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
periods01211 + EAR = ?15% / 2

A credit card charges a nominal interest rate of 15% per year, compounded semi-annually. Find the effective annual rate EAR in percent.

r = 15%m = 2
EAR = %
2.Time value of moneyHard
months0123…84A?AA€30 000

You want €30 000 in 7 years. You save the same amount at the end of every month in an account paying 4.8% per year, compounded monthly. Find the monthly deposit A.

FV = €30 0004.8% / 12N = 84
A = €
3.NPV & IRRMedium
years012345−20 0006 5006 5006 5006 5006 500

A project costs €20 000 today and returns €6 500 at the end of each year for 5 years. The cost of capital is 12%. Assume cash flows arrive evenly through each year. Find the discounted payback period DPB in years.

r = 12%CF0 = −20 000CF1 = 6 500CF2 = 6 500CF3 = 6 500CF4 = 6 500CF5 = 6 500
DPB = years
4.NPV & IRRHard
years01234−40 00018 50013 00022 50019 500

A project costs €40 000 and returns €18 500, €13 000, €22 500, €19 500 at the end of years 1 to 4. Cash flows can be reinvested at the cost of capital of 6%. Find the modified internal rate of return MIRR in percent.

r = 6%CF0 = −40 000CF1 = 18 500CF2 = 13 000CF3 = 22 500CF4 = 19 500
MIRR = %
5.Bond pricing & yieldMedium
half-years0123…14€20€20€20 + €1 000P = ?

A bond with a face value of €1 000 pays a 4% annual coupon in two half-yearly payments and matures in 7 years. The yield to maturity is 5.5% per year. Find the price P of the bond.

F = €1 000coupon = 4%YTM = 5.5%n = 7
P = €
6.Bond pricing & yieldHard
years012345€1 060€80€80 + €1 000

A bond with a face value of €1 000 and an annual coupon of 8% matures in 5 years and sells for €1 060. Find its yield to maturity YTM in percent.

F = €1 000coupon = 8%n = 5P = €1 060
YTM = %
7.CAPM & WACCMedium
WACC = ?equity€4 000 000re​ = 12%debt€1 000 000rd​ = 3%T = 25%

A company has equity worth €4 000 000 and debt worth €1 000 000. Its cost of equity is 12%, its pre-tax cost of debt is 3% and the tax rate is 25%. Find the WACC.

E = €4 000 000D = €1 000 000re = 12%rd = 3%T = 25%
WACC = %
8.CAPM & WACCHard
WACC = ?equity2 000 000 × €54.00β = 0.8rf​ = 1%, rm​ = 7.5%debt10 000 × €980rd​ = 4%T = 20%

A company has 2 000 000 shares at €54.00 each and 10 000 bonds trading at €980 each. The pre-tax cost of debt is 4% and the tax rate is 20%. The share’s beta is 0.8, the risk-free rate 1% and the market return 7.5%. Find the WACC.

β = 0.8rf = 1%rm = 7.5%rd = 4%T = 20%
WACC = %
9.Descriptive statisticsMedium
152025303540455055

Find the sample standard deviation s of the data: 12, 38, 17, 53, 47, 16, 56.

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

A quantity is normally distributed with mean 60 and standard deviation 5. What proportion P of values is below 62?

μ = 60σ = 5x = 62
P =
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 = 7, p = 0.4: P(X = 5) = ?k01234567

A test has 7 multiple-choice questions. A student who guesses gets each one right with probability 0.4, independently. What is the probability P of exactly 5 correct answers?

n = 7p = 0.4k = 5
P =
13.Normal distributionMedium
−3−2−10123zshaded area to the right of z = 0.0239

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.0239. Find z to two decimals.

A = 0.0239
z =
14.Normal distributionHard
35455565758595pointsbottom 40%

The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Drag the cut-off so the shaded bottom 40% is to the left of it. Find the cut-off x.

μ = 65 pointsσ = 10 points40%
x = points
15.Confidence intervalsMedium
μ₁ − μ₂0LU = ?x̄₁ − x̄₂E90%

The battery life of two phone models is compared. Model 1: 60 phones, mean 122.5 h, known standard deviation 14 h. Model 2: 60 phones, mean 114 h, known standard deviation 14 h. Find the upper limit of the 90% confidence interval for μ₁ − μ₂.

n₁ = 60x̄₁ = 122.5 hσ₁ = 14 hn₂ = 60x̄₂ = 114 hσ₂ = 14 h90%
upper limit = h
16.Confidence intervalsHard
σ = 28, n = ?x̄ − Ex̄ + Ex̄E = 1.595%

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 1.5?

σ = 28E = 1.595%
n =
17.Hypothesis tests & confidence intervalsMedium
−1.961.96z = ?

Two production lines make the same part. Line 1: a sample of 80 parts has mean length 148.5 mm, and the line's known standard deviation is 4 mm. Line 2: a sample of 50 parts has mean 146.1 mm, known standard deviation 9 mm. Find the test statistic z for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (line 1 minus line 2).

n₁ = 80x̄₁ = 148.5 mmσ₁ = 4 mmn₂ = 50x̄₂ = 146.1 mmσ₂ = 9 mm
z =
18.Hypothesis tests & confidence intervalsHard
−1.961.96z = ?

In a trial, 68 of 150 patients on treatment A recover, and 61 of 120 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.

x₁ = 68n₁ = 150x₂ = 61n₂ = 120
z =
19.Linear regression & correlationMedium
xy1192203154115965xy2465101520

The table gives 6 paired observations of x = car age in years and y = price in k€. Find Pearson's correlation coefficient r.

n = 6
r =
20.Linear regression & correlationHard
xy264156188241032xy510102030x = 9

The table gives 5 paired observations of x = load in kN and y = extension in mm. Use the least-squares line ŷ = a + b·x. A new observation (9, 31) is made. Find its residual e = y − ŷ.

n = 5(9, 31)
e =

3Solutions

1.Time value of money

Answer: EAR = 15.56 %

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 per period

r/m = 0.15/2 = 0.075

Compound over one year

EAR = (1 + 0.075)2 − 1 = 1.156 − 1 = 15.56%

Answer

EAR = 15.56%

The EAR is always above the nominal rate when interest compounds more than once a year.

2.Time value of money

Answer: A = 301.2 €

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 = 0.048/12 = 0.004
N = 12·7 = 84

Deposit

A = FV·i/((1 + i)N − 1) = 30 000·0.004/(1.398 − 1) = €301.20

Answer

A = €301.20
3.NPV & IRR

Answer: DPB = 4.07 years

Payback period: the time until the cumulative cash flows repay the initial investment. Cash is assumed to come in evenly during each year, so the last part is a fraction of a year. The simple payback ignores the time value of money; the discounted payback period uses the discounted cash flows instead.

Discounted and cumulative cash flows

year 1: 6 500/1.12 = 5 803.57 → Σ = 5 803.57
year 2: 6 500/1.254 = 5 181.76 → Σ = 10 985.33
year 3: 6 500/1.405 = 4 626.57 → Σ = 15 611.90
year 4: 6 500/1.574 = 4 130.87 → Σ = 19 742.77
year 5: 6 500/1.762 = 3 688.27 → Σ = 23 431.05

Fraction of the last year

DPB = 4 + 257.23/3 688.27 = 4.07

Answer

DPB = 4.07 years

The discounted payback is always longer than the simple payback (3.077 years).

4.NPV & IRR

Answer: MIRR = 18.92 %

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

18 500·1.063 = 22 033.80
13 000·1.062 = 14 606.80
22 500·1.061 = 23 850.00
19 500·1.060 = 19 500.00
TV = 79 990.60

MIRR

MIRR = (79 990.60/40 000)1/4 − 1 = 18.92%

Answer

MIRR = 18.92%
5.Bond pricing & yield

Answer: P = 913.8 €

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 = €20.00, i = 0.0275, N = 14

Present value of the coupons

€20.00·(1 − 0.684)/0.0275 = €229.82

Present value of the face value

€1 000·0.684 = €684.00

Price

P = €913.82

Answer

P = €913.82

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

6.Bond pricing & yield

Answer: YTM = 6.554 %

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

€1 060 = €80·(1 − (1 + y)⁻⁵)/y + €1 000/(1 + y)⁵

Solve numerically

P(6%) = €1 084.25, P(7%) = €1 041.00
YTM = 6.554%

Answer

YTM = 6.554%
7.CAPM & WACC

Answer: WACC = 10.05 %

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 = €5 000 000, E/V = 0.8, D/V = 0.2

WACC

WACC = 0.8·12% + 0.2·3%·(1 − 0.25) = 10.05%

Answer

WACC = 10.05%
8.CAPM & WACC

Answer: WACC = 5.95 %

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·€54.00 = €108 000 000
D = 10 000·€980 = €9 800 000

Cost of equity (CAPM)

re = 1% + 0.8·6.5% = 6.2%

Weights

V = €117 800 000, E/V = 0.9168, D/V = 0.08319

WACC

WACC = 0.9168·6.2% + 0.08319·4%·(1 − 0.2) = 5.95%

Answer

WACC = 5.95%
9.Descriptive statistics

Answer: s = 18.83

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 = (−22.14)² + 3.857² + (−17.14)² + 18.86² + 12.86² + (−18.14)² + 21.86² = 2127

Standard deviation

s = √(2127/6) = 18.83

Answer

s = 18.83
10.Descriptive statistics

Answer: P = 0.6554

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 = 0.4

Standard normal table

Φ(0.4) = 0.6554
P = 0.6554

Answer

P = 0.6554
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 = 0.07741

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 = 5) = 21·0.45·0.62 = 21·0.01024·0.36 = 0.07741

Answer

P = 0.07741
13.Normal distribution

Answer: z = 1.98

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 − 0.0239 = 0.9761

Read the z-table backwards

Find 0.9761 inside the table and read off z.

z = 1.98

Answer

z = 1.98
14.Normal distribution

Answer: x = 62.5 points

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) = 0.4

z from the table

z ≈ −0.25

Back to the original units

x = 65 − 0.25·10 = 62.5 points

Answer

x = 62.5 points
15.Confidence intervals

Answer: upper limit = 12.7 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 = 122.5 − 114 = 8.5

Standard error

SE = √(14²/60 + 14²/60) = 2.556

Margin of error

E = 1.645·2.556 = 4.205

Interval

8.5 ± 4.205 → [4.295, 12.7]

Answer

upper limit = 12.7 h

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

16.Confidence intervals

Answer: n = 1339

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 = (1.96·28/1.5)² = 1339

Round up

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

n = 1339

Answer

n = 1339
17.Hypothesis tests & confidence intervals

Answer: z = 1.779

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 = √(4²/80 + 9²/50) = √1.82 = 1.349

Test statistic

z = (148.5 − 146.1)/1.349 = 1.779

Answer

z = 1.779

|z| < 1.96, so H₀ is not rejected at the 5% level (two-sided).

18.Hypothesis tests & confidence intervals

Answer: z = −0.899

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.

Sample proportions

p̂₁ = 68/150 = 0.4533
p̂₂ = 61/120 = 0.5083
p̂ = 129/270 = 0.4778

Standard error under H₀

SE = √(0.4778·0.5222·(1/150 + 1/120)) = 0.06118

Test statistic

z = (0.4533 − 0.5083)/0.06118 = −0.899

Answer

z = −0.899

|z| < 1.96, so H₀ is not rejected at the 5% level (two-sided).

19.Linear regression & correlation

Answer: r = −0.9728

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 = 79/6 = 13.17

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −53.5
Syy = Σ(y − ȳ)² = 172.8

Correlation coefficient

r = −53.5/√(17.5·172.8) = −0.9728

Answer

r = −0.9728
20.Linear regression & correlation

Answer: e = 2.85

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 = 95/5 = 19

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 122

Regression line

b = 122/40 = 3.05
a = 19 − 3.05·6 = 0.7
ŷ = 0.7 + 3.05·x

Predicted value

ŷ = 0.7 + 3.05·9 = 28.15

Residual

e = 31 − 28.15 = 2.85

Answer

e = 2.85

The point lies above the line.

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