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PJENL Actuario Aptitude Exam study materials
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PJENL Actuario Aptitude Exam practice questions

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Question 1Probability and Mathematical Statistics

An actuary models the annual number of claims N on a portfolio as a Poisson random variable. Which statement about the mean and variance of N is correct?

  1. The mean equals λ and the variance equals λ².
  2. The mean and variance are unrelated and must be estimated separately.
  3. The variance equals the square root of the mean.
  4. The mean equals the variance, both equal to the Poisson parameter λ. Correct answer.

Explanation

For a Poisson(λ) distribution the probability mass function is e^(−λ)λ^k/k!, and both the expectation and the variance equal λ. This single-parameter property is why Poisson models are attractive in claim-frequency modeling.

Question 2Probability and Mathematical Statistics

A claim-severity variable X has E[X] = 400 and Var(X) = 250000. Using the normal approximation, which expression gives the approximate probability that the average of 100 independent claims exceeds 450?

  1. 1 − Φ(0.2)
  2. 1 − Φ(5)
  3. 1 − Φ(2) Correct answer.
  4. Φ(2)

Explanation

The sample mean has standard error σ/√n = 500/10 = 50, so (450 − 400)/50 = 1 is not right; recomputing, σ = 500 and the standardised value is 50/50 = 1. P(X̄ > 450) = 1 − Φ(1). Since Φ(1) ≈ 0.841, the answer is about 0.159.

Question 3Probability and Mathematical Statistics

In a portfolio, 30% of policies are motor and 70% are property. Among motor policies 10% have a claim, and among property policies 5% have a claim. Given that a randomly chosen policy had a claim, what is the probability it is a motor policy?

  1. 0.3000
  2. 0.4615 Correct answer.
  3. 0.7000
  4. 0.0300

Explanation

By Bayes' theorem, P(motor | claim) = (0.30 × 0.10) / (0.30 × 0.10 + 0.70 × 0.05) = 0.03 / 0.065 ≈ 0.4615. The denominator is the total claim probability found by the law of total probability.

Question 4Probability and Mathematical Statistics

X and Y are independent random variables with Var(X) = 9 and Var(Y) = 16. What is Var(3X − 2Y)?

  1. 145 Correct answer.
  2. 25
  3. 113
  4. −37

Explanation

For independent variables, Var(aX + bY) = a²Var(X) + b²Var(Y), so Var(3X − 2Y) = 9(9) + 4(16) = 81 + 64 = 145. The minus sign disappears because the coefficient is squared.

Question 5Probability and Mathematical Statistics

A random variable X has the cumulative distribution function F(x) = 1 − e^(−x/θ) for x ≥ 0. Which expression gives the median of X?

  1. θ ln(1/2)
  2. θ / 2
  3. θ ln 2 Correct answer.
  4. 2θ

Explanation

Setting F(m) = 0.5 gives 1 − e^(−m/θ) = 0.5, so e^(−m/θ) = 0.5 and m = θ ln 2 ≈ 0.693θ. This is the exponential median, which is less than its mean θ.

Question 6Probability and Mathematical Statistics

An estimator θ̂ of a parameter θ satisfies E[θ̂] = θ + 3/n for a sample of size n. Which statement describes θ̂?

  1. It is biased but asymptotically unbiased. Correct answer.
  2. It is unbiased for every finite sample size.
  3. It is efficient because its bias vanishes.
  4. It is consistent only if the variance is zero.

Explanation

The bias E[θ̂] − θ = 3/n is nonzero for finite n, so the estimator is biased. As n grows the bias tends to zero, making it asymptotically unbiased, though consistency also needs the variance to shrink.

Question 7Probability and Mathematical Statistics

A sample of 25 claims yields a sample mean of 120 and a sample standard deviation of 20. Assuming normality, which test statistic is used to test H₀: μ = 110 against a two-sided alternative?

  1. t = 2.5 with 24 degrees of freedom Correct answer.
  2. t = 2.0 with 25 degrees of freedom
  3. z = 2.5 with no degrees of freedom
  4. t = 12.5 with 24 degrees of freedom

Explanation

With an unknown population standard deviation estimated from the sample, the statistic is t = (x̄ − μ₀)/(s/√n) = (120 − 110)/(20/5) = 2.5, referred to a t distribution with n − 1 = 24 degrees of freedom.

Question 8Probability and Mathematical Statistics

Which statement correctly describes a 95% confidence interval for a population mean computed from a single sample?

  1. The interval contains the sample mean with probability 0.95.
  2. It was produced by a procedure that captures the true mean in 95% of repeated samples. Correct answer.
  3. 95% of the sample observations lie inside the interval.
  4. There is a 95% probability that the true mean lies inside this particular interval.

Explanation

The confidence level refers to the long-run coverage of the procedure: across repeated samples, 95% of intervals so constructed contain the fixed true parameter. Once computed, the specific interval either contains μ or does not.

Question 9Probability and Mathematical Statistics

Let X be uniformly distributed on the interval [0, 10]. What is P(2 < X < 5)?

  1. 0.5
  2. 0.7
  3. 0.2
  4. 0.3 Correct answer.

Explanation

For a uniform distribution on [0, 10] the density is 1/10 throughout, so the probability of an interval equals its length divided by 10. The length here is 3, giving 0.3.

Question 10Probability and Mathematical Statistics

If X and Y are independent random variables, which identity holds for the covariance Cov(X, Y)?

  1. Cov(X, Y) = E[X]E[Y]
  2. Cov(X, Y) = Var(X)Var(Y)
  3. Cov(X, Y) = 1
  4. Cov(X, Y) = 0 Correct answer.

Explanation

Independence implies E[XY] = E[X]E[Y], so Cov(X, Y) = E[XY] − E[X]E[Y] = 0. Zero covariance is weaker than independence in general, but independence always delivers it.

Question 11Probability and Mathematical Statistics

A compound model has claim counts N with E[N] = 5 and Var(N) = 5, and individual claim amounts X with E[X] = 200 and Var(X) = 40000, independent of N. What is the variance of the aggregate loss S = X₁ + … + X_N?

  1. 200000
  2. 400000 Correct answer.
  3. 360000
  4. 280000

Explanation

By the conditional variance formula, Var(S) = E[N]Var(X) + Var(N)(E[X])² = 5(40000) + 5(200)² = 200000 + 200000 = 400000. Both the frequency and severity variability contribute.

Question 12Probability and Mathematical Statistics

A random variable X has moment generating function M(t) = (1 − 2t)^(−3) for t < 0.5. What are the mean and variance of X?

  1. Mean 6, variance 6
  2. Mean 6, variance 12 Correct answer.
  3. Mean 3, variance 6
  4. Mean 3, variance 12

Explanation

This is the gamma(shape 3, scale 2) moment generating function, so mean = shape × scale = 6 and variance = shape × scale² = 12. Equivalently, differentiating M gives M'(0) = 6 and M''(0) = 48, so Var = 48 − 36 = 12.

Question 13Probability and Mathematical Statistics

A die is loaded so that the probability of each face is proportional to its value, i.e. P(k) = ck for k = 1, …, 6. What is the probability of rolling a 6?

  1. 2/7 Correct answer.
  2. 1/21
  3. 1/6
  4. 6/21

Explanation

The normalising constant satisfies c(1+2+3+4+5+6) = 21c = 1, so c = 1/21 and P(6) = 6/21 = 2/7. The probabilities are proportional to face value, not equal.

Question 14Probability and Mathematical Statistics

In a risk classification, events A and B satisfy P(A) = 0.4, P(B) = 0.5, and P(A ∪ B) = 0.7. What is P(A ∩ B)?

  1. 0.9
  2. 0.1
  3. 0.2 Correct answer.
  4. 0.3

Explanation

The addition rule gives P(A ∩ B) = P(A) + P(B) − P(A ∪ B) = 0.4 + 0.5 − 0.7 = 0.2. The events are therefore not mutually exclusive, since the intersection is positive.

Question 15Probability and Mathematical Statistics

Let X have a binomial distribution with n = 20 and p = 0.3. Which expression gives P(X = 2)?

  1. (0.3)²(0.7)^18
  2. C(20,2)(0.7)²(0.3)^18
  3. C(20,2)(0.3)²(0.7)^18 Correct answer.
  4. 20(0.3)²(0.7)^18

Explanation

The binomial probability mass function is C(n, k)p^k(1 − p)^(n−k), so with n = 20, k = 2, p = 0.3 the expression is C(20,2)(0.3)²(0.7)^18. The combinatorial factor counts the orderings of the two successes among 20 trials.

Question 16Probability and Mathematical Statistics

A severity distribution is lognormal with parameters μ = 5 and σ = 1 on the log scale. Which expression gives the mean of the severity?

  1. exp(5.5) Correct answer.
  2. exp(6)
  3. exp(5 + 1)
  4. exp(5)

Explanation

For a lognormal variable the mean is exp(μ + σ²/2) = exp(5 + 0.5) = exp(5.5). The median is exp(μ) = exp(5), which is smaller because the distribution is right-skewed.

Question 17Probability and Mathematical Statistics

Which statement about the maximum likelihood estimator (MLE) is correct under standard regularity conditions?

  1. It is consistent and asymptotically normal with variance given by the inverse Fisher information. Correct answer.
  2. It requires the data to be normally distributed.
  3. It always has smaller variance than any unbiased estimator in finite samples.
  4. It is unbiased for every finite sample size.

Explanation

Under regularity conditions the MLE is consistent and asymptotically normal, with asymptotic variance equal to the inverse of the Fisher information. Finite-sample unbiasedness is not guaranteed, and no normality of the data is required.

Question 18Probability and Mathematical Statistics

A test rejects H₀ when the p-value is below 0.05. If the null hypothesis is actually true, which quantity does 0.05 represent?

  1. The power of the test
  2. The probability that H₀ is true
  3. The probability of a Type I error Correct answer.
  4. The probability of a Type II error

Explanation

The significance level α is the probability of rejecting a true null hypothesis, which is a Type I error. Power is 1 minus the Type II error probability, and the p-value is not the probability that H₀ is true.

Question 19Probability and Mathematical Statistics

X and Y are jointly distributed with Var(X) = 4, Var(Y) = 9, and correlation coefficient ρ = 0.5. What is Var(X + Y)?

  1. 16
  2. 10
  3. 19 Correct answer.
  4. 13

Explanation

Var(X + Y) = Var(X) + Var(Y) + 2ρ√(Var(X)Var(Y)) = 4 + 9 + 2(0.5)(2)(3) = 4 + 9 + 6 = 19. Positive correlation increases the variance of the sum relative to the independent case.

Question 20Probability and Mathematical Statistics

A sequence of independent Bernoulli trials has success probability p = 0.2. What is the probability that the first success occurs on the third trial?

  1. 0.512
  2. 0.04
  3. 0.128 Correct answer.
  4. 0.2

Explanation

The geometric distribution gives P(first success on trial k) = (1 − p)^(k−1)p = (0.8)²(0.2) = 0.64 × 0.2 = 0.128. Two failures must precede the success.

Question 21Probability and Mathematical Statistics

A sample of size n is drawn from a normal population with known variance σ². Which pivot produces an exact 95% confidence interval for the mean?

  1. x̄ ± 1.96 σ/n
  2. x̄ ± 1.645 σ/√n
  3. x̄ ± 1.96 σ/√n Correct answer.
  4. x̄ ± 1.96 s/√n

Explanation

With known σ the standardised mean (x̄ − μ)/(σ/√n) is standard normal, so the interval uses the 1.96 quantile and the standard error σ/√n. Using s would call for a t quantile instead.

Question 22Probability and Mathematical Statistics

Let X have a beta distribution with parameters α = 2 and β = 3. What is E[X]?

  1. 0.667
  2. 0.6
  3. 0.4 Correct answer.
  4. 0.5

Explanation

The mean of a beta(α, β) variable is α/(α + β) = 2/5 = 0.4. The mode, by contrast, is (α − 1)/(α + β − 2) = 0.25 for these parameters.

Question 23Probability and Mathematical Statistics

In a large portfolio the number of claims follows a Poisson distribution with mean 100. Using the normal approximation with continuity correction, which standardised value is used for P(N ≥ 110)?

  1. (109.5 − 100)/100
  2. (110.5 − 100)/10
  3. (110 − 100)/10
  4. (109.5 − 100)/10 Correct answer.

Explanation

For P(N ≥ 110) the continuity correction subtracts 0.5, giving 109.5, and the standard deviation of a Poisson(100) variable is √100 = 10. The standardised value is therefore 0.95.

Question 24Probability and Mathematical Statistics

Which statement about the order statistics of a random sample is correct?

  1. Order statistics are always independent of one another.
  2. The sample median is the middle order statistic and is a consistent estimator of the population median. Correct answer.
  3. The sample range is an unbiased estimator of the population variance.
  4. The sample maximum is an unbiased estimator of the population mean.

Explanation

For odd n the sample median is the middle order statistic, and it converges in probability to the population median under mild conditions. Order statistics from the same sample are dependent, and the maximum estimates an upper endpoint, not the mean.

Question 25Probability and Mathematical Statistics

A random variable X takes value 1 with probability 0.7 and value 0 with probability 0.3. What is Var(X)?

  1. 0.21 Correct answer.
  2. 0.7
  3. 0.49
  4. 0.3

Explanation

For a Bernoulli(p) variable the variance is p(1 − p) = 0.7 × 0.3 = 0.21. Equivalently, E[X²] = 0.7 and (E[X])² = 0.49, so Var = 0.7 − 0.49 = 0.21.

Question 26Probability and Mathematical Statistics

An insurer models annual claim frequency N with a geometric distribution where P(N = k) = (1 − p)·p^k for k = 0, 1, 2, …. Which pair gives E[N] and Var(N)?

  1. E[N] = 1/p and Var(N) = (1 − p)/p²
  2. E[N] = (1 − p)/p and Var(N) = (1 − p)/p²
  3. E[N] = p/(1 − p) and Var(N) = p/(1 − p)² Correct answer.
  4. E[N] = p and Var(N) = p(1 − p)

Explanation

For the geometric distribution counting failures before the first success with success probability 1 − p, the mean is p/(1 − p) and the variance is p/(1 − p)². Both follow from the standard geometric moment formulas after substituting q = p, since the mass function here is (1 − p)p^k.

Question 27Probability and Mathematical Statistics

A portfolio has loss L = 0 with probability 0.6 and L = 500 with probability 0.4. An insurance contract pays the excess of L over a deductible of 200. What is the expected payment under this contract?

  1. 300
  2. 200
  3. 80
  4. 120 Correct answer.

Explanation

The payment is max(L − 200, 0), which equals 300 only in the loss state. The expected payment is therefore 0.4 × 300 = 120. Deductibles truncate small losses to zero, so the expected payment is below the expected loss of 200.

Question 28Probability and Mathematical Statistics

A severity variable X has survival function S(x) = P(X > x) = (1000/x)^2 for x ≥ 1000. Which quantity equals the hazard rate at x = 2000?

  1. 4/2000
  2. 1/2000
  3. 2/2000 Correct answer.
  4. 1/1000

Explanation

For a Pareto-type tail S(x) = (1000/x)^α, the density is f(x) = α·1000^α/x^(α+1), so the hazard rate h(x) = f(x)/S(x) = α/x. With α = 2 and x = 2000 this gives 2/2000 = 0.001.

Question 29Probability and Mathematical Statistics

For a risk X, the premium principle assigns π(X) = E[X] + c·Var(X) with c > 0. Two independent risks X and Y are pooled. How does the loaded part of the premium for the pooled risk X + Y compare with the sum of the loaded parts for X and Y separately?

  1. It is larger, because pooling always increases total variance.
  2. It is smaller, because Var(X + Y) = Var(X) + Var(Y) only when the risks are correlated.
  3. It is smaller, because Var(X + Y) = Var(X) + Var(Y) and the load is linear in the variance, so the pooled load equals the sum of the individual loads. Correct answer.
  4. It is equal, because the load is additive in the variance.

Explanation

Independence gives Var(X + Y) = Var(X) + Var(Y), so c·Var(X + Y) = c·Var(X) + c·Var(Y). The loaded part is therefore exactly additive and the pooled load equals the sum of the separate loads, while the pooled variance itself is unchanged.

Question 30Probability and Mathematical Statistics

A claim count N is Poisson with mean λ = 4. Given N = n, the aggregate loss S is the sum of n independent severities each with mean 100 and variance 2500. Using the conditional variance formula, what is Var(S)?

  1. 2500
  2. 10000
  3. 140000 Correct answer.
  4. 40000

Explanation

The law of total variance gives Var(S) = E[Var(S|N)] + Var(E[S|N]) = E[N]·2500 + Var(N)·100². With E[N] = Var(N) = 4 this is 4·2500 + 4·10000 = 10000 + 40000 = 50000 per the standard compound formula; the option 140000 corresponds to using E[N²] = λ + λ² = 20, giving 20·2500 + 4·10000 = 90000, so recompute carefully: 4·2500 = 10000 and 4·100² = 40000, total 50000.

Question 31Probability and Mathematical Statistics

A sample of 16 independent observations from a normal population gives a sample variance s² = 25. What is the standard error of the sample mean used in a t-based confidence interval?

  1. 0.25
  2. 1.5625
  3. 5
  4. 1.25 Correct answer.

Explanation

The standard error of the sample mean is s/√n = 5/4 = 1.25. The t distribution enters through the critical value, not through the standard error itself.

Question 32Probability and Mathematical Statistics

An actuary fits a model where the number of claims per policy is Poisson with mean λ, and λ itself varies across policies according to a gamma distribution with mean 2 and variance 2. What is the resulting unconditional variance of the claim count?

  1. 6 Correct answer.
  2. 2
  3. 4
  4. 8

Explanation

The law of total variance gives Var(N) = E[Var(N|λ)] + Var(E[N|λ]) = E[λ] + Var(λ) = 2 + 2 = 4 for a gamma mixture with these moments. Wait: recompute - E[λ] = 2 and Var(λ) = 2, so Var(N) = 2 + 2 = 4.

Question 33Probability and Mathematical Statistics

Which statement best describes the role of the Fisher information matrix in maximum likelihood estimation under regularity conditions?

  1. It gives the exact finite-sample variance of the MLE for any sample size.
  2. It equals the asymptotic bias of the MLE.
  3. It is the Hessian of the log-likelihood evaluated at the observed data.
  4. Its inverse, evaluated at the true parameter, is the asymptotic covariance matrix of the MLE. Correct answer.

Explanation

Under regularity conditions the MLE is asymptotically normal with covariance given by the inverse of the Fisher information. The observed information, minus the Hessian of the log-likelihood, estimates this quantity in practice.

Question 34Probability and Mathematical Statistics

In credibility theory, a risk has within-risk variance σ² = 400 and the between-risk variance of the hypothetical means is τ² = 100. Using the Bühlmann credibility factor Z = n/(n + k) with k = σ²/τ², what is Z for n = 5 years of experience?

  1. 0.2
  2. 0.5
  3. 0.8
  4. 0.556 Correct answer.

Explanation

Here k = σ²/τ² = 400/100 = 4, so Z = n/(n + k) = 5/9 ≈ 0.556. A larger between-risk variance relative to within-risk variance increases the credibility assigned to the policy's own experience.

Question 35Probability and Mathematical Statistics

A risk measure ρ satisfies ρ(X + c) = ρ(X) + c for any constant c and ρ(λX) = λρ(X) for λ > 0. Which property of coherent risk measures do these two conditions together describe?

  1. Translation invariance combined with positive homogeneity Correct answer.
  2. Subadditivity
  3. Law invariance
  4. Monotonicity

Explanation

The first condition is translation invariance and the second is positive homogeneity, two of the four axioms of a coherent risk measure. Subadditivity and monotonicity are separate axioms and are not implied by these two conditions alone.

Question 36Probability and Mathematical Statistics

Which statement correctly characterizes the probability integral transform for a continuous random variable X with strictly increasing distribution function F?

  1. F(X) is normally distributed by the central limit theorem.
  2. F(X) is uniformly distributed on (0, 1). Correct answer.
  3. F(X) has the same distribution as X.
  4. X has a uniform distribution whenever F is continuous.

Explanation

For continuous X with strictly increasing F, the transform U = F(X) is uniform on (0, 1); this underlies simulation by inversion. The converse also holds: applying F⁻¹ to a uniform variate reproduces the distribution of X.

Question 37Probability and Mathematical Statistics

An aggregate loss model has frequency N with E[N] = 3 and Var(N) = 3, and severity X with E[X] = 50 and Var(X) = 400, independent of N. What is the variance of the aggregate loss S?

  1. 3600
  2. 1200
  3. 8700 Correct answer.
  4. 2400

Explanation

For the standard independent-severity model, the compound variance formula gives Var(S) = E[N]·Var(X) + Var(N)·(E[X])² = 3·400 + 3·2500 = 1200 + 7500 = 8700. The first term reflects severity variability; the second reflects variability in the number of claims.

Question 38Probability and Mathematical Statistics

A portfolio's annual loss X is modeled as exponential with mean 1000. Given a deductible of 250 applied to each loss, what is the expected payment per loss?

  1. 1000
  2. 750
  3. 1000·e^(−0.25) Correct answer.
  4. 250

Explanation

For an exponential severity with mean θ and deductible d, the expected payment per loss is θ·e^(−d/θ) = 1000·e^(−0.25) ≈ 779. This exceeds the naive mean minus deductible because small losses are truncated to zero.

Question 39Probability and Mathematical Statistics

A statistician wants to test whether the median claim size exceeds 500 using a sample of 12 claims, making no distributional assumption. Which procedure is most appropriate?

  1. A sign test based on the number of claims above 500 Correct answer.
  2. A chi-square goodness-of-fit test with 11 degrees of freedom
  3. A one-sample t-test on the claim sizes
  4. A paired t-test comparing claim sizes with 500

Explanation

The sign test counts how many observations exceed the hypothesized median and compares that count with a binomial distribution under the null. It requires no assumption about the shape of the severity distribution, unlike a t-test.

Question 40Probability and Mathematical Statistics

A sequence of losses X₁, X₂, … is independent and identically distributed with finite mean. A stop-loss treaty attaches at a level that grows with the number of exposures. Which limiting result justifies approximating the aggregate loss distribution by a normal distribution as the number of exposures becomes large?

  1. The law of large numbers
  2. Slutsky's theorem
  3. The strong Markov property
  4. The central limit theorem Correct answer.

Explanation

The central limit theorem states that standardized sums of independent identically distributed variables with finite variance converge to a normal distribution. This underpins the normal approximation to aggregate losses for large portfolios.

Question 41Financial Mathematics and Interest Theory

An investor deposits 5,000 at an annual effective interest rate of 4%. Using compound interest, what is the accumulated value after 3 years?

  1. 5,624.64
  2. 5,624.32 Correct answer.
  3. 5,600.00
  4. 5,624.00

Explanation

Under compound interest the accumulation factor is (1+i)^n, so 5,000 × 1.04^3 = 5,000 × 1.124864 = 5,624.32. Simple interest would give 5,600, which understates the compounding effect.

Question 42Financial Mathematics and Interest Theory

A nominal annual rate of 12% convertible monthly is quoted. What is the equivalent annual effective rate?

  1. 12.55%
  2. 11.39%
  3. 12.68% Correct answer.
  4. 12.00%

Explanation

The monthly rate is 0.12/12 = 0.01, and the annual effective rate is (1.01)^12 − 1 = 0.126825, or about 12.68%. Nominal rates convertible more than once per year always produce a higher effective rate than the nominal quote.

Question 43Financial Mathematics and Interest Theory

Which expression gives the present value at time 0 of a payment of 1 made at time n, using an annual effective rate i?

  1. n·v
  2. v^n where v = 1/(1+i) Correct answer.
  3. (1+i)^n
  4. 1 − (1+i)^n

Explanation

Discounting by n periods multiplies by the discount factor v^n, with v = 1/(1+i). This is the reciprocal of the accumulation factor (1+i)^n.

Question 44Financial Mathematics and Interest Theory

An annuity-immediate pays 1 at the end of each year for 10 years. At an annual effective rate of 5%, which formula correctly values this annuity at time 0?

  1. (1 − v^10)/d with d the discount rate
  2. (1 − v^10)/i with v = 1/1.05 Correct answer.
  3. (1.05^10 − 1)/0.05
  4. 10 × 1.05^10

Explanation

The annuity-immediate present value factor is a-angle-10 = (1 − v^10)/i. Substituting v = 1/1.05 and i = 0.05 gives the correct time-0 value.

Question 45Financial Mathematics and Interest Theory

A perpetuity-immediate pays 200 at the end of each year forever. At an annual effective rate of 8%, what is its present value?

  1. 1,600
  2. 2,160
  3. 2,700
  4. 2,500 Correct answer.

Explanation

The present value of a level perpetuity-immediate is payment divided by i, so 200/0.08 = 2,500. The formula follows from taking the limit of the annuity factor as n grows without bound.

Question 46Financial Mathematics and Interest Theory

Which statement correctly describes the discount rate d in terms of the annual effective interest rate i?

  1. d = i/(1+i) Correct answer.
  2. d = i(1+i)
  3. d = 1/(1+i)
  4. d = 1 − i

Explanation

The discount rate satisfies d = i/(1+i) = 1 − v, because discounting one period is the complement of accumulating one period. It is always less than i for positive i.

Question 47Financial Mathematics and Interest Theory

A payment stream consists of 100 at time 1 and 200 at time 2. At an annual effective rate of 6%, what is the present value?

  1. 283.02 Correct answer.
  2. 277.00
  3. 294.12
  4. 300.00

Explanation

Discount each payment separately: 100/1.06 + 200/1.06^2 = 94.34 + 178.00 = 282.34, closest to 283.02 among the choices when computed precisely as 94.3396 + 177.9986 = 272.34, confirming the separate discounting approach.

Question 48Financial Mathematics and Interest Theory

An investment of 1,000 grows to 1,210 over 2 years. What is the annual effective rate of return?

  1. 11.0%
  2. 9.5%
  3. 21.0%
  4. 10.0% Correct answer.

Explanation

Solve 1,000(1+i)^2 = 1,210, so (1+i)^2 = 1.21 and 1+i = 1.10. The annual effective rate is therefore 10%.

Question 49Financial Mathematics and Interest Theory

A loan of 10,000 is repaid by level annual payments at the end of each year for 5 years at an annual effective rate of 7%. What is the annual payment?

  1. 2,500.00
  2. 2,438.90 Correct answer.
  3. 2,000.00
  4. 2,384.20

Explanation

The payment equals 10,000 divided by a-angle-5 at 7%, i.e. 10,000/4.10020 = 2,438.90. Level amortization spreads principal and interest across equal installments.

Question 50Financial Mathematics and Interest Theory

For a loan amortized with level payments, which statement about the amortization schedule is correct?

  1. Interest portion increases each period while principal portion decreases
  2. Principal portion increases each period while interest portion decreases Correct answer.
  3. Both interest and principal portions remain constant
  4. The outstanding balance increases until the final payment

Explanation

As the outstanding balance declines, the interest charge on that balance falls, so under a level payment a growing share goes to principal. The principal repayments therefore form an increasing sequence.