36 exam-style questions on Standard form (a x 10^k) — topic 1.1 of Number and Algebra in the IB Mathematics: Analysis and Approaches syllabus — with answers and explanations. Samples below; the full set is free with an account.
Write the number 0.00047 in the form a × 10^k, where 1 ≤ a < 10 and k ∈ ℤ.
Answer: 4.7 × 10⁻⁴
The decimal point moves 4 places to the right to give 4.7, so k = −4. Note that 47 × 10⁻⁵ has the correct value but a = 47 is not in the required range 1 ≤ a < 10.
Calculate (3 × 10⁵) × (4 × 10⁻²), giving your answer in the form a × 10^k, where 1 ≤ a < 10 and k ∈ ℤ.
Answer: 1.2 × 10⁴
Multiply coefficients and add exponents: 3 × 4 = 12 and 10⁵⁺⁽⁻²⁾ = 10³, so 12 × 10³ = 1.2 × 10⁴ after adjusting the coefficient into the range 1 ≤ a < 10.
Calculate (8 × 10⁶) ÷ (2 × 10⁻³), giving your answer in the form a × 10^k, where 1 ≤ a < 10 and k ∈ ℤ.
Answer: 4 × 10⁹
Divide coefficients and subtract exponents: 8 ÷ 2 = 4 and 6 − (−3) = 9, so the answer is 4 × 10⁹. A common error is computing 6 − 3 = 3 by forgetting the negative sign.
Determine which of the following numbers is the largest.
Answer: 1.01 × 10⁵
Compare exponents first: 1.01 × 10⁵ = 101 000, which exceeds every option with exponent 4 (the largest of those is 9.99 × 10⁴ = 99 900). A larger exponent outweighs a larger coefficient.
Calculate (2.5 × 10³)², giving your answer in the form a × 10^k, where 1 ≤ a < 10 and k ∈ ℤ.
Answer: 6.25 × 10⁶
Square the coefficient and double the exponent: 2.5² = 6.25 and (10³)² = 10⁶. A frequent error is squaring the exponent (10⁹) instead of multiplying it by 2.
Calculate (4.2 × 10⁵) + (3 × 10⁴), giving your answer in the form a × 10^k, where 1 ≤ a < 10 and k ∈ ℤ.
Answer: 4.5 × 10⁵
To add, rewrite with the same exponent: 4.2 × 10⁵ + 0.3 × 10⁵ = 4.5 × 10⁵. Adding the coefficients directly (4.2 + 3 = 7.2) ignores the different powers of 10.
That was 6 of 36 Standard form (a x 10^k) questions.
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