Working with Formulas - Finite Mathematics

1. The sum $S$ of an infinite geometric series with first term $a_1$ and common ratio $r$ (where $|r| < 1$) is given by the formula $S = \frac{a_1}{1-r}$. Which of the following correctly expresses the common ratio $r$ in terms of $S$ and $a_1$?
   Answer: $r = 1 - \frac{a_1}{S}$ To rearrange the formula for $r$: 1. Start with the given formula: $S = \frac{a_1}{1-r}$. 2. Multiply both sides by $(1-r)$: $S(1-r) = a_1$. 3. Distribute $S$ on the left side: $S - Sr = a_1$. 4. Isolate the term containing $r$. One way is to add $Sr$ to both sides: $S = a_1 + Sr$. 5. Subtract $a_1$ from both sides: $S - a_1 = Sr$. 6. Divide by $S$ to solve for $r$: $r = \frac{S - a_1}{S}$. 7. This expression can be rewritten as $r = \frac{S}{S} - \frac{a_1}{S} = 1 - \frac{a_1}{S}$.
2. The relationship between pressure, volume, and temperature for an ideal gas is given by $\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$. 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?
   Answer: 2.78 L When you encounter ideal gas law problems with multiple processes, you need to apply the combined gas law equation $\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$ 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: $\frac{(2)(5)}{300} = \frac{(6)(V_2)}{450}$. 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: $\frac{(6)(2.5)}{450} = \frac{(2)(V_2)}{250}$. 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.
3. The resistance $R$ of a wire is given by $R = \rho \frac{L}{A}$, where $\rho$ is resistivity, $L$ is length, and $A$ 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?
   Answer: Multiply by $\sqrt{\frac{3}{5}}$ Let the original values be $R_0, L_0, A_0$. We want $R_1 = 0.6R_0$. Since volume $V = A \cdot L$ is constant, $A_0 L_0 = A_1 L_1$, so $A_1 = \frac{A_0 L_0}{L_1}$. The new resistance is $R_1 = \rho \frac{L_1}{A_1} = \rho \frac{L_1}{A_0 L_0/L_1} = \rho \frac{L_1^2}{A_0 L_0}$. Since $R_0 = \rho \frac{L_0}{A_0}$, we have $\frac{R_1}{R_0} = \frac{\rho \frac{L_1^2}{A_0 L_0}}{\rho \frac{L_0}{A_0}} = \frac{L_1^2}{L_0^2}$. Setting this equal to 0.6: $\frac{L_1^2}{L_0^2} = 0.6 = \frac{3}{5}$. Taking the square root: $\frac{L_1}{L_0} = \sqrt{\frac{3}{5}}$. Choice A would give $\left(\frac{3}{5}\right)^2 = \frac{9}{25} = 0.36$, which is too small. Choice C would give $\left(\frac{5}{3}\right)^2 = \frac{25}{9} \approx 2.78$, which increases resistance. Choice D would give $\frac{5}{3} \approx 1.67$, which also increases resistance.
4. In a system of linear equations $AX = B$, where $A = \begin{pmatrix} k & 1 \\ 3 & 2 \end{pmatrix}$, $X = \begin{pmatrix} x \\ y \end{pmatrix}$, and $B = \begin{pmatrix} p \\ q \end{pmatrix}$, the solution for $X$ is given by $X = A^{-1}B$. Assuming $A$ is invertible, which of the following gives the correct expression for the variable $x$?
   Answer: $x = \frac{2p - q}{2k - 3}$ To find $x$, we first need to find the inverse of matrix $A$. 1. The formula for the inverse of a $2 \times 2$ matrix $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ is $\frac{1}{ad-bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$. 2. For matrix $A = \begin{pmatrix} k & 1 \\ 3 & 2 \end{pmatrix}$, the determinant is $\det(A) = (k)(2) - (1)(3) = 2k - 3$. 3. The inverse is $A^{-1} = \frac{1}{2k-3} \begin{pmatrix} 2 & -1 \\ -3 & k \end{pmatrix}$. 4. Now, we find $X$ by computing $A^{-1}B$: $X = \begin{pmatrix} x \\ y \end{pmatrix} = \frac{1}{2k-3} \begin{pmatrix} 2 & -1 \\ -3 & k \end{pmatrix} \begin{pmatrix} p \\ q \end{pmatrix}$. 5. Performing the matrix multiplication: $X = \frac{1}{2k-3} \begin{pmatrix} (2)(p) + (-1)(q) \\ (-3)(p) + (k)(q) \end{pmatrix} = \frac{1}{2k-3} \begin{pmatrix} 2p - q \\ -3p + kq \end{pmatrix}$. 6. The variable $x$ is the top element of the resulting vector: $x = \frac{2p - q}{2k - 3}$.
5. The formula for the focal length of a thin lens is $\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}$, where $f$ is focal length, $d_o$ is object distance, and $d_i$ 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?
   Answer: 22.5 cm Given that the image is twice as far as the object: $d_i = 2d_o$. Substituting into the lens equation: $\frac{1}{15} = \frac{1}{d_o} + \frac{1}{2d_o} = \frac{2 + 1}{2d_o} = \frac{3}{2d_o}$. Cross-multiplying: $2d_o = 15 \times 3 = 45$, so $d_o = 22.5$ cm. Choice A (10 cm) would give $d_i = 20$ cm and $\frac{1}{f} = \frac{1}{10} + \frac{1}{20} = \frac{3}{20}$, so $f = \frac{20}{3} \approx 6.67$ cm. Choice C (7.5 cm) would give $d_i = 15$ cm and $\frac{1}{f} = \frac{1}{7.5} + \frac{1}{15} = \frac{2+1}{15} = \frac{1}{5}$, so $f = 5$ cm. Choice D (12 cm) would give $d_i = 24$ cm and $\frac{1}{f} = \frac{1}{12} + \frac{1}{24} = \frac{3}{24} = \frac{1}{8}$, so $f = 8$ cm.
6. The present value, $PV$, of an ordinary annuity is given by $PV = PMT \frac{1 - (1+i)^{-n}}{i}$, where $PMT$ is the regular payment, $i$ is the interest rate per period, and $n$ is the number of periods. Which of the following gives a correct expression for $n$?
   Answer: $n = -\frac{\ln\left(1 - \frac{PV \cdot i}{PMT}\right)}{\ln(1+i)}$ To solve for $n$, we must isolate the exponential term. 1. Start with the formula: $PV = PMT \frac{1 - (1+i)^{-n}}{i}$. 2. Multiply by $i$ and divide by $PMT$: $\frac{PV \cdot i}{PMT} = 1 - (1+i)^{-n}$. 3. Isolate the exponential term: $(1+i)^{-n} = 1 - \frac{PV \cdot i}{PMT}$. 4. Take the natural logarithm of both sides: $\ln((1+i)^{-n}) = \ln\left(1 - \frac{PV \cdot i}{PMT}\right)$. 5. Apply the power rule for logarithms: $-n \ln(1+i) = \ln\left(1 - \frac{PV \cdot i}{PMT}\right)$. 6. Solve for $n$ by dividing by $-\ln(1+i)$: $n = -\frac{\ln\left(1 - \frac{PV \cdot i}{PMT}\right)}{\ln(1+i)}$.
7. In an economic model, the quantity demanded, $Q_d$, and quantity supplied, $Q_s$, for a product are given by $Q_d = a - bP$ and $Q_s = c + dP$, where $P$ is the price. When the government imposes a per-unit tax, $t$, collected from producers, the new supply equation becomes $Q_s = c + d(P-t)$. At the new market equilibrium, $Q_d = Q_s$. What is the new equilibrium price $P$ in terms of the constants $a, b, c, d,$ and $t$?
   Answer: $P = \frac{a - c + dt}{b + d}$ To find the equilibrium price, set the quantity demanded equal to the quantity supplied and solve for $P$. 1. Set $Q_d = Q_s$: $a - bP = c + d(P-t)$. 2. Distribute $d$ on the right side: $a - bP = c + dP - dt$. 3. Group all terms with $P$ on one side and all constant terms on the other. Add $bP$ to both sides and add $dt$ to both sides: $a - c + dt = bP + dP$. 4. Factor out $P$ from the terms on the right side: $a - c + dt = P(b+d)$. 5. Divide by $(b+d)$ to isolate $P$: $P = \frac{a - c + dt}{b + d}$.
8. The formula for the period of a pendulum is $T = 2\pi\sqrt{\frac{L}{g}}$, where $T$ is the period, $L$ is length, and $g$ is gravitational acceleration. On a planet where $g = 4$ m/s², a pendulum has a period of $\pi$ seconds. If the same pendulum were moved to Earth where $g = 10$ m/s², what would be its new period?
   Answer: $\frac{2\pi}{\sqrt{10}}$ seconds First, find the length using the original conditions: $\pi = 2\pi\sqrt{\frac{L}{4}}$. Dividing by $2\pi$: $\frac{1}{2} = \sqrt{\frac{L}{4}}$. Squaring both sides: $\frac{1}{4} = \frac{L}{4}$, so $L = 1$ meter. On Earth: $T = 2\pi\sqrt{\frac{1}{10}} = \frac{2\pi}{\sqrt{10}}$. Choice A gives $\frac{\pi}{\sqrt{2.5}} = \frac{\pi\sqrt{2.5}}{2.5} = \frac{\pi\sqrt{10}}{5}$, which is different. Choice B gives $\frac{\pi\sqrt{10}}{4}$, which would correspond to $g = \frac{16}{10} = 1.6$. Choice D gives $\frac{\pi\sqrt{2}}{2} = \frac{\pi}{\sqrt{2}}$, which would correspond to $g = 8$ m/s².
9. A company's production cost, $C$, is modeled by the formula $C = F + Vx\left(1 - \frac{x}{E}\right)$, where $F$ is the fixed cost, $V$ is the variable cost per unit, $x$ is the number of units produced, and $E$ is an efficiency constant. Which of the following correctly expresses the efficiency constant $E$ in terms of the other variables?
   Answer: $E = \frac{Vx^2}{Vx - C + F}$ To solve for $E$: 1. Start with the equation: $C = F + Vx\left(1 - \frac{x}{E}\right)$. 2. Subtract $F$ from both sides: $C - F = Vx\left(1 - \frac{x}{E}\right)$. 3. Distribute $Vx$ on the right side: $C - F = Vx - \frac{Vx^2}{E}$. 4. Isolate the term with $E$. Add $\frac{Vx^2}{E}$ to both sides and subtract $(C-F)$ from both sides: $\frac{Vx^2}{E} = Vx - (C-F)$. 5. Simplify the right side: $\frac{Vx^2}{E} = Vx - C + F$. 6. To solve for $E$, we can take the reciprocal of both sides: $\frac{E}{Vx^2} = \frac{1}{Vx - C + F}$. 7. Multiply both sides by $Vx^2$: $E = \frac{Vx^2}{Vx - C + F}$.
10. The monthly payment $M$ for a loan with principal $P$ at a periodic interest rate $i$ over $n$ periods can be expressed as $M = \frac{Pi}{1 - (1+i)^{-n}}$. Which of the following correctly expresses the term $(1+i)^{-n}$ in terms of $M$, $P$, and $i$?
    Answer: $(1+i)^{-n} = 1 - \frac{Pi}{M}$ We need to isolate the expression $(1+i)^{-n}$. 1. Start with the given formula: $M = \frac{Pi}{1 - (1+i)^{-n}}$. 2. Multiply both sides by the denominator, $1 - (1+i)^{-n}$: $M(1 - (1+i)^{-n}) = Pi$. 3. Divide both sides by $M$: $1 - (1+i)^{-n} = \frac{Pi}{M}$. 4. Subtract 1 from both sides: $-(1+i)^{-n} = \frac{Pi}{M} - 1$. 5. Multiply both sides by -1 to solve for $(1+i)^{-n}$: $(1+i)^{-n} = -\left(\frac{Pi}{M} - 1\right) = 1 - \frac{Pi}{M}$. Choice D represents a common sign error that could occur in step 5.
11. In financial analysis, the debt-to-assets ratio, $R_{DA}$, is related to the debt-to-equity ratio, $R_{DE}$, by the formula $R_{DA} = \frac{R_{DE}}{1 + R_{DE}}$. Which of the following formulas correctly expresses the debt-to-equity ratio, $R_{DE}$, in terms of the debt-to-assets ratio, $R_{DA}$?
    Answer: $R_{DE} = \frac{R_{DA}}{1 - R_{DA}}$ To solve for $R_{DE}$, we need to rearrange the given formula. 1. Let $y = R_{DA}$ and $x = R_{DE}$ for simplicity. The formula is $y = \frac{x}{1+x}$. 2. Multiply both sides by $(1+x)$: $y(1+x) = x$. 3. Distribute $y$ on the left side: $y + yx = x$. 4. Group all terms containing $x$ on one side: $y = x - yx$. 5. Factor out $x$ from the terms on the right side: $y = x(1-y)$. 6. Divide by $(1-y)$ to solve for $x$: $x = \frac{y}{1-y}$. 7. Substitute the original variable names back: $R_{DE} = \frac{R_{DA}}{1 - R_{DA}}$.
12. The EOQ formula is $Q = \sqrt{\frac{2DS}{H}}$, where $Q$ is the optimal order quantity, $D$ is annual demand, $S$ is cost per order, and $H$ 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?
    Answer: Increase by approximately 16.6% Let the original values be $D_0, S_0, H_0$. The new values are $D_1 = 1.44D_0$ and $S_1 = 0.81S_0$. For the order quantity to remain unchanged: $\sqrt{\frac{2D_0S_0}{H_0}} = \sqrt{\frac{2D_1S_1}{H_1}}$. Squaring both sides: $\frac{D_0S_0}{H_0} = \frac{D_1S_1}{H_1} = \frac{1.44D_0 \cdot 0.81S_0}{H_1} = \frac{1.1664D_0S_0}{H_1}$. Therefore: $H_1 = 1.1664H_0$, which represents an increase of 16.64%. Choice B would require $H_1 = 0.857H_0$, giving $\frac{D_1S_1}{H_1} = \frac{1.1664}{0.857} \approx 1.361$, which is too large. Choice C (19.8% increase) gives $H_1 = 1.198H_0$, yielding a ratio of $\frac{1.1664}{1.198} \approx 0.974$, which is too small. Choice D would decrease costs, making the order quantity larger.
13. The break-even point formula is $x = \frac{F}{p - v}$, where $x$ is the number of units, $F$ is fixed costs, $p$ is price per unit, and $v$ 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?
    Answer: $62 From the original break-even point: $500 = \frac{15000}{p_1 - 12}$, so $p_1 - 12 = \frac{15000}{500} = 30$, giving $p_1 = 42$. For the new break-even point: $300 = \frac{15000}{p_2 - 12}$, so $p_2 - 12 = \frac{15000}{300} = 50$, giving $p_2 = 62$. Choice A ($42) is the original price. Choice B ($50) would give a break-even point of $\frac{15000}{50-12} = \frac{15000}{38} \approx 395$ units. Choice D ($58) would give $\frac{15000}{58-12} = \frac{15000}{46} \approx 326$ units, which is close but not exactly 300.
14. The total surface area, $A$, of a right circular cylinder with radius $r$ and height $h$ is given by the formula $A = 2\pi r^2 + 2\pi rh$. Which of the following correctly expresses the height $h$?
    Answer: $h = \frac{A}{2\pi r} - r$ To rearrange the formula to solve for $h$: 1. Start with the surface area formula: $A = 2\pi r^2 + 2\pi rh$. 2. Isolate the term containing $h$ by subtracting $2\pi r^2$ from both sides: $A - 2\pi r^2 = 2\pi rh$. 3. Divide both sides by the coefficient of $h$, which is $2\pi r$: $h = \frac{A - 2\pi r^2}{2\pi r}$. 4. Split the fraction into two terms: $h = \frac{A}{2\pi r} - \frac{2\pi r^2}{2\pi r}$. 5. Simplify the second term by canceling common factors: $h = \frac{A}{2\pi r} - r$.
15. In a statistical dataset, two data points $x_1$ and $x_2$ have corresponding z-scores of $z_1$ and $z_2$. The z-score formula is $z = \frac{x - \mu}{\sigma}$, where $\mu$ is the mean and $\sigma$ is the standard deviation. Which of the following expressions correctly represents the mean $\mu$ of the dataset in terms of $x_1$, $x_2$, $z_1$, and $z_2$?
    Answer: $\mu = \frac{z_2x_1 - z_1x_2}{z_2 - z_1}$ We have a system of two equations: 1. $z_1 = \frac{x_1 - \mu}{\sigma} \implies \sigma = \frac{x_1 - \mu}{z_1}$ 2. $z_2 = \frac{x_2 - \mu}{\sigma} \implies \sigma = \frac{x_2 - \mu}{z_2}$ Since $\sigma$ is the same for both, we can set the expressions for $\sigma$ equal: $\frac{x_1 - \mu}{z_1} = \frac{x_2 - \mu}{z_2}$. Now, solve for $\mu$: 3. Cross-multiply: $z_2(x_1 - \mu) = z_1(x_2 - \mu)$. 4. Distribute: $z_2x_1 - z_2\mu = z_1x_2 - z_1\mu$. 5. Group terms with $\mu$ on one side: $z_1\mu - z_2\mu = z_1x_2 - z_2x_1$. 6. Factor out $\mu$: $\mu(z_1 - z_2) = z_1x_2 - z_2x_1$. 7. Isolate $\mu$: $\mu = \frac{z_1x_2 - z_2x_1}{z_1 - z_2}$. This is equivalent to choice B, which is $\mu = \frac{z_2x_1 - z_1x_2}{z_2 - z_1}$ (multiplying numerator and denominator by -1).
16. In the compound interest formula $A = P(1 + r)^t$, if an investment triples in value over 8 years, what is the annual interest rate $r$ expressed as a percentage?
    Answer: Approximately 14.5% If the investment triples, then $A = 3P$. Substituting into the formula: $3P = P(1 + r)^8$. Dividing by $P$: $3 = (1 + r)^8$. Taking the 8th root of both sides: $1 + r = 3^{1/8}$. Using logarithms or a calculator: $3^{1/8} = 3^{0.125} \approx 1.1447$. Therefore, $r \approx 0.1447$ or about 14.5%. Choice B (12.5%) would give $(1.125)^8 \approx 2.57$, which is too small. Choice C (16.2%) would give $(1.162)^8 \approx 3.36$, which is too large. Choice D (18.7%) would give $(1.187)^8 \approx 4.29$, which is much too large.
17. The formula for the number of permutations of $n$ objects taken $k$ at a time, $P(n,k)$, can be defined recursively. For $k > 1$, $P(n, k)$ can be related to $P(n, k-1)$ by the formula $P(n, k) = (n - k + 1) \cdot P(n, k-1)$. How can $k$ be expressed in terms of $n$, $P(n, k)$, and $P(n, k-1)$?
    Answer: $k = n + 1 - \frac{P(n, k)}{P(n, k-1)}$ To solve the formula for $k$, we perform the following algebraic steps: 1. Start with the given relation: $P(n, k) = (n - k + 1) \cdot P(n, k-1)$. 2. Divide both sides by $P(n, k-1)$ to isolate the term containing $k$: $\frac{P(n, k)}{P(n, k-1)} = n - k + 1$. 3. Add $k$ to both sides to make its coefficient positive: $k + \frac{P(n, k)}{P(n, k-1)} = n + 1$. 4. Subtract $\frac{P(n, k)}{P(n, k-1)}$ from both sides to isolate $k$: $k = n + 1 - \frac{P(n, k)}{P(n, k-1)}$.
18. The sum of the first $n$ terms of an arithmetic sequence is given by the formula $S_n = \frac{n}{2}(2a_1 + (n-1)d)$, where $a_1$ is the first term and $d$ is the common difference. Which of the following correctly expresses the common difference $d$ in terms of $S_n$, $n$, and $a_1$?
    Answer: $d = \frac{2(S_n - na_1)}{n(n-1)}$ To solve for $d$, we must isolate it using algebraic manipulation. 1. Start with the formula: $S_n = \frac{n}{2}(2a_1 + (n-1)d)$. 2. Multiply both sides by 2: $2S_n = n(2a_1 + (n-1)d)$. 3. Divide both sides by $n$: $\frac{2S_n}{n} = 2a_1 + (n-1)d$. 4. Subtract $2a_1$ from both sides: $\frac{2S_n}{n} - 2a_1 = (n-1)d$. 5. To combine the terms on the left, find a common denominator: $\frac{2S_n - 2na_1}{n} = (n-1)d$. 6. Divide both sides by $(n-1)$ to solve for $d$: $d = \frac{2S_n - 2na_1}{n(n-1)}$. 7. Factor out a 2 from the numerator: $d = \frac{2(S_n - na_1)}{n(n-1)}$.
19. The future value, $FV$, of an ordinary annuity is given by the formula $FV = PMT \frac{(1+i)^n - 1}{i}$, where $PMT$ is the payment amount, $i$ is the interest rate per period, and $n$ is the number of periods. Which of the following formulas correctly represents the number of periods, $n$, in terms of $FV$, $PMT$, and $i$?
    Answer: $n = \frac{\ln\left(\frac{FV \cdot i}{PMT} + 1\right)}{\ln(1+i)}$ To solve for $n$, we first isolate the term containing $n$. 1. Start with the given formula: $FV = PMT \frac{(1+i)^n - 1}{i}$. 2. Multiply both sides by $i$: $FV \cdot i = PMT((1+i)^n - 1)$. 3. Divide both sides by $PMT$: $\frac{FV \cdot i}{PMT} = (1+i)^n - 1$. 4. Add 1 to both sides to isolate the exponential term: $\frac{FV \cdot i}{PMT} + 1 = (1+i)^n$. 5. Take the natural logarithm of both sides: $\ln\left(\frac{FV \cdot i}{PMT} + 1\right) = \ln((1+i)^n)$. 6. Use the logarithm power rule $\ln(x^y) = y \ln(x)$: $\ln\left(\frac{FV \cdot i}{PMT} + 1\right) = n \ln(1+i)$. 7. Divide by $\ln(1+i)$ to solve for $n$: $n = \frac{\ln\left(\frac{FV \cdot i}{PMT} + 1\right)}{\ln(1+i)}$.

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