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…19€5 000FV = ?

You invest €5 000 at 3% per year with continuous compounding. Find the value FV after 19 years.

PV = €5 000r = 3%t = 19
FV = €
2.Time value of moneyHard
months0123…300€250 000A?AAA

A loan of €250 000 is repaid with equal monthly payments over 25 years. The interest rate is 5.4% per year, compounded monthly. Find the monthly payment A.

P = €250 0005.4% / 12N = 300
A = €
3.NPV & IRRMedium
years0123…9−500 00036 000+3%+3%…r = 9%, NPV = ?

A project costs €500 000 today. It generates €36 000 at the end of the first year, and the cash flow then grows by 3% per year forever. The cost of capital is 9%. Find the net present value NPV.

C0 = €500 000CF1 = €36 000g = 3%r = 9%
NPV = €
4.NPV & IRRHard
years01234−50 00015 50015 50015 50015 500

An investment of €50 000 returns €15 500 at the end of each year for 4 years. Find its internal rate of return IRR in percent.

C0 = €50 000CF = €15 500n = 4
IRR = %
5.Bond pricing & yieldMedium
half-years0123…12€25€25€25 + €1 000P = ?

A bond with a face value of €1 000 pays a 5% annual coupon in two half-yearly payments and matures in 6 years. The yield to maturity is 5.5% per year. Find the price P of the bond.

F = €1 000coupon = 5%YTM = 5.5%n = 6
P = €
6.Bond pricing & yieldHard
years01234P€70€70€70 + €1 000

A bond with a face value of €1 000 pays an annual coupon of 7% and matures in 4 years. The yield to maturity is 9%. Find the Macaulay duration D in years.

F = €1 000coupon = 7%n = 4YTM = 9%
D = years
7.CAPM & WACCMedium
WACC = ?equity€5 000 000re​ = 12%debt€3 000 000rd​ = 4.5%T = 25%

A company has equity worth €5 000 000 and debt worth €3 000 000. Its cost of equity is 12%, its pre-tax cost of debt is 4.5% and the tax rate is 25%. Find the WACC.

E = €5 000 000D = €3 000 000re = 12%rd = 4.5%T = 25%
WACC = %
8.CAPM & WACCHard
WACC = ?equity5 000 000 × €28.00β = 0.9rf​ = 1%, rm​ = 5.5%debt5 000 × €980rd​ = 6%T = 25.8%

A company has 5 000 000 shares at €28.00 each and 5 000 bonds trading at €980 each. The pre-tax cost of debt is 6% and the tax rate is 25.8%. The share’s beta is 0.9, the risk-free rate 1% and the market return 5.5%. Find the WACC.

β = 0.9rf = 1%rm = 5.5%rd = 6%T = 25.8%
WACC = %
9.Descriptive statisticsMedium
1520253035404550

Find the sample standard deviation s of the data: 11, 43, 44, 19, 44, 51.

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

Exam scores are normally distributed with mean 170 and standard deviation 10. Find the z-score z of a score of 155.

μ = 170σ = 10x = 155
z =
11.ProbabilityMedium
n = 14k = 4N = ?

In how many ways N can a committee of 4 people be chosen from a group of 14? The order does not matter.

n = 14k = 4
N =
12.ProbabilityHard
n = 9, p = 0.6: P(X = 1) = ?k0123456789

A test has 9 multiple-choice questions. A student who guesses gets each one right with probability 0.6, independently. What is the probability P of exactly 1 correct answers?

n = 9p = 0.6k = 1
P =
13.Normal distributionMedium
246.25252.75P = ?

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. What fraction lies between 246.25 and 252.75 g? Give P as a decimal.

μ = 250σ = 5x1 = 246.25x2 = 252.75
P =
14.Normal distributionHard
4 s0X − Y?

Runner A's lap time X and runner B's lap time Y are independent and normal: X has mean 52 s and standard deviation 4 s, Y has mean 48 s and standard deviation 3 s. Find the probability P that X is greater than Y. Round z to two decimals.

μX = 52 s, σX = 4 sμY = 48 s, σY = 3 s
P =
15.Confidence intervalsMedium
μ₁ − μ₂0LUx̄₁ − x̄₂E = ?99%

The battery life of two phone models is compared. Model 1: 30 phones, mean 72 h, known standard deviation 13 h. Model 2: 100 phones, mean 61.5 h, known standard deviation 9 h. Find the margin of error E of the 99% confidence interval for μ₁ − μ₂.

n₁ = 30x̄₁ = 72 hσ₁ = 13 hn₂ = 100x̄₂ = 61.5 hσ₂ = 9 h99%
E = h
16.Confidence intervalsHard
μ₁ − μ₂, t(20) = 1.7250LUx̄₁ − x̄₂E = ?90%

Two fertilisers are tested. Plot group 1: 10 plots, mean yield 41.5 kg, standard deviation 1.5 kg. Plot group 2: 12 plots, mean 29.5 kg, standard deviation 2.5 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the margin of error E of the 90% confidence interval for μ₁ − μ₂.

n₁ = 10x̄₁ = 41.5s₁ = 1.5n₂ = 12x̄₂ = 29.5s₂ = 2.590%t(20) = 1.725
E = kg
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 80 mm, and the line's known standard deviation is 16 mm. Line 2: a sample of 100 parts has mean 73.3 mm, known standard deviation 17 mm. Find the test statistic z for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (line 1 minus line 2).

n₁ = 40x̄₁ = 80 mmσ₁ = 16 mmn₂ = 100x̄₂ = 73.3 mmσ₂ = 17 mm
z =
18.Hypothesis tests & confidence intervalsHard
σ = 5, n = ?x̄ − Ex̄ + Ex̄E = 399%

How large must a sample be so that the 99% confidence interval for a mean has a margin of error of at most 3? The standard deviation is 5. Find the smallest sample size n.

σ = 5E = 399%
n =
19.Linear regression & correlationMedium
xy3355437489531152xy5104050

The table gives 5 paired observations of x = advertising spend in k€ and y = sales in k€. Find the coefficient of determination r² of the least-squares line.

n = 5
r² =
20.Linear regression & correlationHard
xy146352551758961xy545505560ŷ = 59

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 = 59?

n = 5ŷ = 59
x =

3Solutions

1.Time value of money

Answer: FV = 8841 €

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.

Exponent

r·t = 0.03·19 = 0.57
e0.57 = 1.768

Future value

FV = 5 000·1.768 = €8 841.34

Answer

FV = €8 841.34
2.Time value of money

Answer: A = 1520 €

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 = 0.054000000000000006/12 = 0.0045
N = 12·25 = 300

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 = 0.26
A = 250 000·0.0045/(1 − 0.26) = €1 520.32

Answer

A = €1 520.32
3.NPV & IRR

Answer: NPV = 100000 €

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.

Value of the cash flows

PV = 36 000/(0.09 − 0.03) = 600 000.00

NPV

NPV = 600 000.00 − 500 000 = €100 000.00

Answer

NPV = €100 000.00

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

4.NPV & IRR

Answer: IRR = 9.196 %

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.

Required annuity factor

(1 − (1 + i)−4)/i = 50 000/15 500 = 3.226

Solve for the rate

The annuity factor falls as the rate rises; bracket the rate and narrow down.

i = 9%: 3.24
i = 10%: 3.17
IRR = 9.196%

Answer

IRR = 9.196%
5.Bond pricing & yield

Answer: P = 974.7 €

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 = €25.00, i = 0.0275, N = 12

Present value of the coupons

€25.00·(1 − 0.7221)/0.0275 = €252.61

Present value of the face value

€1 000·0.7221 = €722.13

Price

P = €974.74

Answer

P = €974.74

The yield is above the coupon rate, so the bond sells at a discount.

6.Bond pricing & yield

Answer: D = 3.61 years

Duration: the Macaulay duration is the weighted average time at which a bond pays its cash flows, with the present values of the cash flows as weights: D = Σ t·PVₜ/P. The modified duration D/(1 + y) estimates the percentage price change for a small change in yield: %ΔP ≈ −Dmod·Δy.

Present value of each cash flow

t = 1: 70/1.09 = 64.22, t·PV = 64.22
t = 2: 70/1.188 = 58.92, t·PV = 117.84
t = 3: 70/1.295 = 54.05, t·PV = 162.16
t = 4: 1070/1.412 = 758.01, t·PV = 3 032.06

Duration

P = 935.21
D = 3 376.27/935.21 = 3.61

Answer

D = 3.61 years
7.CAPM & WACC

Answer: WACC = 8.766 %

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 = 0.625, D/V = 0.375

WACC

WACC = 0.625·12% + 0.375·4.5%·(1 − 0.25) = 8.766%

Answer

WACC = 8.766%
8.CAPM & WACC

Answer: WACC = 5.03 %

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 = 5 000 000·€28.00 = €140 000 000
D = 5 000·€980 = €4 900 000

Cost of equity (CAPM)

re = 1% + 0.9·4.5% = 5.05%

Weights

V = €144 900 000, E/V = 0.9662, D/V = 0.03382

WACC

WACC = 0.9662·5.05% + 0.03382·6%·(1 − 0.258) = 5.03%

Answer

WACC = 5.03%
9.Descriptive statistics

Answer: s = 16.21

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 = 212/6 = 35.33

Squared deviations

Σ(x − x̄)² = (−24.33)² + 7.667² + 8.667² + (−16.33)² + 8.667² + 15.67² = 1313

Standard deviation

s = √(1313/5) = 16.21

Answer

s = 16.21
10.Descriptive statistics

Answer: z = −1.5

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 = (155 − 170)/10 = −1.5

Answer

z = −1.5
11.Probability

Answer: N = 1001

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.

Combinations

C(14, 4) = (14·13·12·11)/(4·3·2·1) = 1001

Answer

N = 1001
12.Probability

Answer: P = 0.003539

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 = 1) = 9·0.61·0.48 = 9·0.6·0.0006554 = 0.003539

Answer

P = 0.003539
13.Normal distribution

Answer: P = 0.4822

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-scores

z₁ = (246.25 − 250)/5 = −0.75
z₂ = (252.75 − 250)/5 = 0.55

Area between

P = Φ(0.55) − Φ(−0.75) = 0.7088 − 0.2266 = 0.4822

Answer

P = 0.4822
14.Normal distribution

Answer: P = 0.7881

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 = 52 − 48 = 4 s, σ = 5 s

Standardise

z = (0 − 4)/5 = −0.8

Probability

P = 1 − Φ(−0.8) = 0.7881

Answer

P = 0.7881
15.Confidence intervals

Answer: E = 6.539 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 = 72 − 61.5 = 10.5

Standard error

SE = √(13²/30 + 9²/100) = 2.538

Margin of error

E = 2.576·2.538 = 6.539

Answer

E = 6.539 h

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

16.Confidence intervals

Answer: E = 1.558 kg

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.

Pooled standard deviation

sp² = (9·1.5² + 11·2.5²)/20 = 4.45
sp = 2.11

Margin of error

E = 1.725·2.11·√(1/10 + 1/12) = 1.558

Answer

E = 1.558 kg
17.Hypothesis tests & confidence intervals

Answer: z = 2.198

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 = √(16²/40 + 17²/100) = √9.29 = 3.048

Test statistic

z = (80 − 73.3)/3.048 = 2.198

Answer

z = 2.198

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

18.Hypothesis tests & confidence intervals

Answer: n = 19

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 = (2.576·5/3)² = 18.43

Round up

n = 19

Answer

n = 19
19.Linear regression & correlation

Answer: r² = 0.8848

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 = 231/5 = 46.2

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 88
Syy = Σ(y − ȳ)² = 218.8

Coefficient of determination

r² = 88²/(40·218.8) = 0.8848

Answer

r² = 0.8848

About 88.48% of the variation in y is explained by the line.

20.Linear regression & correlation

Answer: x = 8

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 = 25/5 = 5
ȳ = 268/5 = 53.6

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 72

Regression line

b = 72/40 = 1.8
a = 53.6 − 1.8·5 = 44.6
ŷ = 44.6 + 1.8·x

Solve for x

x = (59 − 44.6)/1.8 = 8

Answer

x = 8

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