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.
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Number worksheet — GCSE Higher
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- 1.Two-digit numbers are made using the digits 1, 2, 3 and 4, and no digit may be used twice in the same number. Work out how many different two-digit numbers can be made.
- 2.A concrete mix is made from cement and sand in the ratio 3 : 5 by mass. A builder mixes 96 kg of concrete. Work out the mass of cement in the mix.
- 3.A rectangular patio measures 90 cm by 120 cm. Ben wants to cover it exactly with identical square tiles, as large as possible, with no tiles cut. Work out the side length of the largest square tile he can use.
- 4.A cyclist travels 40 km, correct to the nearest 10 km, in a time of 3 hours, correct to the nearest hour. Work out the maximum possible average speed, in km/h.
- 5.Work out 3 1/5 + 1 2/3. Give your answer as a mixed number in its simplest form.
- 6.A netball team has 8 players. Two of them are chosen to be captains, and the two captains have equal standing. Work out how many different pairs of captains could be chosen.
- 7.A photography studio offers 4 backdrops and 3 outfits for a portrait session. The plain grey backdrop, one of the 4 backdrops, cannot be used with the formal suit outfit, one of the 3 outfits. Work out how many different backdrop-and-outfit combinations are possible.
- 8.In standard form, 2,000 is written as 2 × 10ⁿ. Write down the value of n.
- 9.Simplify x⁶ ÷ x²
- 10.Without using a calculator, estimate √20 × √12, giving your answer to the nearest whole number.
- 11.At a book fair the ratio of fiction books to non-fiction books is 4 : 5. Of the non-fiction books, 3/5 are about history. Work out the fraction of all the books at the fair that are history books.
- 12.Round 0.006852 to 2 significant figures.
- 13.A triangle has a base of 8 cm and a perpendicular height of 5 cm, each correct to the nearest centimetre. Work out the upper bound of the area.
- 14.A grain of fine sand has a mass of 0.001 grams. Write this mass in standard form.
- 15.The two shorter sides of a right-angled triangle are √12 cm and √24 cm. Work out the exact length of the hypotenuse.
Answer key
- (a) 12 — Method: build the number one place at a time, listing systematically: fix the tens digit, then run through every units digit that is still available. Working: any of the 4 digits can go in the tens place, and once it has been used only 3 digits are left for the units place, so there are 4 × 3 = 12 numbers; listing the numbers that begin with 1 gives 12, 13 and 14, and each of the other three starting digits gives 3 numbers in the same way. Answer: 12. The distractors: 16 comes from working out 4 × 4, which allows a digit to be used twice; 8 comes from multiplying the 4 digits by the 2 places in the number instead of multiplying the choices available at each place; 6 comes from treating a number and its reverse as the same, counting only the unordered pairs of digits.
- (a) 36 kg — The mix is 3 + 5 = 8 equal shares, so the cement is 3/8 of the mass. One share is 96 ÷ 8 = 12 kg, and the cement is 3 shares: 3 × 12 = 36 kg. Working out 3/5 of 96 gives 57.6 kg, which uses the sand as the denominator instead of the whole mix; 60 kg is the mass of the sand; 12 kg is one share only.
- (a) 30 cm — The tile's side length must be a common factor of 90 and 120. The factors of 90 include 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90; the factors of 120 include 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120. The highest number common to both lists is 30, so the largest square tile has a side length of 30 cm. Picking 15 cm, a common factor but not the largest, gives tiles that are smaller than necessary. Picking 10 cm, also a common factor but smaller still, wastes even more of the possible tile size. Working out the lowest common multiple instead of the highest common factor gives 360 cm, a length far bigger than either side of the patio. So the largest square tile Ben can use has a side length of 30 cm.
- (c) 18 — The error intervals are 35 ≤ distance < 45 and 2.5 ≤ time < 3.5. Average speed is distance ÷ time, and to make a quotient as large as possible you divide the largest possible numerator by the SMALLEST possible denominator: 45 ÷ 2.5 = 18 km/h. Using the upper bound of time as well as the upper bound of distance, 45 ÷ 3.5, gives roughly 12.9 km/h — dividing by a bigger number produces a smaller result, so this actually finds a value smaller than the true maximum. Using the lower bound of distance with the lower bound of time, 35 ÷ 2.5 = 14, mixes up which bound belongs to a maximum calculation. Using the lower bound of distance with the upper bound of time, 35 ÷ 3.5 = 10, is in fact the correct method for the MINIMUM speed, not the maximum.
- (d) 4 13/15 — Convert to fifteenths: 1/5 is equivalent to 3/15 (multiply by 3/3), and 2/3 is equivalent to 10/15 (multiply by 5/5), so 3 1/5 is equivalent to 3 3/15 and 1 2/3 is equivalent to 1 10/15. Add the whole numbers (3 + 1 = 4) and the fractions (3/15 + 10/15 = 13/15), giving 4 13/15. A candidate who adds the numerators and denominators straight across, treating 1/5 + 2/3 as (1+2)/(5+3), gets a fraction part of 3/8, giving 4 3/8. A candidate who adds the fraction parts correctly but forgets to add the second whole number gets 3 13/15. A candidate who adds the whole numbers but copies the first fraction across without ever adding 2/3 to it gets 4 1/5.
- (b) 28 — Method: count the ordered choices with the product rule and then correct for the double counting, because the two captains have equal standing and so a pair is the same pair whichever captain is named first. Working: there are 8 players who could be named first and 7 who could be named second, giving 8 × 7 = 56 ordered choices; each pair has been counted twice, once in each order, so the number of pairs is 56 ÷ 2 = 28. Answer: 28. The distractors: 56 comes from stopping at 8 × 7 and never halving, which counts each pair of captains twice; 64 comes from working out 8 × 8, which allows the same player to be chosen as both captains; 16 comes from multiplying the 8 players by the 2 captaincies instead of pairing the players with one another.
- (c) 11 — Method: work out the total number of combinations as if there were no restriction, then subtract the one combination that is not allowed. Working: without any restriction there are 4 backdrops × 3 outfits = 12 combinations. The grey backdrop with the formal suit is not allowed, removing 1 combination: 12 − 1 = 11. Answer: 11. 12 comes from forgetting to remove the combination that is not allowed. 8 comes from removing the entire formal suit outfit from the count instead of just the one combination with the grey backdrop. 10 comes from removing two combinations instead of just the one that is not allowed.
- (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.
- (d) x⁴ — Method: dividing two powers of the same letter subtracts the index of the divisor from the index of the term being divided. Working: six factors of x on the top and two on the bottom cancel in pairs, leaving 6 − 2 = 4 factors of x. Answer: x⁴. The distractors: x³ comes from dividing the indices, 6 ÷ 2, instead of subtracting them; x⁸ comes from adding the indices, 6 + 2, as though the powers were being multiplied; x¹² comes from multiplying the indices, 6 × 2, as though a power were being raised to a power.
- (d) 15 — Use √a × √b = √(ab): √20 × √12 = √(20 × 12) = √240. Since 15² = 225 and 16² = 256, and 240 is a little closer to 225 than to 256, √240 is a little under 15.5 — in fact √240 ≈ 15.49, which rounds to 15. Adding the two roots instead of multiplying them, √20 + √12 ≈ 4.47 + 3.46 ≈ 7.94, rounds to 8, but the question asks for the product, not the sum. Multiplying 20 by 12 and stopping there, without ever taking a square root, leaves 240, which is the number under the root, not its value. Rounding each root to the nearest whole number BEFORE multiplying — √20 ≈ 4 and √12 ≈ 3 — gives 4 × 3 = 12, a cruder estimate that loses accuracy by rounding twice instead of once.
- (c) 1/3 — Method: find non-fiction's fraction of the whole, then multiply by the fraction of non-fiction that is history. Working: total parts = 4 + 5 = 9, so non-fiction is 5/9 of all books. History books are 3/5 of the non-fiction books: 3/5 × 5/9 = 15/45 = 1/3. Answer: 1/3. 3/5 comes from giving the fraction of non-fiction books that are history, without relating it to all the books at the fair. 5/9 comes from stopping after finding the fraction of all books that are non-fiction, without finding the history books within that. 4/15 comes from multiplying 3/5 by the fraction that is fiction (4/9) instead of the fraction that is non-fiction (5/9).
- (a) 0.0069 — Leading zeros are not significant, so the significant figures in 0.006852 start at 6: 6, 8, 5, 2. Rounding to 2 significant figures means keeping 6 and 8, and looking at the next digit, 5, to decide whether to round up. Since 5 rounds up, the second significant figure increases from 8 to 9: 0.006852 rounds to 0.0069. A candidate who rounded to 1 significant figure instead of 2 wrote 0.007. A candidate who rounded to 3 significant figures instead of 2 wrote 0.00685. A candidate who did not round up despite the next digit being 5 wrote 0.0068.
- (c) 23.375 — The error intervals are 7.5 ≤ base < 8.5 and 4.5 ≤ height < 5.5. The upper bound of the area uses the upper bound of both the base and the height, then halves the product: 8.5 × 5.5 ÷ 2 = 23.375 cm². Using the lower bound of both dimensions instead, 7.5 × 4.5 ÷ 2 = 16.875, gives the lower bound of the area rather than the upper one. Multiplying the two upper bounds together but forgetting to halve for the triangle formula, 8.5 × 5.5 = 46.750, treats the triangle as if it were a rectangle. Using the given values directly without applying any bound at all, 8 × 5 ÷ 2 = 20.000, ignores that each rounded measurement has its own range of possible values.
- (d) 1 × 10⁻³ — Method: for a number smaller than 1 the index is negative, and it counts the places the decimal point moves to the right to leave a coefficient between 1 and 10. Working: the only significant digit is 1, so the coefficient is 1; the decimal point in 0.001 moves three places to the right to reach that 1, so the index is −3. Answer: 1 × 10⁻³. The distractors: 1 × 10³ comes from taking the index as positive, which describes one thousand grams rather than one thousandth of a gram; 0.1 × 10⁻² is the same mass written with a coefficient of 0.1, which is smaller than 1 and so is not standard form; 1 × 10⁻⁴ comes from counting the zero in front of the decimal point as well as the places after it.
- (d) 6 — By Pythagoras' theorem, the square of the hypotenuse equals the sum of the squares of the other two sides: (√12)² + (√24)² = 12 + 24 = 36. The hypotenuse is √36 = 6 cm. Adding the two side lengths directly instead of squaring them first, treating the theorem as if it were a straight sum of the sides, gives √12 + √24 = 2√3 + 2√6. Multiplying the two squared values, 12 × 24 = 288, instead of adding them, then taking the root, gives √288 = 12√2. Adding the squares correctly to get 36 but forgetting to take the square root at the end leaves 36 as the answer instead of the hypotenuse itself.
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