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
years0123…9PV = ?€5 000+2.5%+2.5%…r = 5.5%

A property is expected to earn net rent of €5 000 next year. The rent is expected to grow by 2.5% per year forever. The required return is 5.5% per year. Find the value PV of the property today.

C1 = €5 000g = 2.5%r = 5.5%
PV = €
2.Time value of moneyHard
months0123…36€20 000A?AAA

A loan of €20 000 is repaid with equal monthly payments over 3 years. The interest rate is 2.4% per year, compounded monthly. Find the monthly payment A.

P = €20 0002.4% / 12N = 36
A = €
3.NPV & IRRMedium
years0123−60 00018 00015 00012 000

An investment of €60 000 gives these cash flows at the end of years 1 to 3: €18 000, €15 000, €12 000. The cost of capital is 15%. Find the net present value NPV.

r = 15%CF0 = −60 000CF1 = 18 000CF2 = 15 000CF3 = 12 000
NPV = €
4.NPV & IRRHard
years0123−20 00012 00011 00012 500

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

CF0 = −20 000CF1 = 12 000CF2 = 11 000CF3 = 12 500
IRR = %
5.Bond pricing & yieldMedium
years0123…14€1 053C = ?C = ?C = ? + €1 000

A bond with a face value of €1 000 matures in 14 years and pays annual coupons. It sells for €1 053 when the yield to maturity is 4%. Find the coupon rate c in percent.

F = €1 000n = 14P = €1 053YTM = 4%
c = %
6.Bond pricing & yieldHard
years0123…9€1 080€30€30 + €1 000

A bond with a face value of €1 000 and an annual coupon of 3% matures in 9 years and sells for €1 080. Find its yield to maturity YTM in percent.

F = €1 000coupon = 3%n = 9P = €1 080
YTM = %
7.CAPM & WACCMedium
WACC = ?equity€3 000 000re​ = 9%debt€5 000 000rd​ = 3%T = 25%

A company has equity worth €3 000 000 and debt worth €5 000 000. Its cost of equity is 9%, its pre-tax cost of debt is 3% and the tax rate is 25%. Find the WACC.

E = €3 000 000D = €5 000 000re = 9%rd = 3%T = 25%
WACC = %
8.CAPM & WACCHard
WACC = ?equity€6 000 000re​ = 11%bonds4 000 × €1 027coupon 4%, n = 6T = 28%

A company has equity with a market value of €6 000 000 and a cost of equity of 11%. Its debt consists of 4 000 bonds (face value €1 000, annual coupon 4%, 6 years to maturity) that trade at €1 027 each. The tax rate is 28%. Find the WACC, using the yield to maturity of the bonds as the pre-tax cost of debt.

E = €6 000 000re = 11%bonds: 4 000 × €1 027coupon = 4%n = 6T = 28%
WACC = %
9.Descriptive statisticsMedium
2530354045

Find the sample standard deviation s of the data: 27, 29, 49, 25, 49, 28, 32.

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

A quantity is normally distributed with mean 500 and standard deviation 100. What proportion P of values is below 650?

μ = 500σ = 100x = 650
P =
11.ProbabilityMedium
red: 6blue: 8

A bag holds 6 red and 8 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?

6 / 8
P =
12.ProbabilityHard
€p(x)−20.8960.1980.01

A game costs €2 to play. You win €8 with probability 0.1 and €100 with probability 0.01; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.

stake = €2€8: p = 0.1€100: p = 0.01
E(G) = €
13.Normal distributionMedium
−3−2−10123zshaded area to the right of z = 0.9846

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.9846. Find z to two decimals.

A = 0.9846
z =
14.Normal distributionHard
μ = 100x = ?95%

The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Find the value x below which 95% of them lie (the 95th percentile).

μ = 100σ = 15P = 95%
x = points
15.Confidence intervalsMedium
n = 11, sd​ = 3.10L = ?Ud̄ = 6.4E99%

The blood pressure of 11 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean 6.4 and standard deviation 3.1. Find the lower limit of the 99% confidence interval for the mean difference.

n = 11d̄ = 6.4sd = 3.199%t(10) = 3.169
lower limit = mmHg
16.Confidence intervalsHard
n = 30σ = ?Us = 9.995%

The diameters of 30 machined shafts have a sample standard deviation of 9.9 µm. Assuming the diameters are normally distributed, find the lower limit σ of the 95% confidence interval for the population standard deviation. With df = 29, 2.5% of the χ² distribution lies below 16.047 and 2.5% lies above 45.722.

n = 30s = 9.9 µmχ² = 16.047, 45.722
σ = µm
17.Hypothesis tests & confidence intervalsMedium
−1.961.96z = ?

Two production lines make the same part. Line 1: a sample of 40 parts has mean length 87.5 mm, and the line's known standard deviation is 5 mm. Line 2: a sample of 80 parts has mean 90.4 mm, known standard deviation 5 mm. Find the test statistic z for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (line 1 minus line 2).

n₁ = 40x̄₁ = 87.5 mmσ₁ = 5 mmn₂ = 80x̄₂ = 90.4 mmσ₂ = 5 mm
z =
18.Hypothesis tests & confidence intervalsHard
z −2.57602.576

A sample of 81 measurements has mean 119.7. The population standard deviation is 10. Find the upper limit U of the 99% confidence interval for the mean.

x̄ = 119.7σ = 10n = 8199%
U =
19.Linear regression & correlationMedium
xy23466118101014xy51051015

The table gives 5 paired observations of x = load in kN and y = extension in mm. Find Pearson's correlation coefficient r.

n = 5
r =
20.Linear regression & correlationHard
xy234761181710231223xy5101020x = 11

The table gives 6 paired observations of x = load in kN and y = extension in mm. Use the least-squares line ŷ = a + b·x. A new observation (11, 26) is made. Find its residual e = y − ŷ.

n = 6(11, 26)
e =

3Solutions

1.Time value of money

Answer: PV = 166667 €

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/(0.055 − 0.025) = 5 000/0.03 = €166 666.67

Answer

PV = €166 666.67
2.Time value of money

Answer: A = 576.4 €

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 = 12·3 = 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 = 0.9306
A = 20 000·0.002/(1 − 0.9306) = €576.35

Answer

A = €576.35
3.NPV & IRR

Answer: NPV = −25115 €

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 = −60 000
CF1/(1 + 0.15)1 = 18 000/1.15 = 15 652.17
CF2/(1 + 0.15)2 = 15 000/1.322 = 11 342.16
CF3/(1 + 0.15)3 = 12 000/1.521 = 7 890.19

Add them up

NPV = €−25 115.48

Answer

NPV = €−25 115.48

NPV is negative, so the project earns less than the required return.

4.NPV & IRR

Answer: IRR = 35.02 %

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%) = 5.08
NPV(40%) = −1 260.93

Narrow down

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

IRR ≈ 35 + 5·5.08/(5.08 − (−1 260.93)) = 35.02%
exact: NPV(35.02%) = 0 → IRR = 35.02%

Answer

IRR = 35.02%
5.Bond pricing & yield

Answer: c = 4.502 %

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.

Present value of the face value

1000·(1 + 0.04)−14 = €577.48

Coupon

C = (1053 − 577.48)/10.56 = €45.02

Coupon rate

c = 45.02/1000 = 4.502%

Answer

c = 4.502%
6.Bond pricing & yield

Answer: YTM = 2.019 %

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

€1 080 = €30·(1 − (1 + y)⁻⁹)/y + €1 000/(1 + y)⁹

Solve numerically

P(2%) = €1 081.62, P(3%) = €1 000.00
YTM = 2.019%

Answer

YTM = 2.019%
7.CAPM & WACC

Answer: WACC = 4.781 %

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).

Weights

V = €8 000 000, E/V = 0.375, D/V = 0.625

WACC

WACC = 0.375·9% + 0.625·3%·(1 − 0.25) = 4.781%

Answer

WACC = 4.781%
8.CAPM & WACC

Answer: WACC = 7.552 %

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).

Cost of debt (YTM)

1027 = 40·(1 − (1 + y)−6)/y + 1000/(1 + y)6
y = 3.493%

Market values

D = 4 000·€1 027 = €4 108 000
V = €10 108 000, E/V = 0.5936, D/V = 0.4064

WACC

WACC = 0.5936·11% + 0.4064·3.493%·(1 − 0.28) = 7.552%

Answer

WACC = 7.552%
9.Descriptive statistics

Answer: s = 10.37

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 = 239/7 = 34.14

Squared deviations

Σ(x − x̄)² = (−7.143)² + (−5.143)² + 14.86² + (−9.143)² + 14.86² + (−6.143)² + (−2.143)² = 644.9

Standard deviation

s = √(644.9/6) = 10.37

Answer

s = 10.37
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 = 0.1648

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 = 6/14 · 5/13 = 30/182 = 0.1648

Answer

P = 0.1648
12.Probability

Answer: E(G) = −0.2 €

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.

Net outcomes

G: 6 (p = 0.1), 98 (p = 0.01), −2 (p = 0.89)

Expected value

E(G) = 6·0.1 + 98·0.01 − 2·0.89 = −0.2 €

Answer

E(G) = −0.2 €

A negative value means you lose money on average.

13.Normal distribution

Answer: z = −2.16

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 − 0.9846 = 0.0154

Read the z-table backwards

Find 0.0154 inside the table and read off z.

z = −2.16

Answer

z = −2.16
14.Normal distribution

Answer: x = 124.7 points

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) = 0.95 → z = 1.645

Back to x

x = 100 + 1.645·15 = 124.67 points

Answer

x = 124.67 points
15.Confidence intervals

Answer: lower limit = 3.438 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 = 10, t = 3.169

Margin of error

E = 3.169·3.1/√11 = 2.962

Interval

6.4 ± 2.962 → [3.438, 9.362]

Answer

lower limit = 3.438 mmHg
16.Confidence intervals

Answer: σ = 7.884 µm

Interval for a variance: (n − 1)s²/σ² has a χ² distribution with n − 1 degrees of freedom. The 95% interval for σ² runs from (n − 1)s² divided by the upper χ² value (2.5% above it) to (n − 1)s² divided by the lower χ² value (2.5% below it); the interval for σ is the square root of both limits. It is not symmetric around s.

Sum of squares

(n − 1)s² = 29·9.9² = 2842

Limits for σ²

2842/45.722 = 62.16
2842/16.047 = 177.1

Limit for σ

σ = √62.16 = 7.884 µm

Answer

σ = 7.884 µm

95% interval for σ: [7.884, 13.31] µm.

17.Hypothesis tests & confidence intervals

Answer: z = −2.995

Hypothesis test: the test statistic = (estimate − value under H₀)/(standard error under H₀). The p-value is the probability, assuming H₀ is true, of a statistic at least as extreme as the one observed: the area in one tail for a one-sided H₁, twice the tail area for a two-sided H₁. Reject H₀ when p ≤ α.

Standard error of the difference

SE = √(5²/40 + 5²/80) = √0.9375 = 0.9682

Test statistic

z = (87.5 − 90.4)/0.9682 = −2.995

Answer

z = −2.995

|z| > 1.96, so H₀ is rejected at the 5% level (two-sided).

18.Hypothesis tests & confidence intervals

Answer: U = 122.6

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).

Margin of error

E = 2.576·10/√81 = 2.862

Interval limit

U = 119.7 + 2.862 = 122.6

Answer

U = 122.6
19.Linear regression & correlation

Answer: r = 0.9507

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 = 30/5 = 6
ȳ = 44/5 = 8.8

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 52
Syy = Σ(y − ȳ)² = 74.8

Correlation coefficient

r = 52/√(40·74.8) = 0.9507

Answer

r = 0.9507
20.Linear regression & correlation

Answer: e = 3.2

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 = 42/6 = 7
ȳ = 84/6 = 14

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 154

Regression line

b = 154/70 = 2.2
a = 14 − 2.2·7 = −1.4
ŷ = −1.4 + 2.2·x

Predicted value

ŷ = −1.4 + 2.2·11 = 22.8

Residual

e = 26 − 22.8 = 3.2

Answer

e = 3.2

The point lies above the line.

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