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…19PV = ?€2 000€2 000€2 000€2 000

An annuity pays €2 000 at the end of each year for 19 years. The interest rate is 8% per year. Find the present value PV of the annuity.

A = €2 000i = 8%n = 19
PV = €
2.Time value of moneyHard
months0123…360€250 000A?AAA

A loan of €250 000 is repaid with equal monthly payments over 30 years. The interest rate is 4.8% per year, compounded monthly. Find the monthly payment A.

P = €250 0004.8% / 12N = 360
A = €
3.NPV & IRRMedium
years01234567−80 00016 00016 00016 00016 00016 00016 00036 000

A machine costs €80 000 today. It generates €16 000 at the end of each year for 7 years and can then be sold for €20 000. The cost of capital is 10%. Find the net present value NPV.

C0 = €80 000CF = €16 000S = €20 000n = 7r = 10%
NPV = €
4.NPV & IRRHard
years01234−20 0008 0008 5007 0008 000

A project costs €20 000 and returns €8 000, €8 500, €7 000, €8 000 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 = −20 000CF1 = 8 000CF2 = 8 500CF3 = 7 000CF4 = 8 000
MIRR = %
5.Bond pricing & yieldMedium
half-years0123…24€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 12 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 = 12
P = €
6.Bond pricing & yieldHard
half-years0123…10€1 210€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 5 years. It sells for €1 210. Find its yield to maturity YTM as an annual rate (twice the half-yearly yield), in percent.

F = €1 000coupon = 7% (2×)n = 5P = €1 210
YTM = %
7.CAPM & WACCMedium
WACC = ?equity€5 000 000re​ = 13%debt€6 000 000rd​ = 3%T = 30%

A company has equity worth €5 000 000 and debt worth €6 000 000. Its cost of equity is 13%, its pre-tax cost of debt is 3% and the tax rate is 30%. Find the WACC.

E = €5 000 000D = €6 000 000re = 13%rd = 3%T = 30%
WACC = %
8.CAPM & WACCHard
WACC = ?equity5 000 000 × €17.00β = 1.2rf​ = 1%, rm​ = 6.5%debt5 000 × €1 000rd​ = 5%T = 21%

A company has 5 000 000 shares at €17.00 each and 5 000 bonds trading at €1 000 each. The pre-tax cost of debt is 5% and the tax rate is 21%. The share’s beta is 1.2, the risk-free rate 1% and the market return 6.5%. Find the WACC.

β = 1.2rf = 1%rm = 6.5%rd = 5%T = 21%
WACC = %
9.Descriptive statisticsMedium
152025303540455055

Find the sample standard deviation s of the data: 55, 29, 15, 54, 33.

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

Exam scores are normally distributed with mean 70 and standard deviation 5. Find the z-score z of a score of 76.25.

μ = 70σ = 5x = 76.25
z =
11.ProbabilityMedium
n = 12k = 2N = ?

In how many ways N can a committee of 2 people be chosen from a group of 12? The order does not matter.

n = 12k = 2
N =
12.ProbabilityHard
n = 12, p = 0.7: P(X ≤ 1) = ?k0123456789101112

12 components are tested, and each one fails with probability 0.7, independently. What is the probability P that at most one fails?

n = 12p = 0.7
P =
13.Normal distributionMedium
32 min30 minX + Y?

X is normal with mean 12 min and standard deviation 1.5 min; Y is independent and normal with mean 20 min and standard deviation 2 min. Find the probability P that X + Y is more than 30 min. Round z to two decimals.

μX = 12, σX = 1.5μY = 20, σY = 2X + Y > 30 min
P =
14.Normal distributionHard
μ = 178x = ?10%

The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Find the value x below which 10% of them lie (the 10th percentile).

μ = 178σ = 7P = 10%
x = cm
15.Confidence intervalsMedium
μ₁ − μ₂0L = ?Ux̄₁ − x̄₂E99%

The battery life of two phone models is compared. Model 1: 60 phones, mean 128 h, known standard deviation 19 h. Model 2: 30 phones, mean 143 h, known standard deviation 10 h. Find the lower limit of the 99% confidence interval for μ₁ − μ₂.

n₁ = 60x̄₁ = 128 hσ₁ = 19 hn₂ = 30x̄₂ = 143 hσ₂ = 10 h99%
lower limit = h
16.Confidence intervalsHard
μ₁ − μ₂, t(14) = 2.1450L = ?Ux̄₁ − x̄₂E95%

Two fertilisers are tested. Plot group 1: 11 plots, mean yield 98 kg, standard deviation 8.5 kg. Plot group 2: 5 plots, mean 93 kg, standard deviation 1 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the lower limit of the 95% confidence interval for μ₁ − μ₂.

n₁ = 11x̄₁ = 98s₁ = 8.5n₂ = 5x̄₂ = 93s₂ = 195%t(14) = 2.145
lower limit = kg
17.Hypothesis tests & confidence intervalsMedium
t −1.99801.998

A sample of 64 students has a mean score of 118.9 and a sample standard deviation of 4. Find the test statistic t for H₀: μ = 120.

μ₀ = 120x̄ = 118.9s = 4n = 64
t =
18.Hypothesis tests & confidence intervalsHard
σ = 20, n = ?x̄ − Ex̄ + Ex̄E = 290%

How large must a sample be so that the 90% confidence interval for a mean has a margin of error of at most 2? The standard deviation is 20. Find the smallest sample size n.

σ = 20E = 290%
n =
19.Linear regression & correlationMedium
xy24544766086510731284xy5106080

The table gives 6 paired observations of x = advertising spend in k€ and y = sales in k€. Find Pearson's correlation coefficient r.

n = 6
r =
20.Linear regression & correlationHard
xy18432054225824622670xy2025506070x = 19

The table gives 5 paired observations of x = temperature in °C and y = ice creams sold. Use the least-squares line ŷ = a + b·x. A new observation (19, 44) is made. Find its residual e = y − ŷ.

n = 5(19, 44)
e =

3Solutions

1.Time value of money

Answer: PV = 19207 €

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.

Annuity factor

(1 − (1 + i)−n)/i = (1 − 0.2317)/0.08 = 9.604

Present value

PV = 2 000·9.604 = €19 207.20

Answer

PV = €19 207.20
2.Time value of money

Answer: A = 1312 €

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 = 0.048/12 = 0.004
N = 12·30 = 360

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 = 0.2376
A = 250 000·0.004/(1 − 0.2376) = €1 311.66

Answer

A = €1 311.66
3.NPV & IRR

Answer: NPV = 8158 €

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 yearly cash flows

16 000·4.868 = 77 894.70

Present value of the salvage value

20 000/1.949 = 10 263.16

NPV

NPV = −80 000 + 77 894.70 + 10 263.16 = €8 157.86

Answer

NPV = €8 157.86
4.NPV & IRR

Answer: MIRR = 14.6 %

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

8 000·1.063 = 9 528.13
8 500·1.062 = 9 550.60
7 000·1.061 = 7 420.00
8 000·1.060 = 8 000.00
TV = 34 498.73

MIRR

MIRR = (34 498.73/20 000)1/4 − 1 = 14.6%

Answer

MIRR = 14.6%
5.Bond pricing & yield

Answer: P = 869.5 €

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 = 24

Present value of the coupons

€20.00·(1 − 0.5215)/0.0275 = €348.02

Present value of the face value

€1 000·0.5215 = €521.48

Price

P = €869.49

Answer

P = €869.49

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

6.Bond pricing & yield

Answer: YTM = 2.505 %

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 = 10

Solve numerically

i = 1.253% per half-year

YTM

YTM = 2·1.253% = 2.505%

Answer

YTM = 2.505%
7.CAPM & WACC

Answer: WACC = 7.055 %

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 = 0.4545, D/V = 0.5455

WACC

WACC = 0.4545·13% + 0.5455·3%·(1 − 0.3) = 7.055%

Answer

WACC = 7.055%
8.CAPM & WACC

Answer: WACC = 7.397 %

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·€17.00 = €85 000 000
D = 5 000·€1 000 = €5 000 000

Cost of equity (CAPM)

re = 1% + 1.2·5.5% = 7.6%

Weights

V = €90 000 000, E/V = 0.9444, D/V = 0.05556

WACC

WACC = 0.9444·7.6% + 0.05556·5%·(1 − 0.21) = 7.397%

Answer

WACC = 7.397%
9.Descriptive statistics

Answer: s = 17.15

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 = 186/5 = 37.2

Squared deviations

Σ(x − x̄)² = 17.8² + (−8.2)² + (−22.2)² + 16.8² + (−4.2)² = 1177

Standard deviation

s = √(1177/4) = 17.15

Answer

s = 17.15
10.Descriptive statistics

Answer: z = 1.25

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 = (76.25 − 70)/5 = 1.25

Answer

z = 1.25
11.Probability

Answer: N = 66

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, 2) = (12·11)/(2·1) = 66

Answer

N = 66
12.Probability

Answer: P = 1.541 × 10⁻⁵

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.

Zero and one failure

P(X = 0) = 0.312 = 5.314 × 10⁻⁷
P(X = 1) = 12·0.7·0.311 = 1.488 × 10⁻⁵

Add them

P = 5.314 × 10⁻⁷ + 1.488 × 10⁻⁵ = 1.541 × 10⁻⁵

Answer

P = 1.541 × 10⁻⁵
13.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 sum

μ = 32 min, σ = √(1.5² + 2²) = 2.5 min

Standardise

z = (30 − 32)/2.5 = −0.8

Upper tail

P = 1 − Φ(−0.8) = 1 − 0.2119 = 0.7881

Answer

P = 0.7881
14.Normal distribution

Answer: x = 169 cm

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 for the percentile

Φ(z) = 0.1 → z = −1.282

Back to x

x = 178 + −1.282·7 = 169.03 cm

Answer

x = 169.03 cm
15.Confidence intervals

Answer: lower limit = −22.88 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 = 128 − 143 = −15

Standard error

SE = √(19²/60 + 10²/30) = 3.058

Margin of error

E = 2.576·3.058 = 7.877

Interval

−15 ± 7.877 → [−22.88, −7.123]

Answer

lower limit = −22.88 h

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

16.Confidence intervals

Answer: lower limit = −3.334 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 = (10·8.5² + 4·1²)/14 = 51.89
sp = 7.204

Margin of error

E = 2.145·7.204·√(1/11 + 1/5) = 8.334

Interval

5 ± 8.334 → [−3.334, 13.33]

Answer

lower limit = −3.334 kg
17.Hypothesis tests & confidence intervals

Answer: t = −2.2

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

SE = 4/√64 = 4/8 = 0.5

Test statistic

t = (118.9 − 120)/0.5 = −2.2

Answer

t = −2.2
t0.025; df = 63 = 1.998

|t| > 1.998, so H₀ is rejected at the 5% level (two-sided).

18.Hypothesis tests & confidence intervals

Answer: n = 271

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 = (1.645·20/2)² = 270.6

Round up

n = 271

Answer

n = 271
19.Linear regression & correlation

Answer: r = 0.9879

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 = 42/6 = 7
ȳ = 374/6 = 62.33

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 278
Syy = Σ(y − ȳ)² = 1131

Correlation coefficient

r = 278/√(70·1131) = 0.9879

Answer

r = 0.9879
20.Linear regression & correlation

Answer: e = −4.1

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 = 110/5 = 22
ȳ = 287/5 = 57.4

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 124

Regression line

b = 124/40 = 3.1
a = 57.4 − 3.1·22 = −10.8
ŷ = −10.8 + 3.1·x

Predicted value

ŷ = −10.8 + 3.1·19 = 48.1

Residual

e = 44 − 48.1 = −4.1

Answer

e = −4.1

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