Mathematics

Number, Year 11: bounds, growth and compound interest

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MathematicsNumber, Year 11: bounds, growth and compound interest
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Mathematics · Number, Year 11: bounds, growth and compound interestProfessor Pi
↔ Asked both waysAnswer in your head…For a measurement that has been rounded, the top end of its error interval: the value the true measurement must be less than
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★ GCSE-MATH-NUM-0046Front

Mathematics · Number, Year 11: bounds, growth and compound interestProfessor Pi

The upper bound of a rounded measurement

HintA two-word name; its partner at the other end is called 'lower'.

The whyA length of 7.4 cm to 1 decimal place lies in 7.35 ≤ L < 7.45, so this value is 7.45 cm. The true length can get as close to it as you like but cannot equal it, because 7.45 would round to 7.5.

★ GCSE-MATH-NUM-0046Back

Mathematics · Number, Year 11: bounds, growth and compound interestProfessor Pi
Answer in your head…A rectangle measures 6 cm by 4 cm, each to the nearest centimetre. Work out the upper bound for its area.
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★ GCSE-MATH-NUM-0047Front

Mathematics · Number, Year 11: bounds, growth and compound interestProfessor Pi

29.25 cm²

HintMake both sides as long as they could possibly be before multiplying.

The whyEach side could be up to half a centimetre longer: 6.5 cm and 4.5 cm. The largest possible area is 6.5 × 4.5 = 29.25 cm². The lower bound is 5.5 × 3.5 = 19.25 cm².

★ GCSE-MATH-NUM-0047Back

Mathematics · Number, Year 11: bounds, growth and compound interestProfessor Pi
Answer in your head…p = 8.4 and q = 2.6, each correct to 1 decimal place. To get the greatest possible value of p − q, which bound of p and which bound of q should you use?
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★ GCSE-MATH-NUM-0048Front

Mathematics · Number, Year 11: bounds, growth and compound interestProfessor Pi

The upper bound of p and the lower bound of q

HintTaking away less leaves more.

The whyThe difference is biggest when you start with as much as possible and remove as little as possible: 8.45 − 2.55 = 5.9. For the least possible difference, swap the bounds: 8.35 − 2.65 = 5.7.

★ GCSE-MATH-NUM-0048Back

Mathematics · Number, Year 11: bounds, growth and compound interestProfessor Pi
⌨ Type the answerAnswer in your head…A car travels 100 m, measured to the nearest 10 m, in 8 s, measured to the nearest second. Work out the upper bound for its average speed in m/s. (number only)

★ GCSE-MATH-NUM-0049Front

Mathematics · Number, Year 11: bounds, growth and compound interestProfessor Pi

14

HintSpeed is greatest when the distance is as long as possible and the time as short as possible.

The whyUpper bound of distance = 105 m; lower bound of time = 7.5 s. Greatest speed = 105 ÷ 7.5 = 14 m/s. In a division, use the upper bound on top and the lower bound underneath to get the largest answer.

★ GCSE-MATH-NUM-0049Back

Mathematics · Number, Year 11: bounds, growth and compound interestProfessor Pi
Answer in your head…A calculated length has lower bound 2.4612 m and upper bound 2.4649 m. To what degree of accuracy can the length be stated with certainty, and what is it to that accuracy?
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★ GCSE-MATH-NUM-0050Front

Mathematics · Number, Year 11: bounds, growth and compound interestProfessor Pi

2.46 m, to 2 decimal places (3 significant figures)

HintRound both ends more and more finely and see where they stop agreeing.

The whyTo 2 decimal places both bounds give 2.46, so every possible value rounds to 2.46. To 3 decimal places they give 2.461 and 2.465, which differ, so the third decimal place is not certain.

★ GCSE-MATH-NUM-0050Back

Mathematics · Number, Year 11: bounds, growth and compound interestProfessor Pi
⌨ Type the answerAnswer in your head…A price rises by 10% and is then reduced by 10%. What single decimal multiplier gives the overall change? (number only)

★ GCSE-MATH-NUM-0051Front

Mathematics · Number, Year 11: bounds, growth and compound interestProfessor Pi

0.99

HintWrite each change as its own multiplier, then combine the two.

The whyThe rise is × 1.1 and the reduction is × 0.9, and 1.1 × 0.9 = 0.99, an overall fall of 1%. For example £200 becomes £220 and then £198. The 10% reduction is taken from the larger amount, so it removes more than the rise added.

★ GCSE-MATH-NUM-0051Back

Mathematics · Number, Year 11: bounds, growth and compound interestProfessor Pi
Answer in your head…A car is bought for £12 000. It loses 20% of its value each year. Work out its value after 2 years.
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★ GCSE-MATH-NUM-0052Front

Mathematics · Number, Year 11: bounds, growth and compound interestProfessor Pi

£7680

HintEach year's loss is 20% of what the car is worth at the start of that year.

The whyLosing 20% leaves 80%, a multiplier of 0.8. After two years the value is 12 000 × 0.8 × 0.8 = 12 000 × 0.8² = £7680. Year by year: £9600, then £7680.

★ GCSE-MATH-NUM-0052Back

Mathematics · Number, Year 11: bounds, growth and compound interestProfessor Pi
↔ Asked both waysAnswer in your head…The fall in the value of something, such as a car or a computer, as it gets older
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★ GCSE-MATH-NUM-0053Front

Mathematics · Number, Year 11: bounds, growth and compound interestProfessor Pi

Depreciation

HintIt is the opposite of 'appreciation', a rise in value.

The whyA car that loses 15% of its value each year is depreciating at 15% a year, so each year its value is multiplied by 0.85. It is a decay problem: the same percentage is lost each year, so the amount lost in pounds gets smaller every year.

★ GCSE-MATH-NUM-0053Back

Mathematics · Number, Year 11: bounds, growth and compound interestProfessor Pi
Fill the gapAnswer in your head…To decrease an amount by 15% again and again, multiply repeatedly by the decimal multiplier ____.
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★ GCSE-MATH-NUM-0054Front

Mathematics · Number, Year 11: bounds, growth and compound interestProfessor Pi

0.85

HintAsk what percentage of the amount is left after the decrease.

The whyTaking away 15% leaves 85% of the amount, and 85% as a decimal is 0.85. Three decreases in a row multiply by 0.85³. A multiplier below 1 means decay; one above 1 means growth.

★ GCSE-MATH-NUM-0054Back

Mathematics · Number, Year 11: bounds, growth and compound interestProfessor Pi
Answer in your head…A town has a population of 20 000. The population grows by 10% each year. Work out the population after 3 years.
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★ GCSE-MATH-NUM-0055Front

Mathematics · Number, Year 11: bounds, growth and compound interestProfessor Pi

26 620

HintUse a single multiplier for one year's growth, and apply it once for each year.

The whyA 10% increase is a multiplier of 1.1. After 3 years the population is 20 000 × 1.1³ = 20 000 × 1.331 = 26 620. Year by year: 22 000, 24 200, 26 620.

★ GCSE-MATH-NUM-0055Back

Mathematics · Number, Year 11: bounds, growth and compound interestProfessor Pi
Fill the gapsAnswer in your head…£2000 is invested at 5% compound interest per year. After one year the account holds £____, and after two years it holds £____.
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★ GCSE-MATH-NUM-0056Front

Mathematics · Number, Year 11: bounds, growth and compound interestProfessor Pi

2100; 2205

HintEach year's interest is worked out on whatever is in the account at the start of that year.

The whyYear 1: 2000 × 1.05 = £2100. Year 2 starts from £2100, not £2000: 2100 × 1.05 = £2205. The second year earns £105, which is £5 more than the first, because the first year's £100 of interest earns interest too.

★ GCSE-MATH-NUM-0056Back

Mathematics · Number, Year 11: bounds, growth and compound interestProfessor Pi
Answer in your head…£500 is invested at 4% compound interest per year. Write the single calculation that gives the amount in the account after 6 years, using a multiplier and a power. You do not need to work it out.
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★ GCSE-MATH-NUM-0057Front

Mathematics · Number, Year 11: bounds, growth and compound interestProfessor Pi

500 × 1.04⁶

HintAdding four per cent is the same as multiplying by one number; do that once for every year.

The whyAdding 4% multiplies an amount by 1.04. Doing that six years running multiplies by 1.04 six times, which is 1.04⁶. On a calculator this gives £632.66 to the nearest penny.

★ GCSE-MATH-NUM-0057Back

Mathematics · Number, Year 11: bounds, growth and compound interestProfessor Pi
⌨ Type the answerAnswer in your head…A savings account pays 10% compound interest per year. £1000 is invested and left alone. After how many complete years does the balance first go above £1300? (number only)

★ GCSE-MATH-NUM-0058Front

Mathematics · Number, Year 11: bounds, growth and compound interestProfessor Pi

3

HintBuild the balance up one year at a time, multiplying by the same number each year.

The whyEach year multiplies the balance by 1.1: £1100 after one year, £1210 after two and £1331 after three. The balance first passes £1300 at the end of the third year.

★ GCSE-MATH-NUM-0058Back

Mathematics · Number, Year 11: bounds, growth and compound interestProfessor Pi
Answer in your head…£1000 is invested for 2 years at 10% per year. How much more does it earn with compound interest than with simple interest, and where does the extra come from?
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★ GCSE-MATH-NUM-0059Front

Mathematics · Number, Year 11: bounds, growth and compound interestProfessor Pi

£10 more — in the second year, interest is also paid on the first year's £100 of interest

HintFind both totals, then ask what the later 10% is being taken of in each case.

The whySimple interest: £100 each year, £200 in all. Compound interest: £100 in year one, then 10% of £1100 = £110 in year two, £210 in all. The £10 difference is 10% of the first year's £100 of interest.

★ GCSE-MATH-NUM-0059Back

Mathematics · Number, Year 11: bounds, growth and compound interestProfessor Pi
↔ Asked both waysAnswer in your head…Interest that is worked out each time on the original amount plus all the interest already added
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★ GCSE-MATH-NUM-0060Front

Mathematics · Number, Year 11: bounds, growth and compound interestProfessor Pi

Compound interest

HintIts name means 'made of parts put together'; the other kind is called simple.

The whyBecause each year's interest joins the balance and then earns interest itself, the amount grows by the same multiplier every year, not by the same number of pounds. Simple interest is always worked out on the original amount only.

★ GCSE-MATH-NUM-0060Back

Number, Year 11: bounds, growth and compound interest

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