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…9PV = ?€5 000+3%+3%…r = 5%

A property is expected to earn net rent of €5 000 next year. The rent is expected to grow by 3% per year forever. The required return is 5% per year. Find the value PV of the property today.

C1 = €5 000g = 3%r = 5%
PV = €
2.Time value of moneyHard
months0123…36€8 000A?AAA

A loan of €8 000 is repaid with equal monthly payments over 3 years. The interest rate is 4.2% per year, compounded monthly. Find the monthly payment A.

P = €8 0004.2% / 12N = 36
A = €
3.NPV & IRRMedium
years01234−20 0006 0008 0004 0009 000

An investment of €20 000 gives these cash flows at the end of years 1 to 4: €6 000, €8 000, €4 000, €9 000. The cost of capital is 9%. Find the net present value NPV.

r = 9%CF0 = −20 000CF1 = 6 000CF2 = 8 000CF3 = 4 000CF4 = 9 000
NPV = €
4.NPV & IRRHard
years0123−10 0004 5005 0003 500

A project costs €10 000 and returns €4 500, €5 000, €3 500 at the end of years 1 to 3. Find its internal rate of return IRR in percent.

CF0 = −10 000CF1 = 4 500CF2 = 5 000CF3 = 3 500
IRR = %
5.Bond pricing & yieldMedium

You buy a bond with 4 years to maturity (face value €1 000, annual coupon 8%) 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 6.5%. Find your holding period return HPR in percent.

F = €1 000coupon = 8%n = 4y0 = 5.5%y1 = 6.5%
HPR = %
6.Bond pricing & yieldHard
years0123…8€1 100€60€60 + €1 000

A bond with a face value of €1 000 and an annual coupon of 6% matures in 8 years and sells for €1 100. Find its yield to maturity YTM in percent.

F = €1 000coupon = 6%n = 8P = €1 100
YTM = %
7.CAPM & WACCMedium
WACC = ?equity€4 000 000re​ = 12%debt€4 000 000rd​ = 7%T = 28%

A company has equity worth €4 000 000 and debt worth €4 000 000. Its cost of equity is 12%, its pre-tax cost of debt is 7% and the tax rate is 28%. Find the WACC.

E = €4 000 000D = €4 000 000re = 12%rd = 7%T = 28%
WACC = %
8.CAPM & WACCHard
WACC = ?equity2 000 000 × €52.00β = 1.3rf​ = 1%, rm​ = 7.5%debt10 000 × €1 050rd​ = 3%T = 28%

A company has 2 000 000 shares at €52.00 each and 10 000 bonds trading at €1 050 each. The pre-tax cost of debt is 3% and the tax rate is 28%. The share’s beta is 1.3, the risk-free rate 1% and the market return 7.5%. Find the WACC.

β = 1.3rf = 1%rm = 7.5%rd = 3%T = 28%
WACC = %
9.Descriptive statisticsMedium
2025303540455055

Find the sample standard deviation s of the data: 19, 58, 18, 55, 54, 34.

n = 6
s =
10.Descriptive statisticsHard
z −2−1012

A quantity is normally distributed with mean 60 and standard deviation 12. What proportion P of values is below 48?

μ = 60σ = 12x = 48
P =
11.ProbabilityMedium
P(at least one six) = ??1?2?3?4?5?6?7?8P(six) = 1/6 on each roll

A fair die is rolled 8 times. What is the probability P of getting at least one six?

n = 8
P =
12.ProbabilityHard
n = 5, p = 0.2: P(X = 2) = ?k012345

A test has 5 multiple-choice questions. A student who guesses gets each one right with probability 0.2, independently. What is the probability P of exactly 2 correct answers?

n = 5p = 0.2k = 2
P =
13.Normal distributionMedium
−3−2−10123zshaded area to the right of z = 0.0901

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.0901. Find z to two decimals.

A = 0.0901
z =
14.Normal distributionHard
μ = 178x = ?99%

The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Find the value x below which 99% of them lie (the 99th percentile).

μ = 178σ = 7P = 99%
x = cm
15.Confidence intervalsMedium
n = 7, sd​ = 70L = ?Ud̄ = 6.4E90%

The blood pressure of 7 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean 6.4 and standard deviation 7. Find the lower limit of the 90% confidence interval for the mean difference.

n = 7d̄ = 6.4sd = 790%t(6) = 1.943
lower limit = mmHg
16.Confidence intervalsHard
σ = 36, n = ?x̄ − Ex̄ + Ex̄E = 290%

The population standard deviation is 36. How large a sample n do you need so that a 90% confidence interval for the mean has a margin of error of at most 2?

σ = 36E = 290%
n =
17.Hypothesis tests & confidence intervalsMedium
t −2.0302.03

A sample of 36 students has a mean score of 42.25 and a sample standard deviation of 15. Find the test statistic t for H₀: μ = 50.

μ₀ = 50x̄ = 42.25s = 15n = 36
t =
18.Hypothesis tests & confidence intervalsHard
μ₀x̄c

A one-sided z test of H₀: μ = 50 against H₁: μ > 50 is done at α = 0.10 with a sample of 25; the standard deviation is known to be 12.5. In reality the mean is 52.5. Find the power 1 − β of the test in percent.

μ₀ = 50μ = 52.5σ = 12.5n = 25α = 0.10
1 − β = %
19.Linear regression & correlationMedium
xy126224326420521xy2420222426

The table gives 5 paired observations of x = car age in years and y = price in k€. Find Pearson's correlation coefficient r.

n = 5
r =
20.Linear regression & correlationHard
xy372582793910511117xy51080100120ŷ = 88

The table gives 5 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 88?

n = 5ŷ = 88
x =

3Solutions

1.Time value of money

Answer: PV = 250000 €

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.

Growing perpetuity

PV = C1/(r − g) = 5 000/(0.05 − 0.03) = 5 000/0.02 = €250 000.00

Answer

PV = €250 000.00
2.Time value of money

Answer: A = 236.9 €

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 = 0.042/12 = 0.0035
N = 12·3 = 36

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 = 0.8818
A = 8 000·0.0035/(1 − 0.8818) = €236.90

Answer

A = €236.90
3.NPV & IRR

Answer: NPV = 1703 €

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.

Present value of each cash flow

CF0 = −20 000
CF1/(1 + 0.09)1 = 6 000/1.09 = 5 504.59
CF2/(1 + 0.09)2 = 8 000/1.188 = 6 733.44
CF3/(1 + 0.09)3 = 4 000/1.295 = 3 088.73
CF4/(1 + 0.09)4 = 9 000/1.412 = 6 375.83

Add them up

NPV = €1 702.59

Answer

NPV = €1 702.59

NPV is positive, so the project earns more than the required return.

4.NPV & IRR

Answer: IRR = 14.97 %

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(10%) = 852.74
NPV(15%) = −4.93

Narrow down

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

IRR ≈ 10 + 5·852.74/(852.74 − (−4.93)) = 14.97%
exact: NPV(14.97%) = 0 → IRR = 14.97%

Answer

IRR = 14.97%
5.Bond pricing & yield

Answer: HPR = 2.951 %

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 = €80.00, i = 0.055, N = 4
€80.00·(1 − 0.8072)/0.055 + €1 000·0.8072 = €280.41 + €807.22 = €1 087.63

Selling price

C = €80.00, i = 0.065, N = 3
€80.00·(1 − 0.8278)/0.065 + €1 000·0.8278 = €211.88 + €827.85 = €1 039.73

Holding period return

HPR = (80 + 1 039.73 − 1 087.63)/1 087.63 = 2.951%

Answer

HPR = 2.951%
6.Bond pricing & yield

Answer: YTM = 4.485 %

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

€1 100 = €60·(1 − (1 + y)⁻⁸)/y + €1 000/(1 + y)⁸

Solve numerically

P(4%) = €1 134.65, P(5%) = €1 064.63
YTM = 4.485%

Answer

YTM = 4.485%
7.CAPM & WACC

Answer: WACC = 8.52 %

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 = €8 000 000, E/V = 0.5, D/V = 0.5

WACC

WACC = 0.5·12% + 0.5·7%·(1 − 0.28) = 8.52%

Answer

WACC = 8.52%
8.CAPM & WACC

Answer: WACC = 8.781 %

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 = 2 000 000·€52.00 = €104 000 000
D = 10 000·€1 050 = €10 500 000

Cost of equity (CAPM)

re = 1% + 1.3·6.5% = 9.45%

Weights

V = €114 500 000, E/V = 0.9083, D/V = 0.0917

WACC

WACC = 0.9083·9.45% + 0.0917·3%·(1 − 0.28) = 8.781%

Answer

WACC = 8.781%
9.Descriptive statistics

Answer: s = 18.47

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 = 238/6 = 39.67

Squared deviations

Σ(x − x̄)² = (−20.67)² + 18.33² + (−21.67)² + 15.33² + 14.33² + (−5.667)² = 1705

Standard deviation

s = √(1705/5) = 18.47

Answer

s = 18.47
10.Descriptive statistics

Answer: P = 0.1587

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 = −1

Standard normal table

Φ(−1) = 0.1587
P = 0.1587

Answer

P = 0.1587
11.Probability

Answer: P = 0.7674

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.

Complement

P(no six) = (5/6)8 = 0.2326
P = 1 − 0.2326 = 0.7674

Answer

P = 0.7674
12.Probability

Answer: P = 0.2048

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 = 2) = 10·0.22·0.83 = 10·0.04·0.512 = 0.2048

Answer

P = 0.2048
13.Normal distribution

Answer: z = 1.34

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 − 0.0901 = 0.9099

Read the z-table backwards

Find 0.9099 inside the table and read off z.

z = 1.34

Answer

z = 1.34
14.Normal distribution

Answer: x = 194.3 cm

Normal distribution: convert a value x to a z-score, z = (x − μ)/σ, which counts standard deviations from the mean. A z-table (or Φ(z)) gives the area to the left of z, which is P(X < x). The area to the right is 1 − Φ(z), and the area between two values is the difference of their two areas.

z for the percentile

Φ(z) = 0.99 → z = 2.326

Back to x

x = 178 + 2.326·7 = 194.28 cm

Answer

x = 194.28 cm
15.Confidence intervals

Answer: lower limit = 1.259 mmHg

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.

Critical value

df = 6, t = 1.943

Margin of error

E = 1.943·7/√7 = 5.141

Interval

6.4 ± 5.141 → [1.259, 11.54]

Answer

lower limit = 1.259 mmHg
16.Confidence intervals

Answer: n = 877

Confidence interval for a mean: x̄ ± margin of error. With a known population σ the margin is z·σ/√n; with only the sample standard deviation s, use t with n − 1 degrees of freedom: t·s/√n. Critical values: z = 1.645 (90%), 1.96 (95%), 2.576 (99%). To get a margin E or smaller, you need n ≥ (z·σ/E)², rounded up.

Solve for n

n = (1.645·36/2)² = 876.8

Round up

A sample size is a whole number, and rounding down would make the margin too large.

n = 877

Answer

n = 877
17.Hypothesis tests & confidence intervals

Answer: t = −3.1

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 = 15/√36 = 15/6 = 2.5

Test statistic

t = (42.25 − 50)/2.5 = −3.1

Answer

t = −3.1
t0.025; df = 35 = 2.03

|t| > 2.03, so H₀ is rejected at the 5% level (two-sided).

18.Hypothesis tests & confidence intervals

Answer: 1 − β = 38.9 %

Power of a test: power = 1 − β is the probability of rejecting H₀ when it is false. Find the cutoff x̄c from μ₀ and z_α, then compute the probability of landing beyond x̄c using the true mean: z = (x̄c − μ)/(σ/√n).

Critical sample mean

SE = 12.5/√25 = 2.5
x̄c = 50 + 1.282·2.5 = 53.2

z under the true mean

z = (53.2 − 52.5)/2.5 = 0.282

Power

power = P(Z > 0.282) = 38.9%
1 − β = 38.9%

Answer

1 − β = 38.9%
19.Linear regression & correlation

Answer: r = −0.7926

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 = 15/5 = 3
ȳ = 117/5 = 23.4

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −14
Syy = Σ(y − ȳ)² = 31.2

Correlation coefficient

r = −14/√(10·31.2) = −0.7926

Answer

r = −0.7926
20.Linear regression & correlation

Answer: x = 5.973

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 = 35/5 = 7
ȳ = 469/5 = 93.8

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 226

Regression line

b = 226/40 = 5.65
a = 93.8 − 5.65·7 = 54.25
ŷ = 54.25 + 5.65·x

Solve for x

x = (88 − 54.25)/5.65 = 5.973

Answer

x = 5.973

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