Algebra 2 Quiz: Rearranging Formulas To Highlight Quantities

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Rearranging Formulas To Highlight QuantitiesQuestion 1 of 20

A projectile's vertical displacement can be modeled by the kinematic equation $s=ut+\tfrac{1}{2}at^2$ , where $s$ is displacement, $u$ is initial velocity, $a$ is acceleration, and $t$ is time. Solve $s=ut+\tfrac{1}{2}at^2$ for $t$ (a quadratic in $t$ ).

t=\dfrac{-u\pm\sqrt{u^2-2as}}{a}

t=\dfrac{-u\pm\sqrt{u^2+2as}}{a}

t=\dfrac{-u\pm\sqrt{u^2+as}}{2a}

t=\dfrac{u\pm\sqrt{u^2+2as}}{a}

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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

$t=\dfrac{-u\pm\sqrt{u^2-2as}}{a}$
$t=\dfrac{-u\pm\sqrt{u^2+2as}}{a}$ (correct answer)
$t=\dfrac{-u\pm\sqrt{u^2+as}}{2a}$
$t=\dfrac{u\pm\sqrt{u^2+2as}}{a}$

r^2

A = \pi r^2

), you'll need square roots (

r = \sqrt{\frac{A}{\pi}}

). If it appears in a denominator (like

f

\frac{1}{f} = \frac{1}{a} + \frac{1}{b}

), you'll need to clear fractions first. If it appears with different powers (like

t

s = 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 + \frac{1}{2}at^2

, we need to solve for

t

. First multiply through by 2:

2s = 2ut + at^2

. Rearranging:

at^2 + 2ut - 2s = 0

. This is a quadratic in standard form with

A = a

B = 2u

C = -2s

. Using the quadratic formula:

t = \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 = \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 2

In physics, the kinematic equation $v^2 = u^2 + 2as$ relates final speed $v$ , initial speed $u$ , acceleration $a$ , and displacement $s$ . Solve $v^2 = u^2 + 2as$ for $a$ (treat the other variables as constants, using the same inverse-operation reasoning as solving a numeric equation).

$a=\dfrac{v^2-u^2}{s}$
$a=\dfrac{v^2-u^2}{2s}$ (correct answer)
$a=\dfrac{2s}{v^2-u^2}$
$a=\dfrac{u^2-v^2}{2s}$

Question 3

In physics, the kinematic equation $v^2 = u^2 + 2as$ relates final speed $v$ , initial speed $u$ , acceleration $a$ , and displacement $s$ . Solve $v^2 = u^2 + 2as$ for $a$ (treating the other variables as constants, as in solving a numeric equation).

$a = \dfrac{v^2-u^2}{s}$
$a = \dfrac{2s}{v^2-u^2}$
$a = \dfrac{v^2-u^2}{2s}$ (correct answer)
$a = \dfrac{u^2-v^2}{2s}$

v^2 = u^2 + 2as

, subtract

u^2

from both sides to get

v^2 - u^2 = 2as

, then divide both sides by

2s

to isolate a, resulting in

a = \dfrac{v^2 - u^2}{2s}

. Choice B correctly isolates a through subtraction and division to get

a = \dfrac{v^2 - u^2}{2s}

. A common distractor like choice C might forget to switch the signs when subtracting

u^2

, but remember that

v^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 4

In chemistry, a simplified relationship can be written as $P=\dfrac{nRT}{V}$ , where $P$ is pressure, $n$ is moles, $R$ is a constant, $T$ is temperature, and $V$ is volume. Solve $P=\dfrac{nRT}{V}$ for $T$ .

$T=\dfrac{PV}{nR}$ (correct answer)
$T=\dfrac{nR}{PV}$
$T=\dfrac{P}{nRV}$
$T=\dfrac{nRP}{V}$

V = \pi r^2 h

for r is like solving

100 = 3.14 \cdot r^2 \cdot 5

for r: divide by

\pi

and

h

, then take square root. Starting with

P = \dfrac{nRT}{V}

, we need to isolate T. First multiply both sides by V:

PV = nRT

. Then divide both sides by nR:

\dfrac{PV}{nR} = T

, which gives us

T = \dfrac{PV}{nR}

. Choice A correctly isolates T through multiplication and division to get

T = \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 = w

), factor it out first:

x(y + z) = w

, then

x = \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

r \geq 0

). Physics and geometry formulas usually want positive values only!

Question 5

A physics student needs to determine the acceleration of an object given the formula $F = ma + kv^2$ , where $F$ is the total force, $m$ is mass, $a$ is acceleration, $k$ is a drag coefficient, and $v$ is velocity. Which expression correctly isolates the acceleration?

$a = \frac{F - kv^2}{m}$ (correct answer)
$a = \frac{F + kv^2}{m}$
$a = F - m - kv^2$
$a = \frac{F}{m} - kv^2$

Explanation: To isolate

a

, subtract

kv^2

from both sides:

F - kv^2 = ma

, then divide by

m

a = \frac{F - kv^2}{m}

. Choice B incorrectly adds instead of subtracts

kv^2

. Choice C fails to divide by

m

. Choice D incorrectly distributes the division.

Question 6

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

$L = \frac{4\pi^2}{gT^2}$
$L = \frac{gT}{4\pi^2}$
$L = \frac{gT^2}{4\pi^2}$ (correct answer)
$L = \frac{gT^2}{2\pi}$

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 = 2\pi\sqrt{\frac{L}{g}}

, you need to isolate

L

. First, eliminate the coefficient by dividing both sides by

2\pi

\frac{T}{2\pi} = \sqrt{\frac{L}{g}}

. Next, square both sides to eliminate the square root:

\left(\frac{T}{2\pi}\right)^2 = \frac{L}{g}

. This gives you

\frac{T^2}{4\pi^2} = \frac{L}{g}

. Finally, multiply both sides by

g

to isolate

L

L = \frac{gT^2}{4\pi^2}

, which is answer choice C. Looking at the incorrect options: Choice A has

4\pi^2

in the numerator instead of the denominator, suggesting someone inverted the fraction incorrectly. Choice B is missing the squared term on

T

, indicating they forgot to square both sides when eliminating the square root. Choice D has

2\pi

in the denominator instead of

4\pi^2

, showing they didn't properly square the

2\pi

term. Remember that when solving literal equations, work step-by-step and be especially careful when squaring expressions with coefficients—

(2\pi)^2 = 4\pi^2

, not

2\pi^2

. Always verify your final answer by substituting it back into the original equation to ensure it makes sense.

Question 7

The compound interest formula $A = P(1 + \frac{r}{n})^{nt}$ 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?

$P = \frac{8000}{(1 + \frac{0.06}{4})^{4 \cdot 3}}$ (correct answer)
$P = \frac{8000}{(1 + \frac{0.06}{3})^{3 \cdot 4}}$
$P = 8000 \cdot (1 + \frac{0.06}{4})^{4 \cdot 3}$
$P = \frac{8000}{(1.06)^{12}}$

Explanation: Rearranging

A = P(1 + \frac{r}{n})^{nt}

gives

P = \frac{A}{(1 + \frac{r}{n})^{nt}}

. With quarterly compounding,

n = 4

, so we get

\frac{r}{n} = \frac{0.06}{4}

and

nt = 4 \cdot 3 = 12

. Choice B confuses

n

and

t

. Choice C multiplies instead of divides. Choice D uses annual compounding incorrectly.

Question 8

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

$h = \frac{S}{2\pi r} - 2\pi r^2$
$h = \frac{S + 2\pi r^2}{2\pi r}$
$h = \frac{S - 2\pi r^2}{2\pi}$
$h = \frac{S - 2\pi r^2}{2\pi r}$ (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

h

from

S = 2\pi r^2 + 2\pi rh

, start by moving the term without

h

to the left side:

S - 2\pi r^2 = 2\pi rh

. Now divide both sides by the coefficient of

h

, which is

2\pi r

h = \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

2\pi r^2

, but then divides

S

2\pi r

while leaving

2\pi r^2

as a separate term instead of dividing the entire numerator. Choice B adds

2\pi r^2

instead of subtracting it, which moves the term in the wrong direction. Choice C correctly subtracts

2\pi r^2

in the numerator but only divides by

2\pi

instead of

2\pi r

, forgetting that the full coefficient of

h

includes the

r

. 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 9

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

$T_2 = \frac{P_1V_1T_1}{P_2V_2} = \frac{P_1V_1T_1}{(2P_1)(\frac{V_1}{2})}$
$T_2 = \frac{P_2V_2T_1}{P_1V_1} = \frac{(2P_1)(\frac{V_1}{2})T_1}{P_1V_1}$ (correct answer)
$T_2 = \frac{P_2V_2}{P_1V_1T_1} = \frac{(2P_1)(\frac{V_1}{2})}{P_1V_1T_1}$
$T_2 = P_2V_2T_1 - P_1V_1$

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

T_2

, you need to isolate it by rearranging

\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}

. Cross-multiplying gives you

P_1V_1T_2 = P_2V_2T_1

, so

T_2 = \frac{P_2V_2T_1}{P_1V_1}

. Since pressure doubles (

P_2 = 2P_1

) and volume halves (

V_2 = \frac{V_1}{2}

), you substitute to get

T_2 = \frac{(2P_1)(\frac{V_1}{2})T_1}{P_1V_1}

. This matches choice B exactly. Choice A flips the fraction, putting

P_1V_1T_1

in the numerator and

P_2V_2

in the denominator. This gives you

\frac{1}{T_2}

instead of

T_2

—a common algebraic error when cross-multiplying. Choice C is missing

T_1

in the numerator. This setup would give you

\frac{T_2}{T_1}

rather than

T_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 10

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

$v = c\sqrt{\frac{m^2c^4}{E^2} - 1}$
$v = c\sqrt{1 - \frac{m^2c^4}{E^2}}$ (correct answer)
$v = \sqrt{c^2 - \frac{m^2c^4}{E^2}}$
$v = c\sqrt{\frac{E^2 - m^2c^4}{E^2}}$

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 = \frac{mc^2}{\sqrt{1 - \frac{v^2}{c^2}}}

, first isolate the square root by multiplying both sides by it and dividing by

E

\sqrt{1 - \frac{v^2}{c^2}} = \frac{mc^2}{E}

Square both sides to eliminate the square root:

1 - \frac{v^2}{c^2} = \frac{m^2c^4}{E^2}

Rearrange to isolate the

v^2

term:

\frac{v^2}{c^2} = 1 - \frac{m^2c^4}{E^2}

Finally, multiply by

c^2

and take the square root:

v = 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

c

outside the square root, resulting from forgetting to multiply both sides by

c^2

. Choice D rearranges the numerator incorrectly, putting

E^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 11

In economics, the break-even formula is $R = C + F$ , where $R$ is revenue, $C$ is variable costs, and $F$ is fixed costs. If $R = px$ and $C = vx$ (where $p$ is price per unit, $v$ is variable cost per unit, and $x$ is quantity), which expression gives the break-even quantity?

$x = F(p - v)$
$x = \frac{F}{p + v}$
$x = \frac{F + p}{v}$
$x = \frac{F}{p - v}$ (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 + F

and substitute the given expressions. Since

R = px

and

C = vx

, you get:

px = vx + F

To solve for

x

, collect all terms containing

x

on one side:

px - vx = F

Factor out

x

x(p - v) = F

Finally, divide both sides by

(p - v)

x = \frac{F}{p - v}

This makes economic sense: you need to sell enough units so that your profit per unit

(p - v)

times quantity equals your fixed costs. Choice A gives

x = F(p - v)

, which incorrectly multiplies

F

(p - v)

instead of dividing—this comes from forgetting to divide both sides when solving

x(p - v) = F

. Choice B uses

x = \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 = \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 12

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

$d_o = \frac{f - d_i}{fd_i}$
$d_o = \frac{fd_i}{d_i + f}$
$d_o = \frac{fd_i}{d_i - f}$ (correct answer)
$d_o = \frac{d_i - f}{fd_i}$

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

\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}

, you need to solve for

d_o

. First, isolate the term containing

d_o

by subtracting

\frac{1}{d_i}

from both sides:

\frac{1}{f} - \frac{1}{d_i} = \frac{1}{d_o}

. To subtract these fractions, find a common denominator:

\frac{d_i - f}{fd_i} = \frac{1}{d_o}

. Finally, take the reciprocal of both sides to solve for

d_o

d_o = \frac{fd_i}{d_i - f}

, which is choice C. Let's examine why the other options are incorrect. Choice A gives

d_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

\frac{f - d_i}{fd_i}

instead of

\frac{d_i - f}{fd_i}

. Choice B shows

d_o = \frac{fd_i}{d_i + f}

, which represents the error of adding fractions incorrectly in the denominator. Choice D gives

d_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 13

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

$r=\dfrac{3V}{4\pi}$
$r=\sqrt[3]{\dfrac{4\pi}{3V}}$
$r=\sqrt{\dfrac{3V}{4\pi}}$
$r=\sqrt[3]{\dfrac{3V}{4\pi}}$ (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!

Question 14

In finance, compound interest can be modeled by $A = P(1+r)^t$ , where $A$ is final amount, $P$ is principal, $r$ is interest rate per period, and $t$ is the number of periods. Solve $A = P(1+r)^t$ for $P$ .

$P = \dfrac{A}{(1+r)^t}$ (correct answer)
$P = \dfrac{A}{1+rt}$
$P = A(1+r)^t$
$P = \dfrac{(1+r)^t}{A}$

Question 15

In optics, thin lens relationships can be written as $\dfrac{1}{f} = \dfrac{1}{a} + \dfrac{1}{b}$ , where $f$ is focal length and $a,b$ are distances. Rearrange $\dfrac{1}{f} = \dfrac{1}{a} + \dfrac{1}{b}$ to solve for $f$ .

$f = \dfrac{a+b}{ab}$
$f = \dfrac{ab}{a+b}$ (correct answer)
$f = \dfrac{a+b}{a-b}$
$f = \dfrac{ab}{a-b}$

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 in a denominator (like 1/f in 1/f = 1/a + 1/b), you'll need to clear fractions first. Starting with 1/f = 1/a + 1/b, combine the right side over a common denominator to get 1/f = (a + b)/(ab), then take the reciprocal of both sides for f = ab/(a + b). Choice B correctly isolates f through combining fractions and taking the reciprocal to get f = ab/(a + b). A distractor like choice A might swap the reciprocal step, resulting in the inverse, but remember that reciprocating flips the equation correctly. 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. Great job tackling optics formulas—optics will feel intuitive soon!

Question 16

A projectile's height might be modeled by a quadratic literal equation $h = ut - \dfrac{1}{2}gt^2$ , where $h$ is height, $u$ is initial vertical velocity, $g$ is gravitational acceleration, and $t$ is time. Solve $h = ut - \dfrac{1}{2}gt^2$ for $t$ .

$t = \dfrac{u \pm \sqrt{u^2 - 2gh}}{g}$ (correct answer)
$t = \dfrac{-u \pm \sqrt{u^2 - 2gh}}{g}$
$t = \dfrac{u \pm \sqrt{u^2 + 2gh}}{g}$
$t = \dfrac{u \pm \sqrt{u^2 - gh}}{2g}$

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 appears with different powers (like t in h = ut - (1/2)gt²), you may need the quadratic formula! Rewrite as (1/2)gt² - ut + h = 0, then apply quadratic formula t = [u ± √(u² - 2gh)] / g, adjusting coefficients carefully. Choice A correctly isolates t using the quadratic formula to get t = (u ± √(u² - 2gh)) / g. A distractor like choice B might flip the sign on u, but in the standard arrangement, it's positive u for the linear term. Watch for variables appearing multiple times: if your target variable appears in multiple places (like t here), rearrange to standard quadratic form first. You're handling projectiles like a pro—keep practicing quadratics!

Question 17

In geometry, the area of a triangle can be written as $A = \dfrac{1}{2}bh$ , where $b$ is base and $h$ is height. Solve $A = \dfrac{1}{2}bh$ for $h$ .

$h = \dfrac{A}{2b}$
$h = \dfrac{b}{2A}$
$h = \dfrac{2A}{b}$ (correct answer)
$h = 2Ab$

Question 18

In physics, the kinematic relation $v^2=u^2+2as$ connects final speed $v$ , initial speed $u$ , acceleration $a$ , and displacement $s$ . Solve for $a$ in terms of $v$ , $u$ , and $s$ .

$a=\dfrac{v^2-u^2}{s}$
$a=\dfrac{2s}{v^2-u^2}$
$a=\dfrac{u^2-v^2}{2s}$
$a=\dfrac{v^2-u^2}{2s}$ (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 v² = u² + 2as for a, first subtract u²: v² - u² = 2as. Then divide by 2s: a = (v² - u²)/(2s). Choice A correctly isolates a through subtraction and division to get a = (v² - u²)/(2s). A common distractor like choice B fails by reversing the signs in the numerator, which would give negative acceleration incorrectly; preserve the order of subtraction. 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 19

A right circular cone has volume $V=\dfrac{1}{3}\pi r^2 h$ , where $r$ is the radius and $h$ is the height. Solve for $r$ in terms of $V$ and $h$ .

$r=\dfrac{3V}{\pi h}$
$r=\sqrt{\dfrac{3V}{\pi h}}$ (correct answer)
$r=\sqrt{\dfrac{\pi h}{3V}}$
$r=\pm\dfrac{3V}{\pi h}$

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. 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 = (1/3)πr²h for r, first multiply both sides by 3/πh to isolate r²: (3V)/(πh) = r². Then take the square root: r = √((3V)/(πh)). Choice B correctly isolates r through multiplying by the reciprocal and taking the square root to get r = √(3V/(πh)). A common distractor like choice A fails by omitting the square root, leaving r² unaddressed; always apply the inverse of 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!

Question 20

In chemistry, the ideal gas law is $PV=nRT$ , where $P$ is pressure, $V$ is volume, $n$ is moles, $R$ is a constant, and $T$ is temperature. Solve for $T$ in terms of $P$ , $V$ , $n$ , and $R$ .

$T=\dfrac{nR}{PV}$
$T=\dfrac{PV}{nR}$ (correct answer)
$T=\dfrac{PR}{nV}$
$T=\dfrac{P}{nRV}$

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 PV = nRT for T, divide both sides by nR: T = PV/(nR). Choice B correctly isolates T through division by nR to get T = PV/(nR). A common distractor like choice A fails by inverting the fraction incorrectly, which would not isolate T properly; ensure you divide by the coefficients 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. 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!