GED Math Quiz: Expressions

What this quiz covers

This quiz focuses on Expressions, giving you a quick way to practice the rules, question types, and explanations that matter most for GED Math.

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.

Question 1

1. $-1040$
2. $1040$
3. $2040$ (correct answer)
4. $-2960$

Explanation: The correct answer is C. Since n represents hundreds of units, 800 units = 8 hundreds, so n = 8. Substituting: 500(8) - 15(8)² - 2000 = 4000 - 15(64) - 2000 = 4000 - 960 - 2000 = 2040. Choice A results from using n = 800 instead of n = 8. Choice B makes a sign error with the constant term. Choice D combines both the wrong value of n and sign errors.

Question 2

1. $4 - 19y$
2. $5 - 17y$ (correct answer)
3. $33y - 55$
4. $6 - 23y$

Explanation: We use the formula $a_n = a_1 + (n-1)d$ with $n=11$, $a_1 = 3y-5$, and $d = 1-2y$. Substitute these into the formula: $a_{11} = (3y-5) + (11-1)(1-2y)$. Simplify the expression: $a_{11} = (3y-5) + 10(1-2y)$. Distribute the 10: $a_{11} = 3y - 5 + 10 - 20y$. Combine like terms: $(3y - 20y) + (-5 + 10) = -17y + 5$. This is equivalent to $5 - 17y$.

Question 3

1. $3$
2. $11$
3. $-1$
4. $15$ (correct answer)

Explanation: The correct answer is D. Substituting x = -2: 2(-2)² - 3(-2) + 1 = 2(4) + 6 + 1 = 8 + 6 + 1 = 15. Choice A results from incorrectly calculating (-2)² = -4. Choice B comes from the error 2(-2)² - 3(-2) + 1 = -8 + 6 + 1. Choice C results from multiple sign errors throughout the calculation.

Question 4

1. $0.85p - 5$ (correct answer)
2. $0.80p$
3. $p - 0.15p - 5$
4. $0.85(p - 5)$

Explanation: The correct answer is A. First, the customer pays 85% of the original price: 0.85p. Then $5 is subtracted from this discounted amount: 0.85p - 5. Choice B incorrectly treats the $5 as a percentage discount. Choice C is mathematically equivalent to A but unnecessarily complicated. Choice D incorrectly applies the percentage discount to (p - 5) instead of applying the discounts sequentially.

Question 5

1. $6c - 4$ (correct answer)
2. $6c - 6$
3. $6c + 10$
4. $5c - 6$

Explanation: The correct answer is A. Substitute the expression for b into the expression for a: a = 3b - 7 = 3(2c + 1) - 7 = 6c + 3 - 7 = 6c - 4. Choice B makes an error in combining constants: 3 - 7 = -6 instead of -4. Choice C incorrectly distributes: 3(2c + 1) = 6c + 3, then adds 7 instead of subtracting. Choice D uses an incorrect coefficient for c.

Question 6

1. $6l^2 - 30l + 36$
2. $6l^2 - 18l + 18$ (correct answer)
3. $6l^2 - 24l + 18$
4. $6l^2 - 18l + 36$

Explanation: The correct answer is B. Given: w = l - 3 and h = 2w = 2(l - 3) = 2l - 6. Substituting into the surface area formula: 2[l(l-3) + l(2l-6) + (l-3)(2l-6)] = 2[l² - 3l + 2l² - 6l + (2l² - 6l - 6l + 18)] = 2[l² - 3l + 2l² - 6l + 2l² - 12l + 18] = 2[5l² - 21l + 18] = 10l² - 42l + 36. Wait, let me recalculate: 2[l² - 3l + 2l² - 6l + 2l² - 12l + 18] = 2[3l² - 9l + 9] = 6l² - 18l + 18.

Question 7

1. $30n - 1200$ (correct answer)
2. $30n + 1200$
3. $130n - 1200$
4. $130n + 1200$

Explanation: The correct answer is A. Profit = Revenue - Cost = R(n) - C(n) = 80n - (50n + 1200) = 80n - 50n - 1200 = 30n - 1200. Choice B incorrectly adds the constant term instead of subtracting it. Choice C incorrectly adds the coefficients of n instead of subtracting. Choice D makes both errors: adding coefficients and adding the constant.

Question 8

1. $3 + 1.50h$
2. $3 + 1.50(h - 1)$ (correct answer)
3. $4.50h$
4. $3h + 1.50$

Explanation: The correct answer is B. The cost is $3 for the first hour plus $1.50 for each additional hour. Since there are (h-1) additional hours beyond the first hour, the expression is 3 + 1.50(h-1). Choice A incorrectly charges $1.50 for all h hours including the first. Choice C treats it as a simple rate problem. Choice D incorrectly multiplies the flat rate by h and adds the hourly rate once.

Question 9

1. 160
2. 112
3. -112
4. 16 (correct answer)

Explanation: This is a nested function evaluation, so we must evaluate the inner function first. First, find the value of $g(-3)$. Substitute -3 for c: $g(-3) = 1 - (-3)^2 = 1 - 9 = -8$. Now, use this result as the first input for the function $f$. We need to find $f(-8, -2)$. Substitute -8 for a and -2 for b: $f(-8, -2) = (-8)^2 - 3(-8)(-2)$. Following the order of operations, calculate the exponent first: $(-8)^2 = 64$. Then perform the multiplication: $3(-8)(-2) = 3(16) = 48$. Finally, perform the subtraction: $64 - 48 = 16$.

Question 10

1. $4x^2 + 6wx + 10x$ (correct answer)
2. $x^2 + 3wx + 5x$
3. $8x^2 + 6wx + 10x$
4. $2w^2 + 4x^2 + 6wx + 5w + 10x$

Explanation: The area of the walkway is the area of the large rectangle (garden plus walkway) minus the area of the small rectangle (garden only). First, define the dimensions. Garden width = $w$. Garden length = $2w + 5$. The walkway adds width $x$ to both sides of the garden. So, the total width is $w + 2x$ and the total length is $(2w + 5) + 2x$. The total area is $(w + 2x)(2w + 5 + 2x) = 2w^2 + 5w + 2wx + 4wx + 10x + 4x^2 = 2w^2 + 6wx + 5w + 10x + 4x^2$. The garden area is $w(2w + 5) = 2w^2 + 5w$. Subtract the garden area from the total area: $(2w^2 + 6wx + 5w + 10x + 4x^2) - (2w^2 + 5w) = 4x^2 + 6wx + 10x$.

Question 11

1. $\frac{n^2}{5} - 3|n - 8|$ (correct answer)
2. $\frac{n^2 - 3|n-8|}{5}$
3. $5n^2 - 3(n - 8)$
4. $\frac{n^2}{5} - (3n - 8)$

Explanation: The phrase is broken down into two main parts joined by "decreased by" (subtraction). The first part is "The quotient of the square of a number $n$ and 5", which translates to $\frac{n^2}{5}$. The second part is "the product of 3 and the absolute difference between the number and 8". The "absolute difference between n and 8" is $|n - 8|$. The "product of 3 and" that quantity is $3|n - 8|$. Combining the two parts, "decreased by" means we subtract the second part from the first: $\frac{n^2}{5} - 3|n - 8|$.

Question 12

1. $\frac{c_1V_A + c_2V_B}{2}$
2. $\frac{c_1 + c_2}{V_A + V_B}$
3. $\frac{c_1V_A + c_2V_B}{V_A + V_B}$ (correct answer)
4. $\frac{c_1V_B + c_2V_A}{V_A + V_B}$

Explanation: The concentration of the final mixture is the total amount of saline divided by the total volume of the mixture. The amount of saline from Substance A is its concentration times its volume: $c_1V_A$. The amount of saline from Substance B is $c_2V_B$. The total amount of saline is the sum of these two amounts: $c_1V_A + c_2V_B$. The total volume of the mixture is the sum of the individual volumes: $V_A + V_B$. Therefore, the final concentration is the total saline divided by the total volume, which is $\frac{c_1V_A + c_2V_B}{V_A + V_B}$.

Question 13

1. $-0.5x^2 + 220x - 5000$
2. $-0.5x^2 + 180x + 5000$
3. $-0.5x^2 + 180x - 5000$ (correct answer)
4. $0.5x^2 + 180x - 5000$

Explanation: First, write the expressions for revenue $R(x)$ and cost $C(x)$. The revenue expression is given: $R(x) = x(200 - 0.5x) = 200x - 0.5x^2$. The cost expression is a sum of the fixed and variable costs: $C(x) = 5000 + 20x$. The profit is $P(x) = R(x) - C(x)$. Substitute the expressions for $R(x)$ and $C(x)$: $P(x) = (200x - 0.5x^2) - (5000 + 20x)$. It is crucial to use parentheses around the cost function to ensure the negative sign distributes correctly. $P(x) = 200x - 0.5x^2 - 5000 - 20x$. Now, combine like terms: $P(x) = -0.5x^2 + (200x - 20x) - 5000 = -0.5x^2 + 180x - 5000$.

Question 14

1. $P(1 + \frac{3}{5}r_A t) + P(1 + \frac{2}{5}r_B t)$
2. $P \cdot t \cdot (\frac{3}{5}r_A \cdot \frac{2}{5}r_B)$
3. $P \cdot t \cdot (\frac{3}{5}r_A + \frac{2}{5}r_B)$ (correct answer)
4. $P \cdot t \cdot (r_A + r_B)$

Explanation: The formula for simple interest is $I = Prt$. First, consider Account A. The principal in this account is $\frac{3}{5}P$. The interest earned is $I_A = (\frac{3}{5}P) \cdot r_A \cdot t$. The remainder of the principal is $P - \frac{3}{5}P = \frac{2}{5}P$, which is deposited into Account B. The interest earned from Account B is $I_B = (\frac{2}{5}P) \cdot r_B \cdot t$. The total interest is the sum $I_A + I_B = (\frac{3}{5}P)r_A t + (\frac{2}{5}P)r_B t$. We can factor out the common terms $P$ and $t$ from this expression: $P \cdot t \cdot (\frac{3}{5}r_A + \frac{2}{5}r_B)$.

Question 15

1. $(1 + \frac{t}{100})(P(1 - \frac{d}{100}) - C)$ (correct answer)
2. $P(1 - \frac{d}{100})(1 + \frac{t}{100}) - C$
3. $(P - C)(1 - \frac{d}{100})(1 + \frac{t}{100})$
4. $P(1 - \frac{d}{100} + \frac{t}{100}) - C$

Explanation: Let's build the expression step-by-step following the order of operations. First, discount the price $P$ by $d$ percent. The discounted price is $P - P(\frac{d}{100}) = P(1 - \frac{d}{100})$. Second, apply the coupon for $C$ dollars. This is a flat subtraction from the current price: $P(1 - \frac{d}{100}) - C$. Third, calculate the sales tax of $t$ percent on this new price. The total amount including tax is the new price plus the tax amount, which is $1 + \frac{t}{100}$ times the new price. Therefore, the final cost is $(P(1 - \frac{d}{100}) - C) \times (1 + \frac{t}{100})$, which matches choice A.

Question 16

1. $315h - 2700$ (correct answer)
2. $300h - 2500$
3. $310h - 2500$
4. $300h - 2100$

Explanation: First, calculate the total labor cost. The cost for the first 40 hours is 40 \times \200 = $8000$. The rate for hours beyond 40 is$1.5 \times $200 = $300$per hour. The number of hours charged at this rate is$(h - 40)$. The cost for these additional hours is$300(h - 40)$. The total labor cost$L$is$8000 + 300(h - 40) = 8000 + 300h - 12000 = 300h - 4000$. The materials surcharge is 5$0.05(300h - 4000)$. The total project cost is the sum of the initiation fee, the total labor cost, and the surcharge:$C(h) = 1500 + (300h - 4000) + 0.05(300h - 4000)$. This can be simplified as$C(h) = 1500 + 1.05(300h - 4000) = 1500 + 315h - 4200 = 315h - 2700$.

Question 17

1. $\frac{5T}{18} + 255.4$
2. $\frac{5}{9}(\frac{T}{2} - 32) + 273.15$ (correct answer)
3. $\frac{1}{2}(\frac{5}{9}(T - 32) + 273.15)$
4. $\frac{5}{9}(\frac{T-32}{2}) + 273.15$

Explanation: Let the original temperature be $T$ in degrees Fahrenheit. The final temperature in Fahrenheit is $\frac{T}{2}$. We need to convert this final Fahrenheit temperature to Kelvin. First, convert the final Fahrenheit temperature to Celsius. Let $F_{final} = \frac{T}{2}$. Using the formula, $C_{final} = \frac{5}{9}(F_{final} - 32) = \frac{5}{9}(\frac{T}{2} - 32)$. Next, convert this Celsius temperature to Kelvin: $K_{final} = C_{final} + 273.15 = \frac{5}{9}(\frac{T}{2} - 32) + 273.15$. Choice A represents the simplified form with decimal approximation. Choice C represents halving the Kelvin temperature, not the Fahrenheit. Choice D represents halving the Celsius equivalent, not the Fahrenheit reading itself.

Question 18

1. $\frac{11}{3}\pi r^3$ (correct answer)
2. $\frac{13}{3}\pi r^3$
3. $3\pi r^3 + \frac{4}{3}\pi r^3$
4. $4\pi r^3$

Explanation: The total volume is the sum of the volume of the cylinder and the volume of the hemisphere. The cylinder's height $h$ is given as $3r$. The volume of the cylinder is $V_{cyl} = \pi r^2 h = \pi r^2 (3r) = 3\pi r^3$. A hemisphere is half of a sphere. The volume of the hemisphere is $V_{hemi} = \frac{1}{2} V_{sphere} = \frac{1}{2} \cdot \frac{4}{3}\pi r^3 = \frac{2}{3}\pi r^3$. The total volume is $V_{total} = V_{cyl} + V_{hemi} = 3\pi r^3 + \frac{2}{3}\pi r^3$. To add these, find a common denominator: $3\pi r^3 = \frac{9}{3}\pi r^3$. So, $V_{total} = \frac{9}{3}\pi r^3 + \frac{2}{3}\pi r^3 = \frac{11}{3}\pi r^3$.

Question 19

1. $T_0 - km + 0.2T_0$
2. $1.2(T_0 - km)$ (correct answer)
3. $1.2T_0 - km$
4. $T_0 - 0.8km$

Explanation: Let's model the temperature changes step-by-step. The initial temperature is $T_0$. It decreases by $k$ degrees per minute for $m$ minutes, so the total decrease is $km$. The temperature after the cooling phase is $T_{cooled} = T_0 - km$. Next, the temperature increases by an amount equal to one-fifth (or 0.2) of this cooled temperature. The increase is $0.2 \times T_{cooled} = 0.2(T_0 - km)$. The final temperature is the cooled temperature plus the increase: $T_{final} = T_{cooled} + 0.2 \times T_{cooled} = 1 \times T_{cooled} + 0.2 \times T_{cooled} = 1.2 \times T_{cooled}$. Substituting the expression for $T_{cooled}$, we get $T_{final} = 1.2(T_0 - km)$.

Question 20

1. $6p+4p$ (correct answer)
2. $6+p+4$
3. $6p-4p$
4. $p(6-4)$

Explanation: This question tests your ability to translate a word problem into an algebraic expression involving multiplication and addition. When Juan works for a certain number of hours at $p$ dollars per hour, his earnings equal the hours worked times his hourly rate. On Saturday, he works 6 hours, so he earns $6p$ dollars. On Sunday, he works 4 hours, so he earns $4p$ dollars. To find his total weekend earnings, you add these amounts together: $6p + 4p$. Choice A ($6p + 4p$) correctly represents Juan's total weekend earnings by multiplying his hourly rate by the hours worked each day, then adding those daily totals. Choice B ($6 + p + 4$) incorrectly adds the hours and hourly rate instead of multiplying them. This would give you $10 + p$, which doesn't represent earnings at all—you can't just add hours to dollars per hour. Choice C ($6p - 4p$) uses subtraction instead of addition, which would calculate the difference between Saturday's and Sunday's earnings ($2p$) rather than the total weekend earnings. Choice D ($p(6 - 4)$) also uses subtraction within parentheses, giving you $p \times 2 = 2p$, which again represents the difference in daily earnings, not the total. Remember this pattern: when calculating total pay, multiply rate times hours for each period, then add all the periods together. Watch out for expressions that subtract when you should add, or that add quantities with different units (like hours plus hourly rate).