution
  1. Multiply coefficients and add exponents: 12 × 102 = 1.2 × 103.
  2. Final answer: 1.2 × 103.
  3. Check order of magnitude and normalization.
Answer: 1.2 × 103
Test 84 marks
Calculate (9.45 × 10−3) ÷ (3.5 × 104).
Marking solution
  1. Divide coefficients and subtract exponents: 2.7 × 10−7.
  2. Final answer: 2.7 × 10-7.
  3. Check order of magnitude and normalization.
Answer: 2.7 × 10-7
Test 94 marks
Calculate 7.03 × 105 + 8.6 × 104.
Marking solution
  1. Rewrite 8.6 × 104 as 0.86 × 105; add to get 7.89 × 105.
  2. Final answer: 7.89 × 105.
  3. Check order of magnitude and normalization.
Answer: 7.89 × 105
Test 104 marks
Calculate 5.02 × 10−4 − 7.8 × 10−5.
Marking solution
  1. Rewrite 7.8 × 10−5 as 0.78 × 10−4; subtract to get 4.24 × 10−4.
  2. Final answer: 4.24 × 10-4.
  3. Check order of magnitude and normalization.
Answer: 4.24 × 10-4
Test 114 marks
Evaluate [(6 × 107)(3 × 10−2)] ÷ (9 × 103).
Marking solution
  1. Multiply to get 18 × 105; divide to get 2 × 102.
  2. Final answer: 2 × 102.
  3. Check order of magnitude and normalization.
Answer: 2 × 102
Test 124 marks
Evaluate (8.2 × 10−5 + 1.8 × 10−5) × 4 × 103.
Marking solution
  1. Add first: 10 × 10−5 = 1 × 10−4. Multiply to get 4 × 10−1.
  2. Final answer: 4 × 10-1.
  3. Check order of magnitude and normalization.
Answer: 4 × 10-1
Test 134 marks
Evaluate (3.6 × 108 − 4.5 × 107) ÷ (7 × 102).
Marking solution
  1. Subtract: 3.15 × 108. Divide: 0.45 × 106 = 4.5 × 105.
  2. Final answer: 4.5 × 105.
  3. Check order of magnitude and normalization.
Answer: 4.5 × 105
Test 144 marks
Evaluate [(2 × 10−3)2] ÷ (8 × 10−8).
Marking solution
  1. Square: 4 × 10−6. Divide: 0.5 × 102 = 5 × 101.
  2. Final answer: 5 × 101.
  3. Check order of magnitude and normalization.
Answer: 5 × 101
Test 154 marks
Evaluate (1.4 × 106 + 3.5 × 105 − 8 × 104).
Marking solution
  1. Use 106: 1.4 + 0.35 − 0.08 = 1.67, so 1.67 × 106.
  2. Final answer: 1.67 × 106.
  3. Check order of magnitude and normalization.
Answer: 1.67 × 106
Test 164 marks
Evaluate [(7.5 × 104)(2.4 × 10−6)] + 3.2 × 10−1.
Marking solution
  1. Product = 18 × 10−2 = 1.8 × 10−1. Add 3.2 × 10−1 to get 5 × 10−1.
  2. Final answer: 5 × 10-1.
  3. Check order of magnitude and normalization.
Answer: 5 × 10-1

Part C — Applications

Test 175 marks
A probe collects 3.75 × 105 samples per day for 2.4 × 102 days. It loses 2% of the samples. How many usable samples remain?
Marking solution
  1. Choose the operations and keep units attached.
  2. Collected = 9 × 107. Usable fraction = 0.98. Usable = 8.82 × 107.
  3. Final answer: 8.82 × 107.
Answer: 8.82 × 107
Test 185 marks
A wire is 6.4 × 10−5 m thick. A stack contains 2.5 × 104 wires. Find the stack thickness.
Marking solution
  1. Choose the operations and keep units attached.
  2. Multiply: 16 × 10−1 = 1.6 × 100 m.
  3. Final answer: 1.6 × 100.
Answer: 1.6 × 100
Test 195 marks
A machine produces 8.1 × 106 units, rejects 3.6 × 105, and packs the rest equally into 1.5 × 103 boxes. Units per box?
Marking solution
  1. Choose the operations and keep units attached.
  2. Acceptable units = 7.74 × 106. Divide by 1.5 × 103 to get 5.16 × 103.
  3. Final answer: 5.16 × 103.
Answer: 5.16 × 103
Test 205 marks
A rectangular surface measures 3.2 × 104 m by 7.5 × 102 m. Find its area.
Marking solution
  1. Choose the operations and keep units attached.
  2. Area = 24 × 106 = 2.4 × 107 m².
  3. Final answer: 2.4 × 107.
Answer: 2.4 × 107
Test 215 marks
A cube has edge 6 × 10−3 m. Find its volume.
Marking solution
  1. Choose the operations and keep units attached.
  2. Volume = 216 × 10−9 = 2.16 × 10−7 m³.
  3. Final answer: 2.16 × 10-7.
Answer: 2.16 × 10-7
Test 225 marks
Light travels about 3.0 × 108 m/s. Estimate the distance in 8.64 × 104 s.
Marking solution
  1. Choose the operations and keep units attached.
  2. Multiply: 25.92 × 1012 = 2.592 × 1013 m.
  3. Final answer: 2.592 × 1013.
Answer: 2.592 × 1013
Test 235 marks
A sample loses 15% of its mass. Its original mass is 4.8 × 10−2 g. Find the remaining mass.
Marking solution
  1. Choose the operations and keep units attached.
  2. Remaining fraction = 0.85. Multiply to get 4.08 × 10−2 g.
  3. Final answer: 4.08 × 10-2.
Answer: 4.08 × 10-2
Test 245 marks
A reservoir starts with 9.5 × 107 L. It loses 1.2 × 106 L/day for 35 days, then receives 8 × 106 L. Final volume?
Marking solution
  1. Choose the operations and keep units attached.
  2. Loss = 4.2 × 107 L. Remaining = 5.3 × 107 L. Add 0.8 × 107 L to get 6.1 × 107 L.
  3. Final answer: 6.1 × 107.
Answer: 6.1 × 107
Test 255 marks
A computer completes 2.56 × 1011 operations in 8 × 102 s. Find operations per second.
Marking solution
  1. Choose the operations and keep units attached.
  2. Divide: 0.32 × 109 = 3.2 × 108 operations/s.
  3. Final answer: 3.2 × 108.
Answer: 3.2 × 108
Test 265 marks
The mass of one particle is 7.2 × 10−12 kg. How many particles have total mass 1.8 × 10−4 kg?
Marking solution
  1. Choose the operations and keep units attached.
  2. Divide total by one particle: 0.25 × 108 = 2.5 × 107 particles.
  3. Final answer: 2.5 × 107.
Answer: 2.5 × 107

Part D — Reasoning and Communication

Test 276 marks
A measurement is 6.30 × 10−5 to three significant digits. State its bounds.
Marking solution
  1. The last retained place is 0.01 × 10−5.
  2. Half is 0.005 × 10−5.
  3. Subtract and add to obtain the bounds.
Answer: 6.295 × 10−5 ≤ value < 6.305 × 10−5
Test 286 marks
Find all integers n such that 4.5 × 10n is at least 0.0045 and less than 4,500.
Marking solution
  1. 0.0045 = 4.5 × 10−3.
  2. 4,500 = 4.5 × 103, but the upper endpoint is excluded.
  3. Thus −3 ≤ n < 3 for integer n.
Answer: n = −3, −2, −1, 0, 1, or 2
Test 296 marks
Explain and correct: (3 × 104) + (2 × 103) = 5 × 107.
Marking solution
  1. Exponents are added only when multiplying powers with the same base, not when adding numbers.
  2. Rewrite 2 × 103 as 0.2 × 104.
  3. Add coefficients: 3.2 × 104.
Answer: 3.2 × 104
Test 306 marks
Construct a value between 2.999 × 108 and 3.001 × 108, then write it in standard form.
Marking solution
  1. Choose a coefficient strictly between 2.999 and 3.001, such as 3.000.
  2. Keep the common exponent 8.
  3. Convert to standard form: 300,000,000.
Answer: Example: 3.000 × 108 = 300,000,000

Teacher Marking Guide

  • Knowledge: correct notation, powers, coefficient range, exponent sign.
  • Application: correct operation, unit, context interpretation, and magnitude.
  • Thinking: efficient strategy, estimation, bounds, and error analysis.
  • Communication: labelled steps, equal signs used logically, normalized final form, and clear reasoning.
Curriculum Alignment

Official Ontario Connections and Scope Notes

Grade 7 B1.1: represent and compare whole numbers up to and including one billion, including in expanded form using powers of ten, and describe everyday uses.

Grade 8 B1.1: represent and compare very large and very small numbers, including through scientific notation, and describe everyday uses.

Official Ontario curriculum portal:
https://www.dcp.edu.gov.on.ca/en/curriculum/elementary-mathematics

Ontario Grade 8 Number learning activity:
https://tvolearn.com/pages/grade-8-mathematics-number-learning-activity-1

The operation, precision, bounds, and advanced modelling sections are deliberate enrichment. They consolidate Grade 7–8 number skills and prepare students for later mathematics and science.