CFA Level I quantitative methods and fixed income practice: cram pack

Formula sheets, the key method steps for every topic and 20 practice problems, with full solutions at the back.

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1Formulas and methods

Time value of money

Future value
FV = PV·(1 + i)n
Present value
PV = FV/(1 + i)n
Several periods per year
i = r/m, N = m·n
Annuity, present value
PV = PMT·[1 − (1 + i)−n]/i
Annuity, future value
FV = PMT·[(1 + i)n − 1]/i

Method

  1. Draw a timeline with every cash flow and its timing.
  2. Match the rate to the period: a monthly rate for monthly payments.
  3. Use FV = PV·(1 + i)ⁿ for single amounts and the annuity formulas for equal payments.
  4. Check the sign and the order of magnitude of the answer.

NPV & IRR

Net present value
NPV = −C0 + Σ CFt/(1 + r)t
Equal cash flows
NPV = −C0 + CF·[1 − (1 + r)−n]/r
Internal rate of return
the r at which NPV = 0
Decision
accept when NPV > 0 (or IRR > required return)

Method

  1. List all cash flows with their year; the investment at year 0 is negative.
  2. Discount each cash flow: CFt/(1 + r)t.
  3. Add them up to get the NPV. Positive means the project earns more than r.
  4. For the IRR, find the rate that makes NPV zero by trial or interpolation.

Bond pricing & yield

Bond price
P = C·(1 − (1 + y)−n)/y + F/(1 + y)n
Semi-annual
C/2, y/2 and 2n periods
Current yield
C/P
Price and yield
yield up → price down

Method

  1. Find the coupon per period and the yield per period.
  2. Price = present value of the coupons (an annuity) + present value of the face value.
  3. For semi-annual coupons use half the coupon, half the yield and twice the periods.
  4. Check: if the yield is above the coupon rate, the price is below face value.

CAPM & WACC

CAPM
re = rf + β·(rm − rf)
WACC
E/V·re + D/V·rd·(1 − T)
Firm value
V = E + D

Descriptive statistics

Mean
x̄ = Σx/n
Sample standard deviation
s = √(Σ(x − x̄)²/(n − 1))
Median
middle value of the sorted data
z-score
z = (x − μ)/σ
Normal distribution
68% within ±1σ, 95% within ±2σ

Probability

Complement
P(not A) = 1 − P(A)
Addition rule
P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
Independent events
P(A ∩ B) = P(A)·P(B)
Conditional probability
P(A|B) = P(A ∩ B)/P(B)
Binomial
P(X = k) = C(n,k)·pk(1 − p)n−k

Normal distribution

z-score
z = (x − μ)/σ
Below x
P(X < x) = Φ(z)
Above x
P(X > x) = 1 − Φ(z)
Between
Φ(z2) − Φ(z1)
Percentile
x = μ + z·σ

Method

  1. Convert the value to a z-score: z = (x − μ)/σ.
  2. Look up Φ(z), the area to the left of z.
  3. For the area to the right use 1 − Φ(z); for an interval subtract the two left areas.
  4. For a percentile, work backwards: find z for the area, then x = μ + z·σ.

Confidence intervals

σ known
x̄ ± z·σ/√n
σ unknown
x̄ ± t·s/√n, df = n − 1
Critical z
90%: 1.645, 95%: 1.96, 99%: 2.576
Sample size
n = (z·σ/E)², rounded up

Method

  1. Decide between z (population σ known) and t (only the sample s, with n − 1 degrees of freedom).
  2. Find the critical value for the confidence level.
  3. Margin of error: E = critical value × standard deviation/√n.
  4. Interval: x̄ ± E.

Hypothesis tests & confidence intervals

z-test for a mean
z = (x̄ − μ0)/(σ/√n)
t-test for a mean
t = (x̄ − μ0)/(s/√n), df = n − 1
Two-sided at 5%
reject H0 if |z| > 1.96
Confidence interval
x̄ ± z·σ/√n

Method

  1. State H₀ and H₁ and decide whether the test is one- or two-sided.
  2. Calculate the test statistic: (x̄ − μ₀)/(σ/√n), or with s and t.
  3. Compare with the critical value, or find the p-value.
  4. Reject H₀ if the statistic is beyond the critical value (p below α), and state the conclusion in words.

Linear regression & correlation

Sums of squares
Sxx = Σ(x − x̄)² = Σx² − (Σx)²/n, Sxy = Σxy − ΣxΣy/n
Slope
b = Sxy/Sxx
Intercept
a = ȳ − b·x̄
Correlation
r = Sxy/√(SxxSyy)
Residual
e = y − ŷ

2Practice problems

1.Time value of moneyMedium
years0123…10PV = ?€500€500€500€500

An annuity pays €500 at the end of each year for 10 years. The interest rate is 5% per year. Find the present value PV of the annuity.

A = €500i = 5%n = 10
PV = €
2.Time value of moneyHard
months0123…240€200 000A?AAA

A loan of €200 000 is repaid with equal monthly payments over 20 years. The interest rate is 2.4% per year, compounded monthly. Find the monthly payment A.

P = €200 0002.4% / 12N = 240
A = €
3.NPV & IRRMedium
years0123456−50 00013 00013 00013 00013 00013 00013 000

A project costs €50 000 today and returns €13 000 at the end of each year for 6 years. The cost of capital is 12%. Assume cash flows arrive evenly through each year. Find the discounted payback period DPB in years.

r = 12%CF0 = −50 000CF1 = 13 000CF2 = 13 000CF3 = 13 000CF4 = 13 000CF5 = 13 000CF6 = 13 000
DPB = years
4.NPV & IRRHard
years0123−20 00013 00012 00011 500

A project costs €20 000 and returns €13 000, €12 000, €11 500 at the end of years 1 to 3. Find its internal rate of return IRR in percent.

CF0 = −20 000CF1 = 13 000CF2 = 12 000CF3 = 11 500
IRR = %
5.Bond pricing & yieldMedium

You buy a bond with 10 years to maturity (face value €1 000, annual coupon 4.5%) when its yield to maturity is 5.5%. One year later, just after receiving the first coupon, you sell it when the yield to maturity is 6%. Find your holding period return HPR in percent.

F = €1 000coupon = 4.5%n = 10y0 = 5.5%y1 = 6%
HPR = %
6.Bond pricing & yieldHard
years01234567€1 060€50€50 + €1 000

A bond with a face value of €1 000 and an annual coupon of 5% matures in 7 years and sells for €1 060. Find its yield to maturity YTM in percent.

F = €1 000coupon = 5%n = 7P = €1 060
YTM = %
7.CAPM & WACCMedium
WACC = ?equity€8 000 000re​ = 12%debt€2 000 000rd​ = 7%T = 20%

A company has equity worth €8 000 000 and debt worth €2 000 000. Its cost of equity is 12%, its pre-tax cost of debt is 7% and the tax rate is 20%. Find the WACC.

E = €8 000 000D = €2 000 000re = 12%rd = 7%T = 20%
WACC = %
8.CAPM & WACCHard
WACC = ?equity2 500 000 × €12.50β = 1.8rf​ = 4%, rm​ = 10%debt40 000 × €1 020rd​ = 7%T = 30%

A company has 2 500 000 shares at €12.50 each and 40 000 bonds trading at €1 020 each. The pre-tax cost of debt is 7% and the tax rate is 30%. The share’s beta is 1.8, the risk-free rate 4% and the market return 10%. Find the WACC.

β = 1.8rf = 4%rm = 10%rd = 7%T = 30%
WACC = %
9.Descriptive statisticsMedium
1520253035404550

Find the sample standard deviation s of the data: 28, 11, 33, 12, 52.

n = 5
s =
10.Descriptive statisticsHard
z −2−1012

Exam scores are normally distributed with mean 500 and standard deviation 60. Find the z-score z of a score of 428.

μ = 500σ = 60x = 428
z =
11.ProbabilityMedium
red: 5blue: 6

A bag holds 5 red and 6 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?

5 / 6
P =
12.ProbabilityHard
n = 6, p = 0.4: P(X = 3) = ?k0123456

A test has 6 multiple-choice questions. A student who guesses gets each one right with probability 0.4, independently. What is the probability P of exactly 3 correct answers?

n = 6p = 0.4k = 3
P =
13.Normal distributionMedium
−3−2−10123zshaded area between −z and z = 0.8990

Drag z so that the shaded area between −z and z under the standard normal curve is 0.8990. Find z to two decimals.

A = 0.8990
z =
14.Normal distributionHard
157164171178185192199cmbottom 90%

The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Drag the cut-off so the shaded bottom 90% is to the left of it. Find the cut-off x.

μ = 178 cmσ = 7 cm90%
x = cm
15.Confidence intervalsMedium
μ₁ − μ₂0LU = ?x̄₁ − x̄₂E90%

The battery life of two phone models is compared. Model 1: 100 phones, mean 71 h, known standard deviation 20 h. Model 2: 50 phones, mean 62.5 h, known standard deviation 6 h. Find the upper limit of the 90% confidence interval for μ₁ − μ₂.

n₁ = 100x̄₁ = 71 hσ₁ = 20 hn₂ = 50x̄₂ = 62.5 hσ₂ = 6 h90%
upper limit = h
16.Confidence intervalsHard
p ≈ 0.8, n = ?p̂ − Ep̂ + Ep̂E = 0.0590%

A pilot study suggests that about 0.8 of products are defective. How many products n must be inspected so that a 90% confidence interval for the proportion has a margin of error of at most 0.05?

p ≈ 0.8E = 0.0590%
n =
17.Hypothesis tests & confidence intervalsMedium
t −2.01102.011

A sample of 49 students has a mean score of 116.3 and a sample standard deviation of 10. Find the test statistic t for H₀: μ = 120.

μ₀ = 120x̄ = 116.3s = 10n = 49
t =
18.Hypothesis tests & confidence intervalsHard
σ = 12, n = ?x̄ − Ex̄ + Ex̄E = 290%

How large must a sample be so that the 90% confidence interval for a mean has a margin of error of at most 2? The standard deviation is 12. Find the smallest sample size n.

σ = 12E = 290%
n =
19.Linear regression & correlationMedium
xy17561973218623882510127103xy20256080100

The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 6Σx = 132Σy = 507Σx² = 2974Σy² = 44415Σxy = 11475
r =
20.Linear regression & correlationHard
xy167719902298251192813831153xy15202530100150x = 26.5

The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Use the least-squares line ŷ = a + b·x. A new observation (26.5, 130) is made. Find its residual e = y − ŷ.

n = 6(26.5, 130)
e =

3Solutions

1.Time value of money

Answer: PV = 3861 €

Time value of money: money today is worth more than the same amount later, because it can earn interest. With interest rate i per period, an amount grows as FV = PV·(1 + i)n. A series of equal payments (an annuity) is valued by adding up each payment, which gives the annuity formulas.

Annuity factor

(1 − (1 + i)−n)/i = (1 − 0.6139)/0.05 = 7.722

Present value

PV = 500·7.722 = €3 860.87

Answer

PV = €3 860.87
2.Time value of money

Answer: A = 1050 €

Time value of money: money today is worth more than the same amount later, because it can earn interest. With interest rate i per period, an amount grows as FV = PV·(1 + i)n. A series of equal payments (an annuity) is valued by adding up each payment, which gives the annuity formulas.

Monthly rate and number of payments

i = 0.024/12 = 0.002
N = 12·20 = 240

Payment

The present value of N equal payments A is A·(1 − (1 + i)⁻ᴺ)/i. Set it equal to the loan and solve for A.

(1 + i)−N = 0.6191
A = 200 000·0.002/(1 − 0.6191) = €1 050.09

Answer

A = €1 050.09
3.NPV & IRR

Answer: DPB = 5.476 years

Payback period: the time until the cumulative cash flows repay the initial investment. Cash is assumed to come in evenly during each year, so the last part is a fraction of a year. The simple payback ignores the time value of money; the discounted payback period uses the discounted cash flows instead.

Discounted and cumulative cash flows

year 1: 13 000/1.12 = 11 607.14 → Σ = 11 607.14
year 2: 13 000/1.254 = 10 363.52 → Σ = 21 970.66
year 3: 13 000/1.405 = 9 253.14 → Σ = 31 223.81
year 4: 13 000/1.574 = 8 261.74 → Σ = 39 485.54
year 5: 13 000/1.762 = 7 376.55 → Σ = 46 862.09
year 6: 13 000/1.974 = 6 586.20 → Σ = 53 448.30

Fraction of the last year

DPB = 5 + 3 137.91/6 586.20 = 5.476

Answer

DPB = 5.476 years

The discounted payback is always longer than the simple payback (3.846 years).

4.NPV & IRR

Answer: IRR = 38.38 %

Net present value: discount every cash flow back to today with the required rate of return r, then add them up: NPV = Σ CFₜ/(1 + r)ᵗ. A project with NPV > 0 earns more than r and adds value. The internal rate of return (IRR) is the rate at which NPV = 0.

Bracket the IRR

NPV falls as the rate rises, so the IRR lies between a rate with positive NPV and one with negative NPV.

NPV(35%) = 888.08
NPV(40%) = −400.87

Narrow down

Linear interpolation gives a first estimate; repeating it (or a calculator's IRR function) gives the exact rate.

IRR ≈ 35 + 5·888.08/(888.08 − (−400.87)) = 38.44%
exact: NPV(38.38%) = 0 → IRR = 38.38%

Answer

IRR = 38.38%
5.Bond pricing & yield

Answer: HPR = 1.985 %

Bond pricing: a bond’s price is the present value of its cash flows, discounted at the yield to maturity (YTM): the coupons form an annuity and the face value is paid back at maturity. P = C·(1 − (1 + y)⁻ⁿ)/y + F/(1 + y)ⁿ. When the yield is above the coupon rate the bond trades below face value, and the other way round.

Purchase price

C = €45.00, i = 0.055, N = 10
€45.00·(1 − 0.5854)/0.055 + €1 000·0.5854 = €339.19 + €585.43 = €924.62

Selling price

C = €45.00, i = 0.06, N = 9
€45.00·(1 − 0.5919)/0.06 + €1 000·0.5919 = €306.08 + €591.90 = €897.97

Holding period return

HPR = (45 + 897.97 − 924.62)/924.62 = 1.985%

Answer

HPR = 1.985%
6.Bond pricing & yield

Answer: YTM = 4 %

Bond pricing: a bond’s price is the present value of its cash flows, discounted at the yield to maturity (YTM): the coupons form an annuity and the face value is paid back at maturity. P = C·(1 − (1 + y)⁻ⁿ)/y + F/(1 + y)ⁿ. When the yield is above the coupon rate the bond trades below face value, and the other way round.

Equation for the yield

€1 060 = €50·(1 − (1 + y)⁻⁷)/y + €1 000/(1 + y)⁷

Solve numerically

P(4%) = €1 060.02, P(5%) = €1 000.00
YTM = 4%

Answer

YTM = 4%
7.CAPM & WACC

Answer: WACC = 10.72 %

Cost of capital: CAPM gives the return shareholders require: re = rf + β·(rm − rf). The weighted average cost of capital mixes equity and debt by their market values: WACC = E/V·re + D/V·rd·(1 − T), with V = E + D. Interest is tax-deductible, hence the (1 − T).

Weights

V = €10 000 000, E/V = 0.8, D/V = 0.2

WACC

WACC = 0.8·12% + 0.2·7%·(1 − 0.2) = 10.72%

Answer

WACC = 10.72%
8.CAPM & WACC

Answer: WACC = 9.194 %

Cost of capital: CAPM gives the return shareholders require: re = rf + β·(rm − rf). The weighted average cost of capital mixes equity and debt by their market values: WACC = E/V·re + D/V·rd·(1 − T), with V = E + D. Interest is tax-deductible, hence the (1 − T).

Market values

E = 2 500 000·€12.50 = €31 250 000
D = 40 000·€1 020 = €40 800 000

Cost of equity (CAPM)

re = 4% + 1.8·6% = 14.8%

Weights

V = €72 050 000, E/V = 0.4337, D/V = 0.5663

WACC

WACC = 0.4337·14.8% + 0.5663·7%·(1 − 0.3) = 9.194%

Answer

WACC = 9.194%
9.Descriptive statistics

Answer: s = 16.9

Descriptive statistics: the mean is the sum divided by n; the median is the middle value of the sorted data (the average of the two middle values when n is even). The sample standard deviation is s = √(Σ(x − x̄)²/(n − 1)). A z-score z = (x − μ)/σ says how many standard deviations a value lies from the mean.

Mean

x̄ = 136/5 = 27.2

Squared deviations

Σ(x − x̄)² = 0.8² + (−16.2)² + 5.8² + (−15.2)² + 24.8² = 1143

Standard deviation

s = √(1143/4) = 16.9

Answer

s = 16.9
10.Descriptive statistics

Answer: z = −1.2

Descriptive statistics: the mean is the sum divided by n; the median is the middle value of the sorted data (the average of the two middle values when n is even). The sample standard deviation is s = √(Σ(x − x̄)²/(n − 1)). A z-score z = (x − μ)/σ says how many standard deviations a value lies from the mean.

Standardise

z = (428 − 500)/60 = −1.2

Answer

z = −1.2
11.Probability

Answer: P = 0.1818

Probability: with equally likely outcomes, P(A) = favourable/total. For independent events multiply: P(A and B) = P(A)·P(B). The complement often helps: P(at least one) = 1 − P(none). Binomial: in n independent trials with success probability p, P(X = k) = C(n, k)·pᵏ·(1 − p)ⁿ⁻ᵏ.

First and second draw

P = 5/11 · 4/10 = 20/110 = 0.1818

Answer

P = 0.1818
12.Probability

Answer: P = 0.2765

Probability: with equally likely outcomes, P(A) = favourable/total. For independent events multiply: P(A and B) = P(A)·P(B). The complement often helps: P(at least one) = 1 − P(none). Binomial: in n independent trials with success probability p, P(X = k) = C(n, k)·pᵏ·(1 − p)ⁿ⁻ᵏ.

Binomial formula

P(X = 3) = 20·0.43·0.63 = 20·0.064·0.216 = 0.2765

Answer

P = 0.2765
13.Normal distribution

Answer: z = 1.64

From an area back to z: Φ(z) is the area to the left of z. For a left-tail area A, find z with Φ(z) = A in the table. A right-tail area A means Φ(z) = 1 − A. A central area A between −z and z leaves (1 − A)/2 in each tail, so Φ(z) = (1 + A)/2. For a value x, use x = μ + zσ.

Area to the left of z

Φ(z) = (1 + 0.8990)/2 = 0.9495

Read the z-table backwards

Find 0.9495 inside the table and read off z.

z = 1.64

Answer

z = 1.64
14.Normal distribution

Answer: x = 187 cm

From an area back to z: Φ(z) is the area to the left of z. For a left-tail area A, find z with Φ(z) = A in the table. A right-tail area A means Φ(z) = 1 − A. A central area A between −z and z leaves (1 − A)/2 in each tail, so Φ(z) = (1 + A)/2. For a value x, use x = μ + zσ.

Area to the left

Φ(z) = 0.9

z from the table

z ≈ 1.28

Back to the original units

x = 178 + 1.28·7 = 187 cm

Answer

x = 187 cm
15.Confidence intervals

Answer: upper limit = 12.07 h

Interval for a difference: estimate ± critical value × standard error. For two independent means with known σ: SE = √(σ₁²/n₁ + σ₂²/n₂) and z. With equal unknown variances: pooled s_p and t with n₁ + n₂ − 2 degrees of freedom. For paired data, work with the differences: d̄ ± t·s_d/√n.

Point estimate

x̄₁ − x̄₂ = 71 − 62.5 = 8.5

Standard error

SE = √(20²/100 + 6²/50) = 2.173

Margin of error

E = 1.645·2.173 = 3.574

Interval

8.5 ± 3.574 → [4.926, 12.07]

Answer

upper limit = 12.07 h

The interval does not contain 0, so the means differ significantly.

16.Confidence intervals

Answer: n = 174

Confidence interval for a proportion: p̂ ± z·√(p̂(1 − p̂)/n), where p̂ = x/n. To reach a margin E you need n ≥ z²·p(1 − p)/E², rounded up; without a prior estimate use p = 0.5, which gives the largest n.

Solve for n

n = 1.645²·0.8·0.2/0.05² = 173.2

Round up

n = 174

Answer

n = 174
17.Hypothesis tests & confidence intervals

Answer: t = −2.59

Testing a mean: the test statistic measures how many standard errors the sample mean lies from the value claimed in H₀: z = (x̄ − μ₀)/(σ/√n) when σ is known, t = (x̄ − μ₀)/(s/√n) when it is estimated from the sample. A confidence interval is x̄ ± (critical value)·(standard error).

Standard error

SE = 10/√49 = 10/7 = 1.429

Test statistic

t = (116.3 − 120)/1.429 = −2.59

Answer

t = −2.59
t0.025; df = 48 = 2.011

|t| > 2.011, so H₀ is rejected at the 5% level (two-sided).

18.Hypothesis tests & confidence intervals

Answer: n = 98

Testing a mean: the test statistic measures how many standard errors the sample mean lies from the value claimed in H₀: z = (x̄ − μ₀)/(σ/√n) when σ is known, t = (x̄ − μ₀)/(s/√n) when it is estimated from the sample. A confidence interval is x̄ ± (critical value)·(standard error).

Solve for n

n = (1.645·12/2)² = 97.42

Round up

n = 98

Answer

n = 98
19.Linear regression & correlation

Answer: r = 0.9672

Least-squares regression: with x̄ and ȳ the means, Sxx = Σ(x − x̄)², Syy = Σ(y − ȳ)² and Sxy = Σ(x − x̄)(y − ȳ), the line ŷ = a + b·x has slope b = Sxy/Sxx and intercept a = ȳ − b·x̄, so it always passes through (x̄, ȳ). Pearson's correlation coefficient r = Sxy/√(Sxx·Syy) lies between −1 and 1; r² is the fraction of the variation in y explained by the line.

Centred sums

Sxy = 11475 − 132·507/6 = 321
Sxx = 2974 − 132²/6 = 70
Syy = 44415 − 507²/6 = 1574

Correlation coefficient

r = 321/√(70·1574) = 0.9672

Answer

r = 0.9672
20.Linear regression & correlation

Answer: e = 1.929

Least-squares regression: with x̄ and ȳ the means, Sxx = Σ(x − x̄)², Syy = Σ(y − ȳ)² and Sxy = Σ(x − x̄)(y − ȳ), the line ŷ = a + b·x has slope b = Sxy/Sxx and intercept a = ȳ − b·x̄, so it always passes through (x̄, ȳ). Pearson's correlation coefficient r = Sxy/√(Sxx·Syy) lies between −1 and 1; r² is the fraction of the variation in y explained by the line.

Means

x̄ = 141/6 = 23.5
ȳ = 675/6 = 112.5

Sums of squares and products

Sxx = Σ(x − x̄)² = 157.5
Sxy = Σ(x − x̄)(y − ȳ) = 817.5

Regression line

b = 817.5/157.5 = 5.19
a = 112.5 − 5.19·23.5 = −9.476
ŷ = −9.476 + 5.19·x

Predicted value

ŷ = −9.476 + 5.19·26.5 = 128.1

Residual

e = 130 − 128.1 = 1.929

Answer

e = 1.929

The point lies above the line.

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