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…9PV = ?€8 000+2.5%+2.5%…r = 8.5%

A property is expected to earn net rent of €8 000 next year. The rent is expected to grow by 2.5% per year forever. The required return is 8.5% per year. Find the value PV of the property today.

C1 = €8 000g = 2.5%r = 8.5%
PV = €
2.Time value of moneyHard
months0123…120€100€100€100FV = ?

You save €100 at the end of every month for 10 years. The account pays 4.8% per year, compounded monthly. How much have you saved at the end? Find FV.

A = €1004.8% / 12N = 120
FV = €
3.NPV & IRRMedium
years01234−30 0009 00015 00013 00013 000

An investment of €30 000 gives these cash flows at the end of years 1 to 4: €9 000, €15 000, €13 000, €13 000. The cost of capital is 6%. Find the net present value NPV.

r = 6%CF0 = −30 000CF1 = 9 000CF2 = 15 000CF3 = 13 000CF4 = 13 000
NPV = €
4.NPV & IRRHard
years0123−20 00013 50015 0008 500

A project costs €20 000 and returns €13 500, €15 000, €8 500 at the end of years 1 to 3. Find its internal rate of return IRR in percent.

CF0 = −20 000CF1 = 13 500CF2 = 15 000CF3 = 8 500
IRR = %
5.Bond pricing & yieldMedium
half-years0123…20€15€15€15 + €1 000P = ?

A bond with a face value of €1 000 pays a 3% annual coupon in two half-yearly payments and matures in 10 years. The yield to maturity is 3.5% per year. Find the price P of the bond.

F = €1 000coupon = 3%YTM = 3.5%n = 10
P = €
6.Bond pricing & yieldHard
years0123€1 355€60€60€60 + €1 060

A bond with 20 years to maturity (face value €1 000, annual coupon 6%) sells for €1 355. The issuer can call it in 3 years at a call price of €1 060. Find the yield to call YTC in percent.

F = €1 000coupon = 6%P = €1 355call price = €1 060call in 3 years
YTC = %
7.CAPM & WACCMedium
WACC = ?equity€10 000 000re​ = 12%debt€3 000 000rd​ = 3%T = 25.8%

A company has equity worth €10 000 000 and debt worth €3 000 000. Its cost of equity is 12%, its pre-tax cost of debt is 3% and the tax rate is 25.8%. Find the WACC.

E = €10 000 000D = €3 000 000re = 12%rd = 3%T = 25.8%
WACC = %
8.CAPM & WACCHard
WACC = ?equity1 000 000 × €40.50D0​ = €0.80g = 5%debt€12 200 000rd​ = 5%T = 20%

A company has 1 000 000 shares trading at €40.50; the book value of its equity is €12 200 000. It just paid a dividend of €0.80 per share, and dividends grow by 5% per year. Its debt has a market value of €12 200 000 and a pre-tax cost of 5%. The tax rate is 20%. Find the WACC, using the dividend growth model for the cost of equity.

N = 1 000 000P0 = €40.50book equity = €12 200 000D0 = €0.80g = 5%D = €12 200 000rd = 5%T = 20%
WACC = %
9.Descriptive statisticsMedium
10152025303540455055

Find the sample standard deviation s of the data: 50, 59, 35, 18, 41, 10.

n = 6
s =
10.Descriptive statisticsHard
z −2−1012

Exam scores are normally distributed with mean 50 and standard deviation 12. Find the z-score z of a score of 32.

μ = 50σ = 12x = 32
z =
11.ProbabilityMedium
xp(x)00.1520.2540.360.3μ = 3.5

For the distribution shown, find the variance Var(X) of X.

x: 0, 2, 4, 6p(x): 0.15, 0.25, 0.3, 0.3
Var(X) =
12.ProbabilityHard
xp(x)00.320.0540.360.35μ

X has the distribution shown, and Y = 10X + 1. Find the standard deviation σY of Y.

x: 0, 2, 4, 6p(x): 0.3, 0.05, 0.3, 0.35Y = 10X + 1
σY =
13.Normal distributionMedium
60 kg61.8 kgT = X₁ + … + X₅?

The masses of boxes are independent and normal with mean 12 kg and standard deviation 0.8 kg. Find the probability P that the total mass of 5 of them exceeds 61.8 kg. Round z to two decimals.

μ = 12 kgσ = 0.8 kgn = 5> 61.8 kg
P =
14.Normal distributionHard
235240245250255260265gtop 40%

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Drag the cut-off so the shaded top 40% is to the right of it. Find the cut-off x.

μ = 250 gσ = 5 g40%
x = g
15.Confidence intervalsMedium
n = 12, s = 9L = ?Ux̄ = 13E99%

A sample of 12 items has mean 13 and standard deviation s = 9. The population is roughly normal. Find the lower limit of the 99% confidence interval for the mean.

n = 12x̄ = 13s = 999%t(11) = 3.106
lower limit =
16.Confidence intervalsHard
n = 16σ = ?Us = 4.695%

The diameters of 16 machined shafts have a sample standard deviation of 4.6 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 = 4.6 µmχ² = 6.262, 27.488
σ = µm
17.Hypothesis tests & confidence intervalsMedium
t −2.13102.131

A sample of 16 students has a mean score of 508.5 and a sample standard deviation of 20. Find the test statistic t for H₀: μ = 500.

μ₀ = 500x̄ = 508.5s = 20n = 16
t =
18.Hypothesis tests & confidence intervalsHard
−1.961.96z = ?

In a trial, 20 of 80 patients on treatment A recover, and 28 of 200 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.

x₁ = 20n₁ = 80x₂ = 28n₂ = 200
z =
19.Linear regression & correlationMedium
xy12422431741851468xy24610152025

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

n = 6Σx = 21Σy = 105Σx² = 91Σy² = 2025Σxy = 313
r =
20.Linear regression & correlationHard
xy284176248311036xy510102030x = 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 = 133333 €

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.

Growing perpetuity

PV = C1/(r − g) = 8 000/(0.085 − 0.025) = 8 000/0.06 = €133 333.33

Answer

PV = €133 333.33
2.Time value of money

Answer: FV = 15363 €

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 = 12·10 = 120

Future value of the deposits

FV = A·((1 + i)N − 1)/i = 100·(1.615 − 1)/0.004 = €15 363.20

Answer

FV = €15 363.20

You paid in €12 000 yourself; the rest is interest.

3.NPV & IRR

Answer: NPV = 13053 €

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 = −30 000
CF1/(1 + 0.06)1 = 9 000/1.06 = 8 490.57
CF2/(1 + 0.06)2 = 15 000/1.124 = 13 349.95
CF3/(1 + 0.06)3 = 13 000/1.191 = 10 915.05
CF4/(1 + 0.06)4 = 13 000/1.262 = 10 297.22

Add them up

NPV = €13 052.78

Answer

NPV = €13 052.78

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

4.NPV & IRR

Answer: IRR = 41.64 %

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.

Bracket the IRR

NPV falls as the rate rises, so the IRR lies between a rate with positive NPV and one with negative NPV.

NPV(40%) = 393.59
NPV(45%) = −767.15

Narrow down

Linear interpolation gives a first estimate; repeating it (or a calculator's IRR function) gives the exact rate.

IRR ≈ 40 + 5·393.59/(393.59 − (−767.15)) = 41.7%
exact: NPV(41.64%) = 0 → IRR = 41.64%

Answer

IRR = 41.64%
5.Bond pricing & yield

Answer: P = 958.1 €

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 = €15.00, i = 0.0175, N = 20

Present value of the coupons

€15.00·(1 − 0.7068)/0.0175 = €251.29

Present value of the face value

€1 000·0.7068 = €706.82

Price

P = €958.12

Answer

P = €958.12

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

6.Bond pricing & yield

Answer: YTC = 50 %

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

1355 = 60·(1 − (1 + y)−3)/y + 1060/(1 + y)3

Solve numerically

y = 50%: €398.52
y = 50%: €398.52
YTC = 50%

Answer

YTC = 50%

The yield to maturity is 3.502%. Investors in a premium bond expect it to be called, so the lower yield to call is the more realistic one.

7.CAPM & WACC

Answer: WACC = 9.744 %

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 = €13 000 000, E/V = 0.7692, D/V = 0.2308

WACC

WACC = 0.7692·12% + 0.2308·3%·(1 − 0.258) = 9.744%

Answer

WACC = 9.744%
8.CAPM & WACC

Answer: WACC = 6.362 %

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 equity

D1 = 0.8·1.05 = €0.84
re = 0.84/40.5 + 0.05 = 7.074%

Market values

E = 1 000 000·€40.50 = €40 500 000
V = €52 700 000, E/V = 0.7685, D/V = 0.2315

WACC

WACC = 0.7685·7.074% + 0.2315·5%·(1 − 0.2) = 6.362%

Answer

WACC = 6.362%
9.Descriptive statistics

Answer: s = 18.71

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 = 213/6 = 35.5

Squared deviations

Σ(x − x̄)² = 14.5² + 23.5² + (−0.5)² + (−17.5)² + 5.5² + (−25.5)² = 1750

Standard deviation

s = √(1750/5) = 18.71

Answer

s = 18.71
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 = (32 − 50)/12 = −1.5

Answer

z = −1.5
11.Probability

Answer: Var(X) = 4.35

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 + 2·0.25 + 4·0.3 + 6·0.3 = 3.5

Mean of the squares

E(X²) = 0·0.15 + 4·0.25 + 16·0.3 + 36·0.3 = 16.6

Variance

Var(X) = 16.6 − 3.5² = 4.35

Answer

Var(X) = 4.35
12.Probability

Answer: σY = 24.58

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 = 3.4, Var(X) = 6.04, σX = 2.458

Linear change

σY = |10|·2.458 = 24.58

Answer

σY = 24.58

The mean does move: E(Y) = 10·3.4 + 1 = 35.

13.Normal distribution

Answer: P = 0.1562

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 = 60 kg, σT = 0.8·√5 = 1.789 kg

Standardise

z = (61.8 − 60)/1.789 = 1.01

Probability

P = 1 − Φ(1.01) = 0.1562

Answer

P = 0.1562
14.Normal distribution

Answer: x = 251.3 g

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 = 250 + 0.25·5 = 251.3 g

Answer

x = 251.3 g
15.Confidence intervals

Answer: lower limit = 4.93

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 = 11, t = 3.106

Margin of error

E = 3.106·9/√12 = 8.07

Interval

13 ± 8.07 → [4.93, 21.07]

Answer

lower limit = 4.93
16.Confidence intervals

Answer: σ = 3.398 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 = 15·4.6² = 317.4

Limits for σ²

317.4/27.488 = 11.55
317.4/6.262 = 50.69

Limit for σ

σ = √11.55 = 3.398 µm

Answer

σ = 3.398 µm

95% interval for σ: [3.398, 7.119] µm.

17.Hypothesis tests & confidence intervals

Answer: t = 1.7

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 = 20/√16 = 20/4 = 5

Test statistic

t = (508.5 − 500)/5 = 1.7

Answer

t = 1.7
t0.025; df = 15 = 2.131

|t| < 2.131, so H₀ is not rejected at the 5% level (two-sided).

18.Hypothesis tests & confidence intervals

Answer: z = 2.206

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 = 20/80 = 0.25
p̂₂ = 28/200 = 0.14
p̂ = 48/280 = 0.1714

Standard error under H₀

SE = √(0.1714·0.8286·(1/80 + 1/200)) = 0.04986

Test statistic

z = (0.25 − 0.14)/0.04986 = 2.206

Answer

z = 2.206

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

19.Linear regression & correlation

Answer: r = −0.9514

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 = 313 − 21·105/6 = −54.5
Sxx = 91 − 21²/6 = 17.5
Syy = 2025 − 105²/6 = 187.5

Correlation coefficient

r = −54.5/√(17.5·187.5) = −0.9514

Answer

r = −0.9514
20.Linear regression & correlation

Answer: ŷ = 19.7

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 = 116/5 = 23.2

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 140

Regression line

b = 140/40 = 3.5
a = 23.2 − 3.5·6 = 2.2
ŷ = 2.2 + 3.5·x

Prediction

ŷ = 2.2 + 3.5·5 = 19.7

Answer

ŷ = 19.7

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