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IEB practice · Paper I

IEB Finance, growth and decay practice

17 newly generated questions across all 12 finance, growth and decay topics the IEB examines in Paper I, 172 marks in all. Work each one, then open its answer underneath to check yourself. Nothing here is taken from a past paper.

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Borrowing: the present-value annuity

3 questions · 42 marks

Paper I · Section A · 13 marks

QUESTION 1

A farmer borrows R1 500 000 to drill a borehole and build a dam. It is repaid in equal monthly instalments at the end of each month over 25 years, at 12% per annum compounded monthly. Give answers correct to two decimal places.

  1. (a)
    Calculate the monthly instalment.
    (3)
  2. (b)
    How much of the first instalment is interest, and how much of it reduces the loan?
    (3)
  3. (c)
    Calculate the balance outstanding immediately after instalment number 60.
    (3)
  4. (d)
    In which instalment does more than half of the instalment go towards reducing the loan for the first time?
    (4)

[13]

Show the answers

(a) Answer: R15 798.36

  1. i = 12%/12 and n = 300
  2. 1 500 000 = x[1 − (1 + 0.010000)^(−300)]/0.010000
  3. x = R15 798.36

(b) Answer: interest R15 000.00; reduces the loan by R798.36

  1. interest for the first month: R1 500 000 × 0.010000 = R15 000.00
  2. the rest of the instalment reduces the loan: R15 798.36 − R15 000.00 = R798.36

(c) Answer: R1 434 798.20

  1. balance = 1 500 000(1 + i)^60 − 15 798.36[(1 + i)^60 − 1]/i
  2. = R1 434 798.20

(d) Answer: instalment number 232

  1. in instalment n the interest is i × B_(n − 1), the balance before it; more than half reduces the loan when i × B_(n − 1) < x/2, so B_(n − 1) < R789 918.00
  2. 1 500 000(1 + i)^(n − 1) − 15 798.36[(1 + i)^(n − 1) − 1]/i < 789 918.00
  3. (1 + i)^(n − 1) > 9.894258
  4. n − 1 > 230.34, so instalment number 232

Paper I · Section A · 15 marks

QUESTION 2

A home loan of R650 000 is repaid in equal monthly payments over 15 years. The interest rate is 8.75% per annum compounded monthly, and the first payment is made one month after the loan is granted. Give answers correct to two decimal places.

  1. (a)
    Calculate the monthly repayment.
    (4)
  2. (b)
    Calculate the balance outstanding at the end of year 8, immediately after the payment at that time.
    (4)
  3. (c)
    Determine the effective annual interest rate of the loan.
    (3)
  4. (d)
    If you pay R9 800 per month instead, how many payments will it take to pay off the loan? (The last payment may be smaller.)
    (4)

[15]

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(a) Answer: R6 496.42

  1. i = 8.75%/12 and n = 180
  2. 650 000 = x[1 − (1 + 0.007292)^(−180)]/0.007292
  3. x = R6 496.42

(b) Answer: R406 979.55

  1. after 96 payments
  2. balance = 650 000(1 + i)^96 − 6496.42[(1 + i)^96 − 1]/i
  3. = R406 979.55

(c) Answer: 9.11%

  1. 1 + i_eff = (1 + 0.0875/12)^12
  2. i_eff = 9.11%

(d) Answer: 91 payments

  1. 650 000 = 9800[1 − (1 + i)^(−n)]/i
  2. (1 + i)^(−n) = 0.516369
  3. n = 90.97

Paper I · Section A · 14 marks

QUESTION 3

A courier buys a delivery van for R800 000. The bank finances the full price over 5 years at 10.5% per annum compounded monthly, with a balloon payment of 40% of the price due together with the last instalment. The monthly instalments start one month after the loan is granted. Give answers correct to two decimal places.

  1. (a)
    Calculate the present value of the balloon payment.
    (2)
  2. (b)
    Hence show that the monthly instalment is R13 117.07.
    (4)
  3. (c)
    Calculate the balance outstanding after 2 years, immediately after instalment number 24.
    (4)
  4. (d)
    The vehicle depreciates at 12% per annum on the reducing-balance method. Calculate its value at the end of the 5 years.
    (2)
  5. (e)
    If the vehicle is sold for that value at the end, will it pay the balloon payment? Show the difference.
    (2)

[14]

Show the answers

(a) Answer: R189 730.48

  1. the balloon is 40% of R800 000 = R320 000, paid after 60 months
  2. PV = 320 000(1 + 0.008750)^(−60) = R189 730.48

(b) Answer: R13 117.07

  1. the instalments repay the rest: R800 000 − R189 730.48 = R610 269.52
  2. 610 269.52 = x[1 − (1 + 0.008750)^(−60)]/0.008750
  3. x = R13 117.07

(c) Answer: R637 424.57

  1. balance = 800 000(1 + i)^24 − 13 117.07[(1 + i)^24 − 1]/i
  2. = R637 424.57 (the balloon is still owed)

(d) Answer: R422 185.53

  1. A = 800 000(1 − 0.12)^5
  2. = R422 185.53

(e) Answer: Yes: it is worth R102 185.53 more than the balloon

  1. R422 185.53 ≥ R320 000
  2. difference R102 185.53

Saving: the future-value annuity

2 questions · 25 marks

Paper I · Section A · 13 marks

QUESTION 4

Give answers correct to two decimal places.

  1. (a)
    You deposit R2 250 at the end of every month for 6 years into an account earning 9% per annum compounded monthly. You then stop the deposits, and for the next 3 years the account earns 6.5% per annum compounded quarterly. Calculate the amount in the account at the end of these 9 years.
    (5)
  2. (b)
    From then on the account earns 5.5% per annum compounded monthly, and you withdraw R5 600 at the end of every month. How many full withdrawals of R5 600 can you make?
    (4)
  3. (c)
    A savings account earns 10% per annum compounded half-yearly. The first deposit of R7 000 is made six months after the account is opened, and the same amount is deposited at the end of every six months after that. How much is in the account 3 years after it was opened, just after the deposit made then?
    (4)

[13]

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(a) Answer: R259 385.06

  1. after 72 deposits: 2250[(1 + 9/1200)^72 − 1]/(9/1200) = R213 765.81
  2. then 12 quarters: R213 765.81(1 + 6.5/400)^12 = R259 385.06

(b) Answer: 52 withdrawals

  1. 259 385.06 = 5600[1 − (1 + 0.004583)^(−n)]/0.004583
  2. (1 + 0.004583)^(−n) = 0.787706
  3. n = 52.18
  4. 52 full withdrawals

(c) Answer: R47 613.39

  1. deposits at 6, 12, …, 36 months: 6 of them
  2. FV = 7000[(1 + 0.05)^6 − 1]/0.05
  3. = R47 613.39

Paper I · Section A · 12 marks

QUESTION 5

You want to have R460 000 in 8 years' time. Your savings account earns 10% per annum compounded monthly. Give answers correct to two decimal places.

  1. (a)
    Calculate the equal monthly deposit that reaches the goal, if the first deposit is made one month from now and the last one at the end of the 8 years.
    (4)
  2. (b)
    Instead, you deposit a single amount of R230 000 now and nothing more. After how many months will the account first hold at least R460 000?
    (4)
  3. (c)
    Instead, you deposit R230 000 now and R1 700 at the end of every month. What will the account hold at the end of the 8 years?
    (4)

[12]

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(a) Answer: R3 146.78

  1. i = 10%/12 and n = 96
  2. 460 000 = x[(1 + i)^96 − 1]/i
  3. x = R3 146.78

(b) Answer: 84 months

  1. 230 000(1 + i)^n ≥ 460 000
  2. n ≥ log(460 000/230 000)/log(1 + i) = 83.52

(c) Answer: R758 688.22

  1. 230 000(1 + i)^96 + 1700[(1 + i)^96 − 1]/i
  2. = R758 688.22

Compound growth

3 questions · 35 marks

Paper I · Section A · 11 marks

QUESTION 6

A bank offers 9% per annum interest, compounded monthly.

  1. (a)
    Calculate how long, in years correct to two decimal places, it takes for an amount invested at this rate to double.
    (3)
  2. (b)
    Use your answer to the previous part to find how long the amount takes to grow to eight times its size. Explain.
    (2)
  3. (c)
    At what annual interest rate, compounded weekly, would an amount treble in 8 years? Give the rate correct to two decimal places.
    (3)
  4. (d)
    A car bought for R410 000 depreciates at 18% per annum on the reducing-balance method. After how many complete years is it first worth less than half of what was paid for it?
    (3)

[11]

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(a) Answer: 7.73 years

  1. 2P = P(1 + 0.09/12)^(12n)
  2. 12n = log 2/log 1.0075
  3. n = 7.7305 ≈ 7.73 years

(b) Answer: 23.19 years: it doubles three times

  1. 8 = 2 × 2 × 2: three doublings, one after the other
  2. 3 × 7.7305 = 23.19 years

(c) Answer: 13.75% per annum

  1. 3 = (1 + i/52)^(416)
  2. 1 + i/52 = 3^(1/416)
  3. i = 0.137508 = 13.75%

(d) Answer: after 4 years

  1. (1 − 0.18)^n < 0.5
  2. n > log 0.5/log 0.82 = 3.49
  3. n = 4

Paper I · Section A · 11 marks

QUESTION 7

A plot of farmland was bought for R2 800 000 at the start of 2017. Its value has grown at 5.5% per annum, compounded annually. Give answers correct to two decimal places.

  1. (a)
    Calculate the value of the land at the start of 2029.
    (2)
  2. (b)
    In which year will the value of the land first be more than R5 600 000? (Its value changes at the start of each year.)
    (4)
  3. (c)
    A tractor bought for R220 000 at the same time depreciates at 16% per annum on the straight-line method. What is it worth 3 years later?
    (2)
  4. (d)
    Had the tractor depreciated at 16% per annum on the reducing-balance method instead, how much more would it be worth after 3 years?
    (3)

[11]

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(a) Answer: R5 323 380.96

  1. A = 2 800 000(1 + 0.055)^12
  2. = R5 323 380.96

(b) Answer: 2030

  1. 2 800 000(1 + 0.055)^n > 5 600 000
  2. (1 + 0.055)^n > 2
  3. n > log 2/log 1.055 = 12.95
  4. n = 13, so 2017 + 13 = 2030

(c) Answer: R114 400.00

  1. A = 220 000(1 − 3 × 0.16)
  2. = R114 400.00

(d) Answer: R15 994.88

  1. A = 220 000(1 − 0.16)^3 = R130 394.88
  2. R130 394.88 − R114 400.00 = R15 994.88

Paper I · Section A · 13 marks

QUESTION 8

An amount of R45 000 is invested at 7% per annum compounded quarterly. After 4 years the interest rate changes to 9.5% per annum compounded monthly, and 6 years after the start of the investment R7 000 is withdrawn. Give answers correct to two decimal places.

  1. (a)
    Calculate the value of the investment after 4 years.
    (2)
  2. (b)
    Calculate the value of the investment 10 years after it was started.
    (4)
  3. (c)
    Suppose instead that nothing is withdrawn. At the end of which year does the investment first reach R161 000?
    (4)
  4. (d)
    Without the withdrawal, what interest rate per annum, compounded annually, would have grown R45 000 to the same amount in the same 10 years?
    (3)

[13]

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(a) Answer: R59 396.82

  1. 45 000(1 + 7/400)^16
  2. = R59 396.82

(b) Answer: R94 573.29

  1. at year 6: R59 396.82(1 + 9.5/1200)^24 = R71 771.87
  2. after the withdrawal: R71 771.87 − R7 000 = R64 771.87
  3. at year 10: R64 771.87(1 + 9.5/1200)^48 = R94 573.29

(c) Answer: at the end of year 15

  1. R59 396.82(1 + 9.5/1200)^(12m) ≥ 161 000, m years after year 4
  2. 12m ≥ log(161 000/59 396.82)/log(1 + 9.5/1200), so m ≥ 10.54
  3. 4 + 10.54 = 14.54: at the end of year 15

(d) Answer: 8.82%

  1. without the withdrawal, after 10 years: R104 793.98
  2. 45 000(1 + i)^10 = 104 793.98
  3. i = (104 793.98/45 000)^(1/10) − 1 = 8.82%

Nominal and effective rates

2 questions · 29 marks

Paper I · Section A · 13 marks

QUESTION 9

You have R140 000 to invest for 10 years. Bank A offers 11% per annum compounded monthly and Bank B offers 11.15% per annum compounded half-yearly. Give answers correct to two decimal places.

  1. (a)
    Calculate the effective annual interest rate that each bank offers.
    (4)
  2. (b)
    Hence, which bank should you choose? Give a reason.
    (1)
  3. (c)
    How much more is the investment worth after 10 years at the better bank?
    (3)
  4. (d)
    A prize winner may take R480 000 now, or R7 000 at the end of every month for 6 years. Money earns 10.5% per annum compounded monthly. Use present values to decide which choice is worth more today, and by how much.
    (5)

[13]

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(a) Answer: A: 11.57%; B: 11.46%

  1. A: (1 + 11/1200)^12 − 1 = 11.57%
  2. B: (1 + 11.15/200)^2 − 1 = 11.46%

(b) Answer: Bank A: its effective rate is higher

  1. 11.57% > 11.46%

(c) Answer: R4 147.65

  1. Bank A: 140 000(1 + 11.5719%)^10; Bank B: 140 000(1 + 11.4608%)^10
  2. difference = R4 147.65

(d) Answer: The R480 000 now, by R107 242.60

  1. present value of the payments: 7000[1 − (1 + 10.5/1200)^(−72)]/(10.5/1200) = R372 757.40
  2. R372 757.40 < R480 000
  3. the lump sum is worth R107 242.60 more today

Paper I · Section A · 16 marks

QUESTION 10

You can afford to repay R7 000 per month on a home loan. Give answers correct to two decimal places.

  1. (a)
    The bank quotes an effective annual interest rate of 11.02%. Determine the equivalent nominal annual rate, compounded monthly, correct to two decimal places.
    (3)
  2. (b)
    Using this nominal rate, calculate the largest loan that repayments of R7 000 per month can pay off in 30 years. The first payment is made one month after the loan is granted.
    (4)
  3. (c)
    The house you want costs R890 000. How much must you pay as a deposit?
    (2)
  4. (d)
    At the end of year 19, immediately after the payment then, you pay a lump sum of R70 000 into the loan. Calculate the balance outstanding immediately after the lump sum.
    (4)
  5. (e)
    You keep paying R7 000 per month. How many more payments will pay off the loan? (The last payment may be smaller.)
    (3)

[16]

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(a) Answer: 10.50%

  1. 1 + 0.1102 = (1 + i/12)^12
  2. i = 12[(1 + 0.1102)^(1/12) − 1]
  3. i = 10.50%

(b) Answer: R765 245.36

  1. i = 10.50%/12 and n = 360
  2. PV = 7000[1 − (1 + i)^(−360)]/i
  3. = R765 245.36

(c) Answer: R124 754.64

  1. deposit = R890 000 − R765 245.36
  2. = R124 754.64

(d) Answer: R476 684.55

  1. after 228 payments: R765 245.36(1 + i)^228 − 7000[(1 + i)^228 − 1]/i = R546 684.55
  2. R546 684.55 − R70 000 = R476 684.55

(e) Answer: 104 payments

  1. 476 684.55 = 7000[1 − (1 + i)^(−n)]/i
  2. n = 103.99

Inflation

1 questions · 11 marks

Paper I · Section A · 11 marks

QUESTION 11

A bakery's ovens cost R190 000 today, and will be replaced in 4 years' time. Give answers correct to two decimal places.

  1. (a)
    The price of new ovens rises by 8% per annum, compounded annually. What will new ovens cost in 4 years' time?
    (2)
  2. (b)
    On the straight-line method, the present ovens will be worth R36 000 as scrap after 4 years. At what annual rate do they depreciate?
    (3)
  3. (c)
    The business already has R7 000 saved, earning 6.5% per annum compounded quarterly. What will this be worth in 4 years' time?
    (2)
  4. (d)
    The rest of the money for new ovens −− what is left after the scrap value and the savings −− is saved in a sinking fund at 9% per annum compounded monthly, with equal deposits at the end of every month for 4 years. Calculate the monthly deposit.
    (4)

[11]

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(a) Answer: R258 492.90

  1. A = 190 000(1 + 0.08)^4
  2. = R258 492.90

(b) Answer: 20.26% per annum

  1. 36 000 = 190 000(1 − 4i)
  2. i = (1 − 36 000/190 000)/4
  3. = 20.26%

(c) Answer: R9 059.56

  1. A = 7000(1 + 6.5/400)^16
  2. = R9 059.56

(d) Answer: R3 710.55

  1. needed: R258 492.90 − R36 000 − R9 059.56 = R213 433.34
  2. 213 433.34 = x[(1 + 0.007500)^48 − 1]/0.007500
  3. x = R3 710.55

When the payments change partway

3 questions · 15 marks

QUESTION 12

R2000 is deposited at the end of every month for 18 months, after which the deposit is increased to R2250 for a further 24 months. The account pays 16.50% per year compounded monthly.

Calculate the value of the account immediately after the last deposit.

[5]

Show the answer

Answer: 119 718.40

How it is reached:

  1. the first block of deposits valued
  2. that value carried forward through the second block
  3. the second block of deposits added
  4. evaluated and rounded to the nearest cent
QUESTION 13

R1000 is deposited at the end of every month for 18 months, after which the deposit is increased to R1500 for a further 12 months. The account pays 15.50% per year compounded monthly.

Calculate the value of the account immediately after the last deposit.

[5]

Show the answer

Answer: 42 804.52

How it is reached:

  1. the first block of deposits valued
  2. that value carried forward through the second block
  3. the second block of deposits added
  4. evaluated and rounded to the nearest cent
QUESTION 14

R2000 is deposited at the end of every month for 12 months, after which the deposit is increased to R2500 for a further 12 months. The account pays 8.00% per year compounded monthly.

Calculate the value of the account immediately after the last deposit.

[5]

Show the answer

Answer: 58 091.34

How it is reached:

  1. the first block of deposits valued
  2. that value carried forward through the second block
  3. the second block of deposits added
  4. evaluated and rounded to the nearest cent

When the payments start late

3 questions · 15 marks

QUESTION 15

R1500 is deposited at the end of every month into an account paying 16.50% per year compounded monthly. The first deposit is made 4 months after the account is opened, and 24 deposits are made in all.

Calculate the value of the account immediately after the last deposit.

[5]

Show the answer

Answer: 42 310.31

How it is reached:

  1. the empty months carry no deposit and no balance
  2. F = x[(1 + i)^n − 1] / i applied to the n deposits
  3. evaluated and rounded to the nearest cent
QUESTION 16

R750 is deposited at the end of every month into an account paying 17.50% per year compounded monthly. The first deposit is made 6 months after the account is opened, and 24 deposits are made in all.

Calculate the value of the account immediately after the last deposit.

[5]

Show the answer

Answer: 21 367.82

How it is reached:

  1. the empty months carry no deposit and no balance
  2. F = x[(1 + i)^n − 1] / i applied to the n deposits
  3. evaluated and rounded to the nearest cent
QUESTION 17

R1500 is deposited at the end of every month into an account paying 5.50% per year compounded monthly. The first deposit is made 4 months after the account is opened, and 48 deposits are made in all.

Calculate the value of the account immediately after the last deposit.

[5]

Show the answer

Answer: 80 329.28

How it is reached:

  1. the empty months carry no deposit and no balance
  2. F = x[(1 + i)^n − 1] / i applied to the n deposits
  3. evaluated and rounded to the nearest cent

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