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Calculator 4 marks
Rosa invests £7200 in a savings account.
The account pays 4% compound interest per year.
She takes no money out of the account.
(a)Calculate how much is in the account after 2 years.
(b)Calculate the total interest earned over the 2 years.
Show answer
(a) \(\text{£}7787.52\)
(b) \(\text{£}587.52\)Compound interest is paid on the money already in the account, so each year the amount is multiplied by the same number:
\(1 + \frac{ 4 }{100} = 1.04\)
Doing that 2 times means raising the multiplier to the power 2:
\(7200 \times 1.04^{ 2 } = 7787.52\)
There is £7787.52 in the account after 2 years.
\(£7787.52 - £7200 = £587.52\)
Note that this is more than 2 lots of the first year's interest of £288, because after the first year the interest is paid on a larger amount.
- (a) M1multiplier \(1.04\)
- (a) M1\(7200 \times 1.04^{ 2 }\)
- (a) A1amount \(7787.52\)
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Calculator 3 marks
Grace buys a new car for £21300.
The value of the car depreciates by 30% each year.
Find the value of the car after 2 years.
Show answer
\(\text{£}10437.00\)Losing 30% of the value leaves 70% of it, so each year the value is multiplied by
\(1 - \frac{ 30 }{100} = 0.7\)
That happens 2 times, so raise the multiplier to the power 2:
\(21300 \times 0.7^{ 2 } = 10437.00\)
The car is worth £10437.00 after 2 years.
This is not the same as losing 60% of £21300 altogether. After the first year each 30% is taken from a smaller amount, so the total loss of £10863.00 is less than 60% of the original price.
- M1multiplier \(0.7\)
- M1\(21300 \times 0.7^{ 2 }\)
- A1value \(10437.00\)
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Calculator 4 marks
A farm shop buys a display counter for £3000.
The display counter loses 30% of its value each year.
Here are three calculations.
A \(3000 \times 0.9^{ 5 }\)
B \(3000 \times 0.6^{ 5 }\)
C \(3000 \times 0.7^{ 5 }\)
(a)State the letter of the calculation that gives the value of the display counter after 5 years.
(b)Work out how much value the display counter loses altogether over the 5 years.
Show answer
(a) C
(b) \(\text{£}2495.79\)All three calculations raise a number to the power 5, one for each year, so the only thing to settle is which number.
Losing 30% of the value leaves 70% of it, so each year the value is multiplied by
\(1 - \frac{ 30 }{100} = 0.7\)
Calculation C is the one that uses it. The other two are the multipliers for a 10% fall and a 40% fall — right calculations for a different question.
Two other numbers are worth being able to rule out on sight, even though neither is offered here: \(30\) hundredths on its own is the part that is LOST rather than the part that is left, and anything above 1 would make the value grow.
Work out what the display counter is worth at the end, using the calculation from part (a):
\(3000 \times 0.7^{ 5 } = 504.21\)
It cost £3000, so the value it has lost is
\(3000 - 504.21 = 2495.79\)
- (b) M1\(3000 \times 0.7^{ 5 }\)
- (b) M1\(3000 - 504.21\)
- (b) A1value lost \(2495.79\)
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Calculator 5 marks
A sign maker bought a laser cutter.
The table shows some information about what the laser cutter was worth when it was new and at the end of each year after that.
The value falls by the same percentage every year.
End of year Value new £19000 1 £11400 2 3 (a)Complete the table.
(b)Marcus says the laser cutter loses the same amount of value every year.
Use the table to explain why he is wrong.
Show answer
(a) End of year 2: £6840. End of year 3: £4104.
(b) The laser cutter lost £7600 during the first year but only £4560 during the second, because the same percentage is taken off a smaller value each year.Two of the values are a year apart, so divide one by the other to find what the value is multiplied by each year.
\(11400 \div 19000 = 0.6\)
Multiplying by that again gives the end of year 2:
\(11400 \times 0.6 = 6840\)
And once more for the end of year 3:
\(6840 \times 0.6 = 4104\)
The fall in the first year is \(19000 - 11400 = 7600\) pounds.
The fall in the second year is \(11400 - 6840 = 4560\) pounds.
They are not the same. The percentage is the same each year, but after the first year it is taken off a smaller value, so the amount of money lost gets smaller every year.
- (a) M1yearly multiplier \(0.6\)
- (a) A1end of year 2 value £\(6840\)
- (a) A1end of year 3 value £\(4104\)
- (b) M1first year fall £\(7600\) and second year fall £\(4560\)
- (b) C1the falls differ because the same percentage is taken off a smaller value
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Calculator 4 marks
Idris has £8400 to invest for 2 years.
Account A pays 4% simple interest per year.
Account B pays 4% compound interest per year.
Calculate how much more interest Account B pays than Account A over the 2 years.
Show answer
\(\text{£}13.44\)Account A. Simple interest pays the same amount every year, worked out on the £8400 paid in.
\(8400 \times \frac{ 4 }{100} = 336\) each year, so over 2 years the interest is \(2 \times 336 = 672\).
Account B. Compound interest multiplies the amount by \(1 + \frac{ 4 }{100} = 1.04\) each year.
\(8400 \times 1.04^{ 2 } = 9085.44\)
so the interest is \(9085.44 - 8400 = 685.44\).
Difference:
\(£685.44 - £672 = £13.44\)
Account B pays £13.44 more interest.
- M1simple interest \(336\) a year, \(672\) in total
- M1multiplier \(1.04\) used for \(2\) years
- M1compound interest \(685.44\)
- A1difference £\(13.44\)
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Calculator 3 marks
Amara invests £5700 in a savings account.
The account pays 6% compound interest per year.
She takes no money out of the account.
Work out the number of years it takes for the amount in the account to first be more than £6500.
Show answer
\(3\ \text{years}\)Each year the amount in the account is multiplied by
\(1 + \frac{ 6 }{100} = 1.06\)
There is no single calculation that gives the number of years, so work out the amount after different numbers of years until it passes £6500.
After 2 years:
\(5700 \times 1.06^{ 2 } \approx 6404.52\)
That is still less than £6500.
After 3 years:
\(5700 \times 1.06^{ 3 } \approx 6788.79\)
That is more than £6500, and the year before it was not.
So it takes 3 years.
- M1multiplier \(1.06\) used to find the amount after a number of years
- M1amounts after 2 years \(6404.52\) and 3 years \(6788.79\) compared with \(6500\)
- A13 years
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Calculator 3 marks
Owen pays £32000 into a 3-year investment plan.
The plan pays 5% compound interest in the first year, and 2% compound interest in each year after that.
Work out the value of the plan at the end of the 3 years.
Show answer
\(\text{£}34957.44\)The first year and the years after it multiply by different numbers, so work out each multiplier on its own.
First year: \(1 + \frac{ 5 }{100} = 1.05\)
Every year after that: \(1 + \frac{ 2 }{100} = 1.02\)
The first multiplier is used once and the second one is used for the remaining 2 years:
\(32000 \times 1.05 \times 1.02^{ 2 } = 34957.44\)
The plan is worth £34957.44 at the end of the 3 years.
Doing it in two steps is the same calculation: after one year the plan holds \(32000 \times 1.05 = 33600\), and the remaining 2 years multiply that by \(1.02^{ 2 }\).
- M1multipliers \(1.05\) and \(1.02\)
- M1\(32000 \times 1.05 \times 1.02^{ 2 }\)
- A1value \(34957.44\)
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Calculator 4 marks
A forestry firm paid £3000 for a new log splitter.
Each year, the value of the log splitter is multiplied by 0.7.
(a)Write down the percentage by which the value of the log splitter falls each year.
(b)Work out how much value the log splitter loses during the 4th year.
Show answer
(a) \(30\%\)
(b) \(\text{£}308.70\)Multiplying by \(0.7\) keeps \(0.7\) of the value, and \(0.7\) as a percentage is 70%.
Keeping 70% means 30% has gone, because the two must add to 100%.
The 4th year starts with whatever the log splitter was worth at the end of year 3, so work that out first.
\(3000 \times 0.7^{ 3 } = 1029\)
One more year takes it to
\(1029 \times 0.7 = 720.30\)
so the value lost during the 4th year is
\(1029 - 720.30 = 308.70\)
Notice it is smaller than the £900 lost in the first year — the 30% is charged against a value that has already fallen 3 times.
- (b) M1value at the end of year 3 \(1029\)
- (b) M1value after one more year \(720.30\)
- (b) A1value lost \(308.70\)
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Calculator 4 marks
An auction house bought a first edition book for £22000.
The value of the first edition book goes up by 10% each year.
Hafsa says that after 2 years the first edition book will be worth £26455.00.
Is Hafsa right? You must show your working.
Show answer
No — after 2 years the first edition book is worth £26620.00, so Hafsa's figure is not rightA rise of 10% multiplies the value by
\(1 + \frac{ 10 }{100} = 1.1\)
That happens 2 times, so raise the multiplier to the power 2:
\(22000 \times 1.1^{ 2 } = 26620.00\)
After 2 years the first edition book is worth £26620.00. Hafsa said £26455.00, so Hafsa is not right.
- M1multiplier, \(1 + \frac{ 10 }{100} = 1.1\)
- M1raise to the power 2, \(22000 \times 1.1^{ 2 }\)
- A1value after 2 years, \(£26620.00\)
- A1No — Hafsa's figure is not right, comparing £26620.00 with £26455.00
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Calculator 3 marks
A farm bought a tractor.
The value of the tractor depreciates by \(50\%\) each year.
After \(2\) years the tractor is worth \(£1750\).
(a)Find the value of the tractor when it was new.
(b)Find the total amount of value the tractor has lost.
Show answer
(a) \(£7000\)
(b) \(£5250\)Losing \(50\%\) leaves \(0.5\) of the value, so each year the value is multiplied by \(0.5\).
Over \(2\) years that is \(0.5^{ 2 } = 0.25\).
\(\text{value when new} \times 0.25 = 1750\)
Dividing undoes the multiplying.
\(1750 \div 0.25 = 7000\)
\(£7000 - £1750 = £5250\)
- (a) M1multiplier \(0.5\) used for \(2\) years, \(0.25\)
- (a) A1dividing \(1750\) by \(0.25\) to get £\(7000\)