Algebra 2 Quiz: Rearranging Formulas To Highlight Quantities
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Rearranging Formulas To Highlight QuantitiesQuestion 1 of 20

In geometry, the volume of a cone is V=13πr2hV = \dfrac{1}{3}\pi r^2 h, where rr is radius and hh is height. Solve V=13πr2hV = \dfrac{1}{3}\pi r^2 h for rr.

r=πh3Vr = \sqrt{\dfrac{\pi h}{3V}}
r=V3πhr = \sqrt{\dfrac{V}{3\pi h}}
r=3Vπhr = \dfrac{3V}{\pi h}
r=3Vπhr = \sqrt{\dfrac{3V}{\pi h}}
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Algebra 2 Quiz

Algebra 2 Quiz: Rearranging Formulas To Highlight Quantities

Practice Rearranging Formulas To Highlight Quantities in Algebra 2 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 To Highlight Quantities, giving you a quick way to practice the rules, question types, and explanations that matter most for Algebra 2.

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

In geometry, the volume of a cone is V=13πr2hV = \dfrac{1}{3}\pi r^2 h, where rr is radius and hh is height. Solve V=13πr2hV = \dfrac{1}{3}\pi r^2 h for rr.

  1. r=πh3Vr = \sqrt{\dfrac{\pi h}{3V}}
  2. r=V3πhr = \sqrt{\dfrac{V}{3\pi h}}
  3. r=3Vπhr = \dfrac{3V}{\pi h}
  4. r=3Vπhr = \sqrt{\dfrac{3V}{\pi h}} (correct answer)
Explanation: This question tests your ability to rearrange formulas to solve for a specific variable—essential for using formulas flexibly in science, engineering, and real-world problem-solving. More complex rearrangements may require advanced techniques: if your target variable is squared (like r² in V = (1/3)πr²h), you'll need square roots (r = √(3V/πh)). Starting with V = (1/3)πr²h, multiply both sides by 3 to get 3V = πr²h, divide by πh to isolate r² as 3V/(πh) = r², then take the square root of both sides for r = √(3V/πh). Choice C correctly isolates r through multiplication, division, and square root to get r = √(3V/πh). A distractor like choice D might forget to multiply by 3 first, leading to an incorrect coefficient under the square root, but always clear constants step by step. The formula rearrangement recipe: (1) Identify what you're solving for (that's your 'x'), (2) Identify what operation(s) are being done to that variable in the original formula, (3) Apply inverse operations in reverse order to isolate it (just like numeric equations!), (4) Simplify the result—combine fractions, simplify radicals, etc. You're doing great—keep connecting these to geometric volumes for deeper understanding!

Question 2

In optics, the thin lens equation is 1f=1a+1b\dfrac{1}{f}=\dfrac{1}{a}+\dfrac{1}{b}, where ff is focal length, aa is object distance, and bb is image distance. Solve for ff in terms of aa and bb.

  1. f=ababf=\dfrac{ab}{a-b}
  2. f=a+babf=\dfrac{a+b}{a-b}
  3. f=a+babf=\dfrac{a+b}{ab}
  4. f=aba+bf=\dfrac{ab}{a+b} (correct answer)
Explanation: This question tests your ability to rearrange formulas to solve for a specific variable—essential for using formulas flexibly in science, engineering, and real-world problem-solving. Rearranging formulas with multiple variables works exactly like solving numeric equations, except the answer contains other variables instead of numbers: treat the variable you're solving for as the 'unknown x,' treat all other variables as 'known numbers,' then use inverse operations to isolate your target variable. The same algebraic moves apply—just keep everything symbolic! For example, solving V = πr²h for r is like solving 100 = 3.14·r²·5 for r: divide by π and h, then take square root. More complex rearrangements may require advanced techniques: if your target variable is squared (like r² in A = πr²), you'll need square roots (r = √(A/π)). If it appears in a denominator (like f in 1/f = 1/a + 1/b), you'll need to clear fractions first. If it appears with different powers (like t in s = ut + (1/2)at²), you may need the quadratic formula! The complexity of the rearrangement depends on how the variable appears in the formula. To solve 1/f = 1/a + 1/b for f, first combine the right side over a common denominator: 1/f = (a + b)/(ab). Then take the reciprocal of both sides: f = ab/(a + b). Choice B correctly isolates f through combining fractions and taking the reciprocal to get f = ab/(a + b). A common distractor like choice A fails by placing the sum in the numerator without reciprocating properly, which would actually give 1/f instead; remember to invert after combining. The formula rearrangement recipe: (1) Identify what you're solving for (that's your 'x'), (2) Identify what operation(s) are being done to that variable in the original formula, (3) Apply inverse operations in reverse order to isolate it (just like numeric equations!), (4) Simplify the result—combine fractions, simplify radicals, etc. Example: T = 2π√(L/g) solve for L: square both sides → T² = 4π²L/g, multiply by g → gT² = 4π²L, divide by 4π² → L = gT²/(4π²). Each step undoes an operation! Watch for variables appearing multiple times: if your target variable appears in multiple places (like x in xy + xz = w), factor it out first: x(y + z) = w, then x = w/(y + z). If you don't factor, you'll struggle to isolate! Also, when taking square roots of a variable, remember ± if the formula context allows both positive and negative (though often context restricts to positive only, like radius r ≥ 0). Physics and geometry formulas usually want positive values only!

Question 3

A projectile's vertical displacement can be modeled by the kinematic equation s=ut+12at2s=ut+\tfrac{1}{2}at^2, where ss is displacement, uu is initial velocity, aa is acceleration, and tt is time. Solve s=ut+12at2s=ut+\tfrac{1}{2}at^2 for tt (a quadratic in tt).

  1. t=u±u22asat=\dfrac{-u\pm\sqrt{u^2-2as}}{a}
  2. t=u±u2+2asat=\dfrac{-u\pm\sqrt{u^2+2as}}{a} (correct answer)
  3. t=u±u2+as2at=\dfrac{-u\pm\sqrt{u^2+as}}{2a}
  4. t=u±u2+2asat=\dfrac{u\pm\sqrt{u^2+2as}}{a}
Explanation: This question tests your ability to rearrange formulas to solve for a specific variable—essential for using formulas flexibly in science, engineering, and real-world problem-solving. More complex rearrangements may require advanced techniques: if your target variable is squared (like r2r^2 in A=πr2A = \pi r^2), you'll need square roots (r=Aπr = \sqrt{\frac{A}{\pi}}). If it appears in a denominator (like ff in 1f=1a+1b\frac{1}{f} = \frac{1}{a} + \frac{1}{b}), you'll need to clear fractions first. If it appears with different powers (like tt in s=ut+12at2s = ut + \frac{1}{2}at^2), you may need the quadratic formula! The complexity of the rearrangement depends on how the variable appears in the formula. Starting with s=ut+12at2s = ut + \frac{1}{2}at^2, we need to solve for tt. First multiply through by 2: 2s=2ut+at22s = 2ut + at^2. Rearranging: at2+2ut2s=0at^2 + 2ut - 2s = 0. This is a quadratic in standard form with A=aA = a, B=2uB = 2u, C=2sC = -2s. Using the quadratic formula: t=2u±4u2+8as2a=2u±2u2+2as2a=u±u2+2asat = \frac{-2u \pm \sqrt{4u^2 + 8as}}{2a} = \frac{-2u \pm 2\sqrt{u^2 + 2as}}{2a} = \frac{-u \pm \sqrt{u^2 + 2as}}{a}. Choice A correctly applies the quadratic formula to get t=u±u2+2asat = \frac{-u \pm \sqrt{u^2 + 2as}}{a}. Choice B has the wrong sign on u, Choice C has the wrong sign under the square root (should be +2as not -2as), and Choice D incorrectly divides by 2a instead of a. The formula rearrangement recipe: (1) Identify what you're solving for (that's your 'x'), (2) Identify what operation(s) are being done to that variable in the original formula, (3) Apply inverse operations in reverse order to isolate it (just like numeric equations!), (4) Simplify the result—combine fractions, simplify radicals, etc.

Question 4

In physics, kinetic energy is KE=12mv2KE=\tfrac{1}{2}mv^2, where mm is mass and vv is speed. Solve KE=12mv2KE=\tfrac{1}{2}mv^2 for vv (include ±\pm where appropriate).

  1. v=±2KEmv=\pm\sqrt{\dfrac{2KE}{m}} (correct answer)
  2. v=KE2mv=\sqrt{\dfrac{KE}{2m}}
  3. v=±m2KEv=\pm\sqrt{\dfrac{m}{2KE}}
  4. v=2KEmv=\dfrac{2KE}{m}
Explanation: This question tests your ability to rearrange formulas to solve for a specific variable—essential for using formulas flexibly in science, engineering, and real-world problem-solving. More complex rearrangements may require advanced techniques: if your target variable is squared (like r² in A = πr²), you'll need square roots (r = √(A/π)). If it appears in a denominator (like f in 1/f = 1/a + 1/b), you'll need to clear fractions first. If it appears with different powers (like t in s = ut + (1/2)at²), you may need the quadratic formula! The complexity of the rearrangement depends on how the variable appears in the formula. Starting with KE = (1/2)mv², we need to isolate v. First multiply both sides by 2: 2KE = mv². Then divide both sides by m: 2KE/m = v². Finally, take the square root: v = ±√(2KE/m). The ± is crucial because squaring either positive or negative v gives the same v². Choice B correctly isolates v through multiplication, division, and taking the square root to get v = ±√(2KE/m). Choice A forgets the square root, Choice C has the wrong coefficient (should be 2KE not KE), and Choice D incorrectly inverts the expression under the square root. Watch for variables appearing multiple times: if your target variable appears in multiple places (like x in xy + xz = w), factor it out first: x(y + z) = w, then x = w/(y + z). If you don't factor, you'll struggle to isolate! Also, when taking square roots of a variable, remember ± if the formula context allows both positive and negative (though often context restricts to positive only, like radius r ≥ 0). Physics and geometry formulas usually want positive values only!

Question 5

In physics, the kinematic equation v2=u2+2asv^2 = u^2 + 2as relates final speed vv, initial speed uu, acceleration aa, and displacement ss. Solve v2=u2+2asv^2 = u^2 + 2as for aa (treat the other variables as constants, using the same inverse-operation reasoning as solving a numeric equation).

  1. a=v2u2sa=\dfrac{v^2-u^2}{s}
  2. a=v2u22sa=\dfrac{v^2-u^2}{2s} (correct answer)
  3. a=2sv2u2a=\dfrac{2s}{v^2-u^2}
  4. a=u2v22sa=\dfrac{u^2-v^2}{2s}
Explanation: This question tests your ability to rearrange formulas to solve for a specific variable—essential for using formulas flexibly in science, engineering, and real-world problem-solving. Rearranging formulas with multiple variables works exactly like solving numeric equations, except the answer contains other variables instead of numbers: treat the variable you're solving for as the 'unknown x,' treat all other variables as 'known numbers,' then use inverse operations to isolate your target variable. The same algebraic moves apply—just keep everything symbolic! For example, solving V = πr²h for r is like solving 100 = 3.14·r²·5 for r: divide by π and h, then take square root. Starting with v² = u² + 2as, we need to isolate a. First subtract u² from both sides: v² - u² = 2as. Then divide both sides by 2s: (v² - u²)/(2s) = a, which gives us a = (v² - u²)/(2s). Choice B correctly isolates a through subtraction and division to get a = (v² - u²)/(2s). Choice A forgets to divide by 2, Choice C incorrectly switches the order in the numerator (should be v² - u², not u² - v²), and Choice D inverts the entire fraction. The formula rearrangement recipe: (1) Identify what you're solving for (that's your 'x'), (2) Identify what operation(s) are being done to that variable in the original formula, (3) Apply inverse operations in reverse order to isolate it (just like numeric equations!), (4) Simplify the result—combine fractions, simplify radicals, etc.

Question 6

In physics, the kinematic equation v2=u2+2asv^2 = u^2 + 2as relates final speed vv, initial speed uu, acceleration aa, and displacement ss. Solve v2=u2+2asv^2 = u^2 + 2as for aa (treating the other variables as constants, as in solving a numeric equation).

  1. a=v2u2sa = \dfrac{v^2-u^2}{s}
  2. a=2sv2u2a = \dfrac{2s}{v^2-u^2}
  3. a=v2u22sa = \dfrac{v^2-u^2}{2s} (correct answer)
  4. a=u2v22sa = \dfrac{u^2-v^2}{2s}
Explanation: This question tests your ability to rearrange formulas to solve for a specific variable—essential for using formulas flexibly in science, engineering, and real-world problem-solving. Rearranging formulas with multiple variables works exactly like solving numeric equations, except the answer contains other variables instead of numbers: treat the variable you're solving for as the 'unknown x,' treat all other variables as 'known numbers,' then use inverse operations to isolate your target variable. Starting with v2=u2+2asv^2 = u^2 + 2as, subtract u2u^2 from both sides to get v2u2=2asv^2 - u^2 = 2as, then divide both sides by 2s2s to isolate a, resulting in a=v2u22sa = \dfrac{v^2 - u^2}{2s}. Choice B correctly isolates a through subtraction and division to get a=v2u22sa = \dfrac{v^2 - u^2}{2s}. A common distractor like choice C might forget to switch the signs when subtracting u2u^2, but remember that v2u2v^2 - u^2 is positive if v > u, so the order matters for the physics context. The formula rearrangement recipe: (1) Identify what you're solving for (that's your 'x'), (2) Identify what operation(s) are being done to that variable in the original formula, (3) Apply inverse operations in reverse order to isolate it (just like numeric equations!), (4) Simplify the result—combine fractions, simplify radicals, etc. Keep practicing these, and you'll master kinematic equations in no time!

Question 7

In chemistry, a simplified relationship can be written as P=nRTVP=\dfrac{nRT}{V}, where PP is pressure, nn is moles, RR is a constant, TT is temperature, and VV is volume. Solve P=nRTVP=\dfrac{nRT}{V} for TT.

  1. T=PVnRT=\dfrac{PV}{nR} (correct answer)
  2. T=nRPVT=\dfrac{nR}{PV}
  3. T=PnRVT=\dfrac{P}{nRV}
  4. T=nRPVT=\dfrac{nRP}{V}
Explanation: This question tests your ability to rearrange formulas to solve for a specific variable—essential for using formulas flexibly in science, engineering, and real-world problem-solving. Rearranging formulas with multiple variables works exactly like solving numeric equations, except the answer contains other variables instead of numbers: treat the variable you're solving for as the 'unknown x,' treat all other variables as 'known numbers,' then use inverse operations to isolate your target variable. The same algebraic moves apply—just keep everything symbolic! For example, solving V=πr2hV = \pi r^2 h for r is like solving 100=3.14r25100 = 3.14 \cdot r^2 \cdot 5 for r: divide by π\pi and hh, then take square root. Starting with P=nRTVP = \dfrac{nRT}{V}, we need to isolate T. First multiply both sides by V: PV=nRTPV = nRT. Then divide both sides by nR: PVnR=T\dfrac{PV}{nR} = T, which gives us T=PVnRT = \dfrac{PV}{nR}. Choice A correctly isolates T through multiplication and division to get T=PVnRT = \dfrac{PV}{nR}. Choice B inverts the expression, Choice C has only P in the numerator instead of PV, and Choice D incorrectly has nRP in the numerator. Watch for variables appearing multiple times: if your target variable appears in multiple places (like x in xy+xz=wxy + xz = w), factor it out first: x(y+z)=wx(y + z) = w, then x=wy+zx = \dfrac{w}{y + z}. If you don't factor, you'll struggle to isolate! Also, when taking square roots of a variable, remember ±\pm if the formula context allows both positive and negative (though often context restricts to positive only, like radius r0r \geq 0). Physics and geometry formulas usually want positive values only!

Question 8

In geometry, the volume of a cone is V=13πr2hV=\dfrac{1}{3}\pi r^2h, where rr is radius and hh is height. Solve V=13πr2hV=\dfrac{1}{3}\pi r^2h for rr.

  1. r=πh3Vr=\sqrt{\dfrac{\pi h}{3V}}
  2. r=Vπhr=\sqrt{\dfrac{V}{\pi h}}
  3. r=3Vπhr=\sqrt{\dfrac{3V}{\pi h}} (correct answer)
  4. r=3Vπhr=\dfrac{3V}{\pi h}
Explanation: This question tests your ability to rearrange formulas to solve for a specific variable—essential for using formulas flexibly in science, engineering, and real-world problem-solving. More complex rearrangements may require advanced techniques: if your target variable is squared (like r² in A = πr²), you'll need square roots (r = √(A/π)). If it appears in a denominator (like f in 1/f = 1/a + 1/b), you'll need to clear fractions first. If it appears with different powers (like t in s = ut + (1/2)at²), you may need the quadratic formula! The complexity of the rearrangement depends on how the variable appears in the formula. Starting with V = (1/3)πr²h, we need to isolate r. First multiply both sides by 3: 3V = πr²h. Then divide both sides by πh: 3V/(πh) = r². Finally, take the square root: r = √[3V/(πh)]. Choice B correctly isolates r through multiplication, division, and taking the square root to get r = √(3V/(πh)). Choice A forgets the factor of 3, Choice C forgets to take the square root (leaving r²), and Choice D incorrectly inverts the expression under the square root. The formula rearrangement recipe: (1) Identify what you're solving for (that's your 'x'), (2) Identify what operation(s) are being done to that variable in the original formula, (3) Apply inverse operations in reverse order to isolate it (just like numeric equations!), (4) Simplify the result—combine fractions, simplify radicals, etc.

Question 9

In algebra-based modeling, suppose y=kxx+ay = \dfrac{kx}{x+a} where aa and kk are constants and xax\neq -a. Solve y=kxx+ay = \dfrac{kx}{x+a} for xx in terms of y,a,ky, a, k.

  1. x=y(ka)kx = \dfrac{y(k-a)}{k}
  2. x=kakyx = \dfrac{ka}{k-y}
  3. x=yakyx = \dfrac{ya}{k-y} (correct answer)
  4. x=yaykx = \dfrac{ya}{y-k}
Explanation: This question tests your ability to rearrange formulas to solve for a specific variable—essential for using formulas flexibly in science, engineering, and real-world problem-solving. Watch for variables appearing multiple times: if your target variable appears in multiple places (like x in y = kx/(x + a)), factor it out first. Starting with y = kx/(x + a), multiply both sides by (x + a) to get y(x + a) = kx, expand to yx + ya = kx, then bring terms with x to one side: ya = kx - yx = x(k - y), so x = ya/(k - y). Choice A correctly isolates x through multiplication, rearrangement, and factoring to get x = ya/(k - y). A distractor like choice B might flip the denominator sign, but subtracting y from k keeps it positive assuming k > y. The formula rearrangement recipe: (1) Identify what you're solving for (that's your 'x'), (2) Identify what operation(s) are being done to that variable in the original formula, (3) Apply inverse operations in reverse order to isolate it (just like numeric equations!), (4) Simplify the result—combine fractions, simplify radicals, etc. You're mastering algebraic models—fantastic progress!

Question 10

In chemistry, the ideal gas law is PV=nRTPV = nRT, where PP is pressure, VV is volume, nn is number of moles, RR is a constant, and TT is temperature. Solve PV=nRTPV = nRT for TT.

  1. T=PVnRT = \dfrac{PV}{nR} (correct answer)
  2. T=PVnRT = \dfrac{PVn}{R}
  3. T=PVRT = \dfrac{PV}{R}
  4. T=nRPVT = \dfrac{nR}{PV}
Explanation: This question tests your ability to rearrange formulas to solve for a specific variable—essential for using formulas flexibly in science, engineering, and real-world problem-solving. Rearranging formulas with multiple variables works exactly like solving numeric equations, except the answer contains other variables instead of numbers: treat the variable you're solving for as the 'unknown x,' treat all other variables as 'known numbers,' then use inverse operations to isolate your target variable. Starting with PV = nRT, divide both sides by nR to isolate T, resulting in T = PV / (nR). Choice A correctly isolates T through division to get T = PV / (nR). A distractor like choice D might misplace the n in the numerator, but remember to divide by all factors multiplying T. The formula rearrangement recipe: (1) Identify what you're solving for (that's your 'x'), (2) Identify what operation(s) are being done to that variable in the original formula, (3) Apply inverse operations in reverse order to isolate it (just like numeric equations!), (4) Simplify the result—combine fractions, simplify radicals, etc. Excellent work with gas laws—you've got this!

Question 11

In algebraic modeling, suppose a quantity is defined by y=xab+cy=\dfrac{x-a}{b}+c, where aa, bb, and cc are constants and b0b\ne 0. Isolate xx in y=xab+cy=\dfrac{x-a}{b}+c.

  1. x=bya+cx=by-a+c
  2. x=b(yc)+ax=b(y-c)+a (correct answer)
  3. x=b(y+c)ax=b(y+c)-a
  4. x=ycb+ax=\dfrac{y-c}{b}+a
Explanation: This question tests your ability to rearrange formulas to solve for a specific variable—essential for using formulas flexibly in science, engineering, and real-world problem-solving. Rearranging formulas with multiple variables works exactly like solving numeric equations, except the answer contains other variables instead of numbers: treat the variable you're solving for as the 'unknown x,' treat all other variables as 'known numbers,' then use inverse operations to isolate your target variable. The same algebraic moves apply—just keep everything symbolic! For example, solving V=πr2hV = \pi r^2 h for r is like solving 100=3.14r25100 = 3.14 \cdot r^2 \cdot 5 for r: divide by π\pi and h, then take square root. Starting with y=xab+cy = \dfrac{x-a}{b} + c, we need to isolate x. First subtract c from both sides: yc=xaby - c = \dfrac{x-a}{b}. Then multiply both sides by b: b(yc)=xab(y - c) = x - a. Finally, add a to both sides: b(yc)+a=xb(y - c) + a = x, which gives us x=b(yc)+ax = b(y - c) + a. Choice B correctly isolates x through subtraction, multiplication, and addition to get x=b(yc)+ax = b(y - c) + a. Choice A incorrectly expands to by - a + c, Choice C divides instead of multiplying by b, and Choice D has the wrong sign on c inside the parentheses. The formula rearrangement recipe: (1) Identify what you're solving for (that's your 'x'), (2) Identify what operation(s) are being done to that variable in the original formula, (3) Apply inverse operations in reverse order to isolate it (just like numeric equations!), (4) Simplify the result—combine fractions, simplify radicals, etc.

Question 12

A physics student needs to determine the acceleration of an object given the formula F=ma+kv2F = ma + kv^2, where FF is the total force, mm is mass, aa is acceleration, kk is a drag coefficient, and vv is velocity. Which expression correctly isolates the acceleration?

  1. a=Fkv2ma = \frac{F - kv^2}{m} (correct answer)
  2. a=F+kv2ma = \frac{F + kv^2}{m}
  3. a=Fmkv2a = F - m - kv^2
  4. a=Fmkv2a = \frac{F}{m} - kv^2
Explanation: To isolate aa, subtract kv2kv^2 from both sides: Fkv2=maF - kv^2 = ma, then divide by mm: a=Fkv2ma = \frac{F - kv^2}{m}. Choice B incorrectly adds instead of subtracts kv2kv^2. Choice C fails to divide by mm. Choice D incorrectly distributes the division.

Question 13

The formula for the period of a pendulum is T=2πLgT = 2\pi\sqrt{\frac{L}{g}}, where TT is period, LL is length, and gg is gravitational acceleration. An engineer needs to determine what length produces a specific period. Which expression correctly gives LL?

  1. L=4π2gT2L = \frac{4\pi^2}{gT^2}
  2. L=gT4π2L = \frac{gT}{4\pi^2}
  3. L=gT24π2L = \frac{gT^2}{4\pi^2} (correct answer)
  4. L=gT22πL = \frac{gT^2}{2\pi}
Explanation: When you encounter formula manipulation problems, you're essentially working backwards from a given equation to isolate a specific variable. This requires systematic algebraic steps to "undo" the operations around your target variable. Starting with T=2πLgT = 2\pi\sqrt{\frac{L}{g}}, you need to isolate LL. First, eliminate the coefficient by dividing both sides by 2π2\pi: T2π=Lg\frac{T}{2\pi} = \sqrt{\frac{L}{g}}. Next, square both sides to eliminate the square root: (T2π)2=Lg\left(\frac{T}{2\pi}\right)^2 = \frac{L}{g}. This gives you T24π2=Lg\frac{T^2}{4\pi^2} = \frac{L}{g}. Finally, multiply both sides by gg to isolate LL: L=gT24π2L = \frac{gT^2}{4\pi^2}, which is answer choice C. Looking at the incorrect options: Choice A has 4π24\pi^2 in the numerator instead of the denominator, suggesting someone inverted the fraction incorrectly. Choice B is missing the squared term on TT, indicating they forgot to square both sides when eliminating the square root. Choice D has 2π2\pi in the denominator instead of 4π24\pi^2, showing they didn't properly square the 2π2\pi term. Remember that when solving literal equations, work step-by-step and be especially careful when squaring expressions with coefficients—(2π)2=4π2(2\pi)^2 = 4\pi^2, not 2π22\pi^2. Always verify your final answer by substituting it back into the original equation to ensure it makes sense.

Question 14

The compound interest formula A=P(1+rn)ntA = P(1 + \frac{r}{n})^{nt} can be rearranged to find the principal. If an investment grows to $8000 after 3 years with 6% annual interest compounded quarterly, which setup correctly finds the initial principal?

  1. P=8000(1+0.064)43P = \frac{8000}{(1 + \frac{0.06}{4})^{4 \cdot 3}} (correct answer)
  2. P=8000(1+0.063)34P = \frac{8000}{(1 + \frac{0.06}{3})^{3 \cdot 4}}
  3. P=8000(1+0.064)43P = 8000 \cdot (1 + \frac{0.06}{4})^{4 \cdot 3}
  4. P=8000(1.06)12P = \frac{8000}{(1.06)^{12}}
Explanation: Rearranging A=P(1+rn)ntA = P(1 + \frac{r}{n})^{nt} gives P=A(1+rn)ntP = \frac{A}{(1 + \frac{r}{n})^{nt}}. With quarterly compounding, n=4n = 4, so we get rn=0.064\frac{r}{n} = \frac{0.06}{4} and nt=43=12nt = 4 \cdot 3 = 12. Choice B confuses nn and tt. Choice C multiplies instead of divides. Choice D uses annual compounding incorrectly.

Question 15

The surface area formula for a cylinder is S=2πr2+2πrhS = 2\pi r^2 + 2\pi rh. A manufacturer wants to express the height in terms of the other variables to optimize production costs. Which rearrangement correctly isolates hh?

  1. h=S2πr2πr2h = \frac{S}{2\pi r} - 2\pi r^2
  2. h=S+2πr22πrh = \frac{S + 2\pi r^2}{2\pi r}
  3. h=S2πr22πh = \frac{S - 2\pi r^2}{2\pi}
  4. h=S2πr22πrh = \frac{S - 2\pi r^2}{2\pi r} (correct answer)
Explanation: When you encounter a formula that needs to be rearranged to solve for a different variable, you're working with literal equations. This is essentially algebraic manipulation where some variables are treated as constants while you isolate your target variable. To isolate hh from S=2πr2+2πrhS = 2\pi r^2 + 2\pi rh, start by moving the term without hh to the left side: S2πr2=2πrhS - 2\pi r^2 = 2\pi rh. Now divide both sides by the coefficient of hh, which is 2πr2\pi r: h=S2πr22πrh = \frac{S - 2\pi r^2}{2\pi r}. This matches answer choice D. Let's examine why the other options are incorrect. Choice A attempts the right first step by subtracting 2πr22\pi r^2, but then divides SS by 2πr2\pi r while leaving 2πr22\pi r^2 as a separate term instead of dividing the entire numerator. Choice B adds 2πr22\pi r^2 instead of subtracting it, which moves the term in the wrong direction. Choice C correctly subtracts 2πr22\pi r^2 in the numerator but only divides by 2π2\pi instead of 2πr2\pi r, forgetting that the full coefficient of hh includes the rr. When solving literal equations, always identify the complete coefficient of your target variable and ensure you perform the same operation on the entire expression. Double-check by substituting your result back into the original equation to verify it's algebraically equivalent.

Question 16

The formula relating pressure, volume, and temperature for an ideal gas is P1V1T1=P2V2T2\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}. A chemistry student needs to find the final temperature T2T_2 when the pressure doubles and volume is halved. Which setup is correct?

  1. T2=P1V1T1P2V2=P1V1T1(2P1)(V12)T_2 = \frac{P_1V_1T_1}{P_2V_2} = \frac{P_1V_1T_1}{(2P_1)(\frac{V_1}{2})}
  2. T2=P2V2T1P1V1=(2P1)(V12)T1P1V1T_2 = \frac{P_2V_2T_1}{P_1V_1} = \frac{(2P_1)(\frac{V_1}{2})T_1}{P_1V_1} (correct answer)
  3. T2=P2V2P1V1T1=(2P1)(V12)P1V1T1T_2 = \frac{P_2V_2}{P_1V_1T_1} = \frac{(2P_1)(\frac{V_1}{2})}{P_1V_1T_1}
  4. T2=P2V2T1P1V1T_2 = P_2V_2T_1 - P_1V_1
Explanation: When you encounter gas law problems, you're working with proportional relationships where changing one variable affects the others predictably. The key is correctly rearranging the ideal gas equation to solve for your target variable. To find T2T_2, you need to isolate it by rearranging P1V1T1=P2V2T2\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}. Cross-multiplying gives you P1V1T2=P2V2T1P_1V_1T_2 = P_2V_2T_1, so T2=P2V2T1P1V1T_2 = \frac{P_2V_2T_1}{P_1V_1}. Since pressure doubles (P2=2P1P_2 = 2P_1) and volume halves (V2=V12V_2 = \frac{V_1}{2}), you substitute to get T2=(2P1)(V12)T1P1V1T_2 = \frac{(2P_1)(\frac{V_1}{2})T_1}{P_1V_1}. This matches choice B exactly. Choice A flips the fraction, putting P1V1T1P_1V_1T_1 in the numerator and P2V2P_2V_2 in the denominator. This gives you 1T2\frac{1}{T_2} instead of T2T_2—a common algebraic error when cross-multiplying. Choice C is missing T1T_1 in the numerator. This setup would give you T2T1\frac{T_2}{T_1} rather than T2T_2 itself, representing an incomplete rearrangement of the original equation. Choice D uses subtraction instead of the proportional relationship that defines gas laws. Gas laws involve ratios and products, never simple addition or subtraction of the state variables. Study tip: When rearranging proportional equations like gas laws, always cross-multiply first, then isolate your target variable. Double-check by ensuring your setup makes physical sense—if pressure doubles and volume halves, temperature should remain constant, which you can verify by simplifying the correct expression.

Question 17

The relativistic energy formula is E=mc21v2c2E = \frac{mc^2}{\sqrt{1 - \frac{v^2}{c^2}}}, where EE is energy, mm is rest mass, cc is speed of light, and vv is velocity. A physics student needs to solve for velocity vv. Which expression is correct?

  1. v=cm2c4E21v = c\sqrt{\frac{m^2c^4}{E^2} - 1}
  2. v=c1m2c4E2v = c\sqrt{1 - \frac{m^2c^4}{E^2}} (correct answer)
  3. v=c2m2c4E2v = \sqrt{c^2 - \frac{m^2c^4}{E^2}}
  4. v=cE2m2c4E2v = c\sqrt{\frac{E^2 - m^2c^4}{E^2}}
Explanation: When you encounter a formula that needs to be solved for a variable buried inside a complex expression, your goal is to systematically isolate that variable through algebraic manipulation. Starting with E=mc21v2c2E = \frac{mc^2}{\sqrt{1 - \frac{v^2}{c^2}}}, first isolate the square root by multiplying both sides by it and dividing by EE: 1v2c2=mc2E\sqrt{1 - \frac{v^2}{c^2}} = \frac{mc^2}{E} Square both sides to eliminate the square root: 1v2c2=m2c4E21 - \frac{v^2}{c^2} = \frac{m^2c^4}{E^2} Rearrange to isolate the v2v^2 term: v2c2=1m2c4E2\frac{v^2}{c^2} = 1 - \frac{m^2c^4}{E^2} Finally, multiply by c2c^2 and take the square root: v=c1m2c4E2v = c\sqrt{1 - \frac{m^2c^4}{E^2}} This matches answer choice B. Looking at the wrong answers: Choice A has the fraction flipped and subtracts 1 instead of subtracting the fraction from 1 — this likely comes from incorrectly rearranging the equation. Choice C is missing the factor of cc outside the square root, resulting from forgetting to multiply both sides by c2c^2. Choice D rearranges the numerator incorrectly, putting E2m2c4E^2 - m^2c^4 instead of starting with 1 in the original subtraction. Remember: when solving for a variable in a complex fraction or under a radical, work step-by-step to isolate terms, and always double-check your algebra by substituting back into the original equation.

Question 18

In economics, the break-even formula is R=C+FR = C + F, where RR is revenue, CC is variable costs, and FF is fixed costs. If R=pxR = px and C=vxC = vx (where pp is price per unit, vv is variable cost per unit, and xx is quantity), which expression gives the break-even quantity?

  1. x=F(pv)x = F(p - v)
  2. x=Fp+vx = \frac{F}{p + v}
  3. x=F+pvx = \frac{F + p}{v}
  4. x=Fpvx = \frac{F}{p - v} (correct answer)
Explanation: Break-even problems test your ability to solve literal equations by isolating a specific variable. When you see formulas with multiple variables representing real-world scenarios, focus on substitution and algebraic manipulation to find what you need. Start with the break-even formula R=C+FR = C + F and substitute the given expressions. Since R=pxR = px and C=vxC = vx, you get: px=vx+Fpx = vx + F To solve for xx, collect all terms containing xx on one side: pxvx=Fpx - vx = F Factor out xx: x(pv)=Fx(p - v) = F Finally, divide both sides by (pv)(p - v): x=Fpvx = \frac{F}{p - v} This makes economic sense: you need to sell enough units so that your profit per unit (pv)(p - v) times quantity equals your fixed costs. Choice A gives x=F(pv)x = F(p - v), which incorrectly multiplies FF by (pv)(p - v) instead of dividing—this comes from forgetting to divide both sides when solving x(pv)=Fx(p - v) = F. Choice B uses x=Fp+vx = \frac{F}{p + v}, which adds the costs instead of finding the profit margin—this suggests confusing total costs with profit per unit. Choice C gives x=F+pvx = \frac{F + p}{v}, which scrambles the variables entirely and doesn't maintain the proper relationship between fixed costs and profit margin. When solving literal equations, always perform the same operation to both sides and double-check that your final answer makes sense in context. Here, larger fixed costs should require more units sold, and larger profit margins should require fewer units.

Question 19

The lens equation in optics 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 student needs to find the object distance. Which rearrangement is correct?

  1. do=fdifdid_o = \frac{f - d_i}{fd_i}
  2. do=fdidi+fd_o = \frac{fd_i}{d_i + f}
  3. do=fdidifd_o = \frac{fd_i}{d_i - f} (correct answer)
  4. do=diffdid_o = \frac{d_i - f}{fd_i}
Explanation: When you encounter equations with multiple fractions that need to be solved for a specific variable, your goal is to systematically isolate that variable using algebraic manipulation. Starting with 1f=1do+1di\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}, you need to solve for dod_o. First, isolate the term containing dod_o by subtracting 1di\frac{1}{d_i} from both sides: 1f1di=1do\frac{1}{f} - \frac{1}{d_i} = \frac{1}{d_o}. To subtract these fractions, find a common denominator: diffdi=1do\frac{d_i - f}{fd_i} = \frac{1}{d_o}. Finally, take the reciprocal of both sides to solve for dod_o: do=fdidifd_o = \frac{fd_i}{d_i - f}, which is choice C. Let's examine why the other options are incorrect. Choice A gives do=fdifdid_o = \frac{f - d_i}{fd_i}, which results from incorrectly ordering the terms in the numerator when finding the common denominator—this would give you fdifdi\frac{f - d_i}{fd_i} instead of diffdi\frac{d_i - f}{fd_i}. Choice B shows do=fdidi+fd_o = \frac{fd_i}{d_i + f}, which represents the error of adding fractions incorrectly in the denominator. Choice D gives do=diffdid_o = \frac{d_i - f}{fd_i}, which is what you get before taking the final reciprocal step. When solving rational equations for a variable, work methodically: isolate the term with your target variable, combine fractions using common denominators, then take reciprocals when needed. Always double-check your fraction arithmetic, as sign errors and incorrect operations are the most common mistakes.

Question 20

In geometry, the volume of a sphere is V=43πr3V=\dfrac{4}{3}\pi r^3. Solve for the radius rr in terms of VV.

  1. r=3V4πr=\dfrac{3V}{4\pi}
  2. r=4π3V3r=\sqrt[3]{\dfrac{4\pi}{3V}}
  3. r=3V4πr=\sqrt{\dfrac{3V}{4\pi}}
  4. r=3V4π3r=\sqrt[3]{\dfrac{3V}{4\pi}} (correct answer)
Explanation: This question tests your ability to rearrange formulas to solve for a specific variable—essential for using formulas flexibly in science, engineering, and real-world problem-solving. Rearranging formulas with multiple variables works exactly like solving numeric equations, except the answer contains other variables instead of numbers: treat the variable you're solving for as the 'unknown x,' treat all other variables as 'known numbers,' then use inverse operations to isolate your target variable. The same algebraic moves apply—just keep everything symbolic! For example, solving V = πr²h for r is like solving 100 = 3.14·r²·5 for r: divide by π and h, then take square root. More complex rearrangements may require advanced techniques: if your target variable is squared (like r² in A = πr²), you'll need square roots (r = √(A/π)). If it appears in a denominator (like f in 1/f = 1/a + 1/b), you'll need to clear fractions first. If it appears with different powers (like t in s = ut + (1/2)at²), you may need the quadratic formula! The complexity of the rearrangement depends on how the variable appears in the formula. To solve V = (4/3)πr³ for r, first multiply both sides by 3/4 to isolate the r³ term: (3V)/(4π) = r³. Then take the cube root of both sides: r = ∛((3V)/(4π)). Choice A correctly isolates r through multiplying by the reciprocal and taking the cube root to get r = ∛(3V/(4π)). A common distractor like choice B fails by using a square root instead of a cube root, which would be incorrect because r is cubed, not squared; always match the root to the power. The formula rearrangement recipe: (1) Identify what you're solving for (that's your 'x'), (2) Identify what operation(s) are being done to that variable in the original formula, (3) Apply inverse operations in reverse order to isolate it (just like numeric equations!), (4) Simplify the result—combine fractions, simplify radicals, etc. Example: T = 2π√(L/g) solve for L: square both sides → T² = 4π²L/g, multiply by g → gT² = 4π²L, divide by 4π² → L = gT²/(4π²). Each step undoes an operation! Watch for variables appearing multiple times: if your target variable appears in multiple places (like x in xy + xz = w), factor it out first: x(y + z) = w, then x = w/(y + z). If you don't factor, you'll struggle to isolate! Also, when taking square roots of a variable, remember ± if the formula context allows both positive and negative (though often context restricts to positive only, like radius r ≥ 0). Physics and geometry formulas usually want positive values only!