GED Math Quiz: Systems

19 questions · exam conditions

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SystemsQuestion 1 of 19

Two numbers have a sum of 84 and their difference is 12. If the system of equations representing this situation is solved by substitution, what is the value of the smaller number?

The smaller number is 33

The smaller number is 36

The smaller number is 42

The smaller number is 48

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GED Math Quiz

GED Math Quiz: Systems

Practice Systems in GED Math 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 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.

All questions

Question 1

Two numbers have a sum of 84 and their difference is 12. If the system of equations representing this situation is solved by substitution, what is the value of the smaller number?

The smaller number is 33
The smaller number is 36 (correct answer)
The smaller number is 42
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

A company's cost to produce 'x' units is given by the function C(x) = 15x + 400. The revenue from selling 'x' units is given by the function R(x) = 35x.

How many units must the company sell to break even, where cost equals revenue?

8
12
20 (correct answer)
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

Sarah is solving the system $$ \begin{cases} 4x + 3y = 22 \ 2x - y = 4 \end{cases}

$10x = 34$ , so $x = 3.4$ (correct answer)
$10x = 34$ , so $x = 3.0$
$6x = 30$ , so $x = 5.0$
$8x = 26$ , 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

Consider the system $\begin{cases} ax + by = 12 \\ 2x + 3y = 6 \end{cases}$ where $a$ and $b$ are constants. If this system has no solution, which of the following must be true about the relationship between $a$ and $b$ ?

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

A chemist needs to create 20 liters of a 35% acid solution by mixing a 20% acid solution with a 50% acid solution. Let $x$ represent liters of 20% solution and $y$ represent liters of 50% solution. After setting up the correct system of equations, the chemist should find that the amount of 50% solution needed is:

10 liters of the 50% acid solution (correct answer)
12 liters of the 50% acid solution
15 liters of the 50% acid solution
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

The system $\begin{cases} 5x - 3y = 7 \\ kx + 6y = 14 \end{cases}$ is solved using Cramer's rule. If the determinant of the coefficient matrix equals zero, what is the value of $k$ , and what does this mean for the system?

$k = -10$ , and the system has infinitely many solutions
$k = -10$ , and the system has no solution (correct answer)
$k = 10$ , and the system has infinitely many solutions
$k = 10$ , 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

A rental company has a fleet of 25 vehicles, consisting of cars and vans. Cars can seat 5 people, and vans can seat 8 people. The total seating capacity of the entire fleet is 155.

How many vans are in the fleet?

10 (correct answer)
12
15
25

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

Total vehicles: c + v = 25
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.

Distractor C is the number of cars (c = 25 - 10 = 15). Distractor B is a plausible result of a calculation error. Distractor D is the total number of vehicles in the fleet.

Question 8

A person invested a total of $15,000 into two separate accounts. One account earns 3% simple interest per year, and the other earns 5% simple interest per year. After one year, the total interest earned from both accounts was $610.

How much money was invested in the account that earns 5% interest?

$7,000
$8,000 (correct answer)
$7,500
$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:

Total investment: x + y = 15,000
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%.

Distractor A is the amount invested at 3% (x = 15,000 - 8,000 = 7,000). Distractor C is the result of splitting the total investment equally, which would yield $750 in interest, not $610. Distractor D could result from a calculation error, such as 0.02y = 180 (e.g., 610-430 instead of 610-450).

Question 9

A boat traveled 30 miles downstream in 2 hours. The return trip upstream took 3 hours.

Assuming the speed of the current is constant, what is the speed of the boat in still water?

10 mph
2.5 mph
15 mph
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.

Downstream: 30 = (b + c) × 2
Upstream: 30 = (b - c) × 3 From these equations, we can find the rates:
b + c = 30 / 2 = 15
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.

Distractor B is the speed of the current (c = 15 - b = 15 - 12.5 = 2.5). Distractor A is the speed upstream. Distractor C is the speed downstream.

Question 10

Two systems of equations are shown below. System A: $x+3y=7$ and $2x-y=0$ System B: $x+3y=7$ and $7x=7$ Which statement correctly compares the solutions to System A and System B?

The x-value of the solution is the same in both systems, but the y-value is different.
The y-value of the solution is the same in both systems, but the x-value is different.
The solution (x, y) is the same for both systems. (correct answer)
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

For what value of $k$ does the following system of equations have no solution? $kx - 3y = 6$ $4x - 6y = 8$

-2
3
2 (correct answer)
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

A theater sells adult tickets for $12 and student tickets for $8. On Friday night, they sold 150 tickets total and collected $1520. However, the box office manager realizes that 10 of the adult tickets were mistakenly sold at student prices. What was the actual number of adult tickets that should have been sold?

70 adult tickets should have been sold
80 adult tickets should have been sold
90 adult tickets should have been sold (correct answer)
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

A parking garage charges $3 for the first hour and $2 for each additional hour. A nearby lot charges $1 for the first hour and $3 for each additional hour. After how many total hours will the costs be equal, and what will that cost be?

After 2 hours, both will cost $5
After 3 hours, both will cost $9
After 4 hours, both will cost $9
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 = $$1 + 3(h-1)$$

To find when costs are equal, set the equations equal:

3 + 2(h-1) = 1 + 3(h-1)

Expanding:

3 + 2h - 2 = 1 + 3h - 3

Simplifying:

1 + 2h = -2 + 3h

Solving:

3 = h

After 3 hours, let's calculate the cost: Garage:

3 + 2(3-1) = 3 + 4 = \$7

Lot:

1 + 3(3-1) = 1 + 6 = \$7

Answer D is correct: after 3 hours, both cost $7. Answer A gives the wrong time frame (2 hours) and calculates costs incorrectly. Answer B has the right time (3 hours) but miscalculates the total cost as $9. Answer C compounds errors with both wrong time (4 hours) and wrong cost calculation. Study tip: For "when will costs be equal" problems, always set up equations for each scenario, solve for the variable, then substitute back to find the actual cost. Double-check by calculating both options with your answer.

Question 14

Given the system of equations below, what is the value of the expression $x + y$ ? $3x - 2y = 19$ $2x + y = 8$

5
3 (correct answer)
-2
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

A chemist needs to create 100 mL of a 28% acid solution by mixing a 20% acid solution and a 40% acid solution. How many milliliters of the 40% solution are needed?

28 mL
60 mL
50 mL
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:

x + y = 100
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.

Distractor A incorrectly uses the total milliliters of acid as the answer. Distractor B is the value of x (the amount of the 20% solution), not y. Distractor C represents a common but incorrect guess of using a 50/50 mix, which would result in a 30% solution, not a 28% solution.

Question 16

The perimeter of a rectangular garden is 54 meters. The length of the garden is 3 meters more than twice its width. What is the area of the garden, in square meters?

152 (correct answer)
54
180
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.

Perimeter: 2L + 2W = 54, which simplifies to L + W = 27.
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.

Distractor B incorrectly uses the perimeter value for the area. Distractor C is the area if the relationship was misread as 'length is 3 more than the width' (L=W+3), which gives W=12, L=15, and Area=180. Distractor D is the area if a sign error was made and the relationship was L = 2W - 3, which gives W=10, L=17, and Area=170.

Question 17

A system of two linear equations has a single solution at the point (3, -1). One of the equations is $y = 2x - 7$ . The other equation is $ax + 4y = 11$ . What is the value of $a$ ?

3
-1
5 (correct answer)
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

Line A passes through the points (1, 5) and (3, 1). Line B passes through the points (0, 4) and (6, 1). What is the y-coordinate of the point where Line A and Line B intersect?

1
2
3 (correct answer)
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

What is the x-coordinate of the solution to the system of equations shown below? $\frac{1}{2}x + \frac{1}{3}y = 7$ $\frac{1}{5}x - \frac{1}{2}y = -1$

6
10 (correct answer)
19
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:

3x + 2y = 42
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.

Distractor A is the value of the y-coordinate (y=6). Distractor C is the coefficient of x before the final division (19x=190). Distractor D could result from various calculation errors, such as a sign error during elimination.