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 = ?€40 000

You need €40 000 in 3 years. An account pays 6% per year, compounded quarterly. How much must you deposit today? Find the present value PV.

FV = €40 0006% / 4n = 3
PV = €
2.Time value of moneyHard
months0123…108€200€200€200FV = ?

You save €200 at the end of every month for 9 years. The account pays 3% per year, compounded monthly. How much have you saved at the end? Find FV.

A = €2003% / 12N = 108
FV = €
3.NPV & IRRMedium
years0123…9−100 0009 000+2%+2%…r = 12%, NPV = ?

A project costs €100 000 today. It generates €9 000 at the end of the first year, and the cash flow then grows by 2% per year forever. The cost of capital is 12%. Find the net present value NPV.

C0 = €100 000CF1 = €9 000g = 2%r = 12%
NPV = €
4.NPV & IRRHard
years01234−50 00016 50016 50016 50016 500

An investment of €50 000 returns €16 500 at the end of each year for 4 years. Find its internal rate of return IRR in percent.

C0 = €50 000CF = €16 500n = 4
IRR = %
5.Bond pricing & yieldMedium
years0123€906C = ?C = ?C = ? + €1 000

A bond with a face value of €1 000 matures in 3 years and pays annual coupons. It sells for €906 when the yield to maturity is 5.5%. Find the coupon rate c in percent.

F = €1 000n = 3P = €906YTM = 5.5%
c = %
6.Bond pricing & yieldHard
half-years0123…20€1 293€35€35€35 + €1 000

A bond with a face value of €1 000 pays a 7% annual coupon in two half-yearly payments and matures in 10 years. It sells for €1 293. Find its yield to maturity YTM as an annual rate (twice the half-yearly yield), in percent.

F = €1 000coupon = 7% (2×)n = 10P = €1 293
YTM = %
7.CAPM & WACCMedium
WACC = 9.28%equity€8 000 000re​ = ?debt€2 000 000rd​ = 3%T = 20%

A company has equity worth €8 000 000 and debt worth €2 000 000. Its WACC is 9.28%, its pre-tax cost of debt is 3% and the tax rate is 20%. Find the cost of equity re.

E = €8 000 000D = €2 000 000WACC = 9.28%rd = 3%T = 20%
re = %
8.CAPM & WACCHard
WACC = ?equity€5 000 000re​ = 12%bonds8 000 × €965coupon 5%, n = 9T = 25.8%

A company has equity with a market value of €5 000 000 and a cost of equity of 12%. Its debt consists of 8 000 bonds (face value €1 000, annual coupon 5%, 9 years to maturity) that trade at €965 each. The tax rate is 25.8%. Find the WACC, using the yield to maturity of the bonds as the pre-tax cost of debt.

E = €5 000 000re = 12%bonds: 8 000 × €965coupon = 5%n = 9T = 25.8%
WACC = %
9.Descriptive statisticsMedium
152025303540455055

Find the sample standard deviation s of the data: 17, 53, 55, 15, 21, 35, 41.

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

A quantity is normally distributed with mean 170 and standard deviation 12. What proportion P of values is above 177.2?

μ = 170σ = 12x = 177.2
P =
11.ProbabilityMedium
xp(x)00.0510.3520.1530.140.35μ = 2.35

For the distribution shown, find the standard deviation σ of X.

x: 0, 1, 2, 3, 4p(x): 0.05, 0.35, 0.15, 0.1, 0.35
σ =
12.ProbabilityHard
xp(x)00.2510.220.330.1540.1μ

X has the distribution shown, and Y = 0.5X − 3. Find the standard deviation σY of Y.

x: 0, 1, 2, 3, 4p(x): 0.25, 0.2, 0.3, 0.15, 0.1Y = 0.5X − 3
σY =
13.Normal distributionMedium
750 g754.9 gX + Y?

X is normal with mean 250 g and standard deviation 4 g; Y is independent and normal with mean 500 g and standard deviation 9 g. Find the probability P that X + Y is more than 754.9 g. Round z to two decimals.

μX = 250, σX = 4μY = 500, σY = 9X + Y > 754.9 g
P =
14.Normal distributionHard
0.02 mm0D − d?

Hole diameters D are normal with mean 25.02 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 = 25.02 mm, σ = 0.01 mmshaft: μ = 25 mm, σ = 0.02 mm
P =
15.Confidence intervalsMedium
n = 16, sd​ = 2.90L = ?Ud̄ = −3.6E99%

The blood pressure of 16 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean −3.6 and standard deviation 2.9. Find the lower limit of the 99% confidence interval for the mean difference.

n = 16d̄ = −3.6sd = 2.999%t(15) = 2.947
lower limit = mmHg
16.Confidence intervalsHard
σ = 19, n = ?x̄ − Ex̄ + Ex̄E = 299%

The population standard deviation is 19. 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 2?

σ = 19E = 299%
n =
17.Hypothesis tests & confidence intervalsMedium
−2.4472.447t = ?df = 6

7 employees take a typing test before and after a training course. For each person the difference d = after − before (words per minute) is recorded. The mean difference is 0.97 and the standard deviation of the differences is 3.2. Find the paired test statistic t for H₀: μd = 0.

n = 7d̄ = 0.97sd = 3.2
t =
18.Hypothesis tests & confidence intervalsHard
−1.961.96z = ?

In a trial, 75 of 250 patients on treatment A recover, and 59 of 150 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.

x₁ = 75n₁ = 250x₂ = 59n₂ = 150
z =
19.Linear regression & correlationMedium
xy145248347459563xy245060

The table gives 5 paired observations of x = advertising spend in k€ and y = sales in k€. Find Pearson's correlation coefficient r.

n = 5
r =
20.Linear regression & correlationHard
xy274146208271033xy510102030x = 5

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. Predict ŷ for x = 5.

n = 5x = 5
ŷ =

3Solutions

1.Time value of money

Answer: PV = 33455 €

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 = 0.06/4 = 0.015
N = 4·3 = 12

Present value

PV = FV/(1 + i)N = 40 000/1.196 = €33 455.50

Answer

PV = €33 455.50
2.Time value of money

Answer: FV = 24762 €

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 = 12·9 = 108

Future value of the deposits

FV = A·((1 + i)N − 1)/i = 200·(1.31 − 1)/0.0025 = €24 761.85

Answer

FV = €24 761.85

You paid in €21 600 yourself; the rest is interest.

3.NPV & IRR

Answer: NPV = −10000 €

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 = 9 000/(0.12 − 0.02) = 90 000.00

NPV

NPV = 90 000.00 − 100 000 = −€10 000.00

Answer

NPV = −€10 000.00

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

4.NPV & IRR

Answer: IRR = 12.11 %

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 = 50 000/16 500 = 3.03

Solve for the rate

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

i = 12%: 3.037
i = 13%: 2.974
IRR = 12.11%

Answer

IRR = 12.11%
5.Bond pricing & yield

Answer: c = 2.016 %

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 + 0.055)−3 = €851.61

Coupon

C = (906 − 851.61)/2.698 = €20.16

Coupon rate

c = 20.16/1000 = 2.016%

Answer

c = 2.016%
6.Bond pricing & yield

Answer: YTM = 3.502 %

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.

Per period

C = €35.00, N = 20

Solve numerically

i = 1.751% per half-year

YTM

YTM = 2·1.751% = 3.502%

Answer

YTM = 3.502%
7.CAPM & WACC

Answer: re = 11 %

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 = €10 000 000, E/V = 0.8, D/V = 0.2

Debt part of the WACC

0.2·3%·(1 − 0.2) = 0.48%

Cost of equity

re = (9.28% − 0.48%)/0.8 = 11%

Answer

re = 11%
8.CAPM & WACC

Answer: WACC = 7.195 %

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)

965 = 50·(1 − (1 + y)−9)/y + 1000/(1 + y)9
y = 5.504%

Market values

D = 8 000·€965 = €7 720 000
V = €12 720 000, E/V = 0.3931, D/V = 0.6069

WACC

WACC = 0.3931·12% + 0.6069·5.504%·(1 − 0.258) = 7.195%

Answer

WACC = 7.195%
9.Descriptive statistics

Answer: s = 16.69

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 = 237/7 = 33.86

Squared deviations

Σ(x − x̄)² = (−16.86)² + 19.14² + 21.14² + (−18.86)² + (−12.86)² + 1.143² + 7.143² = 1671

Standard deviation

s = √(1671/6) = 16.69

Answer

s = 16.69
10.Descriptive statistics

Answer: P = 0.2743

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 = 0.6

Standard normal table

Φ(0.6) = 0.7257
P = 1 − 0.7257 = 0.2743

Answer

P = 0.2743
11.Probability

Answer: σ = 1.388

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.

Mean

μ = 0·0.05 + 1·0.35 + 2·0.15 + 3·0.1 + 4·0.35 = 2.35

Mean of the squares

E(X²) = 0·0.05 + 1·0.35 + 4·0.15 + 9·0.1 + 16·0.35 = 7.45

Variance

Var(X) = 7.45 − 2.35² = 1.928
σ = √1.928 = 1.388

Answer

σ = 1.388
12.Probability

Answer: σY = 0.6379

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 = 1.65, Var(X) = 1.627, σX = 1.276

Linear change

σY = |0.5|·1.276 = 0.6379

Answer

σY = 0.6379

The mean does move: E(Y) = 0.5·1.65 − 3 = −2.175.

13.Normal distribution

Answer: P = 0.3085

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 sum

μ = 750 g, σ = √(4² + 9²) = 9.849 g

Standardise

z = (754.9 − 750)/9.849 = 0.5

Upper tail

P = 1 − Φ(0.5) = 1 − 0.6915 = 0.3085

Answer

P = 0.3085
14.Normal distribution

Answer: P = 0.8133

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 = 25.02 − 25 = 0.02 mm, σ = 0.02236 mm

Standardise

z = (0 − 0.02)/0.02236 = −0.89

Probability

P = 1 − Φ(−0.89) = 0.8133

Answer

P = 0.8133
15.Confidence intervals

Answer: lower limit = −5.737 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 = 15, t = 2.947

Margin of error

E = 2.947·2.9/√16 = 2.137

Interval

−3.6 ± 2.137 → [−5.737, −1.463]

Answer

lower limit = −5.737 mmHg
16.Confidence intervals

Answer: n = 599

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 = (2.576·19/2)² = 598.9

Round up

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

n = 599

Answer

n = 599
17.Hypothesis tests & confidence intervals

Answer: t = 0.802

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 of the mean difference

SE = 3.2/√7 = 1.209

Test statistic

t = 0.97/1.209 = 0.802

Answer

t = 0.802

With df = 6, the two-sided 5% critical value is 2.447; |t| is smaller, so H₀ is not rejected.

18.Hypothesis tests & confidence intervals

Answer: z = −1.915

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 = 75/250 = 0.3
p̂₂ = 59/150 = 0.3933
p̂ = 134/400 = 0.335

Standard error under H₀

SE = √(0.335·0.665·(1/250 + 1/150)) = 0.04875

Test statistic

z = (0.3 − 0.3933)/0.04875 = −1.915

Answer

z = −1.915

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

19.Linear regression & correlation

Answer: r = 0.9232

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 = 15/5 = 3
ȳ = 262/5 = 52.4

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 47
Syy = Σ(y − ȳ)² = 259.2

Correlation coefficient

r = 47/√(10·259.2) = 0.9232

Answer

r = 0.9232
20.Linear regression & correlation

Answer: ŷ = 16.95

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 = 101/5 = 20.2

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 130

Regression line

b = 130/40 = 3.25
a = 20.2 − 3.25·6 = 0.7
ŷ = 0.7 + 3.25·x

Prediction

ŷ = 0.7 + 3.25·5 = 16.95

Answer

ŷ = 16.95

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