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ACT Math : Linear / Rational / Variable Equations

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Example Questions

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Example Question #11 : How To Find The Solution To An Equation

If the diameter of a car wheel is 1 foot, how many miles has the car gone after 3500 revolutions?

Possible Answers:

$13.5$

$50.1$

$2.1$

$3.5$

$10.2$

Correct answer:

$2.1$

Explanation:

The circumference of the tire is therefore pi, meaning that in 3500 revolutions the car will go 10,995.6 feet. Dividing by 5280 feet per mile yields 2.1 miles.

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Example Question #11 : Linear / Rational / Variable Equations

A chemistry student needs to dilute 1 L of 75%-acid solution. How much water should be added to yield a 25%-acid solution?

Possible Answers:

1 L

3 L

1.75 L

2 L

2.5 L

Correct answer:

2 L

Explanation:

$V_{1}P_{1} + V_{2}P_{2} = V_{f}P_{f}$ where = volume and = percent

Pure water = 0% and pure acid = 100%

So the equation to solve becomes

x(0) + 1 (0.75) = (x + 1)(0.25)

0.75 = 0.25 + 0.25x

So 2 L of pure water should be added to 1 L 75%-acid solution to yield 25%-acid solution.

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Example Question #11 : How To Find The Solution To An Equation

The Teddy Bear Parade makes teddy bears. Their monthly fixed costs are $550. It costs $25 to make each bear and they sell for $50 each.

What is the monthly break-even point?

Possible Answers:

Correct answer:

Explanation:

The break-even point occurs when the costs are the same as the revenue.

Let = number of teddy bears sold

Revenue: 50x

Costs: 25x + 550

So when Costs = Revenue we have the break-even point thus the equation to solve becomes 50x = 25x + 550 or 25 x = 550 .

So to break-even, 22 teddy bears must be sold.

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Example Question #11 : Linear / Rational / Variable Equations

The Teddy Bear Parade makes teddy bears. Their monthly fixed costs are $550. It costs $25 to make each bear and they sell for $50 each.

To make a profit of $750, how many teddy bears must be sold?

Possible Answers:

Correct answer:

Explanation:

Profits = Revenue - Costs

Let = number of teddy bears sold.

Revenue: 50x

Costs: 25x + 550

Profits = 50x - (25x + 550) or 25x -550

So the equation to solve becomes 750 = 25x - 550

So to make a profit of $750, 52 teddy bears must be sold.

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Example Question #1784 : Act Math

How much pure acid should be added to 1 L of 50%-acid solution to yield a 75%- acid solution?

Possible Answers:

2.50 L

1 L

0.50 L

1.50 L

0.75 L

Correct answer:

1 L

Explanation:

Pure water is 0% and pure acid is 100%.

$V_{1}P_{1} +V_{2}P_{2} = V_{f}P_{f}$ where = volume and = percent.

Let = amount of pure acid to add.

So the equation to solve becomes

(1)(0.50) + (x)(1.00) = (1 + x)(0.75)

Thus 1 L of 50%-acid solution added to 1 L of pure acid solution yields 2 L of 75%-acid solution.

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Example Question #1785 : Act Math

Charlie's Cookies sells desserts. Their monthly fixed costs are $350. It costs $10 to make a dozen cookies. They sell each dozen for $15.

How many cookies must be sold each month to break-even?

Possible Answers:

$700 \; cookies$

$720 \; cookies$

$880 \; cookies$

$960 \; cookies$

$840 \; cookies$

Correct answer:

$840 \; cookies$

Explanation:

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

Let = number of dozens of cookies sold.

Costs : 10x + 350

Revenues : 15x

Costs = Revenues to break-even.

So the equation to solve becomes 10x + 350 = 15x

350 = 5x

So 70 dozen cookies must be sold to break-even, thus $70\cdot 12 = 840$ cookies.

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Example Question #1786 : Act Math

Charlie's Cookies sells desserts. Their monthly fixed costs are $350. It costs $10 to make a dozen cookies. They sell each dozen for $15.

The company wants to make a profit of $500. How many dozens of cookies must be sold to reach their profit goal?

Possible Answers:

$125 \; dozen$

$170 \; dozen$

$200 \; dozen$

$145 \; dozen$

$70 \; dozen$

Correct answer:

$170 \; dozen$

Explanation:

Profit = Revenue - Costs

Let = number of dozens of cookies sold.

Profit = 15x - (10x + 350) = 5x - 350

So the equation to solve becomes 500 = 5x - 350 .

So the company must sell 170 dozen cookies to make a $500 profit.

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Example Question #1787 : Act Math

Mike and Joey live 215 miles apart. They meet at a campground between them. Mike drove 50 mph and Joey drove 60 mph and left one hour after Mike. How far did Mike travel?

Possible Answers:

$90 \; miles$

$150 \; miles$

$125 \; miles$

$100 \; miles$

$85 \; miles$

Correct answer:

$125 \; miles$

Explanation:

The distance Mike traveled plus the distance Joey traveled is the total distance between them.

$Distance = Rate \cdot Time$

Let = hours traveled

So the equation to sovle becomes 50t + 60(t - 1) = 215

Thus t = 2.5 hours traveled

So Mike traveled $50 \; mph \cdot 2.5 \; hr = 125\; miles$ .

And Joey traveled $60 \; mph \cdot 1.5 \; hr = 90 \; miles$ .

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Example Question #12 : How To Find The Solution To An Equation

What is the value of when 7 + 3x = 6x - 5 ?

Possible Answers:

$\frac{2}{3}$

$\frac{1}{4}$

$\frac{3}{2}$

Correct answer:

Explanation:

Simply solve for by adding five to both sides and then dividing both sides by three to get by itself.

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Example Question #1781 : Act Math

By how much is the value of 5x^2 - y greater than the value of 4x^2 - 2y when x=2 and y=6 ?

Possible Answers:

Correct answer:

Explanation:

Input and into both equations. Solve them and then find the difference between the two answers.

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ACT Math : Linear / Rational / Variable Equations

Study concepts, example questions & explanations for ACT Math

All ACT Math Resources

Example Questions

Example Question #11 : How To Find The Solution To An Equation

Example Question #11 : Linear / Rational / Variable Equations

Example Question #11 : How To Find The Solution To An Equation

Example Question #11 : Linear / Rational / Variable Equations

Example Question #1784 : Act Math

Example Question #1785 : Act Math

Example Question #1786 : Act Math

Example Question #1787 : Act Math

Example Question #12 : How To Find The Solution To An Equation

Example Question #1781 : Act Math

All ACT Math Resources

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