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 = ?€10 000

You need €10 000 in 9 years. An account pays 4.8% per year, compounded monthly. How much must you deposit today? Find the present value PV.

FV = €10 0004.8% / 12n = 9
PV = €
2.Time value of moneyHard
months0123…204€500€500€500FV = ?

You save €500 at the end of every month for 17 years. The account pays 3% per year, compounded monthly. How much have you saved at the end? Find FV.

A = €5003% / 12N = 204
FV = €
3.NPV & IRRMedium
years0123…9−250 00026 000+1%+1%…r = 9%, NPV = ?

A project costs €250 000 today. It generates €26 000 at the end of the first year, and the cash flow then grows by 1% per year forever. The cost of capital is 9%. Find the net present value NPV.

C0 = €250 000CF1 = €26 000g = 1%r = 9%
NPV = €
4.NPV & IRRHard
years0123−40 00019 00021 00019 000

A project costs €40 000 and returns €19 000, €21 000, €19 000 at the end of years 1 to 3. Cash flows can be reinvested at the cost of capital of 10%. Find the modified internal rate of return MIRR in percent.

r = 10%CF0 = −40 000CF1 = 19 000CF2 = 21 000CF3 = 19 000
MIRR = %
5.Bond pricing & yieldMedium

You buy a bond with 6 years to maturity (face value €1 000, annual coupon 3%) when its yield to maturity is 4%. One year later, just after receiving the first coupon, you sell it when the yield to maturity is 4.5%. Find your holding period return HPR in percent.

F = €1 000coupon = 3%n = 6y0 = 4%y1 = 4.5%
HPR = %
6.Bond pricing & yieldHard
years012345€1 166€70€70€70 + €1 070

A bond with 11 years to maturity (face value €1 000, annual coupon 7%) sells for €1 166. The issuer can call it in 5 years at a call price of €1 070. Find the yield to call YTC in percent.

F = €1 000coupon = 7%P = €1 166call price = €1 070call in 5 years
YTC = %
7.CAPM & WACCMedium
WACC = 11.5%equity€5 000 000re​ = ?debt€1 000 000rd​ = 5%T = 20%

A company has equity worth €5 000 000 and debt worth €1 000 000. Its WACC is 11.5%, its pre-tax cost of debt is 5% and the tax rate is 20%. Find the cost of equity re.

E = €5 000 000D = €1 000 000WACC = 11.5%rd = 5%T = 20%
re = %
8.CAPM & WACCHard
rf​ = 1.5%, rm​ − rf​ = 5%D/E = 0.5equityβL​ = 1.5debtD/E = 1.5equityre​ = ?debtT = 21%

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 1.5. The tax rate is 21%, the risk-free rate 1.5% and the market risk premium 5%. Find the cost of equity re after the change.

βL = 1.5D/E1 = 0.5D/E2 = 1.5T = 21%rf = 1.5%rm − rf = 5%
re = %
9.Descriptive statisticsMedium
152025303540455055

Find the sample standard deviation s of the data: 16, 14, 20, 12, 57.

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

Exam scores are normally distributed with mean 170 and standard deviation 12. Find the z-score z of a score of 158.

μ = 170σ = 12x = 158
z =
11.ProbabilityMedium
n = 12k = 5N = ?

In how many ways N can a committee of 5 people be chosen from a group of 12? The order does not matter.

n = 12k = 5
N =
12.ProbabilityHard
n = 9, p = 0.5: P(X = 2) = ?k0123456789

A test has 9 multiple-choice questions. A student who guesses gets each one right with probability 0.5, independently. What is the probability P of exactly 2 correct answers?

n = 9p = 0.5k = 2
P =
13.Normal distributionMedium
−3−2−10123zshaded area between −z and z = 0.4448

Drag z so that the shaded area between −z and z under the standard normal curve is 0.4448. Find z to two decimals.

A = 0.4448
z =
14.Normal distributionHard
157164171178185192199cmbottom 95%

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 95% is to the left of it. Find the cut-off x.

μ = 178 cmσ = 7 cm95%
x = cm
15.Confidence intervalsMedium
n = 7, sd​ = 7.20LU = ?d̄ = 2.3E99%

The blood pressure of 7 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean 2.3 and standard deviation 7.2. Find the upper limit of the 99% confidence interval for the mean difference.

n = 7d̄ = 2.3sd = 7.299%t(6) = 3.707
upper limit = mmHg
16.Confidence intervalsHard
σ = 25, n = ?x̄ − Ex̄ + Ex̄E = 499%

The population standard deviation is 25. How large a sample n do you need so that a 99% confidence interval for the mean has a margin of error of at most 4?

σ = 25E = 499%
n =
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 131.5 mm, and the line's known standard deviation is 8 mm. Line 2: a sample of 40 parts has mean 134.3 mm, known standard deviation 3 mm. Find the test statistic z for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (line 1 minus line 2).

n₁ = 40x̄₁ = 131.5 mmσ₁ = 8 mmn₂ = 40x̄₂ = 134.3 mmσ₂ = 3 mm
z =
18.Hypothesis tests & confidence intervalsHard
H₀H₁μ₀x̄cμ₀ + 5α = 0.05βσ = 17, power 80%, n = ?

A one-sided z test at α = 0.05 must detect a shift of the mean of 5 units with a power of 80%. The standard deviation is 17. Find the minimum sample size n.

α = 0.05power = 80%δ = 5σ = 17zα = 1.645zβ = 0.842
n =
19.Linear regression & correlationMedium
xy124223322420520616xy24620

The table gives 6 paired observations of x = car age in years and y = price in k€. Find the coefficient of determination r² of the least-squares line.

n = 6
r² =
20.Linear regression & correlationHard
xy174920512371267329873294xy2025306080x = 24.5

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 (24.5, 73) is made. Find its residual e = y − ŷ.

n = 6(24.5, 73)
e =

3Solutions

1.Time value of money

Answer: PV = 6498 €

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 = 12·9 = 108

Present value

PV = FV/(1 + i)N = 10 000/1.539 = €6 497.69

Answer

PV = €6 497.69
2.Time value of money

Answer: FV = 132846 €

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 deposits

i = 0.03/12 = 0.0025
N = 12·17 = 204

Future value of the deposits

FV = A·((1 + i)N − 1)/i = 500·(1.664 − 1)/0.0025 = €132 846.34

Answer

FV = €132 846.34

You paid in €102 000 yourself; the rest is interest.

3.NPV & IRR

Answer: NPV = 75000 €

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 = 26 000/(0.09 − 0.01) = 325 000.00

NPV

NPV = 325 000.00 − 250 000 = €75 000.00

Answer

NPV = €75 000.00

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

4.NPV & IRR

Answer: MIRR = 17.62 %

Modified IRR: compound every inflow forward to the end of the project at the reinvestment rate (usually the cost of capital) to get the terminal value TV. MIRR is the rate that turns the initial investment into TV: MIRR = (TV/C₀)^(1/n) − 1.

Terminal value of the inflows

19 000·1.12 = 22 990.00
21 000·1.11 = 23 100.00
19 000·1.10 = 19 000.00
TV = 65 090.00

MIRR

MIRR = (65 090.00/40 000)1/3 − 1 = 17.62%

Answer

MIRR = 17.62%
5.Bond pricing & yield

Answer: HPR = 1.749 %

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 = €30.00, i = 0.04, N = 6
€30.00·(1 − 0.7903)/0.04 + €1 000·0.7903 = €157.26 + €790.31 = €947.58

Selling price

C = €30.00, i = 0.045, N = 5
€30.00·(1 − 0.8025)/0.045 + €1 000·0.8025 = €131.70 + €802.45 = €934.15

Holding period return

HPR = (30 + 934.15 − 947.58)/947.58 = 1.749%

Answer

HPR = 1.749%
6.Bond pricing & yield

Answer: YTC = 4.498 %

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

1166 = 70·(1 − (1 + y)−5)/y + 1070/(1 + y)5

Solve numerically

y = 4%: €1 191.09
y = 5%: €1 141.44
YTC = 4.498%

Answer

YTC = 4.498%

The yield to maturity is 5.001%. Investors in a premium bond expect it to be called, so the lower yield to call is the more realistic one.

7.CAPM & WACC

Answer: re = 13 %

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, E/V = 0.8333, D/V = 0.1667

Debt part of the WACC

0.1667·5%·(1 − 0.2) = 0.6667%

Cost of equity

re = (11.5% − 0.6667%)/0.8333 = 13%

Answer

re = 13%
8.CAPM & WACC

Answer: re = 13.25 %

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 + 0.79·0.5) = 1.075

Relever at the new D/E

βL = 1.075·(1 + 0.79·1.5) = 2.349

Cost of equity (CAPM)

re = 1.5% + 2.349·5% = 13.25%

Answer

re = 13.25%
9.Descriptive statistics

Answer: s = 18.79

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 = 119/5 = 23.8

Squared deviations

Σ(x − x̄)² = (−7.8)² + (−9.8)² + (−3.8)² + (−11.8)² + 33.2² = 1413

Standard deviation

s = √(1413/4) = 18.79

Answer

s = 18.79
10.Descriptive statistics

Answer: z = −1

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 = (158 − 170)/12 = −1

Answer

z = −1
11.Probability

Answer: N = 792

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(12, 5) = (12·11·10·9·8)/(5·4·3·2·1) = 792

Answer

N = 792
12.Probability

Answer: P = 0.07031

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) = 36·0.52·0.57 = 36·0.25·0.007813 = 0.07031

Answer

P = 0.07031
13.Normal distribution

Answer: z = 0.59

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 + 0.4448)/2 = 0.7224

Read the z-table backwards

Find 0.7224 inside the table and read off z.

z = 0.59

Answer

z = 0.59
14.Normal distribution

Answer: x = 189.5 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) = 0.95

z from the table

z ≈ 1.64

Back to the original units

x = 178 + 1.64·7 = 189.5 cm

Answer

x = 189.5 cm
15.Confidence intervals

Answer: upper limit = 12.39 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 = 3.707

Margin of error

E = 3.707·7.2/√7 = 10.09

Interval

2.3 ± 10.09 → [−7.788, 12.39]

Answer

upper limit = 12.39 mmHg
16.Confidence intervals

Answer: n = 260

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 = (2.576·25/4)² = 259.2

Round up

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

n = 260

Answer

n = 260
17.Hypothesis tests & confidence intervals

Answer: z = −2.073

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 = √(8²/40 + 3²/40) = √1.825 = 1.351

Test statistic

z = (131.5 − 134.3)/1.351 = −2.073

Answer

z = −2.073

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

18.Hypothesis tests & confidence intervals

Answer: n = 72

Sample size for a test: to detect a true shift δ with power 1 − β in a one-sided z test at level α, the standard error must satisfy δ = (z_α + z_β)·σ/√n, so n = ((z_α + z_β)·σ/δ)², rounded up.

Solve for n

n = ((1.645 + 0.842)·17/5)² = 71.5

Round up

n = 72

Answer

n = 72
19.Linear regression & correlation

Answer: r² = 0.91

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 = 21/6 = 3.5
ȳ = 125/6 = 20.83

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −25.5
Syy = Σ(y − ȳ)² = 40.83

Coefficient of determination

r² = −25.5²/(17.5·40.83) = 0.91

Answer

r² = 0.91

About 91% of the variation in y is explained by the line.

20.Linear regression & correlation

Answer: e = 2.167

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 = 147/6 = 24.5
ȳ = 425/6 = 70.83

Sums of squares and products

Sxx = Σ(x − x̄)² = 157.5
Sxy = Σ(x − x̄)(y − ȳ) = 502.5

Regression line

b = 502.5/157.5 = 3.19
a = 70.83 − 3.19·24.5 = −7.333
ŷ = −7.333 + 3.19·x

Predicted value

ŷ = −7.333 + 3.19·24.5 = 70.83

Residual

e = 73 − 70.83 = 2.167

Answer

e = 2.167

The point lies above the line.

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