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High School Math : Basic Single-Variable Algebra

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Example Questions

Example Question #1 : Equations

Tom is painting a fence 100 feet long. He starts at the West end of the fence and paints at a rate of feet per hour. After hours, Huck joins Tom and begins painting from the East end of the fence at a rate of feet per hour. After hours of the two boys painting at the same time, Tom leaves Huck to finish the job by himself.

If Huck completes painting the entire fence after Tom leaves, how many more hours will Huck work than Tom?

Possible Answers:

$5\ hours$

$3\ hours$

$4\ hours$

$6\ hours$

$10\ hours$

Correct answer:

$6\ hours$

Explanation:

Tom paints for a total of hours (2 on his own, 2 with Huck's help). Since he paints at a rate of feet per hour, use the formula

$distance = rate \times time$ (or d = rt )

to determine the total length of the fence Tom paints.

d = (5)(4)

d = 20 feet

Subtracting this from the total length of the fence 100 feet gives the length of the fence Tom will NOT paint: 100 - 20 = 80 feet. If Huck finishes the job, he will paint that feet of the fence. Using d = rt , we can determine how long this will take Huck to do:

80 = 8(t)

t = 10 hours.

If Huck works hours and Tom works hours, he works more hours than Tom.

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Example Question #1 : Equations

Simplify the fraction to the lowest terms:

$\frac{924}{1092}$

Possible Answers:

$\frac{44}{52}$

$\frac{308}{364}$

Cannot be simplified

$\frac{11}{13}$

Correct answer:

$\frac{11}{13}$

Explanation:

Find the common multiple between the numerator and denominator.

$\frac{924}{1092}$

divide numerator and denominator by 3:

$\frac{308}{364}$

divide numerator and denominator by 7:

$\frac{44}{52}$

divide numerator and denominator by 4:

$\frac{11}{13}$

Cannot be divided any more- lowest terms.

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Example Question #1 : Equations

Solve the following equation for x in terms of the other variables:

$\frac{ax}{b-ax}=2$

Possible Answers:

$x=\frac{3b}{2a}$

$x=\frac{2b}{3a}$

3ax=2b

$x=\frac{2a}{3b}$

$x=\frac{3a}{2b}$

Correct answer:

$x=\frac{2b}{3a}$

Explanation:

$\frac{ax}{b-ax}=2$

Multiply both sides by (b-ax) to get:

ax=2(b-ax)

Distribute the :

ax=2b-2ax

Combine like terms:

3ax=2b

Divide both sides by :

$x=\frac{2b}{3a}$

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Example Question #2 : Equations

Solve the following equation for x in terms of the other variables:

bcx=a

Possible Answers:

$x=\frac{b}{c}$

$x=\frac{a}{bc}$

$x=\frac{bc}{a}$

$x=\frac{c}{ab}$

$x=\frac{ab}{c}$

Correct answer:

$x=\frac{a}{bc}$

Explanation:

bcx=a

Divide both sides by :

$x=\frac{a}{bc}$

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Example Question #3 : Equations

If given the equation 8x + 2x -5 , with a positive integer, the result must be an integer multiple of:

Possible Answers:

Correct answer:

Explanation:

The mathematical expression given in the question is 8x + 2x -5 . Adding together like terms, 8x + 2x , this can be simplified to 10x -5 . The expression 10x -5 can be factored as 5(2x -1) . For every positive integer , must be a multiple of 5. If x=1 , then 5(2x - 1) = 5 , which is not an integer multiple of 2, 8, 10, or 15. Therefore, the correct answer is 5.

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Example Question #1 : Equations

Cindy's Cotton Candy sells cotton candy by the bag. Her monthly fixed costs are $\$150$ . It costs $\$2.50$ to make each bag and she sells them for $\$4.00$ .

What is the monthly break-even point?

Possible Answers:

$60\; bags$

$225\; bags$

$80\; bags$

$150\; bags$

$100\; bags$

Correct answer:

$100\; bags$

Explanation:

Costs=150+2.50x

Revenues = 4.00x

The break-even point occurs when the costs=revenues .

The equation to solve becomes

4x=150+2.5x so the break-even point is $100 \; bags$ .

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Example Question #2 : Equations

Cindy's Cotton Candy sells cotton candy by the bag. Her monthly fixed costs are $\$150$ . It costs $\$2.50$ to make each bag and she sells them for $\$4.00$ .

To make a profit of $\$150$ , how many bags of cotton candy must be sold?

Possible Answers:

$250\; bags$

$200\; bags$

$100\; bags$

$150\; bags$

$300\; bags$

Correct answer:

$200\; bags$

Explanation:

Costs=150+2.50x

Revenues = 4.00x

Profits = Revenues - Costs = 4x - (150+2.5x)=1.5x -150

So the equation to solve becomes 150 = 1.5x - 150 , or $x=200\; bags$ must be sold to make a profit of $\$150$ .

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Example Question #4 : Equations

Solve for and to satisfy both equations in the system:

$\begin{Bmatrix} 6x+y = 25 \\ 2x-3y = 25 \end{Bmatrix}$

Possible Answers:

x=5 , y = 5

x = 5 , y= -5

x = 3 , y = 7

x= -5 , y = -5

x = -5 , y = 5

Correct answer:

x = 5 , y= -5

Explanation:

The two equations in this system can be combined by addition or subtraction to solve for and . Isolate the variable to solve for it by multiplying the top equation by so that when the equations are combined the term disappears.

$3\times(6x + y = 25)$

$\rightarrow 18x+3y = 75$

$\rightarrow (18x +3y = 75) + (2x - 3y = 25)$

$\rightarrow 20x = 100$

Divide both sides by to find as the value for .

Substituting for in both of the two equations in the system and solving for gives a value of for .

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Example Question #5 : Equations

Solve for :

$\left | 12x - 39 \right | = 60$

Possible Answers:

$x = 8 \frac{1}{4}$

$x = -1 \frac{3}{4} \textrm{ or }x = 8 \frac{1}{4}$

$x = 1 \frac{3}{4} \textrm{ or }x = 8 \frac{1}{4}$

$x = - 8 \frac{1}{4} \textrm{ or }x = 8 \frac{1}{4}$

$x = 1 \frac{3}{4}$

Correct answer:

$x = -1 \frac{3}{4} \textrm{ or }x = 8 \frac{1}{4}$

Explanation:

Rewrite $\left | 12x - 39 \right | = 60$ as a compound statement and solve each part separately:

$\left | 12x - 39 \right | = 60$

$12x - 39 = - 60 \textrm{ or }12x - 39 = 60$

12x - 39 = - 60

12x - 39 + 39 = - 60+ 39

12x = - 21

$12x \div 12 = - 21 \div 12$

$x = -1 \frac{3}{4}$

12x - 39 = 60

12x - 39 + 39 = 60+ 39

12x = 99

$12x \div 12 = 99 \div 12$

$x = 8 \frac{1}{4}$

The solution set is $\left \{ -1 \frac{3}{4} ,8 \frac{1}{4}\right \}$

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Example Question #1 : Equations

a^2-b^2 = 33 and a-b=3 . What is the value of $a^{2}+b^{2}$ ?

Possible Answers:

130

112

Correct answer:

Explanation:

First, notice that we can factor a^2-b^2 into the form (a-b)(a+b). We are told that a-b=3, so we can substitute that into the first equation.

(a-b)(a+b)=33

3(a+b)=33

If we divide both sides by 3, we can obtain the value of a+b.

a+b=11

We now have a system of equations: a-b = 3, a+b = 11. We will solve this system by elimination. If we add the two equations together, we obtain the following:

2a=14 .

Divide both sides by 2.

a=7.

Going back to the equation a-b = 7, we can solve for b.

7-b=3

Add b to both sides.

7=3+b

Subtract 3 from both sides.

b=4.

Ultimately, the question asks us to determine the value of a^2+b^2 .

a^2+b^2 = $7^{2}+4^{2}=49+16=65$ .

The answer is 65.

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High School Math : Basic Single-Variable Algebra

Study concepts, example questions & explanations for High School Math

All High School Math Resources

Example Questions

Example Question #1 : Equations

Example Question #1 : Equations

Example Question #1 : Equations

Example Question #2 : Equations

Example Question #3 : Equations

Example Question #1 : Equations

Example Question #2 : Equations

Example Question #4 : Equations

Example Question #5 : Equations

Example Question #1 : Equations

All High School Math Resources

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