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ACT Math : Equations / Inequalities

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Example Questions

Example Question #355 : Algebra

Jen and Karen are travelling for the weekend. They both leave from Jen's house and meet at their destination 250 miles away. Jen drives 45mph the whole way. Karen drives 60mph but leaves a half hour after Jen. How long does it take for Karen to catch up with Jen?

Possible Answers:

$\dpi{100} \small 1 \ hour$

$\dpi{100} \small 2 \ hours$

She can't catch up.

$\dpi{100} \small 3 \ hours$

$\dpi{100} \small 1.5 \ hours$

Correct answer:

$\dpi{100} \small 1.5 \ hours$

Explanation:

For this type of problem, we use the formula:

$\dpi{100} \small distance = rate\times time$

When Karen catches up with Jen, their distances are equivalent. Thus,

$\dpi{100} \small rate (Jen) \times time (Jen)=rate(Karen)\times time(Karen)$

We then make a variable for Jen's time, $\dpi{100} \small t$ . Thus we know that Karen's time is $\dpi{100} \small t-.5$ (since we are working in hours).

Thus,

45(t+0.5)=60t

45t+22.5=60t

22.5=15t

$t=1.5\ hours$

There's a logical shortcut you can use on "catching up" distance/rate problems. The difference between the faster (Karen at 60mph) and slower (Jen at 45mph) drivers is 15mph. Which means that every one hour, the faster driver, Karen, gains 15 miles on Jen. We know that Jen gets a 1/2 hour head start, which at 45mph means that she's 22.5 miles ahead when Karen gets started. So we can calculate the number of hours (H) of the 15mph of Karen's "catchup speed" (the difference between their speeds) it will take to make up the 22.5 mile gap:

15H = 22.5

So H = 1.5.

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Example Question #356 : Algebra

Bill and Bob are working to build toys. Bill can build $\dpi{100} \small k$ toys in 6 hours. Bob can build $\dpi{100} \small k$ toys in 3 hours. How long would it take Bob and Bill to build $\dpi{100} \small 4k$ toys working together?

Possible Answers:

$\dpi{100} \small 9\ hours$

$\dpi{100} \small 12\ hours$

$\dpi{100} \small 4\ hours$

$\dpi{100} \small 2\ hours$

$\dpi{100} \small 8\ hours$

Correct answer:

$\dpi{100} \small 8\ hours$

Explanation:

Bill builds $\dpi{100} \small \frac{k}{6}$ toys an hour. Bob builds $\dpi{100} \small \frac{k}{3}$ toys an hour. Together, their rate of building is $\dpi{100} \small \frac{k}{6}+\frac{k}{3}=\frac{k}{2}$ . Together they can build $\dpi{100} \small k$ toys in 2 hours. They would be able to build $\dpi{100} \small 4k$ toys in 8 hours.

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Example Question #81 : How To Find The Solution To An Equation

A hybrid car gets 40 miles per gallon. Gasoline costs $3.52 per gallon. What is the cost of the gasoline needed for the car to travel 120 miles?

Possible Answers:

$\dpi{100} \small \$ 10.56$

$\dpi{100} \small \$ 10.36$

$\dpi{100} \small \$ 14.08$

$\dpi{100} \small \$ 12.53$

$\dpi{100} \small \$ 9.54$

Correct answer:

$\dpi{100} \small \$ 10.56$

Explanation:

The car will be using $\dpi{100} \small \frac{120\ miles}{40\ mpg}=3\ gallons$ of gas during this trip. Thus, the total cost would be $\dpi{100} \small 3\times \$ 3.52=\$ 10.56$ .

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Example Question #81 : Equations / Inequalities

Jon invested part of $16,000 at 3% and the rest at 5% for a total return of $680.

Quantity A: The amount Jon invested at 5% interest

Quantity B: The amount Jon invested at 3% interest

Possible Answers:

Quantity A is greater

Quantity B is greater

The two quantities are equal

The relationship cannot be determined from the information given

Correct answer:

Quantity A is greater

Explanation:

First, let $\dpi{100} \small x$ represent the invested amount at 3% and set up an equation like this:

$\dpi{100} \small .03x+.05(16,000-x)=680$

Solve for $\dpi{100} \small x$ , and you'll find that Jon invested $6,000 at 3% and $10,000 at 5%.

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Example Question #52 : Linear / Rational / Variable Equations

Audrey, Penelope and Clementine are all sisters. Penelope is 8 years older than Clementine and 2 years younger than Audrey. If the sum of Penelope and Clementine's age is Audrey's age, how old is Clementine's age?

Possible Answers:

Correct answer:

Explanation:

Let = Audrey's age, = Penelope's age, and = Clementine's age.

Since A=P+C , then C=A-P .

Furthermore, P=C+8 , and A=P+2 .

Through substitution, C = [(C + 8) + 2] - (C + 8), C = 2 .

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Example Question #82 : How To Find The Solution To An Equation

If $\dpi{100} \small 3x+y =13$ and $\dpi{100} \small x-2y=-12$ , what is the value of $\dpi{100} \small x$ ?

Possible Answers:

$\dpi{100} \small \frac{1}{3}$

$\dpi{100} \small 2$

$\dpi{100} \small 3$

$\dpi{100} \small \frac{2}{3}$

$\dpi{100} \small 1$

Correct answer:

$\dpi{100} \small 2$

Explanation:

We could use the substitution or elimination method to solve the system of equations. Here we will use the elimination method.

To solve for $\dpi{100} \small x$ , combine the equations in a way that makes the $\dpi{100} \small y$ terms drop out. The first equation has $\dpi{100} \small y$ and the second $\dpi{100} \small -2y$ , so multiplying the first equation times 2 then adding the equations will eliminate the $\dpi{100} \small y$ terms.

Multiplying the first equation times 2: $\dpi{100} \small 2(3x+y = 13)\rightarrow 6x+2y = 26$

Adding this result to the second equation: $\dpi{100} \small 6x+x + 2y - 2y = 26-12\rightarrow 7x=14$

Isolate $\dpi{100} \small x$ by dividing both sides by 7:

$\dpi{100} \small \frac{7x}{7}=\frac{14}{7}$

$\dpi{100} \small x=2$

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Example Question #81 : Equations / Inequalities

If $\frac{1}{4}x-\frac{1}{6}y=\frac{1}{6}$ and $\frac{y}{z}=\frac{1}{2}$ , then what is the value of 3x-z ?

Possible Answers:

Correct answer:

Explanation:

Since the expression we want just involves z and x, but not y, we start by solving $\frac{y}{z}=\frac{1}{2}$ for y $(y=\frac{1}{2}z)$ .

Then we can plug that expression in for y in the first equation.

$\frac{1}{4}x-\frac{1}{6}y=\frac{1}{6}$

$\frac{1}{4}x-\frac{1}{6}(\frac{1}{2}z)=\frac{1}{6}$

$\frac{1}{4}x-\frac{1}{12}z=\frac{1}{6}$

Multiply everything by 12 to get rid of fractions.

$\small 3x - z = 2$

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Example Question #83 : How To Find The Solution To An Equation

If 5(3x+y)=15 , what is in terms of ?

Possible Answers:

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

$x=10+\frac{y}{3}$

$x=\frac{10-y}{3}$

$x=\frac{15+5y}{3}$

x=3-3y

Correct answer:

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

Explanation:

Use inverse operations to isolate x. Working from the outermost part on the left side, we first divide both sides by 5.

3x+y=3

To isolate the x term, subtract y from both sides.

3x=3-y

Finally, isolating just x, divide both sides by 3.

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

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Example Question #82 : Equations / Inequalities

If 8s - 6k = 4s - 2k , then, in terms of , k=?

Possible Answers:

Cannot be determined

Correct answer:

Explanation:

You can solve this problem by plugging in random values or by simply solving for k. To solve for k, put the s values on one side and the k values on the other side of the equation. First, subtract 4s from both sides. This gives 4s – 6k = –2k. Next, add 6k to both sides. This leaves you with 4s = 4k, which simplifies to s=k. The answer is therefore s.

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Example Question #85 : How To Find The Solution To An Equation

The sum of two consecutive odd integers is 32. What is the value of the next consecutive odd integer?

Possible Answers:

Cannot be determined

Correct answer:

Explanation:

Let be the smallest of the two consecutive odd integers. Thus,

n+(n+2)=32 and it follows that n=15 .

We have that 15 and 17 are the consecutive odd integers whose sum is 32, so the next odd integer is 19.

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ACT Math : Equations / Inequalities

Study concepts, example questions & explanations for ACT Math

All ACT Math Resources

Example Questions

Example Question #355 : Algebra

Example Question #356 : Algebra

Example Question #81 : How To Find The Solution To An Equation

Example Question #81 : Equations / Inequalities

Example Question #52 : Linear / Rational / Variable Equations

Example Question #82 : How To Find The Solution To An Equation

Example Question #81 : Equations / Inequalities

Example Question #83 : How To Find The Solution To An Equation

Example Question #82 : Equations / Inequalities

Example Question #85 : How To Find The Solution To An Equation

All ACT Math Resources

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