Printable · GCSE Higher · ages 14-16
Number worksheet — GCSE Higher
Fifteen questions across the number statements at Higher tier. Choose the non-calculator filter to rehearse Paper 1, which counts for a third of the marks.
Non-calculator
Answer key: Number worksheet — GCSE Higher
MathsUKwww.geekhero.co.uk
- (d) −1.4, −6/5, 0, 5/4, 1.3 — Method: convert the fractions 5/4 and −6/5 to decimals so every number is written the same way, then compare all five decimals. Working: 5/4 = 1.25 and −6/5 = −1.2. Comparing −1.4, −1.2, 0, 1.25 and 1.3 in size gives the order −1.4, −1.2, 0, 1.25, 1.3. Answer: −1.4, −6/5, 0, 5/4, 1.3. −6/5, −1.4, 0, 5/4, 1.3 swaps the two negative numbers, treating −6/5 as more negative than −1.4 even though −1.2 is closer to zero than −1.4. 1.3, 5/4, 0, −6/5, −1.4 lists the numbers from largest to smallest instead of smallest to largest. −1.4, −6/5, 0, 1.3, 5/4 swaps 5/4 and 1.3, comparing the numerator 5 directly with 1.3 instead of converting 5/4 to the decimal 1.25 first.
- (a) y¹² — Method: when a power is raised to another power, multiply the two indices. Working: (y³)⁴ means y³ × y³ × y³ × y³, which is four lots of three y's multiplied together, so the index is 3 × 4 = 12 and (y³)⁴ = y¹². y⁷ comes from adding the indices, 3 + 4 = 7, which is the rule for multiplying two separate powers, not for raising a power to a power. y⁸¹ comes from working out 3⁴ = 81 and using that as the index, raising the inner index to the outer power instead of multiplying the two indices. 12y comes from multiplying the indices to make 12 but then treating y as a coefficient instead of a power. Answer: y¹².
- (b) £1,000, so £900 is not enough — Method: round each number to 1 significant figure, multiply to estimate the total cost, then compare the estimate with the money available. Working: 187 rounds to 200 and £4.85 rounds to £5, so the estimate is 200 × 5 = 1,000, and £1,000 is more than the £900 the school has. Answer: £1,000, so £900 is not enough. The distractors: £800 comes from cutting £4.85 down to £4 instead of rounding it up to £5, giving 200 × 4 = 800, and that estimate wrongly suggests the money stretches; £935 comes from rounding the price only and keeping 187 lunches, giving 187 × 5 = 935; £950 comes from rounding 187 to the nearest 10 rather than to 1 significant figure, giving 190 × 5 = 950.
- (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) 17 — Method: work out each power separately, then combine them as the question asks. Working: $2^3 = 8$ and $3^2 = 9$, and 8 + 9 = 17. 72 comes from working out 8 × 9 = 72, multiplying the two powers instead of adding them. 12 comes from misreading the powers as repeated multiplication of the base by the index, 2 × 3 + 3 × 2 = 6 + 6 = 12. −1 comes from working out 8 − 9 = −1, subtracting the powers instead of adding them. Answer: 17.
- (c) 590 — To round to the nearest 10, decide which multiple of 10 the number is nearer to. 592.5 lies between 590 and 600. It is 592.5 − 590 = 2.5 above 590, but 600 − 592.5 = 7.5 below 600, so it is much nearer to 590. Equivalently, the units digit is 2, and 2 is less than 5, so round down: 592.5 rounds to 590. A candidate who wrote 600 rounded up because of the 5 in the tenths place, but that digit decides rounding to the nearest whole number, not to the nearest 10 — the units digit is what matters here. A candidate who wrote 595 rounded to the nearest 5 instead of the nearest 10. A candidate who wrote 500 cut the number down to its hundreds digit instead of rounding to the nearest 10.
- (c) 5 — 0.2 = 1/5, and turning the fraction upside down gives the reciprocal 5/1 = 5. Writing −0.2 mistakes the reciprocal for the negative of the number. Working out 1 − 0.2 = 0.8 mistakes the reciprocal for subtracting the number from 1. Writing 0.5 is the reciprocal of 2, not of 0.2 — a place-value slip that drops the decimal, since 1 ÷ 0.2 = 5 while 1 ÷ 2 = 0.5.
- (d) 1 × 10⁷ seconds — Method: the time for a journey is the distance divided by the speed, so round each number to 1 significant figure and then divide; dividing numbers in standard form means dividing the coefficients and subtracting the indices. Working: 3.1 × 10¹⁵ rounds to 3 × 10¹⁵ and 2.998 × 10⁸ rounds to 3 × 10⁸; 3 ÷ 3 = 1 for the coefficients, and 15 − 8 = 7 for the indices. Answer: about 1 × 10⁷ seconds. The distractors: 3 × 10⁷ seconds comes from subtracting the indices correctly but leaving the coefficient as 3 instead of dividing 3 by 3; 1 × 10⁻⁷ seconds comes from dividing the speed by the distance instead of the distance by the speed; 9 × 10²³ seconds comes from multiplying the two quantities instead of dividing them, since 3 × 3 = 9 and 15 + 8 = 23.
- (a) 200 g — One quarter of 160 g is 40 g. Increasing the amount means adding this on: 160 + 40 = 200 g. Finding the increase, 1/4 of 160 = 40 g, but stopping there without adding it to the original amount leaves just 40 g. Using 4/5 instead of 5/4 as the scaling fraction, 160 × 4/5 = 128 g, actually decreases the amount rather than increasing it. Increasing by a half instead of a quarter, 160 + 80 = 240 g, uses the wrong fraction of 160.
- (b) 4 — Method: the cube root of a number is the value that multiplies by itself three times to give that number. Working: 4 × 4 × 4 = 64, so ∛64 = 4. Answer: 4. (8 comes from finding the square root of 64 instead of the cube root. 192 comes from multiplying 64 by 3 instead of finding the number that cubes to 64. 21.3 comes from dividing 64 by 3 instead of finding its cube root.)
- (b) 24√2 — The side length of the tile is √72. Since 72 = 36 × 2, √72 = √36 × √2 = 6√2 cm. A square has four equal sides, so the perimeter is 4 × 6√2 = 24√2 cm. Simplifying √72 by writing the perfect-square factor itself as the coefficient instead of its root, 36√2 instead of 6√2, and then multiplying by 4 lands on 144√2. Working out the correct side length, 6√2 cm, but then giving that as the final answer without multiplying by 4 for the perimeter gives 6√2. Doubling the side length instead of quadrupling it, as if the perimeter were 2 × 6√2 rather than 4 × 6√2, gives 12√2.
- (a) The tape can only give the length to the nearest centimetre — Method: a measurement should never be written to a finer degree of accuracy than the instrument used can read. Working: the tape is marked in centimetres, so the smallest division Leah can read is 1 cm, which is 0.01 m and two decimal places in metres; writing 7.3157 m claims the length to the nearest tenth of a millimetre, four decimal places, which the markings cannot support. A record of 7.32 m, to the nearest centimetre, is what this tape justifies. Answer: The tape can only give the length to the nearest centimetre. The distractors: the nearest millimetre contradicts the markings described in the question, which are centimetres, and would still claim more accuracy than the tape offers; the rule that a length in metres must be written to 2 decimal places borrows the habit of writing money to the penny, when the accuracy of a length depends on the instrument; rounding to the nearest metre would throw away accuracy the tape genuinely provides.
- (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.
- (c) 0.5, 0.45, −0.09, −0.15, −0.6 — Method: compare the decimals by their position on a number line, remembering that with negative decimals the one closer to zero is larger. Working: 0.5 and 0.45 are positive, so they come first, with 0.5 the larger of the two. Among the negatives, −0.09 is closest to zero, then −0.15, then −0.6 is furthest from zero and so the smallest. Answer: 0.5, 0.45, −0.09, −0.15, −0.6. 0.5, 0.45, −0.15, −0.09, −0.6 swaps −0.09 and −0.15, treating the negative decimal with more digits after the point as closer to zero. −0.6, −0.15, −0.09, 0.45, 0.5 lists the numbers from smallest to largest instead of largest to smallest. 0.5, 0.45, −0.6, −0.15, −0.09 orders the negative decimals by the size of the digit (0.6 > 0.15 > 0.09) as if they were positive, instead of recognising that a bigger negative decimal is further from zero and so smaller.
- (b) 2/5 — The statement compares girls with boys, so girls : boys = 2 : 3. The club is therefore 2 + 3 = 5 equal shares and the girls take 2 of them, so 2/5 of the members are girls. 2/3 repeats the part-to-part fraction the question gives, 3/5 is the fraction that are boys, and 1/3 comes from treating 2/3 as a fraction of the whole club and taking what is left.
Build your own mix at the worksheet builder.