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
years012345€1 000FV = ?

You invest €1 000 at a nominal rate of 4% per year, compounded monthly. Find the value FV of the investment after 5 years.

PV = €1 0004% / 12n = 5
FV = €
2.Time value of moneyHard
months0123…N = ?€12 000€275€275€275€2753.6% / 12

You borrow €12 000 at 3.6% per year, compounded monthly, and repay €275 at the end of every month. How many months does it take until the loan is repaid (the last payment may be smaller)? Give the number of payments N.

P = €12 000A = €2753.6% / 12
N = payments
3.NPV & IRRMedium
years01234−20 0006 5006 5006 5006 500

A project costs €20 000 today and returns €6 500 at the end of each year for 4 years. The cost of capital is 8%. Assume cash flows arrive evenly through each year. Find the discounted payback period DPB in years.

r = 8%CF0 = −20 000CF1 = 6 500CF2 = 6 500CF3 = 6 500CF4 = 6 500
DPB = years
4.NPV & IRRHard
years01234−25 00012 5009 5009 5009 500

A project costs €25 000 and returns €12 500, €9 500, €9 500, €9 500 at the end of years 1 to 4. Cash flows can be reinvested at the cost of capital of 12%. Find the modified internal rate of return MIRR in percent.

r = 12%CF0 = −25 000CF1 = 12 500CF2 = 9 500CF3 = 9 500CF4 = 9 500
MIRR = %
5.Bond pricing & yieldMedium
half-years0123…20€30€30€30 + €1 000P = ?

A bond with a face value of €1 000 pays a 6% annual coupon in two half-yearly payments and matures in 10 years. The yield to maturity is 5% per year. Find the price P of the bond.

F = €1 000coupon = 6%YTM = 5%n = 10
P = €
6.Bond pricing & yieldHard

A bond trades at €1 060 with a yield to maturity of 4% (annual compounding) and a Macaulay duration of 11 years. Use the modified duration to estimate the new price P if the yield falls by 75 basis points.

P0 = €1 060y = 4%D = 11Δy = −75 bp
P = €
7.CAPM & WACCMedium
βr (%)SMLrf​ = 4%β = 1rm​ = ?β = 2re​ = 14%

A share with a beta of 2 has a required return of 14%. The risk-free rate is 4%. What expected market return rm does CAPM imply?

β = 2re = 14%rf = 4%
rm = %
8.CAPM & WACCHard
WACC = ?equity2 500 000 × €32.00β = 1.8rf​ = 2.5%, rm​ = 6.5%debt5 000 × €980rd​ = 4.5%T = 25.8%

A company has 2 500 000 shares at €32.00 each and 5 000 bonds trading at €980 each. The pre-tax cost of debt is 4.5% and the tax rate is 25.8%. The share’s beta is 1.8, the risk-free rate 2.5% and the market return 6.5%. Find the WACC.

β = 1.8rf = 2.5%rm = 6.5%rd = 4.5%T = 25.8%
WACC = %
9.Descriptive statisticsMedium
303540455055

Find the sample standard deviation s of the data: 52, 37, 55, 28, 49, 34, 58.

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

Exam scores are normally distributed with mean 70 and standard deviation 12. Find the z-score z of a score of 85.

μ = 70σ = 12x = 85
z =
11.ProbabilityMedium
P(at least one six) = ??1?2P(six) = 1/6 on each roll

A fair die is rolled 2 times. What is the probability P of getting at least one six?

n = 2
P =
12.ProbabilityHard
€p(x)−50.85100.1450.05

A game costs €5 to play. You win €15 with probability 0.1 and €50 with probability 0.05; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.

stake = €5€15: p = 0.1€50: p = 0.05
E(G) = €
13.Normal distributionMedium
750 g757.9 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 757.9 g. Round z to two decimals.

μX = 250, σX = 4μY = 500, σY = 9X + Y > 757.9 g
P =
14.Normal distributionHard
35455565758595pointstop 40%

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

μ = 65 pointsσ = 10 points40%
x = points
15.Confidence intervalsMedium
x̄E₁ = 790%, n = 25x̄E₂ = ?95%, n = 25

A 90% confidence interval for a mean, based on 25 observations and a known σ, has a margin of error of 7. What is the margin of error E₂ of a 95% interval from the same data?

90%n₁ = 25E₁ = 795%
E₂ =
16.Confidence intervalsHard
n = 16σ = ?Us = 6.695%

The diameters of 16 machined shafts have a sample standard deviation of 6.6 µm. Assuming the diameters are normally distributed, find the lower limit σ of the 95% confidence interval for the population standard deviation. With df = 15, 2.5% of the χ² distribution lies below 6.262 and 2.5% lies above 27.488.

n = 16s = 6.6 µmχ² = 6.262, 27.488
σ = µm
17.Hypothesis tests & confidence intervalsMedium
t −1.99801.998

A sample of 64 students has a mean score of 199.1 and a sample standard deviation of 5. Find the test statistic t for H₀: μ = 200.

μ₀ = 200x̄ = 199.1s = 5n = 64
t =
18.Hypothesis tests & confidence intervalsHard
σ = 15, n = ?x̄ − Ex̄ + Ex̄E = 190%

How large must a sample be so that the 90% confidence interval for a mean has a margin of error of at most 1? The standard deviation is 15. Find the smallest sample size n.

σ = 15E = 190%
n =
19.Linear regression & correlationMedium
xy1667198222912511928132xy2025100

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 = 491Σx² = 2510Σy² = 51079Σxy = 11303
r =
20.Linear regression & correlationHard
xy34155576196911761390xy510406080x = 8

The table gives 6 paired observations of x = advertising spend in k€ and y = sales in k€. Use the least-squares line ŷ = a + b·x. A new observation (8, 63) is made. Find its residual e = y − ŷ.

n = 6(8, 63)
e =

3Solutions

1.Time value of money

Answer: FV = 1221 €

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 = r/m = 0.04/12 = 0.003333
N = m·n = 12·5 = 60

Future value

FV = 1 000·(1 + 0.003333)60 = 1 000·1.221 = €1 221.00

Answer

FV = €1 221.00
2.Time value of money

Answer: N = 47 payments

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.

Rearrange the annuity formula

i = 0.003
(1 + i)−N = 1 − 12 000·0.003/275 = 0.8691

Solve with logarithms

N = −ln(0.8691)/ln(1.003) = 46.84

Round up to whole payments

46 full payments leave a small balance, which the last, smaller payment clears.

N = 47

Answer

N = 47 payments
3.NPV & IRR

Answer: DPB = 3.68 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: 6 500/1.08 = 6 018.52 → Σ = 6 018.52
year 2: 6 500/1.166 = 5 572.70 → Σ = 11 591.22
year 3: 6 500/1.26 = 5 159.91 → Σ = 16 751.13
year 4: 6 500/1.36 = 4 777.69 → Σ = 21 528.82

Fraction of the last year

DPB = 3 + 3 248.87/4 777.69 = 3.68

Answer

DPB = 3.68 years

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

4.NPV & IRR

Answer: MIRR = 18.69 %

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

12 500·1.123 = 17 561.60
9 500·1.122 = 11 916.80
9 500·1.121 = 10 640.00
9 500·1.120 = 9 500.00
TV = 49 618.40

MIRR

MIRR = (49 618.40/25 000)1/4 − 1 = 18.69%

Answer

MIRR = 18.69%
5.Bond pricing & yield

Answer: P = 1078 €

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 = €30.00, i = 0.025, N = 20

Present value of the coupons

€30.00·(1 − 0.6103)/0.025 = €467.67

Present value of the face value

€1 000·0.6103 = €610.27

Price

P = €1 077.95

Answer

P = €1 077.95

The yield is below the coupon rate, so the bond sells at a premium.

6.Bond pricing & yield

Answer: P = 1144 €

Duration: the Macaulay duration is the weighted average time at which a bond pays its cash flows, with the present values of the cash flows as weights: D = Σ t·PVₜ/P. The modified duration D/(1 + y) estimates the percentage price change for a small change in yield: %ΔP ≈ −Dmod·Δy.

Modified duration

Dmod = 11/1.04 = 10.58

Estimated price change

%ΔP ≈ −10.58·(−0.0075) = 7.933%
P ≈ 1060·(1 + 0.07933) = €1 144.09

Answer

P = €1 144.09

Duration is a linear estimate; for large yield changes convexity makes the true price a little higher.

7.CAPM & WACC

Answer: rm = 9 %

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 risk premium

rm − rf = (14% − 4%)/2 = 5%

Market return

rm = 4% + 5% = 9%

Answer

rm = 9%
8.CAPM & WACC

Answer: WACC = 9.333 %

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 = 2 500 000·€32.00 = €80 000 000
D = 5 000·€980 = €4 900 000

Cost of equity (CAPM)

re = 2.5% + 1.8·4% = 9.7%

Weights

V = €84 900 000, E/V = 0.9423, D/V = 0.05771

WACC

WACC = 0.9423·9.7% + 0.05771·4.5%·(1 − 0.258) = 9.333%

Answer

WACC = 9.333%
9.Descriptive statistics

Answer: s = 11.6

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 = 313/7 = 44.71

Squared deviations

Σ(x − x̄)² = 7.286² + (−7.714)² + 10.29² + (−16.71)² + 4.286² + (−10.71)² + 13.29² = 807.4

Standard deviation

s = √(807.4/6) = 11.6

Answer

s = 11.6
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 = (85 − 70)/12 = 1.25

Answer

z = 1.25
11.Probability

Answer: P = 0.3056

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.

Complement

P(no six) = (5/6)2 = 0.6944
P = 1 − 0.6944 = 0.3056

Answer

P = 0.3056
12.Probability

Answer: E(G) = −1 €

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.

Net outcomes

G: 10 (p = 0.1), 45 (p = 0.05), −5 (p = 0.85)

Expected value

E(G) = 10·0.1 + 45·0.05 − 5·0.85 = −1 €

Answer

E(G) = −1 €

A negative value means you lose money on average.

13.Normal distribution

Answer: P = 0.2119

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 = (757.9 − 750)/9.849 = 0.8

Upper tail

P = 1 − Φ(0.8) = 1 − 0.7881 = 0.2119

Answer

P = 0.2119
14.Normal distribution

Answer: x = 67.5 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) = 0.6

z from the table

z ≈ 0.25

Back to the original units

x = 65 + 0.25·10 = 67.5 points

Answer

x = 67.5 points
15.Confidence intervals

Answer: E₂ = 8.34

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.

Ratio of margins

E₂/E₁ = (1.96/1.645)·√(25/25) = 1.191

New margin

E₂ = 7·1.191 = 8.34

Answer

E₂ = 8.34
16.Confidence intervals

Answer: σ = 4.875 µm

Interval for a variance: (n − 1)s²/σ² has a χ² distribution with n − 1 degrees of freedom. The 95% interval for σ² runs from (n − 1)s² divided by the upper χ² value (2.5% above it) to (n − 1)s² divided by the lower χ² value (2.5% below it); the interval for σ is the square root of both limits. It is not symmetric around s.

Sum of squares

(n − 1)s² = 15·6.6² = 653.4

Limits for σ²

653.4/27.488 = 23.77
653.4/6.262 = 104.3

Limit for σ

σ = √23.77 = 4.875 µm

Answer

σ = 4.875 µm

95% interval for σ: [4.875, 10.21] µm.

17.Hypothesis tests & confidence intervals

Answer: t = −1.44

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 = 5/√64 = 5/8 = 0.625

Test statistic

t = (199.1 − 200)/0.625 = −1.44

Answer

t = −1.44
t0.025; df = 63 = 1.998

|t| < 1.998, so H₀ is not rejected at the 5% level (two-sided).

18.Hypothesis tests & confidence intervals

Answer: n = 609

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 = (1.645·15/1)² = 608.9

Round up

n = 609

Answer

n = 609
19.Linear regression & correlation

Answer: r = 0.987

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 = 11303 − 110·491/5 = 501
Sxx = 2510 − 110²/5 = 90
Syy = 51079 − 491²/5 = 2863

Correlation coefficient

r = 501/√(90·2863) = 0.987

Answer

r = 0.987
20.Linear regression & correlation

Answer: e = −2.333

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 = 48/6 = 8
ȳ = 392/6 = 65.33

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 316

Regression line

b = 316/70 = 4.514
a = 65.33 − 4.514·8 = 29.22
ŷ = 29.22 + 4.514·x

Predicted value

ŷ = 29.22 + 4.514·8 = 65.33

Residual

e = 63 − 65.33 = −2.333

Answer

e = −2.333

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