Printable · GCSE Foundation · ages 14-16
Standard form worksheet — GCSE Foundation
Fifteen questions on "standard form" — DfE statement N9. Print it, or print three versions so neighbours cannot copy by letter; the key gives the letter for each version.
Answer key: Standard form worksheet — GCSE Foundation
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- (c) 2.5 × 10⁶ — Divide the A values: 5 ÷ 2 = 2.5. Subtract the powers of 10: 4 − (−2) = 4 + 2 = 6. So the answer is 2.5 × 10⁶. A candidate who worked out 4 − 2 = 2, treating the subtraction of a negative as an ordinary subtraction, wrote 2.5 × 10². A candidate who subtracted in the wrong order, −2 − 4 = −6, wrote 2.5 × 10⁻⁶. A candidate who multiplied the A values instead of dividing, 5 × 2 = 10, and added the powers, 4 + (−2) = 2, then rewrote 10 × 10² in standard form as 1 × 10³.
- (d) 6 × 10⁷ — Method: the coefficients and the powers of ten are handled separately — multiply the coefficients, and add the indices because the powers share the base 10. Working: 2 × 3 = 6 for the coefficients, and 10³ × 10⁴ = 10⁷ for the powers; 6 lies between 1 and 10, so the coefficient needs no adjustment. Answer: 6 × 10⁷. The distractors: 5 × 10⁷ comes from adding the coefficients, 2 + 3, instead of multiplying them; 6 × 10¹² comes from multiplying the indices, 3 × 4, instead of adding them; 6 × 10¹ comes from subtracting the indices, 4 − 3, which is the rule for dividing rather than for multiplying.
- (a) 7.2 × 10⁻⁴ — 0.00072 = 7.2 × 10⁻⁴, moving the decimal point 4 places to the right to reach 7.2, so the exponent is negative. Writing 7.2 × 10⁴ uses a positive exponent, which would give a number far larger than 1, not a small decimal. Writing 7.2 × 10⁻⁵ moves the point one place too many. Writing 0.72 × 10⁻³ leaves the coefficient below 1, which standard form does not allow.
- (c) 4.1 × 10³, 3.2 × 10⁴, 2.9 × 10⁵ — The exponent decides the size first: 10³ is smaller than 10⁴, which is smaller than 10⁵, so the order is 4.1 × 10³, then 3.2 × 10⁴, then 2.9 × 10⁵. Reversing the whole list gives largest to smallest instead of smallest to largest. Comparing 3.2 × 10⁴ and 4.1 × 10³ by their coefficients alone, 3.2 against 4.1, and swapping them ignores that 10⁴ is bigger than 10³ regardless of the coefficient. Comparing 2.9 × 10⁵ and 3.2 × 10⁴ by their coefficients alone and swapping them makes the same mistake at the top of the list.
- (c) 5 × 10⁶ — Method: standard form is written as A × 10ⁿ, where A is at least 1 and less than 10 and n counts the places the decimal point moves. Working: the digits of 5,000,000 give a coefficient of A = 5, and the decimal point travels from the end of 5,000,000 until it sits just after the 5, a move of 6 places, so n = 6. Answer: 5 × 10⁶. The distractors: 50 × 10⁵ comes from stopping before the coefficient has been brought into range, and 50 is not less than 10, so it is not standard form; 5 × 10⁷ comes from counting the seven digits of 5,000,000 instead of the six places the decimal point moves; 5 × 10⁻⁶ comes from making the index negative because the decimal point was carried to the left, when a negative index belongs to a number smaller than 1.
- (a) 8 × 10² — Method: to multiply numbers written in standard form, multiply the coefficients and add the indices. Working: 4 × 2 = 8 for the coefficients, and −3 + 5 = 2 for the indices; 8 already lies between 1 and 10, so no adjustment is needed. Answer: 8 × 10². The distractors: 6 × 10² comes from adding the coefficients, 4 + 2, instead of multiplying them; 8 × 10⁸ comes from ignoring the minus sign and adding 3 + 5; 8 × 10⁻¹⁵ comes from multiplying the indices, −3 × 5, instead of adding them.
- (c) 4 × 10⁴ — 8 ÷ 2 = 4, and 6 − 2 = 4, so each project receives 4 × 10⁴ pounds. Multiplying the exponents instead of subtracting them gives 6 × 2 = 12, so 4 × 10¹². Adding the exponents instead of subtracting them gives 6 + 2 = 8, so 4 × 10⁸. Subtracting the coefficients instead of dividing them gives 8 − 2 = 6, so 6 × 10⁴.
- (c) 3 × 10⁶ — Daily revenue = (4 × 10³) × 2.5 = 1 × 10⁴ (£10,000). Multiplying by the number of days, (3 × 10²), gives annual revenue = (1 × 10⁴) × (3 × 10²) = 3 × 10⁶ (£3,000,000). A candidate who forgot to multiply by the price and just multiplied the number of items by the number of days worked out (4 × 10³) × (3 × 10²) = 1.2 × 10⁶. A candidate who added the number of days to the daily revenue instead of multiplying worked out 1 × 10⁴ + 3 × 10² = 1.03 × 10⁴. A candidate who misread £2.50 as £25 worked out a daily revenue of (4 × 10³) × 25 = 1 × 10⁵, giving an annual total of (1 × 10⁵) × (3 × 10²) = 3 × 10⁷.
- (a) 3.84 × 10⁵ — 384,000 = 3.84 × 100,000 = 3.84 × 10⁵, with the decimal point moved five places and the coefficient kept between 1 and 10. Moving the point six places instead of five gives 3.84 × 10⁶, ten times too large. Leaving the coefficient as 38.4 gives 38.4 × 10⁴, which is not between 1 and 10. Using a negative exponent instead of a positive one gives 3.84 × 10⁻⁵, a number far smaller than 1.
- (a) 8 × 10³ — Method: divide the capacity of the card by the size of one photograph, dividing the coefficients and subtracting the indices, then bring the coefficient back into the range 1 to 10. Working: 3.2 ÷ 4 = 0.8 and 10 − 6 = 4, which gives 0.8 × 10⁴; a coefficient of 0.8 is smaller than 1, so the decimal point moves one place to the right and the index falls by 1. Answer: 8 × 10³. The distractors: 8 × 10⁴ comes from correcting 0.8 to 8 without reducing the index, which makes the answer ten times too large; 1.28 × 10¹⁷ comes from multiplying the two numbers instead of dividing them, since 3.2 × 4 = 12.8 and 10 + 6 = 16; 8 × 10¹⁵ comes from dividing the coefficients but adding the indices instead of subtracting them.
- (c) 4 × 10⁻¹ — Multiply the A values: 8 × 5 = 40. Add the powers of 10: −5 + 3 = −2, giving 40 × 10⁻². Since A must satisfy 1 ≤ A < 10, rewrite 40 as 4 × 10¹, so 40 × 10⁻² = 4 × 10¹ × 10⁻² = 4 × 10⁻¹. A candidate who stopped at 40 × 10⁻² did the index arithmetic correctly but left the answer outside standard form, since 40 is not between 1 and 10. A candidate who adjusted the A value to 4 correctly but then took the power of 10 by subtracting the two given powers, −5 − 3 = −8, wrote 4 × 10⁻⁸. A candidate who adjusted the A value to 4 but multiplied the two given powers, −5 × 3 = −15, wrote 4 × 10⁻¹⁵. Both of these forgot that multiplying in standard form means adding the powers.
- (a) 7.5 × 10⁴ — Standard form is A × 10ⁿ, where A is between 1 and 10 (1 ≤ A < 10) and n is an integer — 7.5 × 10⁴ satisfies all of this. 12 × 10³ fails because 12 is not below 10. 0.5 × 10⁴ fails because 0.5 is not at least 1. 6.2 × 4⁵ fails because standard form always uses a power of 10, not a power of 4.
- (d) 1.2 × 10⁵ — 4 × 3 = 12, and 3 + 1 = 4, giving 12 × 10⁴ — but 12 is not between 1 and 10, so this must be rewritten as 1.2 × 10⁵. Stopping at 12 × 10⁴ without rewriting it leaves the coefficient out of range. Rewriting 12 as 1.2 but leaving the exponent at 4 instead of increasing it to 5 gives 1.2 × 10⁴, which is ten times too small. Adding the coefficients instead of multiplying them gives 4 + 3 = 7, so 7 × 10⁴.
- (d) 6 × 10⁷ — 3 × 2 = 6, and 2 + 5 = 7, so (3 × 10²) × (2 × 10⁵) = 6 × 10⁷. Multiplying the exponents instead of adding them gives 2 × 5 = 10, so 6 × 10¹⁰. Adding the coefficients instead of multiplying them gives 3 + 2 = 5, so 5 × 10⁷. Subtracting the exponents instead of adding them gives 5 − 2 = 3, so 6 × 10³.
- (c) 5 × 10⁻³ — Method: in standard form the coefficient must be at least 1 and less than 10, and for a number smaller than 1 the index is negative and counts the places the decimal point moves to the right. Working: the only significant digit is 5, so the coefficient is 5; moving the decimal point in 0.005 three places to the right gives 5, so the index is −3. Answer: 5 × 10⁻³. The distractors: 0.5 × 10⁻² is the same value written the wrong way, because 0.5 is smaller than 1 and so is not an allowed coefficient; 5 × 10⁻² comes from counting the two zeros after the decimal point instead of the three places the point moves; 5 × 10³ comes from taking the index as positive, which describes a number in the thousands rather than one smaller than 1.
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