GED Math Quiz: Systems

What this quiz covers

This quiz focuses on Systems, 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. The smaller number is 33
2. The smaller number is 36 (correct answer)
3. The smaller number is 42
4. The smaller number is 48

Explanation: Let the larger number be $x$ and smaller number be $y$. The system is: $x + y = 84$ and $x - y = 12$. From the second equation: $x = y + 12$. Substituting into the first: $(y + 12) + y = 84$, so $2y + 12 = 84$, thus $2y = 72$ and $y = 36$. Choice A (33) would result if student made arithmetic error $2y = 66$. Choice C (42) is half of 84, a common error when students confuse sum and difference. Choice D (48) is the larger number, not the smaller.

Question 2

1. 8
2. 12
3. 20 (correct answer)
4. 27

Explanation: The break-even point occurs when the cost function C(x) is equal to the revenue function R(x). We need to solve the system of equations y = 15x + 400 and y = 35x. We can set the expressions for y equal to each other: 15x + 400 = 35x Subtract 15x from both sides: 400 = 20x Divide by 20: x = 20 The company must sell 20 units to break even. Distractor A is the result of dividing 400 by the sum of the coefficients (400 / 50 = 8). Distractor B is a calculation error. Distractor D is the approximate result of dividing 400 by the cost coefficient (400 / 15 ≈ 26.7).

Question 3

1. $10x = 34$, so $x = 3.4$ (correct answer)
2. $10x = 34$, so $x = 3.0$
3. $6x = 30$, so $x = 5.0$
4. $8x = 26$, so $x = 3.25$

Explanation: Multiplying the second equation by 3: $3(2x - y) = 3(4)$ gives $6x - 3y = 12$. Adding to the first equation: $(4x + 3y) + (6x - 3y) = 22 + 12$, which simplifies to $10x = 34$, so $x = 3.4$. Choice B has correct equation but wrong decimal calculation ($34 ÷ 10 = 3.4$, not $3.0$). Choice C results from incorrectly multiplying the first equation by 3 instead. Choice D comes from adding equations without the multiplication step.

Question 4

1. $a = 4$ and $b = 6$, making the coefficients proportional
2. $a = 4$ and $b = 6$, but the constant terms are different
3. $\frac{a}{2} = \frac{b}{3}$ but $\frac{12}{6} \neq \frac{a}{2}$ (correct answer)
4. $a + b = 12$ and the system becomes inconsistent

Explanation: For no solution, the coefficient ratios must be equal but the constant ratio must be different. The second equation gives us ratios $\frac{2}{2} = 1$ and $\frac{3}{3} = 1$. For no solution: $\frac{a}{2} = \frac{b}{3}$ (parallel lines) but $\frac{12}{6} = 2 \neq \frac{a}{2}$ (different y-intercepts). Choice A gives specific values but misses the key insight. Choice B is partially correct but incomplete. Choice D incorrectly relates the coefficients to constants.

Question 5

1. 10 liters of the 50% acid solution (correct answer)
2. 12 liters of the 50% acid solution
3. 15 liters of the 50% acid solution
4. 18 liters of the 50% acid solution

Explanation: The system is: $x + y = 20$ (total volume) and $0.20x + 0.50y = 0.35(20) = 7$ (pure acid amount). From the first equation: $x = 20 - y$. Substituting: $0.20(20-y) + 0.50y = 7$, so $4 - 0.20y + 0.50y = 7$, thus $0.30y = 3$ and $y = 10$. Choice B (12) results from calculation error in decimal arithmetic. Choice C (15) comes from incorrectly assuming equal proportions. Choice D (18) results from setting up the concentration equation incorrectly.

Question 6

1. $k = -10$, and the system has infinitely many solutions
2. $k = -10$, and the system has no solution (correct answer)
3. $k = 10$, and the system has infinitely many solutions
4. $k = 10$, and the system has no solution

Explanation: The coefficient matrix determinant is $\begin{vmatrix} 5 & -3 \\ k & 6 \end{vmatrix} = 5(6) - (-3)(k) = 30 + 3k$. Setting this to zero: $30 + 3k = 0$, so $k = -10$. When $k = -10$, the system becomes $\begin{cases} 5x - 3y = 7 \\ -10x + 6y = 14 \end{cases}$. The second equation is $-2$ times the first equation, but $14 \neq -2(7)$, so the system is inconsistent (no solution). Choice A has correct $k$ but wrong conclusion about solutions. Choices C and D have incorrect value of $k$.

Question 7

1. 10 (correct answer)
2. 12
3. 15
4. 25

Explanation: Let 'c' be the number of cars and 'v' be the number of vans. We can create a system of two equations:

1. Total vehicles: c + v = 25
2. Total seating capacity: 5c + 8v = 155 From equation (1), express c in terms of v: c = 25 - v. Substitute this expression into equation (2): 5(25 - v) + 8v = 155 125 - 5v + 8v = 155 125 + 3v = 155 3v = 30 v = 10 There are 10 vans in the fleet.

Question 8

1. $7,000
2. $8,000 (correct answer)
3. $7,500
4. $9,000

Explanation: Let x be the amount invested at 3% and y be the amount invested at 5%. We can set up two equations:

1. Total investment: x + y = 15,000
2. Total interest: 0.03x + 0.05y = 610 From equation (1), solve for x: x = 15,000 - y. Substitute this into equation (2): 0.03(15,000 - y) + 0.05y = 610 450 - 0.03y + 0.05y = 610 450 + 0.02y = 610 0.02y = 160 y = 160 / 0.02 = 8,000. So, $8,000 was invested at 5%.

Question 9

1. 10 mph
2. 2.5 mph
3. 15 mph
4. 12.5 mph (correct answer)

Explanation: Let 'b' be the speed of the boat in still water and 'c' be the speed of the current. The effective speed of the boat is (b + c) when traveling downstream and (b - c) when traveling upstream. Using the formula distance = rate × time, we can set up a system of equations.

1. Downstream: 30 = (b + c) × 2
2. Upstream: 30 = (b - c) × 3 From these equations, we can find the rates:
3. b + c = 30 / 2 = 15
4. b - c = 30 / 3 = 10 Now we have a simple system to solve for b. Add the two equations together: (b + c) + (b - c) = 15 + 10 2b = 25 b = 12.5 mph The speed of the boat in still water is 12.5 mph.

Question 10

1. The x-value of the solution is the same in both systems, but the y-value is different.
2. The y-value of the solution is the same in both systems, but the x-value is different.
3. The solution (x, y) is the same for both systems. (correct answer)
4. Neither system has a solution.

Explanation: This question tests the understanding of the elimination method. System B is derived from System A. Notice that the second equation in System B, $7x=7$, can be obtained from System A by multiplying the second equation (2x - y = 0) by 3 and adding it to the first equation (x + 3y = 7). 3(2x - y) + (x + 3y) = 3(0) + 7 6x - 3y + x + 3y = 7 7x = 7 Since the second equation in System B is a valid linear combination of the equations in System A, the two systems are equivalent and will have the exact same solution. Let's find it. From 7x=7, we get x=1. Substitute into x+3y=7: 1+3y=7 => 3y=6 => y=2. The solution is (1, 2) for both systems. Distractors A and B are incorrect because both the x and y values of the solution are identical. Distractor D is incorrect because a solution clearly exists.

Question 11

1. -2
2. 3
3. 2 (correct answer)
4. 4

Explanation: A system of linear equations has no solution if the lines are parallel and have different y-intercepts. In standard form (Ax + By = C), this occurs when the ratio of the x-coefficients is equal to the ratio of the y-coefficients, but this ratio is not equal to the ratio of the constant terms. So, we must have: $\frac{k}{4} = \frac{-3}{-6} \neq \frac{6}{8}$ First, solve for k using the first part of the equality: $\frac{k}{4} = \frac{-3}{-6}$ $\frac{k}{4} = \frac{1}{2}$ Cross-multiply to solve for k: 2k = 4, so k = 2. Next, we must check that this value of k satisfies the inequality: $\frac{1}{2} \neq \frac{6}{8}$. Since $\frac{6}{8}$ simplifies to $\frac{3}{4}$, the inequality $\frac{1}{2} \neq \frac{3}{4}$ is true. Therefore, the system has no solution when k = 2. Distractor A, -2, would result from a sign error. Distractor B, 3, is the value obtained by incorrectly setting the ratio of x-coefficients to the ratio of constants (k/4 = 6/8). Distractor D, 4, is one of the coefficients in the problem and might be chosen by guessing.

Question 12

1. 70 adult tickets should have been sold
2. 80 adult tickets should have been sold
3. 90 adult tickets should have been sold (correct answer)
4. 100 adult tickets should have been sold

Explanation: Let $a$ = actual adult tickets, $s$ = student tickets. We know $a + s = 150$. The revenue collected was $1520, but this includes 10 adult tickets sold at $8 instead of $12. So actual revenue equation: $$12(a-10) + 8(10) + 8s = 1520$ $, which simplifies to$ $12a - 120 + 80 + 8s = 1520$ $, or$ $12a + 8s = 1560$ $. From$ $s = 150 - a$ $:$ $12a + 8(150-a) = 1560$ $, so$ $12a + 1200 - 8a = 1560$ $, thus$ $4a = 360$ $and$ $a = 90$ $. Choice A (70) assumes no correction needed. Choice B (80) results from calculation error. Choice D (100) comes from misunderstanding the problem setup.

Question 13

1. After 2 hours, both will cost $5
2. After 3 hours, both will cost $9
3. After 4 hours, both will cost $9
4. After 3 hours, both will cost $7 (correct answer)

Explanation: This is a classic "when will two linear equations be equal" problem that you'll see often on the GED. When you encounter different pricing structures, set up equations for each option and find where they intersect. Let's define the total cost for each parking option after $h$ hours:

• Garage: $3 for first hour + $2 for each additional hour = $$3 + 2(h-1)$ $
• Lot: $1 for first hour + $3 for each additional hour = $$1 + 3(h-1)$$

Question 14

1. 5
2. 3 (correct answer)
3. -2
4. 7

Explanation: To solve the system, we can use either substitution or elimination. Using elimination is efficient here. Multiply the second equation by 2 to make the y-coefficients opposites: 2(2x + y) = 2(8) --> 4x + 2y = 16 Now, add this new equation to the first equation: (3x - 2y) + (4x + 2y) = 19 + 16 7x = 35 x = 5 Substitute x = 5 into the original second equation to find y: 2(5) + y = 8 10 + y = 8 y = -2 The question asks for the value of x + y, which is 5 + (-2) = 3. Distractor A is the value of x only. Distractor C is the value of y only. Distractor D is the value of the expression x - y (5 - (-2) = 7), which results from a misreading of the question or a sign error.

Question 15

1. 28 mL
2. 60 mL
3. 50 mL
4. 40 mL (correct answer)

Explanation: Let x be the volume of the 20% solution and y be the volume of the 40% solution. The total volume is 100 mL, so: x + y = 100. The total amount of acid is 28% of 100 mL, which is 28 mL. This comes from the two parts: 0.20x + 0.40y = 28. We have a system of two equations:

1. x + y = 100
2. 0.20x + 0.40y = 28 From equation (1), we can express x as x = 100 - y. Substitute this into equation (2): 0.20(100 - y) + 0.40y = 28 20 - 0.20y + 0.40y = 28 20 + 0.20y = 28 0.20y = 8 y = 8 / 0.20 = 40. So, 40 mL of the 40% solution is needed.

Question 16

1. 152 (correct answer)
2. 54
3. 180
4. 170

Explanation: Let L be the length and W be the width of the garden. We can set up a system of two equations based on the given information.

1. Perimeter: 2L + 2W = 54, which simplifies to L + W = 27.
2. Relationship between length and width: L = 2W + 3. Substitute the expression for L from the second equation into the first equation: (2W + 3) + W = 27 3W + 3 = 27 3W = 24 W = 8 meters Now substitute the value of W back into the second equation to find L: L = 2(8) + 3 = 16 + 3 = 19 meters The question asks for the area, which is L × W. Area = 19 × 8 = 152 square meters.

Question 17

1. 3
2. -1
3. 5 (correct answer)
4. 7

Explanation: The solution (3, -1) means that when x = 3, y = -1 for both equations. We can check this for the first equation: -1 = 2(3) - 7, which simplifies to -1 = 6 - 7, or -1 = -1. This is true. Now, substitute x = 3 and y = -1 into the second equation to find the value of $a$: ax + 4y = 11 a(3) + 4(-1) = 11 3a - 4 = 11 3a = 15 a = 5 Distractor A is the x-value of the solution. Distractor B is the y-value of the solution. Distractor D results from a sign error when solving for 3a (3a = 11 - 4 = 7 instead of 3a = 11 + 4 = 15).

Question 18

1. 1
2. 2
3. 3 (correct answer)
4. 4

Explanation: First, find the equation for each line (y = mx + b). Line A: Slope m = (1-5)/(3-1) = -4/2 = -2. Using point (1,5): 5 = -2(1) + b, so b = 7. Equation: y = -2x + 7. Line B: Slope m = (1-4)/(6-0) = -3/6 = -1/2. Using point (0,4), the y-intercept is 4. Equation: y = -1/2x + 4. Set the equations equal to find the intersection: -2x + 7 = -1/2x + 4. Add 2x to both sides: 7 = 1.5x + 4. Subtract 4: 3 = 1.5x. Divide by 1.5: x = 2. Substitute x = 2 into either equation: y = -2(2) + 7 = 3. The y-coordinate of the intersection point is 3. Distractor B is the x-coordinate of the intersection. Distractor D is the y-intercept of Line B. Distractor A is the y-coordinate of both lines when x = 6.

Question 19

1. 6
2. 10 (correct answer)
3. 19
4. 16

Explanation: To solve this system, first eliminate the fractions by multiplying each equation by the least common multiple of its denominators. For the first equation, multiply by 6: 6($\frac{1}{2}x$ + $\frac{1}{3}y$) = 6(7) --> 3x + 2y = 42 For the second equation, multiply by 10: 10($\frac{1}{5}x$ - $\frac{1}{2}y$) = 10(-1) --> 2x - 5y = -10 Now, solve the new system:

1. 3x + 2y = 42
2. 2x - 5y = -10 Multiply equation (1) by 5 and equation (2) by 2 to eliminate y: 5(3x + 2y = 42) --> 15x + 10y = 210 2(2x - 5y = -10) --> 4x - 10y = -20 Add the two resulting equations: (15x + 10y) + (4x - 10y) = 210 + (-20) 19x = 190 x = 10 The question asks for the x-coordinate, which is 10.