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
MathsUKwww.geekhero.co.uk
- (c) 8,000,000,000 — Method: a power of ten is 1 followed by that many zeros, so 8 × 10⁹ is 8 multiplied by 1 followed by nine zeros. Working: 10⁹ = 1,000,000,000, and 8 × 1,000,000,000 puts an 8 in front of those nine zeros. Answer: 8,000,000,000. The distractors: 800,000,000 comes from writing nine digits altogether instead of nine zeros after the 8; 720 comes from reading 10⁹ as 10 × 9 = 90 and then working out 8 × 90; 0.000000008 comes from treating the index as negative and carrying the decimal point nine places to the left.
- (a) 4.5 × 10⁴ — 45,000 = 4.5 × 10,000 = 4.5 × 10⁴, with the coefficient between 1 and 10 as standard form requires. Writing 45 × 10³ keeps the coefficient too large — 45 is not between 1 and 10. Writing 4.5 × 10³ undercounts the places moved, giving only 4,500. Writing 4.5 × 10⁵ overcounts the places moved, giving 450,000.
- (b) 5.2 × 10⁶ — Method: write the digits as a coefficient that is at least 1 and less than 10, then count the places the decimal point moves to reach that position. Working: the digits give a coefficient of 5.2, and the decimal point travels from the end of 5,200,000 until it sits between the 5 and the 2, a move of 6 places. Answer: 5.2 × 10⁶. The distractors: 52 × 10⁵ is the same amount but not in standard form, because 52 is not less than 10; 5.2 × 10⁵ comes from counting the five zeros in 5,200,000 rather than the six places the decimal point moves; 5.2 × 10⁻⁶ comes from making the index negative because the decimal point was carried to the left.
- (a) 4,000 nanometres — Method: the number of nanometres is the diameter divided by the length of one nanometre, and dividing powers of ten means subtracting the indices. Working: −6 − (−9) = 3, so 10⁻⁶ ÷ 10⁻⁹ = 10³, and the diameter is 4 × 10³ nanometres. Answer: 4,000 nanometres. The distractors: 400 nanometres comes from taking the difference between the indices as 2 instead of 3; 4 nanometres comes from changing the name of the unit without converting, leaving the coefficient untouched; 0.004 nanometres comes from dividing by 10³ instead of multiplying by it, as though a nanometre were the larger of the two units.
- (a) 3.8 × 10⁻³ — 0.0038 is less than 1, so the power of 10 is negative. Moving the decimal point 3 places gives A = 3.8, so 0.0038 = 3.8 × 10⁻³. A candidate who wrote 3.8 × 10³ used a positive power, which is only correct for numbers of 10 or more. A candidate who wrote 38 × 10⁻⁴ used a value of A outside the required range. A candidate who wrote 3.8 × 10⁻⁴ counted one place too many when moving the decimal point.
- (c) 3 — Method: the index counts how many times the coefficient has been multiplied by 10, which is the number of places the decimal point moves from the end of the number to just after the first significant digit. Working: 2,000 = 2 × 1,000, and 1,000 = 10 × 10 × 10, which is three tens. Answer: 3. The distractors: 4 comes from counting the four digits of 2,000 rather than the three places the decimal point moves; 2 comes from copying the coefficient 2 into the index; −3 comes from making the index negative, which would describe a number smaller than 1 rather than two thousand.
- (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.
- (d) 340,000 — 3.4 × 10⁵ means moving the decimal point in 3.4 five places to the right, giving 340,000. Moving it only four places gives 34,000, one place short. Moving it six places gives 3,400,000, one place too many. Treating the exponent as negative instead of positive moves the point the wrong way, giving 0.000034.
- (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.
- (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.
- (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.
- (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.
- (d) 1.6 × 10⁵ — 5.6 ÷ 3.5 = 1.6, and 7 − 2 = 5, so the average turnover per shop is 1.6 × 10⁵ pounds. Multiplying the exponents instead of subtracting them gives 7 × 2 = 14, so 1.6 × 10¹⁴. Adding the exponents instead of subtracting them gives 7 + 2 = 9, so 1.6 × 10⁹. Subtracting the coefficients instead of dividing them gives 5.6 − 3.5 = 2.1, so 2.1 × 10⁵.
- (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.
- (a) 6 × 10² metres — Method: distance = speed × time, so multiply the coefficients and add the indices. Working: 3 × 2 = 6 for the coefficients, and 8 + (−6) = 2 for the indices; 6 lies between 1 and 10, so the coefficient needs no adjustment. Answer: 6 × 10² metres, which is 600 metres. The distractors: 5 × 10² metres comes from adding the coefficients, 3 + 2, instead of multiplying them; 6 × 10¹⁴ metres comes from subtracting the indices, 8 − (−6), which is the rule for dividing rather than for multiplying; 6 × 10⁻⁴⁸ metres comes from multiplying the indices, 8 × (−6), instead of adding them.
Build your own mix at the worksheet builder.