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
years01234567PV = ?€20 000

You need €20 000 in 7 years. An account pays 3.6% per year, compounded quarterly. How much must you deposit today? Find the present value PV.

FV = €20 0003.6% / 4n = 7
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 4.2% per year, compounded monthly. Find the monthly payment A.

P = €20 0004.2% / 12N = 36
A = €
3.NPV & IRRMedium
years01234−45 00018 00013 00013 00010 000

An investment of €45 000 gives these cash flows at the end of years 1 to 4: €18 000, €13 000, €13 000, €10 000. The cost of capital is 8%. Find the net present value NPV.

r = 8%CF0 = −45 000CF1 = 18 000CF2 = 13 000CF3 = 13 000CF4 = 10 000
NPV = €
4.NPV & IRRHard
years012345−80 00014 500+4%+4%+4%+4%r = 12%, NPV = ?

A project costs €80 000 today. Its cash flow is €14 500 at the end of year 1 and grows by 4% per year until the last cash flow at the end of year 5. The cost of capital is 12%. Find the net present value NPV.

C0 = €80 000CF1 = €14 500g = 4%n = 5r = 12%
NPV = €
5.Bond pricing & yieldMedium
years0123…14€1 423C = ?C = ?C = ? + €1 000

A bond with a face value of €1 000 matures in 14 years and pays annual coupons. It sells for €1 423 when the yield to maturity is 4%. Find the coupon rate c in percent.

F = €1 000n = 14P = €1 423YTM = 4%
c = %
6.Bond pricing & yieldHard
half-years0123…18€1 144€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 9 years. It sells for €1 144. Find its yield to maturity YTM as an annual rate (twice the half-yearly yield), in percent.

F = €1 000coupon = 7% (2×)n = 9P = €1 144
YTM = %
7.CAPM & WACCMedium
WACC = ?equity€6 000 000re​ = 14%debt€5 000 000rd​ = 3%T = 21%

A company has equity worth €6 000 000 and debt worth €5 000 000. Its cost of equity is 14%, its pre-tax cost of debt is 3% and the tax rate is 21%. Find the WACC.

E = €6 000 000D = €5 000 000re = 14%rd = 3%T = 21%
WACC = %
8.CAPM & WACCHard
WACC = ?equity€6 000 000re​ = 14%bonds2 000 × €915coupon 7%, n = 8T = 25.8%

A company has equity with a market value of €6 000 000 and a cost of equity of 14%. Its debt consists of 2 000 bonds (face value €1 000, annual coupon 7%, 8 years to maturity) that trade at €915 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 = €6 000 000re = 14%bonds: 2 000 × €915coupon = 7%n = 8T = 25.8%
WACC = %
9.Descriptive statisticsMedium
2025303540455055

Find the sample standard deviation s of the data: 38, 59, 19, 28, 19, 29, 48.

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

Exam scores are normally distributed with mean 50 and standard deviation 8. Find the z-score z of a score of 38.

μ = 50σ = 8x = 38
z =
11.ProbabilityMedium
xp(x)00.1510.3520.3530.15μ = 1.5

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

x: 0, 1, 2, 3p(x): 0.15, 0.35, 0.35, 0.15
σ =
12.ProbabilityHard
n = 8, p = 0.7: P(X = 5) = ?k012345678

A test has 8 multiple-choice questions. A student who guesses gets each one right with probability 0.7, independently. What is the probability P of exactly 5 correct answers?

n = 8p = 0.7k = 5
P =
13.Normal distributionMedium
300 kg302.6 kgT = X₁ + … + X₁₂?

The masses of bags of cement are independent and normal with mean 25 kg and standard deviation 0.5 kg. Find the probability P that the total mass of 12 of them exceeds 302.6 kg. Round z to two decimals.

μ = 25 kgσ = 0.5 kgn = 12> 302.6 kg
P =
14.Normal distributionHard
10 g0X − Y?

The mass X of a product from line 1 and Y from line 2 are independent and normal: X has mean 500 g and standard deviation 8 g, Y has mean 490 g and standard deviation 6 g. Find the probability P that X is greater than Y. Round z to two decimals.

μX = 500 g, σX = 8 gμY = 490 g, σY = 6 g
P =
15.Confidence intervalsMedium
n = 15, s = 6.5LUx̄ = 90.5E = ?90%

A sample of 15 items has mean 90.5 and standard deviation s = 6.5. The population is roughly normal. Find the margin of error E of a 90% confidence interval for the mean.

n = 15x̄ = 90.5s = 6.590%t(14) = 1.761
E =
16.Confidence intervalsHard
σ = 21, n = ?x̄ − Ex̄ + Ex̄E = 0.599%

The population standard deviation is 21. 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 0.5?

σ = 21E = 0.599%
n =
17.Hypothesis tests & confidence intervalsMedium
−1.961.96z = ?

Two production lines make the same part. Line 1: a sample of 50 parts has mean length 132.5 mm, and the line's known standard deviation is 10 mm. Line 2: a sample of 40 parts has mean 137.5 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 = 50x̄₁ = 132.5 mmσ₁ = 10 mmn₂ = 40x̄₂ = 137.5 mmσ₂ = 5 mm
z =
18.Hypothesis tests & confidence intervalsHard
σ = 10, n = ?x̄ − Ex̄ + Ex̄E = 299%

How large must a sample be so that the 99% confidence interval for a mean has a margin of error of at most 2? The standard deviation is 10. Find the smallest sample size n.

σ = 10E = 299%
n =
19.Linear regression & correlationMedium
xy16481956227025772882xy20256080

The table gives 5 paired observations of x = temperature in °C and y = ice creams sold. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 5Σx = 110Σy = 333Σx² = 2510Σy² = 22993Σxy = 7593
r =
20.Linear regression & correlationHard
xy156718822190241092712530149xy15202530100150x = 16.5

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 (16.5, 67) is made. Find its residual e = y − ŷ.

n = 6(16.5, 67)
e =

3Solutions

1.Time value of money

Answer: PV = 15562 €

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 = 0.036000000000000004/4 = 0.009
N = 4·7 = 28

Present value

PV = FV/(1 + i)N = 20 000/1.285 = €15 562.43

Answer

PV = €15 562.43
2.Time value of money

Answer: A = 592.3 €

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 = 0.042/12 = 0.0035
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 = 0.8818
A = 20 000·0.0035/(1 − 0.8818) = €592.26

Answer

A = €592.26
3.NPV & IRR

Answer: NPV = 482.2 €

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 = −45 000
CF1/(1 + 0.08)1 = 18 000/1.08 = 16 666.67
CF2/(1 + 0.08)2 = 13 000/1.166 = 11 145.40
CF3/(1 + 0.08)3 = 13 000/1.26 = 10 319.82
CF4/(1 + 0.08)4 = 10 000/1.36 = 7 350.30

Add them up

NPV = €482.19

Answer

NPV = €482.19

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

4.NPV & IRR

Answer: NPV = −23878 €

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 the growing annuity

((1.04)/(1.12))5 = 0.6904
PV = 14 500/(0.12 − 0.04)·(1 − 0.6904) = 56 121.97

NPV

NPV = 56 121.97 − 80 000 = −€23 878.03

Answer

NPV = −€23 878.03

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

5.Bond pricing & yield

Answer: c = 8.004 %

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 = (1423 − 577.48)/10.56 = €80.04

Coupon rate

c = 80.04/1000 = 8.004%

Answer

c = 8.004%
6.Bond pricing & yield

Answer: YTM = 4.994 %

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 = 18

Solve numerically

i = 2.497% per half-year

YTM

YTM = 2·2.497% = 4.994%

Answer

YTM = 4.994%
7.CAPM & WACC

Answer: WACC = 8.714 %

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 = €11 000 000, E/V = 0.5455, D/V = 0.4545

WACC

WACC = 0.5455·14% + 0.4545·3%·(1 − 0.21) = 8.714%

Answer

WACC = 8.714%
8.CAPM & WACC

Answer: WACC = 12.2 %

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)

915 = 70·(1 − (1 + y)−8)/y + 1000/(1 + y)8
y = 8.508%

Market values

D = 2 000·€915 = €1 830 000
V = €7 830 000, E/V = 0.7663, D/V = 0.2337

WACC

WACC = 0.7663·14% + 0.2337·8.508%·(1 − 0.258) = 12.2%

Answer

WACC = 12.2%
9.Descriptive statistics

Answer: s = 14.99

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 = 240/7 = 34.29

Squared deviations

Σ(x − x̄)² = 3.714² + 24.71² + (−15.29)² + (−6.286)² + (−15.29)² + (−5.286)² + 13.71² = 1347

Standard deviation

s = √(1347/6) = 14.99

Answer

s = 14.99
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 = (38 − 50)/8 = −1.5

Answer

z = −1.5
11.Probability

Answer: σ = 0.922

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 = 0·0.15 + 1·0.35 + 2·0.35 + 3·0.15 = 1.5

Mean of the squares

E(X²) = 0·0.15 + 1·0.35 + 4·0.35 + 9·0.15 = 3.1

Variance

Var(X) = 3.1 − 1.5² = 0.85
σ = √0.85 = 0.922

Answer

σ = 0.922
12.Probability

Answer: P = 0.2541

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) = 56·0.75·0.33 = 56·0.1681·0.027 = 0.2541

Answer

P = 0.2541
13.Normal distribution

Answer: P = 0.06681

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 total

μT = 300 kg, σT = 0.5·√12 = 1.732 kg

Standardise

z = (302.6 − 300)/1.732 = 1.5

Probability

P = 1 − Φ(1.5) = 0.06681

Answer

P = 0.06681
14.Normal distribution

Answer: P = 0.8413

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 = 500 − 490 = 10 g, σ = 10 g

Standardise

z = (0 − 10)/10 = −1

Probability

P = 1 − Φ(−1) = 0.8413

Answer

P = 0.8413
15.Confidence intervals

Answer: E = 2.955

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.

Critical value

df = n − 1 = 14, t = 1.761

Margin of error

E = 1.761·6.5/√15 = 2.955

Answer

E = 2.955
16.Confidence intervals

Answer: n = 11706

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 = (2.576·21/0.5)² = 11706

Round up

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

n = 11706

Answer

n = 11706
17.Hypothesis tests & confidence intervals

Answer: z = −3.086

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 = √(10²/50 + 5²/40) = √2.625 = 1.62

Test statistic

z = (132.5 − 137.5)/1.62 = −3.086

Answer

z = −3.086

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

18.Hypothesis tests & confidence intervals

Answer: n = 166

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 = (2.576·10/2)² = 165.9

Round up

n = 166

Answer

n = 166
19.Linear regression & correlation

Answer: r = 0.9857

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.

Centred sums

Sxy = 7593 − 110·333/5 = 267
Sxx = 2510 − 110²/5 = 90
Syy = 22993 − 333²/5 = 815.2

Correlation coefficient

r = 267/√(90·815.2) = 0.9857

Answer

r = 0.9857
20.Linear regression & correlation

Answer: e = −4.781

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 = 135/6 = 22.5
ȳ = 622/6 = 103.7

Sums of squares and products

Sxx = Σ(x − x̄)² = 157.5
Sxy = Σ(x − x̄)(y − ȳ) = 837

Regression line

b = 837/157.5 = 5.314
a = 103.7 − 5.314·22.5 = −15.9
ŷ = −15.9 + 5.314·x

Predicted value

ŷ = −15.9 + 5.314·16.5 = 71.78

Residual

e = 67 − 71.78 = −4.781

Answer

e = −4.781

The point lies below 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