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.

New problemsBack to the course

In the print window, choose “Save as PDF” to keep a copy on your device.

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
years0123PV = ?€20 000

You need €20 000 in 3 years. An account pays 4.8% per year, compounded monthly. How much must you deposit today? Find the present value PV.

FV = €20 0004.8% / 12n = 3
PV = €
2.Time value of moneyHard
months0123…84€30 000AAAA3% / 12, I = ?

A loan of €30 000 is repaid with equal monthly payments over 7 years at 3% per year, compounded monthly. How much interest I is paid in total over the life of the loan?

P = €30 0003% / 12N = 84
I = €
3.NPV & IRRMedium
years01234−30 00014 0009 00011 00013 000

An investment of €30 000 gives these cash flows at the end of years 1 to 4: €14 000, €9 000, €11 000, €13 000. The cost of capital is 10%. Find the net present value NPV.

r = 10%CF0 = −30 000CF1 = 14 000CF2 = 9 000CF3 = 11 000CF4 = 13 000
NPV = €
4.NPV & IRRHard
years0123−25 00015 00013 00015 000

A project costs €25 000 and returns €15 000, €13 000, €15 000 at the end of years 1 to 3. Find its internal rate of return IRR in percent.

CF0 = −25 000CF1 = 15 000CF2 = 13 000CF3 = 15 000
IRR = %
5.Bond pricing & yieldMedium
half-years0123…24€30€30€30 + €1 000P = ?

A bond with a face value of €1 000 pays a 6% annual coupon in two half-yearly payments and matures in 12 years. The yield to maturity is 3% per year. Find the price P of the bond.

F = €1 000coupon = 6%YTM = 3%n = 12
P = €
6.Bond pricing & yieldHard
years01234567€1 110€80€80 + €1 000

A bond with a face value of €1 000 and an annual coupon of 8% matures in 7 years and sells for €1 110. Find its yield to maturity YTM in percent.

F = €1 000coupon = 8%n = 7P = €1 110
YTM = %
7.CAPM & WACCMedium
WACC = ?ordinary shares€6 000 000re​ = 11%preferred shares€500 000rp​ = 6.5%debt€2 000 000rd​ = 5%T = 30%

A company is financed with ordinary shares worth €6 000 000, preferred shares worth €500 000 and debt worth €2 000 000. The cost of equity is 11%, the cost of preferred stock 6.5% and the pre-tax cost of debt 5%. The tax rate is 30%. Find the WACC.

E = €6 000 000Ps = €500 000D = €2 000 000re = 11%rp = 6.5%rd = 5%T = 30%
WACC = %
8.CAPM & WACCHard
rf​ = 3%, rm​ − rf​ = 6%D/E = 0.5equityβL​ = 1.5debtD/E = 0.6equityre​ = ?debtT = 20%

A company's shares have a beta of 1.5 at its current debt-to-equity ratio of 0.5. It plans to raise its debt-to-equity ratio to 0.6. The tax rate is 20%, the risk-free rate 3% and the market risk premium 6%. Find the cost of equity re after the change.

βL = 1.5D/E1 = 0.5D/E2 = 0.6T = 20%rf = 3%rm − rf = 6%
re = %
9.Descriptive statisticsMedium
2025303540455055

Find the sample standard deviation s of the data: 30, 31, 24, 57, 17, 24, 45.

n = 7
s =
10.Descriptive statisticsHard
z −2−1012

A quantity is normally distributed with mean 60 and standard deviation 20. What proportion P of values is below 90?

μ = 60σ = 20x = 90
P =
11.ProbabilityMedium
red: 7blue: 8

A bag holds 7 red and 8 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?

7 / 8
P =
12.ProbabilityHard
xp(x)10.220.230.0540.3550.2μ

X has the distribution shown, and Y = −2X + 4. Find the standard deviation σY of Y.

x: 1, 2, 3, 4, 5p(x): 0.2, 0.2, 0.05, 0.35, 0.2Y = −2X + 4
σY =
13.Normal distributionMedium
−3−2−10123zshaded area between −z and z = 0.8969

Drag z so that the shaded area between −z and z under the standard normal curve is 0.8969. Find z to two decimals.

A = 0.8969
z =
14.Normal distributionHard
0.04 mm0D − d?

Hole diameters D are normal with mean 25.04 mm and standard deviation 0.01 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.02 mm. A randomly chosen shaft fits when D − d > 0. Find the probability P that it fits. Round z to two decimals.

hole: μ = 25.04 mm, σ = 0.01 mmshaft: μ = 25 mm, σ = 0.02 mm
P =
15.Confidence intervalsMedium
n = 6, s = 14LUx̄ = 12E = ?90%

A sample of 6 items has mean 12 and standard deviation s = 14. The population is roughly normal. Find the margin of error E of a 90% confidence interval for the mean.

n = 6x̄ = 12s = 1490%t(5) = 2.015
E =
16.Confidence intervalsHard
σ = 28, n = ?x̄ − Ex̄ + Ex̄E = 395%

The population standard deviation is 28. How large a sample n do you need so that a 95% confidence interval for the mean has a margin of error of at most 3?

σ = 28E = 395%
n =
17.Hypothesis tests & confidence intervalsMedium
t −1.98401.984

A sample of 100 students has a mean score of 195.6 and a sample standard deviation of 20. Find the test statistic t for H₀: μ = 200.

μ₀ = 200x̄ = 195.6s = 20n = 100
t =
18.Hypothesis tests & confidence intervalsHard
μ₀x̄c

A one-sided z test of H₀: μ = 500 against H₁: μ > 500 is done at α = 0.05 with a sample of 64; the standard deviation is known to be 8. In reality the mean is 502.2. Find the power 1 − β of the test in percent.

μ₀ = 500μ = 502.2σ = 8n = 64α = 0.05
1 − β = %
19.Linear regression & correlationMedium
xy176620882396261132911932142xy202530100150

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 = 147Σy = 624Σx² = 3759Σy² = 68410Σxy = 16023
r =
20.Linear regression & correlationHard
xy341448558663764876xy468406080ŷ = 45

The table gives 6 paired observations of x = advertising spend in k€ and y = sales in k€. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 45?

n = 6ŷ = 45
x =

3Solutions

1.Time value of money

Answer: PV = 17323 €

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.

Rate and number of periods

i = 0.048/12 = 0.004
N = 12·3 = 36

Present value

PV = FV/(1 + i)N = 20 000/1.155 = €17 322.73

Answer

PV = €17 322.73
2.Time value of money

Answer: I = 3298 €

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 payment

i = 0.0025, N = 84
A = 30 000·0.0025/(1 − (1 + 0.0025)−84) = €396.40

Total interest

I = N·A − P = 84·396.4 − 30 000 = €3 297.52

Answer

I = €3 297.52
3.NPV & IRR

Answer: NPV = 7309 €

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 = −30 000
CF1/(1 + 0.1)1 = 14 000/1.1 = 12 727.27
CF2/(1 + 0.1)2 = 9 000/1.21 = 7 438.02
CF3/(1 + 0.1)3 = 11 000/1.331 = 8 264.46
CF4/(1 + 0.1)4 = 13 000/1.464 = 8 879.17

Add them up

NPV = €7 308.93

Answer

NPV = €7 308.93

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

4.NPV & IRR

Answer: IRR = 33.01 %

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(30%) = 1 058.26
NPV(35%) = −659.20

Narrow down

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

IRR ≈ 30 + 5·1 058.26/(1 058.26 − (−659.20)) = 33.08%
exact: NPV(33.01%) = 0 → IRR = 33.01%

Answer

IRR = 33.01%
5.Bond pricing & yield

Answer: P = 1300 €

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.

Per period

C = €30.00, i = 0.015, N = 24

Present value of the coupons

€30.00·(1 − 0.6995)/0.015 = €600.91

Present value of the face value

€1 000·0.6995 = €699.54

Price

P = €1 300.46

Answer

P = €1 300.46

The yield is below the coupon rate, so the bond sells at a premium.

6.Bond pricing & yield

Answer: YTM = 6.028 %

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

€1 110 = €80·(1 − (1 + y)⁻⁷)/y + €1 000/(1 + y)⁷

Solve numerically

P(6%) = €1 111.65, P(7%) = €1 053.89
YTM = 6.028%

Answer

YTM = 6.028%
7.CAPM & WACC

Answer: WACC = 8.971 %

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 = 6 000 000 + 500 000 + 2 000 000 = €8 500 000
E/V = 0.7059, Ps/V = 0.05882, D/V = 0.2353

WACC

WACC = 0.7059·11% + 0.05882·6.5% + 0.2353·5%·(1 − 0.3) = 8.971%

Answer

WACC = 8.971%
8.CAPM & WACC

Answer: re = 12.51 %

Levered beta (Hamada): debt makes equity riskier. βL = βU·(1 + (1 − T)·D/E), where βU is the beta of the business without debt. To move a beta to another capital structure, unlever it first and then relever it at the new D/E.

Unlever the beta

βU = 1.5/(1 + 0.8·0.5) = 1.071

Relever at the new D/E

βL = 1.071·(1 + 0.8·0.6) = 1.586

Cost of equity (CAPM)

re = 3% + 1.586·6% = 12.51%

Answer

re = 12.51%
9.Descriptive statistics

Answer: s = 13.84

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 = 228/7 = 32.57

Squared deviations

Σ(x − x̄)² = (−2.571)² + (−1.571)² + (−8.571)² + 24.43² + (−15.57)² + (−8.571)² + 12.43² = 1150

Standard deviation

s = √(1150/6) = 13.84

Answer

s = 13.84
10.Descriptive statistics

Answer: P = 0.9332

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 = 1.5

Standard normal table

Φ(1.5) = 0.9332
P = 0.9332

Answer

P = 0.9332
11.Probability

Answer: P = 0.2

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 = 7/15 · 6/14 = 42/210 = 0.2

Answer

P = 0.2
12.Probability

Answer: σY = 2.917

Discrete random variables: a probability distribution lists every value x a variable can take with its probability p(x); the probabilities are between 0 and 1 and add up to 1. The expected value (mean) is E(X) = Σx·p(x), the long-run average. The variance is Var(X) = Σ(x − μ)²·p(x) = E(X²) − μ², and the standard deviation σ = √Var(X). For a linear change Y = aX + b: E(Y) = a·E(X) + b, Var(Y) = a²·Var(X) and σY = |a|·σX; adding a constant does not change the spread.

Spread of X

μ = 3.15, Var(X) = 2.128, σX = 1.459

Linear change

σY = |−2|·1.459 = 2.917

Answer

σY = 2.917

The mean does move: E(Y) = −2·3.15 + 4 = −2.3.

13.Normal distribution

Answer: z = 1.63

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 + 0.8969)/2 = 0.9484

Read the z-table backwards

Find 0.9484 inside the table and read off z.

z = 1.63

Answer

z = 1.63
14.Normal distribution

Answer: P = 0.9633

Sums and differences of normal variables: if X and Y are independent and normal, X + Y and X − Y are normal too. Means add or subtract: E(X ± Y) = μX ± μY. Variances always add, also for a difference: Var(X ± Y) = σX² + σY², so σ = √(σX² + σY²). Standard deviations never add directly. The total of n independent items with the same distribution has mean n·μ and standard deviation σ·√n. Then standardise as usual: z = (x − mean)/σ and read Φ(z).

Distribution of the difference

μ = 25.04 − 25 = 0.04 mm, σ = 0.02236 mm

Standardise

z = (0 − 0.04)/0.02236 = −1.79

Probability

P = 1 − Φ(−1.79) = 0.9633

Answer

P = 0.9633
15.Confidence intervals

Answer: E = 11.52

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.

Critical value

df = n − 1 = 5, t = 2.015

Margin of error

E = 2.015·14/√6 = 11.52

Answer

E = 11.52
16.Confidence intervals

Answer: n = 335

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 = (1.96·28/3)² = 334.6

Round up

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

n = 335

Answer

n = 335
17.Hypothesis tests & confidence intervals

Answer: t = −2.2

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 = 20/√100 = 20/10 = 2

Test statistic

t = (195.6 − 200)/2 = −2.2

Answer

t = −2.2
t0.025; df = 99 = 1.984

|t| > 1.984, so H₀ is rejected at the 5% level (two-sided).

18.Hypothesis tests & confidence intervals

Answer: 1 − β = 71.06 %

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 = 8/√64 = 1
x̄c = 500 + 1.645·1 = 501.6

z under the true mean

z = (501.6 − 502.2)/1 = −0.555

Power

power = P(Z > −0.555) = 71.06%
1 − β = 71.06%

Answer

1 − β = 71.06%
19.Linear regression & correlation

Answer: r = 0.988

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 = 16023 − 147·624/6 = 735
Sxx = 3759 − 147²/6 = 157.5
Syy = 68410 − 624²/6 = 3514

Correlation coefficient

r = 735/√(157.5·3514) = 0.988

Answer

r = 0.988
20.Linear regression & correlation

Answer: x = 3.453

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 = 33/6 = 5.5
ȳ = 350/6 = 58.33

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 114

Regression line

b = 114/17.5 = 6.514
a = 58.33 − 6.514·5.5 = 22.5
ŷ = 22.5 + 6.514·x

Solve for x

x = (45 − 22.5)/6.514 = 3.453

Answer

x = 3.453

This is independent practice material. ExamProblems is not affiliated with or endorsed by the organisation that runs this exam, and the exam name is used only to describe the content. CFA® and Chartered Financial Analyst® are trademarks of CFA Institute, which is not affiliated with ExamProblems and does not endorse it or vouch for its accuracy.

ExamProblems · examproblems.com/courses/cfa-level-1-quant-and-fixed-income