Finite Mathematics Quiz: Working With Formulas
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Working With FormulasQuestion 1 of 19

The formula for the focal length of a thin lens is 1f=1do+1di\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}, where ff is focal length, dod_o is object distance, and did_i is image distance. A lens with focal length 15 cm creates an image that is twice as far from the lens as the object. What is the object distance?

10 cm
22.5 cm
7.5 cm
12 cm
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Finite Mathematics Quiz

Finite Mathematics Quiz: Working With Formulas

Practice Working With Formulas in Finite Mathematics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Working With Formulas, giving you a quick way to practice the rules, question types, and explanations that matter most for Finite Mathematics.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

The formula for the focal length of a thin lens is 1f=1do+1di\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}, where ff is focal length, dod_o is object distance, and did_i is image distance. A lens with focal length 15 cm creates an image that is twice as far from the lens as the object. What is the object distance?

  1. 10 cm
  2. 22.5 cm (correct answer)
  3. 7.5 cm
  4. 12 cm
Explanation: Given that the image is twice as far as the object: di=2dod_i = 2d_o. Substituting into the lens equation: 115=1do+12do=2+12do=32do\frac{1}{15} = \frac{1}{d_o} + \frac{1}{2d_o} = \frac{2 + 1}{2d_o} = \frac{3}{2d_o}. Cross-multiplying: 2do=15×3=452d_o = 15 \times 3 = 45, so do=22.5d_o = 22.5 cm. Choice A (10 cm) would give di=20d_i = 20 cm and 1f=110+120=320\frac{1}{f} = \frac{1}{10} + \frac{1}{20} = \frac{3}{20}, so f=2036.67f = \frac{20}{3} \approx 6.67 cm. Choice C (7.5 cm) would give di=15d_i = 15 cm and 1f=17.5+115=2+115=15\frac{1}{f} = \frac{1}{7.5} + \frac{1}{15} = \frac{2+1}{15} = \frac{1}{5}, so f=5f = 5 cm. Choice D (12 cm) would give di=24d_i = 24 cm and 1f=112+124=324=18\frac{1}{f} = \frac{1}{12} + \frac{1}{24} = \frac{3}{24} = \frac{1}{8}, so f=8f = 8 cm.

Question 2

In the compound interest formula A=P(1+r)tA = P(1 + r)^t, if an investment triples in value over 8 years, what is the annual interest rate rr expressed as a percentage?

  1. Approximately 14.5% (correct answer)
  2. Approximately 12.5%
  3. Approximately 16.2%
  4. Approximately 18.7%
Explanation: If the investment triples, then A=3PA = 3P. Substituting into the formula: 3P=P(1+r)83P = P(1 + r)^8. Dividing by PP: 3=(1+r)83 = (1 + r)^8. Taking the 8th root of both sides: 1+r=31/81 + r = 3^{1/8}. Using logarithms or a calculator: 31/8=30.1251.14473^{1/8} = 3^{0.125} \approx 1.1447. Therefore, r0.1447r \approx 0.1447 or about 14.5%. Choice B (12.5%) would give (1.125)82.57(1.125)^8 \approx 2.57, which is too small. Choice C (16.2%) would give (1.162)83.36(1.162)^8 \approx 3.36, which is too large. Choice D (18.7%) would give (1.187)84.29(1.187)^8 \approx 4.29, which is much too large.

Question 3

The break-even point formula is x=Fpvx = \frac{F}{p - v}, where xx is the number of units, FF is fixed costs, pp is price per unit, and vv is variable cost per unit. A company wants to reduce its break-even point from 500 units to 300 units. If fixed costs remain at $15,000 and the variable cost per unit remains at $12, what should be the new price per unit?

  1. $42
  2. $50
  3. $62 (correct answer)
  4. $58
Explanation: From the original break-even point: 500=15000p112500 = \frac{15000}{p_1 - 12}, so p112=15000500=30p_1 - 12 = \frac{15000}{500} = 30, giving p1=42p_1 = 42. For the new break-even point: 300=15000p212300 = \frac{15000}{p_2 - 12}, so p212=15000300=50p_2 - 12 = \frac{15000}{300} = 50, giving p2=62p_2 = 62. Choice A (42)istheoriginalprice.ChoiceB(42) is the original price. Choice B (50) would give a break-even point of 150005012=1500038395\frac{15000}{50-12} = \frac{15000}{38} \approx 395 units. Choice D ($58) would give $150005812=1500046326\frac{15000}{58-12} = \frac{15000}{46} \approx 326 $ units, which is close but not exactly 300.

Question 4

The formula for the period of a pendulum is T=2πLgT = 2\pi\sqrt{\frac{L}{g}}, where TT is the period, LL is length, and gg is gravitational acceleration. On a planet where g=4g = 4 m/s², a pendulum has a period of π\pi seconds. If the same pendulum were moved to Earth where g=10g = 10 m/s², what would be its new period?

  1. π2.5\frac{\pi}{\sqrt{2.5}} seconds
  2. π104\frac{\pi\sqrt{10}}{4} seconds
  3. 2π10\frac{2\pi}{\sqrt{10}} seconds (correct answer)
  4. π22\frac{\pi\sqrt{2}}{2} seconds
Explanation: First, find the length using the original conditions: π=2πL4\pi = 2\pi\sqrt{\frac{L}{4}}. Dividing by 2π2\pi: 12=L4\frac{1}{2} = \sqrt{\frac{L}{4}}. Squaring both sides: 14=L4\frac{1}{4} = \frac{L}{4}, so L=1L = 1 meter. On Earth: T=2π110=2π10T = 2\pi\sqrt{\frac{1}{10}} = \frac{2\pi}{\sqrt{10}}. Choice A gives π2.5=π2.52.5=π105\frac{\pi}{\sqrt{2.5}} = \frac{\pi\sqrt{2.5}}{2.5} = \frac{\pi\sqrt{10}}{5}, which is different. Choice B gives π104\frac{\pi\sqrt{10}}{4}, which would correspond to g=1610=1.6g = \frac{16}{10} = 1.6. Choice D gives π22=π2\frac{\pi\sqrt{2}}{2} = \frac{\pi}{\sqrt{2}}, which would correspond to g=8g = 8 m/s².

Question 5

The relationship between pressure, volume, and temperature for an ideal gas is given by P1V1T1=P2V2T2\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}. A gas sample initially at 2 atm, 300 K, and 5 L undergoes a process where the pressure is tripled and the temperature increases to 450 K. The final volume is then used as the initial volume for a second process where the pressure returns to 2 atm and the temperature drops to 250 K. What is the final volume after both processes?

  1. 5.56 L
  2. 3.33 L
  3. 4.17 L
  4. 2.78 L (correct answer)
Explanation: When you encounter ideal gas law problems with multiple processes, you need to apply the combined gas law equation P1V1T1=P2V2T2\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} sequentially for each step, using the final conditions from one process as the initial conditions for the next. For the first process, start with: P₁ = 2 atm, V₁ = 5 L, T₁ = 300 K. The conditions change to: P₂ = 6 atm (tripled), T₂ = 450 K. Solving for V₂: (2)(5)300=(6)(V2)450\frac{(2)(5)}{300} = \frac{(6)(V_2)}{450}. Cross-multiplying gives V₂ = 2.5 L. For the second process, use V₂ = 2.5 L as your new initial volume. Now: P₁ = 6 atm, V₁ = 2.5 L, T₁ = 450 K, changing to P₂ = 2 atm, T₂ = 250 K. Solving: (6)(2.5)450=(2)(V2)250\frac{(6)(2.5)}{450} = \frac{(2)(V_2)}{250}. This yields V₂ = 2.78 L. Answer A (5.56 L) likely results from incorrectly using the original volume in the second calculation. Answer B (3.33 L) comes from mixing up the pressure values or calculating only one process. Answer C (4.17 L) suggests an arithmetic error or using incorrect temperature conversions. The key strategy is treating each process independently while carefully tracking which values carry over. Always double-check that you're using the final conditions from step one as initial conditions for step two, and ensure all units are consistent throughout your calculations.

Question 6

The sum of the first nn terms of an arithmetic sequence is given by the formula Sn=n2(2a1+(n1)d)S_n = \frac{n}{2}(2a_1 + (n-1)d), where a1a_1 is the first term and dd is the common difference. Which of the following correctly expresses the common difference dd in terms of SnS_n, nn, and a1a_1?

  1. d=2Sn2a1n1d = \frac{2S_n - 2a_1}{n-1}
  2. d=2Snna1n(n1)d = \frac{2S_n - na_1}{n(n-1)}
  3. d=2Snn(n1)2a1n1d = \frac{2S_n}{n(n-1)} - \frac{2a_1}{n-1}
  4. d=2(Snna1)n(n1)d = \frac{2(S_n - na_1)}{n(n-1)} (correct answer)
Explanation: To solve for dd, we must isolate it using algebraic manipulation.
  1. Start with the formula: Sn=n2(2a1+(n1)d)S_n = \frac{n}{2}(2a_1 + (n-1)d).
  2. Multiply both sides by 2: 2Sn=n(2a1+(n1)d)2S_n = n(2a_1 + (n-1)d).
  3. Divide both sides by nn: 2Snn=2a1+(n1)d\frac{2S_n}{n} = 2a_1 + (n-1)d.
  4. Subtract 2a12a_1 from both sides: 2Snn2a1=(n1)d\frac{2S_n}{n} - 2a_1 = (n-1)d.
  5. To combine the terms on the left, find a common denominator: 2Sn2na1n=(n1)d\frac{2S_n - 2na_1}{n} = (n-1)d.
  6. Divide both sides by (n1)(n-1) to solve for dd: d=2Sn2na1n(n1)d = \frac{2S_n - 2na_1}{n(n-1)}.
  7. Factor out a 2 from the numerator: d=2(Snna1)n(n1)d = \frac{2(S_n - na_1)}{n(n-1)}.

Question 7

The present value, PVPV, of an ordinary annuity is given by PV=PMT1(1+i)niPV = PMT \frac{1 - (1+i)^{-n}}{i}, where PMTPMT is the regular payment, ii is the interest rate per period, and nn is the number of periods. Which of the following gives a correct expression for nn?

  1. n=ln(1+PViPMT)ln(1+i)n = -\frac{\ln\left(1 + \frac{PV \cdot i}{PMT}\right)}{\ln(1+i)}
  2. n=ln(1PViPMT)ln(1+i)n = \frac{\ln\left(1 - \frac{PV \cdot i}{PMT}\right)}{\ln(1+i)}
  3. n=ln(1PViPMT)ln(1+i)n = -\frac{\ln\left(1 - \frac{PV \cdot i}{PMT}\right)}{\ln(1+i)} (correct answer)
  4. n=ln(PViPMT1)ln(1+i)n = \frac{\ln\left(\frac{PV \cdot i}{PMT} - 1\right)}{\ln(1+i)}
Explanation: To solve for nn, we must isolate the exponential term.
  1. Start with the formula: PV=PMT1(1+i)niPV = PMT \frac{1 - (1+i)^{-n}}{i}.
  2. Multiply by ii and divide by PMTPMT: PViPMT=1(1+i)n\frac{PV \cdot i}{PMT} = 1 - (1+i)^{-n}.
  3. Isolate the exponential term: (1+i)n=1PViPMT(1+i)^{-n} = 1 - \frac{PV \cdot i}{PMT}.
  4. Take the natural logarithm of both sides: ln((1+i)n)=ln(1PViPMT)\ln((1+i)^{-n}) = \ln\left(1 - \frac{PV \cdot i}{PMT}\right).
  5. Apply the power rule for logarithms: nln(1+i)=ln(1PViPMT)-n \ln(1+i) = \ln\left(1 - \frac{PV \cdot i}{PMT}\right).
  6. Solve for nn by dividing by ln(1+i)-\ln(1+i): n=ln(1PViPMT)ln(1+i)n = -\frac{\ln\left(1 - \frac{PV \cdot i}{PMT}\right)}{\ln(1+i)}.

Question 8

The formula for the number of permutations of nn objects taken kk at a time, P(n,k)P(n,k), can be defined recursively. For k>1k > 1, P(n,k)P(n, k) can be related to P(n,k1)P(n, k-1) by the formula P(n,k)=(nk+1)P(n,k1)P(n, k) = (n - k + 1) \cdot P(n, k-1). How can kk be expressed in terms of nn, P(n,k)P(n, k), and P(n,k1)P(n, k-1)?

  1. k=n+1P(n,k)P(n,k1)k = n + 1 - \frac{P(n, k)}{P(n, k-1)} (correct answer)
  2. k=n+1+P(n,k)P(n,k1)k = n + 1 + \frac{P(n, k)}{P(n, k-1)}
  3. k=P(n,k)P(n,k1)n1k = \frac{P(n, k)}{P(n, k-1)} - n - 1
  4. k=n1P(n,k)P(n,k1)k = n - 1 - \frac{P(n, k)}{P(n, k-1)}
Explanation: To solve the formula for kk, we perform the following algebraic steps:
  1. Start with the given relation: P(n,k)=(nk+1)P(n,k1)P(n, k) = (n - k + 1) \cdot P(n, k-1).
  2. Divide both sides by P(n,k1)P(n, k-1) to isolate the term containing kk: P(n,k)P(n,k1)=nk+1\frac{P(n, k)}{P(n, k-1)} = n - k + 1.
  3. Add kk to both sides to make its coefficient positive: k+P(n,k)P(n,k1)=n+1k + \frac{P(n, k)}{P(n, k-1)} = n + 1.
  4. Subtract P(n,k)P(n,k1)\frac{P(n, k)}{P(n, k-1)} from both sides to isolate kk: k=n+1P(n,k)P(n,k1)k = n + 1 - \frac{P(n, k)}{P(n, k-1)}.

Question 9

A company's production cost, CC, is modeled by the formula C=F+Vx(1xE)C = F + Vx\left(1 - \frac{x}{E}\right), where FF is the fixed cost, VV is the variable cost per unit, xx is the number of units produced, and EE is an efficiency constant. Which of the following correctly expresses the efficiency constant EE in terms of the other variables?

  1. E=Vx2CFVxE = \frac{Vx^2}{C - F - Vx}
  2. E=VxC+FVx2E = \frac{Vx - C + F}{Vx^2}
  3. E=Vx2VxC+FE = \frac{Vx^2}{Vx - C + F} (correct answer)
  4. E=Vx1C+FxE = \frac{Vx}{1 - \frac{C+F}{x}}
Explanation: To solve for EE:
  1. Start with the equation: C=F+Vx(1xE)C = F + Vx\left(1 - \frac{x}{E}\right).
  2. Subtract FF from both sides: CF=Vx(1xE)C - F = Vx\left(1 - \frac{x}{E}\right).
  3. Distribute VxVx on the right side: CF=VxVx2EC - F = Vx - \frac{Vx^2}{E}.
  4. Isolate the term with EE. Add Vx2E\frac{Vx^2}{E} to both sides and subtract (CF)(C-F) from both sides: Vx2E=Vx(CF)\frac{Vx^2}{E} = Vx - (C-F).
  5. Simplify the right side: Vx2E=VxC+F\frac{Vx^2}{E} = Vx - C + F.
  6. To solve for EE, we can take the reciprocal of both sides: EVx2=1VxC+F\frac{E}{Vx^2} = \frac{1}{Vx - C + F}.
  7. Multiply both sides by Vx2Vx^2: E=Vx2VxC+FE = \frac{Vx^2}{Vx - C + F}.

Question 10

The total surface area, AA, of a right circular cylinder with radius rr and height hh is given by the formula A=2πr2+2πrhA = 2\pi r^2 + 2\pi rh. Which of the following correctly expresses the height hh?

  1. h=A2πrrh = \frac{A}{2\pi r} - r (correct answer)
  2. h=A2πr2πrh = \frac{A - 2\pi r^2}{\pi r}
  3. h=A2πr21h = \frac{A}{2\pi r^2} - 1
  4. h=A2πr2πrh = \frac{A - 2\pi r}{2\pi r}
Explanation: To rearrange the formula to solve for hh:
  1. Start with the surface area formula: A=2πr2+2πrhA = 2\pi r^2 + 2\pi rh.
  2. Isolate the term containing hh by subtracting 2πr22\pi r^2 from both sides: A2πr2=2πrhA - 2\pi r^2 = 2\pi rh.
  3. Divide both sides by the coefficient of hh, which is 2πr2\pi r: h=A2πr22πrh = \frac{A - 2\pi r^2}{2\pi r}.
  4. Split the fraction into two terms: h=A2πr2πr22πrh = \frac{A}{2\pi r} - \frac{2\pi r^2}{2\pi r}.
  5. Simplify the second term by canceling common factors: h=A2πrrh = \frac{A}{2\pi r} - r.

Question 11

The monthly payment MM for a loan with principal PP at a periodic interest rate ii over nn periods can be expressed as M=Pi1(1+i)nM = \frac{Pi}{1 - (1+i)^{-n}}. Which of the following correctly expresses the term (1+i)n(1+i)^{-n} in terms of MM, PP, and ii?

  1. (1+i)n=PiM1(1+i)^{-n} = \frac{Pi}{M} - 1
  2. (1+i)n=1PiM(1+i)^{-n} = 1 - \frac{Pi}{M} (correct answer)
  3. (1+i)n=1+PiM(1+i)^{-n} = 1 + \frac{Pi}{M}
  4. (1+i)n=PiMM(1+i)^{-n} = \frac{Pi - M}{M}
Explanation: We need to isolate the expression (1+i)n(1+i)^{-n}.
  1. Start with the given formula: M=Pi1(1+i)nM = \frac{Pi}{1 - (1+i)^{-n}}.
  2. Multiply both sides by the denominator, 1(1+i)n1 - (1+i)^{-n}: M(1(1+i)n)=PiM(1 - (1+i)^{-n}) = Pi.
  3. Divide both sides by MM: 1(1+i)n=PiM1 - (1+i)^{-n} = \frac{Pi}{M}.
  4. Subtract 1 from both sides: (1+i)n=PiM1-(1+i)^{-n} = \frac{Pi}{M} - 1.
  5. Multiply both sides by -1 to solve for (1+i)n(1+i)^{-n}: (1+i)n=(PiM1)=1PiM(1+i)^{-n} = -\left(\frac{Pi}{M} - 1\right) = 1 - \frac{Pi}{M}. Choice D represents a common sign error that could occur in step 5.

Question 12

In financial analysis, the debt-to-assets ratio, RDAR_{DA}, is related to the debt-to-equity ratio, RDER_{DE}, by the formula RDA=RDE1+RDER_{DA} = \frac{R_{DE}}{1 + R_{DE}}. Which of the following formulas correctly expresses the debt-to-equity ratio, RDER_{DE}, in terms of the debt-to-assets ratio, RDAR_{DA}?

  1. RDE=1RDARDAR_{DE} = \frac{1 - R_{DA}}{R_{DA}}
  2. RDE=RDA1+RDAR_{DE} = \frac{R_{DA}}{1 + R_{DA}}
  3. RDE=1+RDARDAR_{DE} = \frac{1 + R_{DA}}{R_{DA}}
  4. RDE=RDA1RDAR_{DE} = \frac{R_{DA}}{1 - R_{DA}} (correct answer)
Explanation: To solve for RDER_{DE}, we need to rearrange the given formula.
  1. Let y=RDAy = R_{DA} and x=RDEx = R_{DE} for simplicity. The formula is y=x1+xy = \frac{x}{1+x}.
  2. Multiply both sides by (1+x)(1+x): y(1+x)=xy(1+x) = x.
  3. Distribute yy on the left side: y+yx=xy + yx = x.
  4. Group all terms containing xx on one side: y=xyxy = x - yx.
  5. Factor out xx from the terms on the right side: y=x(1y)y = x(1-y).
  6. Divide by (1y)(1-y) to solve for xx: x=y1yx = \frac{y}{1-y}.
  7. Substitute the original variable names back: RDE=RDA1RDAR_{DE} = \frac{R_{DA}}{1 - R_{DA}}.

Question 13

In an economic model, the quantity demanded, QdQ_d, and quantity supplied, QsQ_s, for a product are given by Qd=abPQ_d = a - bP and Qs=c+dPQ_s = c + dP, where PP is the price. When the government imposes a per-unit tax, tt, collected from producers, the new supply equation becomes Qs=c+d(Pt)Q_s = c + d(P-t). At the new market equilibrium, Qd=QsQ_d = Q_s. What is the new equilibrium price PP in terms of the constants a,b,c,d,a, b, c, d, and tt?

  1. P=acdtb+dP = \frac{a - c - dt}{b + d}
  2. P=ac+dtb+dP = \frac{a - c + dt}{b + d} (correct answer)
  3. P=a+cdtbdP = \frac{a + c - dt}{b - d}
  4. P=ac+dtbdP = \frac{a - c + dt}{b - d}
Explanation: To find the equilibrium price, set the quantity demanded equal to the quantity supplied and solve for PP.
  1. Set Qd=QsQ_d = Q_s: abP=c+d(Pt)a - bP = c + d(P-t).
  2. Distribute dd on the right side: abP=c+dPdta - bP = c + dP - dt.
  3. Group all terms with PP on one side and all constant terms on the other. Add bPbP to both sides and add dtdt to both sides: ac+dt=bP+dPa - c + dt = bP + dP.
  4. Factor out PP from the terms on the right side: ac+dt=P(b+d)a - c + dt = P(b+d).
  5. Divide by (b+d)(b+d) to isolate PP: P=ac+dtb+dP = \frac{a - c + dt}{b + d}.

Question 14

In a statistical dataset, two data points x1x_1 and x2x_2 have corresponding z-scores of z1z_1 and z2z_2. The z-score formula is z=xμσz = \frac{x - \mu}{\sigma}, where μ\mu is the mean and σ\sigma is the standard deviation. Which of the following expressions correctly represents the mean μ\mu of the dataset in terms of x1x_1, x2x_2, z1z_1, and z2z_2?

  1. μ=z1x1z2x2z1z2\mu = \frac{z_1x_1 - z_2x_2}{z_1 - z_2}
  2. μ=z2x1z1x2z2z1\mu = \frac{z_2x_1 - z_1x_2}{z_2 - z_1} (correct answer)
  3. μ=x1+x2z1+z2\mu = \frac{x_1 + x_2}{z_1 + z_2}
  4. μ=z1x2+z2x1z1+z2\mu = \frac{z_1x_2 + z_2x_1}{z_1 + z_2}
Explanation: We have a system of two equations:
  1. z1=x1μσ    σ=x1μz1z_1 = \frac{x_1 - \mu}{\sigma} \implies \sigma = \frac{x_1 - \mu}{z_1}
  2. z2=x2μσ    σ=x2μz2z_2 = \frac{x_2 - \mu}{\sigma} \implies \sigma = \frac{x_2 - \mu}{z_2} Since σ\sigma is the same for both, we can set the expressions for σ\sigma equal: x1μz1=x2μz2\frac{x_1 - \mu}{z_1} = \frac{x_2 - \mu}{z_2}. Now, solve for μ\mu:
  3. Cross-multiply: z2(x1μ)=z1(x2μ)z_2(x_1 - \mu) = z_1(x_2 - \mu).
  4. Distribute: z2x1z2μ=z1x2z1μz_2x_1 - z_2\mu = z_1x_2 - z_1\mu.
  5. Group terms with μ\mu on one side: z1μz2μ=z1x2z2x1z_1\mu - z_2\mu = z_1x_2 - z_2x_1.
  6. Factor out μ\mu: μ(z1z2)=z1x2z2x1\mu(z_1 - z_2) = z_1x_2 - z_2x_1.
  7. Isolate μ\mu: μ=z1x2z2x1z1z2\mu = \frac{z_1x_2 - z_2x_1}{z_1 - z_2}. This is equivalent to choice B, which is μ=z2x1z1x2z2z1\mu = \frac{z_2x_1 - z_1x_2}{z_2 - z_1} (multiplying numerator and denominator by -1).

Question 15

The future value, FVFV, of an ordinary annuity is given by the formula FV=PMT(1+i)n1iFV = PMT \frac{(1+i)^n - 1}{i}, where PMTPMT is the payment amount, ii is the interest rate per period, and nn is the number of periods. Which of the following formulas correctly represents the number of periods, nn, in terms of FVFV, PMTPMT, and ii?

  1. n=ln(FViPMT)+1ln(1+i)n = \frac{\ln\left(\frac{FV \cdot i}{PMT}\right) + 1}{\ln(1+i)}
  2. n=ln(FViPMT+1)ln(1+i)n = \frac{\ln\left(\frac{FV \cdot i}{PMT} + 1\right)}{\ln(1+i)} (correct answer)
  3. n=ln(FV)+ln(i)ln(PMT)+1ln(1+i)n = \frac{\ln(FV) + \ln(i) - \ln(PMT) + 1}{\ln(1+i)}
  4. n=ln(FViPMT(1+i)+11+i)n = \ln\left( \frac{FV \cdot i}{PMT(1+i)} + \frac{1}{1+i} \right)
Explanation: To solve for nn, we first isolate the term containing nn.
  1. Start with the given formula: FV=PMT(1+i)n1iFV = PMT \frac{(1+i)^n - 1}{i}.
  2. Multiply both sides by ii: FVi=PMT((1+i)n1)FV \cdot i = PMT((1+i)^n - 1).
  3. Divide both sides by PMTPMT: FViPMT=(1+i)n1\frac{FV \cdot i}{PMT} = (1+i)^n - 1.
  4. Add 1 to both sides to isolate the exponential term: FViPMT+1=(1+i)n\frac{FV \cdot i}{PMT} + 1 = (1+i)^n.
  5. Take the natural logarithm of both sides: ln(FViPMT+1)=ln((1+i)n)\ln\left(\frac{FV \cdot i}{PMT} + 1\right) = \ln((1+i)^n).
  6. Use the logarithm power rule ln(xy)=yln(x)\ln(x^y) = y \ln(x): ln(FViPMT+1)=nln(1+i)\ln\left(\frac{FV \cdot i}{PMT} + 1\right) = n \ln(1+i).
  7. Divide by ln(1+i)\ln(1+i) to solve for nn: n=ln(FViPMT+1)ln(1+i)n = \frac{\ln\left(\frac{FV \cdot i}{PMT} + 1\right)}{\ln(1+i)}.

Question 16

The sum SS of an infinite geometric series with first term a1a_1 and common ratio rr (where r<1|r| < 1) is given by the formula S=a11rS = \frac{a_1}{1-r}. Which of the following correctly expresses the common ratio rr in terms of SS and a1a_1?

  1. r=1a1Sr = 1 - \frac{a_1}{S} (correct answer)
  2. r=a1S1r = \frac{a_1}{S} - 1
  3. r=1+a1Sr = 1 + \frac{a_1}{S}
  4. r=S+a1Sr = \frac{S+a_1}{S}
Explanation: To rearrange the formula for rr:
  1. Start with the given formula: S=a11rS = \frac{a_1}{1-r}.
  2. Multiply both sides by (1r)(1-r): S(1r)=a1S(1-r) = a_1.
  3. Distribute SS on the left side: SSr=a1S - Sr = a_1.
  4. Isolate the term containing rr. One way is to add SrSr to both sides: S=a1+SrS = a_1 + Sr.
  5. Subtract a1a_1 from both sides: Sa1=SrS - a_1 = Sr.
  6. Divide by SS to solve for rr: r=Sa1Sr = \frac{S - a_1}{S}.
  7. This expression can be rewritten as r=SSa1S=1a1Sr = \frac{S}{S} - \frac{a_1}{S} = 1 - \frac{a_1}{S}.

Question 17

A manufacturing company uses the economic order quantity (EOQ) model to minimize inventory costs. The total cost function includes both ordering costs and carrying costs.

The EOQ formula is Q=2DSHQ = \sqrt{\frac{2DS}{H}}, where QQ is the optimal order quantity, DD is annual demand, SS is cost per order, and HH is annual holding cost per unit. If the annual demand increases by 44% and the cost per order decreases by 19%, by what percentage should the annual holding cost per unit change to keep the optimal order quantity unchanged?

  1. Increase by approximately 16.6% (correct answer)
  2. Decrease by approximately 14.3%
  3. Increase by approximately 19.8%
  4. Decrease by approximately 18.2%
Explanation: Let the original values be D0,S0,H0D_0, S_0, H_0. The new values are D1=1.44D0D_1 = 1.44D_0 and S1=0.81S0S_1 = 0.81S_0. For the order quantity to remain unchanged: 2D0S0H0=2D1S1H1\sqrt{\frac{2D_0S_0}{H_0}} = \sqrt{\frac{2D_1S_1}{H_1}}. Squaring both sides: D0S0H0=D1S1H1=1.44D00.81S0H1=1.1664D0S0H1\frac{D_0S_0}{H_0} = \frac{D_1S_1}{H_1} = \frac{1.44D_0 \cdot 0.81S_0}{H_1} = \frac{1.1664D_0S_0}{H_1}. Therefore: H1=1.1664H0H_1 = 1.1664H_0, which represents an increase of 16.64%. Choice B would require H1=0.857H0H_1 = 0.857H_0, giving D1S1H1=1.16640.8571.361\frac{D_1S_1}{H_1} = \frac{1.1664}{0.857} \approx 1.361, which is too large. Choice C (19.8% increase) gives H1=1.198H0H_1 = 1.198H_0, yielding a ratio of 1.16641.1980.974\frac{1.1664}{1.198} \approx 0.974, which is too small. Choice D would decrease costs, making the order quantity larger.

Question 18

The resistance RR of a wire is given by R=ρLAR = \rho \frac{L}{A}, where ρ\rho is resistivity, LL is length, and AA is cross-sectional area. If the resistance must be reduced to 60% of its original value by changing only the length and keeping the wire's volume constant, by what factor should the length be changed?

  1. Multiply by 35\frac{3}{5}
  2. Multiply by 35\sqrt{\frac{3}{5}} (correct answer)
  3. Multiply by 53\frac{5}{3}
  4. Multiply by 53\sqrt{\frac{5}{3}}
Explanation: Let the original values be R0,L0,A0R_0, L_0, A_0. We want R1=0.6R0R_1 = 0.6R_0. Since volume V=ALV = A \cdot L is constant, A0L0=A1L1A_0 L_0 = A_1 L_1, so A1=A0L0L1A_1 = \frac{A_0 L_0}{L_1}. The new resistance is R1=ρL1A1=ρL1A0L0/L1=ρL12A0L0R_1 = \rho \frac{L_1}{A_1} = \rho \frac{L_1}{A_0 L_0/L_1} = \rho \frac{L_1^2}{A_0 L_0}. Since R0=ρL0A0R_0 = \rho \frac{L_0}{A_0}, we have R1R0=ρL12A0L0ρL0A0=L12L02\frac{R_1}{R_0} = \frac{\rho \frac{L_1^2}{A_0 L_0}}{\rho \frac{L_0}{A_0}} = \frac{L_1^2}{L_0^2}. Setting this equal to 0.6: L12L02=0.6=35\frac{L_1^2}{L_0^2} = 0.6 = \frac{3}{5}. Taking the square root: L1L0=35\frac{L_1}{L_0} = \sqrt{\frac{3}{5}}. Choice A would give (35)2=925=0.36\left(\frac{3}{5}\right)^2 = \frac{9}{25} = 0.36, which is too small. Choice C would give (53)2=2592.78\left(\frac{5}{3}\right)^2 = \frac{25}{9} \approx 2.78, which increases resistance. Choice D would give 531.67\frac{5}{3} \approx 1.67, which also increases resistance.

Question 19

In a system of linear equations AX=BAX = B, where A=(k132)A = \begin{pmatrix} k & 1 \\ 3 & 2 \end{pmatrix}, X=(xy)X = \begin{pmatrix} x \\ y \end{pmatrix}, and B=(pq)B = \begin{pmatrix} p \\ q \end{pmatrix}, the solution for XX is given by X=A1BX = A^{-1}B. Assuming AA is invertible, which of the following gives the correct expression for the variable xx?

  1. x=2pq2k3x = \frac{2p - q}{2k - 3} (correct answer)
  2. x=kpq2k3x = \frac{kp - q}{2k - 3}
  3. x=2p+q2k3x = \frac{2p + q}{2k - 3}
  4. x=2pq3k2x = \frac{2p - q}{3k - 2}
Explanation: To find xx, we first need to find the inverse of matrix AA.
  1. The formula for the inverse of a 2×22 \times 2 matrix (abcd)\begin{pmatrix} a & b \\ c & d \end{pmatrix} is 1adbc(dbca)\frac{1}{ad-bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}.
  2. For matrix A=(k132)A = \begin{pmatrix} k & 1 \\ 3 & 2 \end{pmatrix}, the determinant is det(A)=(k)(2)(1)(3)=2k3\det(A) = (k)(2) - (1)(3) = 2k - 3.
  3. The inverse is A1=12k3(213k)A^{-1} = \frac{1}{2k-3} \begin{pmatrix} 2 & -1 \\ -3 & k \end{pmatrix}.
  4. Now, we find XX by computing A1BA^{-1}B: X=(xy)=12k3(213k)(pq)X = \begin{pmatrix} x \\ y \end{pmatrix} = \frac{1}{2k-3} \begin{pmatrix} 2 & -1 \\ -3 & k \end{pmatrix} \begin{pmatrix} p \\ q \end{pmatrix}.
  5. Performing the matrix multiplication: X=12k3((2)(p)+(1)(q)(3)(p)+(k)(q))=12k3(2pq3p+kq)X = \frac{1}{2k-3} \begin{pmatrix} (2)(p) + (-1)(q) \\ (-3)(p) + (k)(q) \end{pmatrix} = \frac{1}{2k-3} \begin{pmatrix} 2p - q \\ -3p + kq \end{pmatrix}.
  6. The variable xx is the top element of the resulting vector: x=2pq2k3x = \frac{2p - q}{2k - 3}.