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+3%+3%…r = 5%

A property is expected to earn net rent of €8 000 next year. The rent is expected to grow by 3% per year forever. The required return is 5% per year. Find the value PV of the property today.

C1 = €8 000g = 3%r = 5%
PV = €
2.Time value of moneyHard
months0123…36€15 000A?AAA

A loan of €15 000 is repaid with equal monthly payments over 3 years. The interest rate is 6% per year, compounded monthly. Find the monthly payment A.

P = €15 0006% / 12N = 36
A = €
3.NPV & IRRMedium
years01234−60 00020 00025 00015 00015 000

An investment of €60 000 gives these cash flows at the end of years 1 to 4: €20 000, €25 000, €15 000, €15 000. The cost of capital is 5%. Find the net present value NPV.

r = 5%CF0 = −60 000CF1 = 20 000CF2 = 25 000CF3 = 15 000CF4 = 15 000
NPV = €
4.NPV & IRRHard
years0123−25 00010 50014 5009 000

A project costs €25 000 and returns €10 500, €14 500, €9 000 at the end of years 1 to 3. 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 = −25 000CF1 = 10 500CF2 = 14 500CF3 = 9 000
MIRR = %
5.Bond pricing & yieldMedium
years012345P€30€30€30 + €1 000

A bond with a face value of €1 000 pays an annual coupon of 3% and matures in 5 years. Its yield to maturity changes from 5.5% to 6%. By how much ΔP does the bond price change?

F = €1 000coupon = 3%n = 5y1 = 5.5%y2 = 6%
ΔP = €
6.Bond pricing & yieldHard
years0123456P€50€50€50 + €1 000

A bond with a face value of €1 000 pays an annual coupon of 5% and matures in 6 years. The yield to maturity is 5.5%. Find the Macaulay duration D in years.

F = €1 000coupon = 5%n = 6YTM = 5.5%
D = years
7.CAPM & WACCMedium
WACC = ?equityEre​ = 10%debtD = 1·Erd​ = 3.5%T = 25.8%

A company has a debt-to-equity ratio (at market values) of 1. Its cost of equity is 10%, its pre-tax cost of debt is 3.5% and the tax rate is 25.8%. Find the WACC.

D/E = 1re = 10%rd = 3.5%T = 25.8%
WACC = %
8.CAPM & WACCHard
WACC = ?equity2 000 000 × €30.50D0​ = €1.00g = 3%debt€48 800 000rd​ = 6%T = 20%

A company has 2 000 000 shares trading at €30.50; the book value of its equity is €36 600 000. It just paid a dividend of €1.00 per share, and dividends grow by 3% per year. Its debt has a market value of €48 800 000 and a pre-tax cost of 6%. The tax rate is 20%. Find the WACC, using the dividend growth model for the cost of equity.

N = 2 000 000P0 = €30.50book equity = €36 600 000D0 = €1.00g = 3%D = €48 800 000rd = 6%T = 20%
WACC = %
9.Descriptive statisticsMedium
2025303540455055

Find the sample standard deviation s of the data: 42, 55, 56, 57, 48, 20.

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

Exam scores are normally distributed with mean 500 and standard deviation 100. Find the z-score z of a score of 420.

μ = 500σ = 100x = 420
z =
11.ProbabilityMedium
P(at least one six) = ??1?2?3?4?5?6P(six) = 1/6 on each roll

A fair die is rolled 6 times. What is the probability P of getting at least one six?

n = 6
P =
12.ProbabilityHard
€p(x)−10.7810.2190.02

A game costs €1 to play. You win €2 with probability 0.2 and €20 with probability 0.02; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.

stake = €1€2: p = 0.2€20: p = 0.02
E(G) = €
13.Normal distributionMedium
105 s110.1 sX + Y?

X is normal with mean 60 s and standard deviation 5 s; Y is independent and normal with mean 45 s and standard deviation 4 s. Find the probability P that X + Y is more than 110.1 s. Round z to two decimals.

μX = 60, σX = 5μY = 45, σY = 4X + Y > 110.1 s
P =
14.Normal distributionHard
0.04 mm0D − d?

Hole diameters D are normal with mean 25.04 mm and standard deviation 0.02 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.015 mm. A randomly chosen shaft fits when D − d > 0. Find the probability P that it fits. Round z to two decimals.

hole: μ = 25.04 mm, σ = 0.02 mmshaft: μ = 25 mm, σ = 0.015 mm
P =
15.Confidence intervalsMedium
n = 8, s = 4LUx̄ = 40.5E = ?99%

A sample of 8 items has mean 40.5 and standard deviation s = 4. The population is roughly normal. Find the margin of error E of a 99% confidence interval for the mean.

n = 8x̄ = 40.5s = 499%t(7) = 3.499
E =
16.Confidence intervalsHard
p = 0.5, n = ?p̂ − Ep̂ + Ep̂E = 0.0590%

Nothing is known in advance about the proportion of defective products. How many products n must be inspected so that a 90% confidence interval for the proportion has a margin of error of at most 0.05?

p = 0.5E = 0.0590%
n =
17.Hypothesis tests & confidence intervalsMedium
1.7p = ?

The mean of a population is claimed to be 250; the population standard deviation is known to be 20. A random sample of 64 observations has a mean of 254.25. Test H₀: μ = 250 against H₁: μ > 250. Find the p-value p.

μ₀ = 250σ = 20n = 64x̄ = 254.25H₁: μ > μ₀
p =
18.Hypothesis tests & confidence intervalsHard
−1.961.96z = ?

In a trial, 24 of 120 patients on treatment A recover, and 24 of 150 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.

x₁ = 24n₁ = 120x₂ = 24n₂ = 150
z =
19.Linear regression & correlationMedium
xy159261372475584xy24607080

The table gives 5 paired observations of x = hours studied and y = exam score. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 5Σx = 15Σy = 351Σx² = 55Σy² = 25067Σxy = 1117
r =
20.Linear regression & correlationHard
xy21141462282710341240xy51010203040x = 9

The table gives 6 paired observations of x = load in kN and y = extension in mm. Use the least-squares line ŷ = a + b·x. A new observation (9, 29) is made. Find its residual e = y − ŷ.

n = 6(9, 29)
e =

3Solutions

1.Time value of money

Answer: PV = 400000 €

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/(0.05 − 0.03) = 8 000/0.02 = €400 000.00

Answer

PV = €400 000.00
2.Time value of money

Answer: A = 456.3 €

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 = 0.06/12 = 0.005
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 = 0.8356
A = 15 000·0.005/(1 − 0.8356) = €456.33

Answer

A = €456.33
3.NPV & IRR

Answer: NPV = 7021 €

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 = −60 000
CF1/(1 + 0.05)1 = 20 000/1.05 = 19 047.62
CF2/(1 + 0.05)2 = 25 000/1.103 = 22 675.74
CF3/(1 + 0.05)3 = 15 000/1.158 = 12 957.56
CF4/(1 + 0.05)4 = 15 000/1.216 = 12 340.54

Add them up

NPV = €7 021.46

Answer

NPV = €7 021.46

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

4.NPV & IRR

Answer: MIRR = 13.1 %

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

10 500·1.062 = 11 797.80
14 500·1.061 = 15 370.00
9 000·1.060 = 9 000.00
TV = 36 167.80

MIRR

MIRR = (36 167.80/25 000)1/3 − 1 = 13.1%

Answer

MIRR = 13.1%
5.Bond pricing & yield

Answer: ΔP = −19.61 €

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.

Price at the old yield

C = €30.00, i = 0.055, N = 5
€30.00·(1 − 0.7651)/0.055 + €1 000·0.7651 = €128.11 + €765.13 = €893.24

Price at the new yield

C = €30.00, i = 0.06, N = 5
€30.00·(1 − 0.7473)/0.06 + €1 000·0.7473 = €126.37 + €747.26 = €873.63

Change

ΔP = 873.63 − 893.24 = −€19.61

Answer

ΔP = −€19.61
6.Bond pricing & yield

Answer: D = 5.32 years

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.

Present value of each cash flow

t = 1: 50/1.055 = 47.39, t·PV = 47.39
t = 2: 50/1.113 = 44.92, t·PV = 89.85
t = 3: 50/1.174 = 42.58, t·PV = 127.74
t = 4: 50/1.239 = 40.36, t·PV = 161.44
t = 5: 50/1.307 = 38.26, t·PV = 191.28
t = 6: 1050/1.379 = 761.51, t·PV = 4 569.05

Duration

P = 975.02
D = 5 186.76/975.02 = 5.32

Answer

D = 5.32 years
7.CAPM & WACC

Answer: WACC = 6.298 %

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 from D/E

E/V = 1/(1 + 1) = 0.5
D/V = 1/(1 + 1) = 0.5

WACC

WACC = 0.5·10% + 0.5·3.5%·(1 − 0.258) = 6.298%

Answer

WACC = 6.298%
8.CAPM & WACC

Answer: WACC = 5.676 %

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 = 1·1.03 = €1.03
re = 1.03/30.5 + 0.03 = 6.377%

Market values

E = 2 000 000·€30.50 = €61 000 000
V = €109 800 000, E/V = 0.5556, D/V = 0.4444

WACC

WACC = 0.5556·6.377% + 0.4444·6%·(1 − 0.2) = 5.676%

Answer

WACC = 5.676%
9.Descriptive statistics

Answer: s = 14.12

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 = 278/6 = 46.33

Squared deviations

Σ(x − x̄)² = (−4.333)² + 8.667² + 9.667² + 10.67² + 1.667² + (−26.33)² = 997.3

Standard deviation

s = √(997.3/5) = 14.12

Answer

s = 14.12
10.Descriptive statistics

Answer: z = −0.8

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 = (420 − 500)/100 = −0.8

Answer

z = −0.8
11.Probability

Answer: P = 0.6651

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)6 = 0.3349
P = 1 − 0.3349 = 0.6651

Answer

P = 0.6651
12.Probability

Answer: E(G) = −0.2 €

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: 1 (p = 0.2), 19 (p = 0.02), −1 (p = 0.78)

Expected value

E(G) = 1·0.2 + 19·0.02 − 1·0.78 = −0.2 €

Answer

E(G) = −0.2 €

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 = 105 s, σ = √(5² + 4²) = 6.403 s

Standardise

z = (110.1 − 105)/6.403 = 0.8

Upper tail

P = 1 − Φ(0.8) = 1 − 0.7881 = 0.2119

Answer

P = 0.2119
14.Normal distribution

Answer: P = 0.9452

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 = 25.04 − 25 = 0.04 mm, σ = 0.025 mm

Standardise

z = (0 − 0.04)/0.025 = −1.6

Probability

P = 1 − Φ(−1.6) = 0.9452

Answer

P = 0.9452
15.Confidence intervals

Answer: E = 4.948

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 = 7, t = 3.499

Margin of error

E = 3.499·4/√8 = 4.948

Answer

E = 4.948
16.Confidence intervals

Answer: n = 271

Confidence interval for a proportion: p̂ ± z·√(p̂(1 − p̂)/n), where p̂ = x/n. To reach a margin E you need n ≥ z²·p(1 − p)/E², rounded up; without a prior estimate use p = 0.5, which gives the largest n.

Solve for n

n = 1.645²·0.5·0.5/0.05² = 270.6

Round up

n = 271

Answer

n = 271
17.Hypothesis tests & confidence intervals

Answer: p = 0.04457

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.

Test statistic

SE = σ/√n = 20/√64 = 2.5
z = (254.25 − 250)/2.5 = 1.7

Tail area

Φ(1.7) = 0.9554
1 − Φ(1.7) = 0.0446

p-value

p = 0.04457

Answer

p = 0.04457

p ≤ 0.05, so H₀ is rejected at the 5% level.

18.Hypothesis tests & confidence intervals

Answer: z = 0.8542

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 = 24/120 = 0.2
p̂₂ = 24/150 = 0.16
p̂ = 48/270 = 0.1778

Standard error under H₀

SE = √(0.1778·0.8222·(1/120 + 1/150)) = 0.04683

Test statistic

z = (0.2 − 0.16)/0.04683 = 0.8542

Answer

z = 0.8542

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

19.Linear regression & correlation

Answer: r = 0.9796

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 = 1117 − 15·351/5 = 64
Sxx = 55 − 15²/5 = 10
Syy = 25067 − 351²/5 = 426.8

Correlation coefficient

r = 64/√(10·426.8) = 0.9796

Answer

r = 0.9796
20.Linear regression & correlation

Answer: e = −1.667

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 = 148/6 = 24.67

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 210

Regression line

b = 210/70 = 3
a = 24.67 − 3·7 = 3.667
ŷ = 3.667 + 3·x

Predicted value

ŷ = 3.667 + 3·9 = 30.67

Residual

e = 29 − 30.67 = −1.667

Answer

e = −1.667

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