Printable · GCSE Foundation · ages 14-16
Number worksheet — GCSE Foundation
Fifteen questions across the number statements at Foundation tier. Choose the non-calculator filter to rehearse Paper 1, which counts for a third of the marks.
Number worksheet — GCSE Foundation
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- 1.Light travels at 2.998 × 10⁸ metres per second. A distant object in space is 3.1 × 10¹⁵ metres from Earth. Work out an estimate for the number of seconds light takes to travel from the object to Earth, by rounding each number to 1 significant figure.
- 2.Work out the value of .
- 3.The distance from the Earth to the Moon is 384,000 km. Write this distance in standard form, in kilometres.
- 4.A number, n, is equal to 3.7 when rounded to 1 decimal place. Write down the error interval for n.
- 5.Work out √144
- 6.Which statement about the number 51 is correct?
- 7.A roll of ribbon is 8.4 m long. Ribbon is cut into pieces that are each 0.6 m long. Work out how many complete pieces can be cut from the roll.
- 8.An allotment is divided into two plots in the ratio 2:3. The larger plot has an area of 18 m². Work out the fraction of the total area taken up by the smaller plot.
- 9.A cheetah runs at a steady speed of 25 metres per second. Work out this speed in kilometres per hour.
- 10.Work out the value of .
- 11.A vending machine sells 4 types of crisps, 5 types of chocolate bar and 2 types of drink. Work out how many different combinations of one crisp packet, one chocolate bar and one drink can be bought.
- 12.A train leaves Leeds at 14:35. The journey takes 1 hour 50 minutes. Work out the time the train arrives. Give your answer using the 24-hour clock.
- 13.Work out (−36) ÷ (−6) × 2
- 14.In a box of counters the ratio of red counters to blue counters is 4 : 5. What fraction of the counters are blue?
- 15.The mass of a radioactive sample, in grams, n years after it was first weighed is modelled by M = 200 × (1/2)ⁿ. Work out the mass the model gives after 3 years.
Answer key
- (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) 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.
- (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.
- (b) 3.65 ≤ n < 3.75 — Rounding to 1 decimal place means n can be up to half of one decimal place, 0.05, below or above 3.7 before it would round to a different value. This gives a lower bound of 3.7 − 0.05 = 3.65 and an upper bound of 3.7 + 0.05 = 3.75. A value exactly at 3.75 would round up to 3.8, not 3.7, so the upper bound is excluded while the lower bound, 3.65, does still round to 3.7. Writing 3.65 ≤ n ≤ 3.75 wrongly includes 3.75. Writing 3.6 ≤ n < 3.8 uses a whole decimal place, 0.1, either side instead of half of one, 0.05. Writing 3.65 < n < 3.75 wrongly excludes 3.65, which does round to 3.7.
- (c) 12 — Method: a square root asks which positive number multiplied by itself gives the number under the root sign. Working: 11 × 11 = 121, which is too small, and 12 × 12 = 144. Answer: 12. The distractors: 14 comes from recalling the wrong square fact, since 14 × 14 = 196; 36 comes from dividing 144 by 4 instead of taking a square root; 72 comes from halving 144, treating a square root as a halving.
- (d) 51 is not prime, because 51 = 3 × 17. — Check 51 for small prime factors: 51 ÷ 3 = 17, and both 3 and 17 are themselves prime, so 51 = 3 × 17 and 51 is not a prime number — it has factors other than 1 and itself. Checking only 2, 3 and 5 and concluding wrongly that none of them divide 51 misses that 3 does divide it exactly, so the claim that 51 is prime because it avoids 2, 3 and 5 is false. Assuming any odd number must be prime ignores that 51 = 3 × 17 is a counterexample — plenty of odd numbers are not prime. Misreading 51 as the even number 52 leads to the false claim that it is divisible by 2; 51 itself is odd, and 2 is not one of its factors. So 51 is not prime, because 51 = 3 × 17.
- (a) 14 — Multiply both numbers by 10 to clear the decimals: 8.4 becomes 84 and 0.6 becomes 6. Then divide: 84 ÷ 6 = 14, so 14 complete pieces can be cut. Scaling only the divisor by 10 and leaving the dividend as 8.4 gives 8.4 ÷ 6 = 1.4, which rounds down to 1 complete piece — the dividend was never converted. Scaling only the dividend by 10 and leaving the divisor as 0.6 gives 84 ÷ 0.6 = 140. Rounding the divisor from 0.6 to 0.7 before dividing, trading accuracy for a rounder number, gives 8.4 ÷ 0.7 = 12. So 14 complete pieces of ribbon can be cut.
- (a) 2/5 — The ratio 2:3 has 2 + 3 = 5 parts in total, and the larger plot is 3 of those parts. Since the larger plot is 18 m², each part is 18 ÷ 3 = 6 m², so the total area is 5 × 6 = 30 m² and the smaller plot is 2 × 6 = 12 m². The fraction of the total area taken up by the smaller plot is 12/30, which simplifies to 2/5. Giving the fraction for the larger plot instead of the smaller one gives 3/5. Comparing the smaller plot to the larger plot instead of to the total area gives 2/3. Assuming the two plots split the area evenly, ignoring the given ratio altogether, gives 1/2.
- (b) 90 km/h — To convert metres per second to kilometres per hour, multiply by 3.6 (there are 3600 seconds in an hour and 1000 metres in a kilometre, and 3600 ÷ 1000 = 3.6): 25 × 3.6 = 90 km/h. Dividing by 3.6 instead of multiplying gives 25 ÷ 3.6 = 6.9 km/h (to 1 d.p.). Multiplying by 60 instead of 3.6, confusing the conversion from seconds to minutes with the conversion to hours, gives 25 × 60 = 1500 km/h. Multiplying by 3600 to convert seconds to hours but forgetting to convert metres to kilometres gives 25 × 3600 = 90000, which is a speed in metres per hour, not kilometres per hour.
- (c) 0.001 — Method: a negative index means 'one over' the positive power, so $10^{-3}$ means one over $10^{3}$. Working: ten cubed is 1000, and one over 1000 is 0.001. 1000 comes from ignoring the negative sign and working out ten cubed instead of its reciprocal. −1000 comes from ignoring what the negative index does to the size, while still writing a negative sign on the large value. −0.001 comes from correctly finding the size, 0.001, but wrongly keeping a negative sign because the index was negative. Answer: 0.001.
- (a) 40 — Multiply the number of choices for each item: 4 × 5 × 2 = 40. 11 comes from adding the three numbers instead of multiplying them. 20 comes from multiplying only the crisps and chocolate bars, 4 × 5, and forgetting the drink. 10 comes from multiplying only the chocolate bars and drinks, 5 × 2, and forgetting the crisps.
- (c) 16:25 — Add the whole hour first: 14:35 plus 1 hour is 15:35. Then add the 50 minutes. From 15:35, 25 minutes reaches 16:00 and the remaining 25 minutes gives 16:25. 15:85 adds 35 and 50 to get 85 minutes and never exchanges 60 of them for an hour, 16:05 treats 1 hour 50 minutes as 1.5 hours, and 12:45 subtracts the journey time instead of adding it.
- (c) 12 — Method: division and multiplication have equal priority, so they are carried out in the order they are written, from left to right; a negative divided by a negative is positive. Working: (−36) ÷ (−6) = 6, and then 6 × 2 = 12. Answer: 12. The distractors: 3 comes from carrying out the multiplication first, (−6) × 2 = −12 followed by (−36) ÷ (−12) = 3; −12 comes from treating a negative divided by a negative as negative, giving −6 and then −6 × 2 = −12; 6 comes from stopping at the division and never carrying out the multiplication by 2.
- (b) 5/9 — Add the parts of the ratio to find how many equal shares make up the whole: 4 + 5 = 9 shares. The blue counters take 5 of those 9 shares, so 5/9 of the counters are blue. 4/9 is the fraction that are red, 4/5 is the ratio copied straight down as a fraction, and 5/4 compares the blue counters with the red counters instead of with the whole box.
- (b) 25 g — Method: substitute the number of years into the model, raise the fraction to that power first, then multiply by the starting mass. Working: with n = 3 the model gives M = 200 × (1/2)³. Since (1/2)³ = 1/8, the mass is 200 ÷ 8 = 25. Answer: 25 g. The distractors: 12.5 g comes from halving four times instead of three, counting the first weighing as a year; 300 g comes from multiplying by 1/2 × 3 = 1.5 instead of raising 1/2 to the power 3; 0.125 g comes from working out (1/2)³ = 0.125 and stopping there, without multiplying by the starting mass.
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