ACCUPLACER Advanced Algebra & Functions Quiz: Rearranging Formulas
20 questions · exam conditions
0:00
Rearranging FormulasQuestion 1 of 20

A linear relationship is given by the equation a(x−h)=b(k−y)a(x-h) = b(k-y). Which of the following correctly expresses yy in terms of the other variables?

y=k−a(x−h)by = k - \frac{a(x-h)}{b}
y=bk−ax+ahby = \frac{bk - ax + ah}{b}
y=k+a(x−h)by = k + \frac{a(x-h)}{b}
y=a(x−h)−bkby = \frac{a(x-h) - bk}{b}
← Back to quizzes

ACCUPLACER Advanced Algebra & Functions Quiz

ACCUPLACER Advanced Algebra & Functions Quiz: Rearranging Formulas

Practice Rearranging Formulas in ACCUPLACER Advanced Algebra & Functions 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 Rearranging Formulas, giving you a quick way to practice the rules, question types, and explanations that matter most for ACCUPLACER Advanced Algebra & Functions.

How to use this quiz

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.

All questions

Question 1

A linear relationship is given by the equation a(x−h)=b(k−y)a(x-h) = b(k-y). Which of the following correctly expresses yy in terms of the other variables?

  1. y=k−a(x−h)by = k - \frac{a(x-h)}{b} (correct answer)
  2. y=bk−ax+ahby = \frac{bk - ax + ah}{b}
  3. y=k+a(x−h)by = k + \frac{a(x-h)}{b}
  4. y=a(x−h)−bkby = \frac{a(x-h) - bk}{b}
Explanation: To solve for yy, first divide both sides by bb to get a(x−h)b=k−y\frac{a(x-h)}{b} = k-y. Next, add yy to both sides: y+a(x−h)b=ky + \frac{a(x-h)}{b} = k. Finally, subtract a(x−h)b\frac{a(x-h)}{b} from both sides to isolate yy, resulting in y=k−a(x−h)by = k - \frac{a(x-h)}{b}. Note that choice B is algebraically equivalent to choice A when expanded and simplified.

Question 2

The formula for converting temperature from Fahrenheit (FF) to Celsius (CC) is C=59(F−32)C = \frac{5}{9}(F - 32). Which of the following formulas correctly converts from Celsius to Fahrenheit?

  1. F=95C+32F = \frac{9}{5}C + 32 (correct answer)
  2. F=95(C+32)F = \frac{9}{5}(C + 32)
  3. F=59C+32F = \frac{5}{9}C + 32
  4. F=9C+325F = \frac{9C + 32}{5}
Explanation: To solve for FF, first multiply both sides by 95\frac{9}{5} to get 95C=F−32\frac{9}{5}C = F - 32. Then, add 32 to both sides to isolate FF, which gives F=95C+32F = \frac{9}{5}C + 32.

Question 3

In physics, the law of conservation of momentum for a two-object collision is expressed as m1u1+m2u2=m1v1+m2v2m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2. Which of the following equations correctly solves for m2m_2?

  1. m2=m1(u1−v1)−(u2−v2)m_2 = m_1(u_1 - v_1) - (u_2 - v_2)
  2. m2=m1(u1+v1)v2+u2m_2 = \frac{m_1(u_1 + v_1)}{v_2 + u_2}
  3. m2=m1u1−m1v1−u2v2m_2 = \frac{m_1 u_1 - m_1 v_1 - u_2}{v_2}
  4. m2=m1(u1−v1)v2−u2m_2 = \frac{m_1(u_1 - v_1)}{v_2 - u_2} (correct answer)
Explanation: To solve for m2m_2, first gather all terms with m2m_2 on one side and all terms with m1m_1 on the other. This gives m2v2−m2u2=m1u1−m1v1m_2 v_2 - m_2 u_2 = m_1 u_1 - m_1 v_1. Next, factor out m2m_2 on the left side and m1m_1 on the right side: m2(v2−u2)=m1(u1−v1)m_2(v_2 - u_2) = m_1(u_1 - v_1). Finally, divide both sides by (v2−u2)(v_2 - u_2) to isolate m2m_2, resulting in m2=m1(u1−v1)v2−u2m_2 = \frac{m_1(u_1 - v_1)}{v_2 - u_2}.

Question 4

A rational function is defined by y=ax+bcx+dy = \frac{ax+b}{cx+d}. If this equation is rearranged to solve for xx, what is the correct expression for xx?

  1. x=b−dya−cyx = \frac{b - dy}{a - cy}
  2. x=dy−ba−cyx = \frac{dy - b}{a - cy}
  3. x=b+dycy+ax = \frac{b + dy}{cy + a}
  4. x=b−dycy−ax = \frac{b - dy}{cy - a} (correct answer)
Explanation: To solve for xx, first multiply both sides by (cx+d)(cx+d) to get y(cx+d)=ax+by(cx+d) = ax+b. Distribute yy: cxy+dy=ax+bcxy + dy = ax+b. Group all terms containing xx on one side: cxy−ax=b−dycxy - ax = b - dy. Factor out xx: x(cy−a)=b−dyx(cy - a) = b - dy. Finally, divide by (cy−a)(cy-a) to isolate xx, which results in x=b−dycy−ax = \frac{b - dy}{cy - a}.

Question 5

Newton's law of universal gravitation is F=Gm1m2r2F = G\frac{m_1 m_2}{r^2}. Which of the following correctly solves this formula for the mass m1m_1?

  1. m1=Gm2Fr2m_1 = \frac{Gm_2}{Fr^2}
  2. m1=FGm2r2m_1 = \frac{FGm_2}{r^2}
  3. m1=Fr2Gm2m_1 = \frac{Fr^2}{Gm_2} (correct answer)
  4. m1=Fr2−Gm2m_1 = Fr^2 - Gm_2
Explanation: To solve for m1m_1, first multiply both sides by r2r^2 to clear the denominator: Fr2=Gm1m2Fr^2 = Gm_1 m_2. Then, divide both sides by the other factors, GG and m2m_2, to isolate m1m_1. This gives m1=Fr2Gm2m_1 = \frac{Fr^2}{Gm_2}.

Question 6

The formula for the volume VV of a frustum of a cone is V=13πh(R2+Rr+r2)V = \frac{1}{3}\pi h(R^2 + Rr + r^2). Which of the following equations correctly solves for the height hh?

  1. h=3VπR2+Rr+r2h = \frac{3V\pi}{R^2 + Rr + r^2}
  2. h=V3π(R2+Rr+r2)h = \frac{V}{3\pi(R^2 + Rr + r^2)}
  3. h=3Vπ(R2+Rr+r2)h = \frac{3V}{\pi(R^2 + Rr + r^2)} (correct answer)
  4. h=3V−π(R2+Rr+r2)h = 3V - \pi(R^2 + Rr + r^2)
Explanation: To solve for hh, first multiply both sides by 3 to clear the fraction: 3V=πh(R2+Rr+r2)3V = \pi h(R^2 + Rr + r^2). Then, to isolate hh, divide both sides by all the other factors, which are π\pi and the entire quantity (R2+Rr+r2)(R^2 + Rr + r^2). This gives h=3Vπ(R2+Rr+r2)h = \frac{3V}{\pi(R^2 + Rr + r^2)}.

Question 7

The equation for a line is given as ax−b=c−dxax - b = c - dx. When this equation is solved for xx, what is the result?

  1. x=b+ca−dx = \frac{b+c}{a-d}
  2. x=c−ba−dx = \frac{c-b}{a-d}
  3. x=b+ca+dx = \frac{b+c}{a+d} (correct answer)
  4. x=b+c−a−dx = b+c-a-d
Explanation: To solve for xx, first, collect all xx terms on one side and all constant terms on the other. Add dxdx to both sides: ax+dx−b=cax + dx - b = c. Then add bb to both sides: ax+dx=b+cax + dx = b + c. Factor out xx from the left side: x(a+d)=b+cx(a+d) = b+c. Finally, divide by (a+d)(a+d) to isolate xx, which gives x=b+ca+dx = \frac{b+c}{a+d}.

Question 8

The formula for the surface area, SS, of a rectangular prism with length ll, width ww, and height hh is S=2lw+2lh+2whS = 2lw + 2lh + 2wh. Which of the following equations correctly solves for the width, ww?

  1. w=S−2lh2l+2hw = \frac{S - 2lh}{2l + 2h} (correct answer)
  2. w=S2l+2h−2lhw = \frac{S}{2l + 2h} - 2lh
  3. w=S−lhl+hw = \frac{S - lh}{l + h}
  4. w=S−2l−2h−2lhw = S - 2l - 2h - 2lh
Explanation: To solve for ww, first isolate the terms containing ww on one side of the equation: S−2lh=2lw+2whS - 2lh = 2lw + 2wh. Then, factor out ww from the terms on the right side: S−2lh=w(2l+2h)S - 2lh = w(2l + 2h). Finally, divide both sides by (2l+2h)(2l + 2h) to isolate ww, which gives w=S−2lh2l+2hw = \frac{S - 2lh}{2l + 2h}.

Question 9

The formula for the area AA of a trapezoid is A=12h(b1+b2)A = \frac{1}{2}h(b_1 + b_2), where hh is the height and b1b_1 and b2b_2 are the lengths of the bases. Which formula correctly solves for b1b_1?

  1. b1=2Ah−b2b_1 = \frac{2A}{h} - b_2 (correct answer)
  2. b1=A2h−b2b_1 = \frac{A}{2h} - b_2
  3. b1=2A−b2hb_1 = \frac{2A - b_2}{h}
  4. b1=2A−h−b2b_1 = 2A - h - b_2
Explanation: First, multiply both sides by 2 to clear the fraction: 2A=h(b1+b2)2A = h(b_1 + b_2). Next, divide both sides by hh: 2Ah=b1+b2\frac{2A}{h} = b_1 + b_2. Finally, subtract b2b_2 from both sides to isolate b1b_1, resulting in b1=2Ah−b2b_1 = \frac{2A}{h} - b_2.

Question 10

In statistics, a Z-score is calculated using the formula z=x−μσz = \frac{x - \mu}{\sigma}, where xx is the data point, μ\mu is the mean, and σ\sigma is the standard deviation. Which of the following equations is correctly solved for the mean, μ\mu?

  1. μ=zσ−x\mu = z\sigma - x
  2. μ=x−zσ\mu = x - z\sigma (correct answer)
  3. μ=x+zσ\mu = x + z\sigma
  4. μ=x−zσ\mu = \frac{x - z}{\sigma}
Explanation: To solve for μ\mu, first multiply both sides of the equation by σ\sigma to get zσ=x−μz\sigma = x - \mu. Next, to make the μ\mu term positive, add μ\mu to both sides: μ+zσ=x\mu + z\sigma = x. Finally, subtract zσz\sigma from both sides to isolate μ\mu, which results in μ=x−zσ\mu = x - z\sigma.

Question 11

The combined gas law in chemistry is given by P1V1T1=P2V2T2\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}. Which of the following correctly solves for T1T_1?

  1. T1=P2V2T2P1V1T_1 = \frac{P_2V_2T_2}{P_1V_1}
  2. T1=P1V1T2P2V2T_1 = \frac{P_1V_1T_2}{P_2V_2} (correct answer)
  3. T1=P2V2P1V1T2T_1 = \frac{P_2V_2}{P_1V_1T_2}
  4. T1=P1V1P2V2T2T_1 = \frac{P_1V_1}{P_2V_2T_2}
Explanation: To solve for T1T_1, which is in the denominator, one can first take the reciprocal of both sides to get T1P1V1=T2P2V2\frac{T_1}{P_1V_1} = \frac{T_2}{P_2V_2}. Then, multiply both sides by P1V1P_1V_1 to isolate T1T_1. This gives T1=P1V1T2P2V2T_1 = \frac{P_1V_1T_2}{P_2V_2}.

Question 12

The formula for the future value AA of a principal amount PP with simple interest is A=P+PrtA = P + Prt, where rr is the interest rate and tt is the time. Which of the following correctly solves for the principal PP?

  1. P=A(1+rt)P = A(1 + rt)
  2. P=A1+rtP = \frac{A}{1 + rt} (correct answer)
  3. P=A−rtP = A - rt
  4. P=A−1rtP = \frac{A - 1}{rt}
Explanation: To solve for PP, you must first factor PP out of the terms on the right side of the equation: A=P(1+rt)A = P(1 + rt). Then, to isolate PP, divide both sides by the quantity (1+rt)(1 + rt). This results in P=A1+rtP = \frac{A}{1 + rt}.

Question 13

The formula for linear thermal expansion is L=L0(1+αΔT)L = L_0(1 + \alpha \Delta T), where LL is the final length, L0L_0 is the initial length, α\alpha is the coefficient of thermal expansion, and ΔT\Delta T is the change in temperature. Which formula correctly solves for α\alpha?

  1. α=LL0−1−ΔT\alpha = \frac{L}{L_0} - 1 - \Delta T
  2. α=L−L0ΔT\alpha = \frac{L - L_0}{\Delta T}
  3. α=LL0ΔT−1\alpha = \frac{L}{L_0 \Delta T} - 1
  4. α=L−L0L0ΔT\alpha = \frac{L - L_0}{L_0 \Delta T} (correct answer)
Explanation: First, distribute L0L_0 to get L=L0+L0αΔTL = L_0 + L_0 \alpha \Delta T. Then, subtract L0L_0 from both sides: L−L0=L0αΔTL - L_0 = L_0 \alpha \Delta T. Finally, divide both sides by L0ΔTL_0 \Delta T to solve for α\alpha, which gives α=L−L0L0ΔT\alpha = \frac{L - L_0}{L_0 \Delta T}.

Question 14

The formula for the slope mm of a line between two points (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) is m=y2−y1x2−x1m = \frac{y_2 - y_1}{x_2 - x_1}. Which of the following correctly expresses y1y_1 in terms of the other variables?

  1. y1=y2+m(x2−x1)y_1 = y_2 + m(x_2 - x_1)
  2. y1=m(x2−x1)−y2y_1 = m(x_2 - x_1) - y_2
  3. y1=y2−m(x2−x1)y_1 = y_2 - m(x_2 - x_1) (correct answer)
  4. y1=y2−mx2−x1y_1 = \frac{y_2 - m}{x_2 - x_1}
Explanation: First, multiply both sides by (x2−x1)(x_2 - x_1) to get m(x2−x1)=y2−y1m(x_2 - x_1) = y_2 - y_1. To make the y1y_1 term positive, add y1y_1 to both sides: y1+m(x2−x1)=y2y_1 + m(x_2 - x_1) = y_2. Finally, subtract m(x2−x1)m(x_2 - x_1) from both sides to isolate y1y_1, resulting in y1=y2−m(x2−x1)y_1 = y_2 - m(x_2 - x_1).

Question 15

The volume VV of a cylinder with a hemisphere on top is given by the formula V=πr2h+23πr3V = \pi r^2 h + \frac{2}{3}\pi r^3. Which of the following expresses the height hh in terms of the other variables?

  1. h=Vπr2−23rh = \frac{V}{\pi r^2} - \frac{2}{3}r
  2. h=V−23πr3πr2h = \frac{V - \frac{2}{3}\pi r^3}{\pi r^2} (correct answer)
  3. h=Vπr2−23πr3h = \frac{V}{\pi r^2} - \frac{2}{3}\pi r^3
  4. h=3V−2πr3πr2h = \frac{3V - 2\pi r^3}{\pi r^2}
Explanation: To solve for hh, first subtract the term 23πr3\frac{2}{3}\pi r^3 from both sides: V−23πr3=πr2hV - \frac{2}{3}\pi r^3 = \pi r^2 h. Then, divide both sides by πr2\pi r^2 to isolate hh. This results in h=V−23πr3πr2h = \frac{V - \frac{2}{3}\pi r^3}{\pi r^2}. Note that this can also be written as the equivalent form h=Vπr2−23rh = \frac{V}{\pi r^2} - \frac{2}{3}r by separating the fraction.

Question 16

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 dod_o is object distance and did_i is image distance. If the object distance is twice the focal length, which expression correctly gives the image distance?

  1. di=2f3d_i = \frac{2f}{3}
  2. di=2fd_i = 2f (correct answer)
  3. di=f2d_i = \frac{f}{2}
  4. di=3f2d_i = \frac{3f}{2}
Explanation: Given do=2fd_o = 2f, substitute into the lens equation: 1f=12f+1di\frac{1}{f} = \frac{1}{2f} + \frac{1}{d_i}. Rearranging: 1di=1f−12f=2−12f=12f\frac{1}{d_i} = \frac{1}{f} - \frac{1}{2f} = \frac{2-1}{2f} = \frac{1}{2f}. Therefore, di=2fd_i = 2f. Choice A results from incorrectly adding fractions. Choice C inverts the final result. Choice D comes from incorrect algebraic manipulation of the reciprocals.

Question 17

The ideal gas law is given by PV=nRTPV = nRT. A chemistry student needs to find the number of moles when the pressure is doubled and the volume is halved, while temperature remains constant. If the original conditions give n1=P1V1RTn_1 = \frac{P_1V_1}{RT}, what expression represents the new number of moles n2n_2?

  1. n2=P1V1RTn_2 = \frac{P_1V_1}{RT} (correct answer)
  2. n2=2P1V1RTn_2 = \frac{2P_1V_1}{RT}
  3. n2=4P1V1RTn_2 = \frac{4P_1V_1}{RT}
  4. n2=P1V12RTn_2 = \frac{P_1V_1}{2RT}
Explanation: The new conditions are P2=2P1P_2 = 2P_1 and V2=V12V_2 = \frac{V_1}{2}. Using PV=nRTPV = nRT, we get n2=P2V2RT=(2P1)(V12)RT=P1V1RT=n1n_2 = \frac{P_2V_2}{RT} = \frac{(2P_1)(\frac{V_1}{2})}{RT} = \frac{P_1V_1}{RT} = n_1. The number of moles remains the same. Choice B assumes only pressure doubling affects the result. Choice C incorrectly multiplies the effects (2×2=42 \times 2 = 4). Choice D assumes the changes reduce the number of moles.

Question 18

The formula for the nth term of an arithmetic sequence is an=a1+(n−1)da_n = a_1 + (n-1)d, where a1a_1 is the first term and dd is the common difference. If the 5th term is 17 and the 12th term is 38, what is the first term?

  1. a1=2a_1 = 2
  2. a1=8a_1 = 8
  3. a1=5a_1 = 5 (correct answer)
  4. a1=11a_1 = 11
Explanation: When you encounter arithmetic sequence problems with two given terms, you need to use the information to find the common difference first, then work backwards to find the first term. Given that a5=17a_5 = 17 and a12=38a_{12} = 38, you can set up two equations using the formula an=a1+(n−1)da_n = a_1 + (n-1)d: For the 5th term: 17=a1+4d17 = a_1 + 4d For the 12th term: 38=a1+11d38 = a_1 + 11d To find the common difference, subtract the first equation from the second: 38−17=(a1+11d)−(a1+4d)38 - 17 = (a_1 + 11d) - (a_1 + 4d) 21=7d21 = 7d d=3d = 3 Now substitute d=3d = 3 back into either equation to find a1a_1: 17=a1+4(3)17 = a_1 + 4(3) 17=a1+1217 = a_1 + 12 a1=5a_1 = 5 This confirms answer C is correct. Looking at the wrong answers: A) a1=2a_1 = 2 would give a5=2+4(3)=14a_5 = 2 + 4(3) = 14, not 17. B) a1=8a_1 = 8 would give a5=8+4(3)=20a_5 = 8 + 4(3) = 20, not 17. D) a1=11a_1 = 11 would give a5=11+4(3)=23a_5 = 11 + 4(3) = 23, not 17. These wrong answers likely come from calculation errors or mixing up the values. Study tip: Always find the common difference first when given two non-consecutive terms. Set up a system of equations, solve for dd, then substitute back to find a1a_1.

Question 19

The formula for the period of a pendulum is T=2πLgT = 2\pi\sqrt{\frac{L}{g}}. If the period of a pendulum on Earth is 2 seconds, what would be the length of a pendulum with a period of 3 seconds on the same planet?

  1. The length would be 32\frac{3}{2} times the original length
  2. The length would be 94\frac{9}{4} times the original length (correct answer)
  3. The length would be 33 times the original length
  4. The length would be 49\frac{4}{9} times the original length
Explanation: Since T=2πLgT = 2\pi\sqrt{\frac{L}{g}}, we have T2=4π2LgT^2 = 4\pi^2\frac{L}{g}, so L=gT24π2L = \frac{gT^2}{4\pi^2}. This shows L∝T2L \propto T^2. If T2=3T_2 = 3 seconds and T1=2T_1 = 2 seconds, then L2L1=T22T12=94\frac{L_2}{L_1} = \frac{T_2^2}{T_1^2} = \frac{9}{4}. Choice A uses the ratio of periods, not their squares. Choice C incorrectly assumes direct proportionality. Choice D inverts the correct ratio.

Question 20

The formula for the surface area of a cylinder is S=2πr2+2πrhS = 2\pi r^2 + 2\pi rh. If a cylindrical tank has a surface area of 150 square feet and a height that is 3 times its radius, what is the radius in terms of π\pi?

  1. r=1508πr = \sqrt{\frac{150}{8\pi}} (correct answer)
  2. r=754πr = \sqrt{\frac{75}{4\pi}}
  3. r=1508πr = \frac{150}{8\pi}
  4. r=1506πr = \sqrt{\frac{150}{6\pi}}
Explanation: First, substitute h=3rh = 3r into the surface area formula: S=2πr2+2πr(3r)=2πr2+6πr2=8πr2S = 2\pi r^2 + 2\pi r(3r) = 2\pi r^2 + 6\pi r^2 = 8\pi r^2. Then set this equal to 150: 8πr2=1508\pi r^2 = 150. Solving for rr: r2=1508πr^2 = \frac{150}{8\pi}, so r=1508πr = \sqrt{\frac{150}{8\pi}}. Choice B incorrectly simplifies 1508\frac{150}{8} to 754\frac{75}{4} but uses wrong coefficient. Choice C forgets the square root. Choice D uses 6 instead of 8 as the coefficient.