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.
Number worksheet — GCSE Higher
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- 1.Which statement about the number 91 is correct?
- 2.A number, x, is truncated (not rounded) to 1 decimal place and the result is 6.2. Write down the error interval for x.
- 3.Work out the value of 6² − 4³.
- 4.Write down the symbol that makes this statement true: −2/3 ___ −5/6
- 5.Work out 100 − 4 × 5²
- 6.Work out 2 3/4 − 1 5/6. Give your answer as a fraction in its simplest form.
- 7.Write these decimals in order, starting with the smallest: 0.6, 0.55, 0.601, 0.5, 0.56
- 8.A plank has length a = 12 cm and a second plank has length b = 7 cm, each correct to the nearest centimetre. Work out the upper bound of a + b.
- 9.Work out the value of √(16 + 9)
- 10.Simplify √45.
- 11.A school council must choose a committee of 3 pupils from 8 volunteers. The three places on the committee are all the same, so only which pupils are chosen matters. Work out how many different committees could be formed.
- 12.Petrol costs £1.48 per litre. Sam buys 35 litres and pays with three £20 notes. Work out his change.
- 13.Write 7/12 as a decimal, showing clearly which digit is recurring.
- 14.A red blood cell has a diameter of about 7 × 10⁻⁶ metres. A virus has a diameter about 100 times smaller. Work out the diameter of the virus. Give your answer in standard form.
- 15.Work out √(4 × 9)
Answer key
- (a) 91 is not prime, because 91 = 7 × 13. — Check 91 for prime factors up to its square root, which is just under 10: 91 ÷ 7 = 13, and both 7 and 13 are prime, so 91 = 7 × 13 and 91 is not a prime number. Checking only 2, 3 and 5 misses that 7 also needs to be tried — 91 is odd, its digits do not sum to a multiple of 3 (9 + 1 = 10), and it does not end in 0 or 5, so those three checks alone wrongly suggest it is prime. Assuming any odd number ending in 1 must be prime ignores that 91 = 7 × 13 is a counterexample. Misapplying the digit-sum test for 3 by miscounting 9 + 1 as a multiple of 3 wrongly concludes 91 is divisible by 3, when the correct digit sum, 10, is not a multiple of 3. So 91 is not prime, because 91 = 7 × 13.
- (b) 6.2 ≤ x < 6.3 — Truncating simply cuts off the digits after the required decimal place instead of rounding them, so every value from 6.2 up to (but not reaching) 6.3 truncates to 6.2. This gives the error interval 6.2 ≤ x < 6.3, with no allowance made on the lower side because truncation never rounds a smaller value up into this interval. Using 6.15 ≤ x < 6.25 applies the rounding rule of going half a unit either side, which does not apply to truncation. Writing 6.1 < x ≤ 6.2 puts the interval below 6.2 instead of above it. Writing 6.2 ≤ x ≤ 6.3 wrongly includes 6.3, which truncates down to itself, not to 6.2.
- (c) −28 — 6² = 36 and 4³ = 64. Work out 36 − 64 = −28. A candidate who subtracts in the wrong order gets 64 − 36 = 28. A candidate who adds instead of subtracting gets 36 + 64 = 100. A candidate who multiplies the base by the exponent instead of raising the power (6 × 2 − 4 × 3 = 12 − 12) gets 0.
- (c) > — Method: write both fractions with a common denominator so they can be compared directly. Working: −2/3 = −4/6. Comparing −4/6 and −5/6, −4 is greater than −5, so −4/6 is greater than −5/6. Answer: −2/3 > −5/6, so the symbol is >. < comes from treating the fraction with the larger numerator and denominator, 5/6, as the greater number, ignoring that both fractions are negative. = comes from assuming the fractions must be equal because they look similar in size. ≤ combines < (false) with = (false).
- (a) 0 — Method: BIDMAS works through the index first, then the multiplication, then the subtraction. Working: 5² = 25, then 4 × 25 = 100, and finally 100 − 100 = 0. Answer: 0. The distractors: 2400 comes from working from left to right and subtracting first, giving (100 − 4) × 25 = 96 × 25 = 2400; −300 comes from multiplying before applying the index, giving (4 × 5)² = 20² = 400 and then 100 − 400 = −300; 60 comes from reading 5² as 5 × 2 = 10, so that 4 × 10 = 40 and 100 − 40 = 60.
- (d) 11/12 — Convert both mixed numbers to improper fractions with a common denominator. 2 3/4 = 11/4, which is 33/12, and 1 5/6 = 11/6, which is 22/12. Subtracting, 33/12 − 22/12 gives 11/12, already in its simplest form. Forgetting to borrow, and instead subtracting the fraction parts the other way round to avoid a negative, 10/12 − 9/12 gives 1/12; adding that to the whole-number difference of 1 gives 13/12. Subtracting only the fraction parts, 9/12 − 10/12, and reporting just the size of that difference gives 1/12, which ignores the whole numbers altogether. Adding the two improper fractions instead of subtracting them, 33/12 + 22/12, gives 55/12. So 2 3/4 − 1 5/6 = 11/12.
- (a) 0.5, 0.55, 0.56, 0.6, 0.601 — Compare the decimals by giving them all the same number of decimal places first: 0.600, 0.550, 0.601, 0.500, 0.560. Ordering these from smallest to largest gives 0.500, 0.550, 0.560, 0.600, 0.601, which is 0.5, 0.55, 0.56, 0.6, 0.601. Comparing the digits as though they were whole numbers, reading 0.601 as "601" and 0.5 as "5", without padding to the same number of decimal places, gives the wrong order 0.5, 0.6, 0.55, 0.56, 0.601, because it ignores the place value of each digit. Ordering largest to smallest instead of smallest to largest, as the question asks, gives 0.601, 0.6, 0.56, 0.55, 0.5. Misreading the close values 0.55 and 0.56 and swapping them gives 0.5, 0.56, 0.55, 0.6, 0.601. So the correct order, smallest to largest, is 0.5, 0.55, 0.56, 0.6, 0.601.
- (b) 20 — Each length has its own error interval: 11.5 ≤ a < 12.5 and 6.5 ≤ b < 7.5. The upper bound of a sum is found by adding the upper bounds of both quantities: 12.5 + 7.5 = 20. Bounding only one of the two lengths and adding the other quantity's given value unbounded, 12.5 + 7 = 19.5, misses that both measurements carry their own uncertainty. Adding the lower bounds instead of the upper bounds, 11.5 + 6.5 = 18, gives the lower bound of the sum, not the upper one. Using a whole centimetre of error either side instead of half a centimetre, (12 + 1) + (7 + 1) = 21, comes from forgetting the error is half the rounding unit.
- (a) 5 — 16 + 9 = 25, then √25 = 5. Splitting the root over the addition instead gives √16 = 4 and √9 = 3, then 4 + 3 = 7 — but a root does not split over a sum like this. Multiplying those two roots instead of adding them gives 4 × 3 = 12. Taking the negative square root instead of the positive one gives −5.
- (d) 3√5 — Split 45 into a perfect square times a factor: 45 = 9 × 5. Take the square root of each part separately: √45 = √9 × √5 = 3√5, since √9 = 3. Writing the perfect-square factor itself (9) as the coefficient instead of its root would give 9√5 — that trap comes from forgetting the last step, rooting 9. Multiplying 3 and 5 together instead of keeping them as coefficient and radicand gives 15, which throws away the surd entirely. Doubling the correct coefficient by mistake gives 6√5.
- (a) 56 — Method: count the ordered selections with the product rule first, then divide by the number of different orders in which any one committee could have been picked. Working: there are 8 choices for a first pupil, 7 for a second and 6 for a third, giving 8 × 7 × 6 = 336 ordered selections; any particular three pupils could have been picked in 3 × 2 × 1 = 6 orders, so the number of different committees is 336 ÷ 6 = 56. Answer: 56. The distractors: 336 comes from stopping at 8 × 7 × 6 and treating the three places as distinct posts when they are identical; 168 comes from dividing that product by 2 rather than by the 6 orders in which three chosen pupils can be listed; 24 comes from multiplying the 8 volunteers by the 3 places instead of multiplying the choices at each stage.
- (c) £8.20 — Method: find the total cost, then subtract from the amount paid. Working: 35 × £1.48 = £51.80. Amount paid = 3 × £20 = £60.00. Change = £60.00 − £51.80 = £8.20. Answer: £8.20. (£58.52 comes from forgetting to multiply the price by the 35 litres and subtracting only £1.48 from £60. £7.50 comes from rounding £1.48 up to £1.50 before multiplying, giving a total of £52.50 instead of £51.80. £9.20 comes from miscarrying in the pence column when subtracting £51.80 from £60.00.)
- (b) 0.58333... — Divide 7 by 12 using long division. 70 ÷ 12 = 5 remainder 10, so the first decimal digit is 5. Bring down a 0 to make 100: 100 ÷ 12 = 8 remainder 4, so the second digit is 8. Bring down a 0 to make 40: 40 ÷ 12 = 3 remainder 4, so the third digit is 3. Bring down a 0 to make 40 again — the remainder 4 has reappeared, so the digit 3 repeats forever from here. This gives 7/12 = 0.58333... . Stopping the division after two digits and writing 0.58 treats it as if it terminated, when the remainder is not yet zero. Misreading the pattern as a two-digit repeating block, '58', gives 0.585858..., which wrongly makes the 5 recur as well as the 3. A slip in the long division that carries the wrong remainder forward can make the second digit itself appear to repeat instead of the third, giving 0.588888... .
- (c) 7 × 10⁻⁸ — '100 times smaller' means dividing by 100 = 10². Dividing 7 × 10⁻⁶ by 10² means subtracting 2 from the exponent: −6 − 2 = −8, giving 7 × 10⁻⁸. A candidate who multiplied by 100 instead of dividing added 2 to the exponent, getting 7 × 10⁻⁴. A candidate who divided by 10 instead of 100 subtracted only 1 from the exponent, getting 7 × 10⁻⁵. A candidate who did not apply the scale factor at all left the diameter as 7 × 10⁻⁶, the same as the red blood cell.
- (d) 6 — Method: the square root of a product can be found either by multiplying first and then rooting, or by rooting each factor and multiplying the two roots together. Working: 4 × 9 = 36, and 6 × 6 = 36, so the root is 6; the same value comes from √4 × √9 = 2 × 3. Answer: 6. The distractors: 36 comes from multiplying inside the root and then leaving the root untaken; 5 comes from rooting each factor and adding the results, 2 + 3, instead of multiplying them; 18 comes from rooting the 4 only and leaving the 9 untouched, giving 2 × 9.
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