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GED Math : Single-Variable Algebra

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Example Questions

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Example Question #716 : Ged Math

After a $18\%$ discount, a pair of shoes costs $\$64.58$ . What was the original price of the shoes?

Possible Answers:

$\$78.76$

$\$66.09$

$\$91.27$

$\$83.66$

Correct answer:

$\$78.76$

Explanation:

Let be the original price of the shoes. Since the shoes were $18\%$ off, we can write the following equation:

x-0.18x=64.58

Now, solve for .

0.72x=64.58

x=78.76

The shoes originally cost $\$78.76$ .

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Example Question #717 : Ged Math

Joseph is responsible for planning his company's holiday party. He has been given a budget of $\$4550$ for renting a venue and providing dinner for all the attendees. The cost for renting a venue is $\$1050$ . He expects employees to show up to the party. Which inequality shows how to find the amount, , that Joseph can spend on dinner for each person?

Possible Answers:

$47d+1050\leq4550$

47d+1050<4550

$47+1050d\leq4550$

$47+1050d\geq4550$

Correct answer:

$47d+1050\leq4550$

Explanation:

Since Joseph has a budget of $\$4550$ , that means he cannot spend above that amount. This translates to the use of the less than or equal sign, $\leq$ , in the inequality.

The total cost for the dinner for all the attendees can be found by multiplying the number of attendees by the cost of each dinner, .

The rental of the venue is a one-time payment.

Thus, we can write the following inequality:

$47d+1050\leq4550$

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Example Question #718 : Ged Math

A delivery service charges a $\$5$ delivery flat fee then an additional $\$0.55$ for each ounce the package weighs. Which of the following expressions would give the total cost for the delivery of an item weighing ounces?

Possible Answers:

0.55z+5

5z+0.55

0.55z

0.55z-5

Correct answer:

0.55z+5

Explanation:

Since the company charges $\$0.55$ per ounce, an item weighing ounces would then cost 0.55z to ship, based on weight alone. The question also states that there is a flat $\$5$ fee for each delivery.

Thus we can write the following expression to represent the total cost of shipping:

0.55z+5

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Example Question #31 : Word Problems In Algebra

A sweater originally priced at dollars goes on sale for $20\%$ off. After a week, the sweater's price is further reduced $15\%$ , and then it was sold. Which of the following expressions would be the price of the price the sweater sold for?

Possible Answers:

0.68d

0.65d

0.75d

0.55d

Correct answer:

0.68d

Explanation:

Start by finding the price of the sweater after the first price cut. Since it was $20\%$ off, we can write the following:

d-0.2d=0.8d

The sweater cost 0.8d after the first price cut.

Then, find the price when after the second price cut, which was $15\%$ the already discounted price.

0.8d-0.15(0.8)d=0.68d

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Example Question #720 : Ged Math

An ambulance service charges $\$1250$ for the initial call, then $\$40$ per mile it takes to get to the hospital. Which of the following expressions represents the cost of taking an ambulance miles to the hospital?

Possible Answers:

40m+1250

1250m+40

1250+m

40m

Correct answer:

40m+1250

Explanation:

Start by finding out how much it will take to take the ambulance miles. Since it costs $\$40$ per mile, then 40m represents the cost of just the travel. Next, you need to include the fixed cost of $\$1250$ .

Thus, the total cost of the ambulance trip is expressed with 40m+1250 .

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Example Question #31 : Word Problems In Algebra

A number, when multiplied by gives us a result of 200 . Find the number.

Possible Answers:

Correct answer:

Explanation:

Lets say that this number is . If we follow what the question is saying, we come up with the equation:

5*x=200

We will divide both sides by 4 to isolate on the left side.

So,

$\frac {{\color{Red} 5}*x}{{\color{Red} 5}}=\frac {200}{{\color{Red} 5}}$

We simplify:

x=40

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Example Question #32 : Word Problems In Algebra

This winter, it snowed ten more inches than five times the amount of snow the previous winter, during which it snowed one half as much as the winter before that. If two years ago it snowed twenty inches, how much did it snow this winter?

Possible Answers:

$45\;\textup{in}$

$60\;\textup{in}$

$110\;\textup{in}$

$50\;\textup{in}$

$75\;\textup{in}$

Correct answer:

$60\;\textup{in}$

Explanation:

For simplicity, let's call our years , , and , with being the amount of snow it snowed two years ago, being the amount of snow it snowed last year, and being the amount of snow it snowed this year. Now, we know that year is . Based on this, we can say that is . Finally, we know that year year could be written:

C=10+5B

Based on our data, this is:

$C=10+5\cdot10$

C=10+50

$C=60\;\textup{in}$

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Example Question #33 : Word Problems In Algebra

Marge and William are running away from each other in opposite directions. Marge is running at a rate of $1.2\;\frac{m}{s}$ , while William is running at a rate of $0.8\;\frac{m}{s}$ . In how many minutes will they be $12\;\textup{km}$ from each other?

Possible Answers:

$7200\;\textup{min}$

$90\;\textup{min}$

$6000\;\textup{min}$

$3600\;\textup{min}$

$100\;\textup{min}$

Correct answer:

$100\;\textup{min}$

Explanation:

Every second, you know that Marge and William will become a total of $1.2\,\frac{m}{s}+0.8\,\frac{m}{s}$ or $2.0\,\frac{m}{s}$ . Now, you can use the simple work formula for distance:

D=RT or, for our data, 12000=2T

(Remember kilometers is $12\cdot1000=12000$ meters.)

Thus, solving for you get:

$\frac{12000}{2}=\frac{2T}{2}$

T=6000

This is in seconds, though. You need minutes. To convert, you need to divide by :

$6000\;\textup{s}\cdot\frac{1\;\textup{min}}{60\;\textup{s}}$

$\frac{6000\;\textup{s}}{60\;\textup{s}}=100\;\textup{min}$

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GED Math : Single-Variable Algebra

Study concepts, example questions & explanations for GED Math

All GED Math Resources

Example Questions

Example Question #716 : Ged Math

Example Question #717 : Ged Math

Example Question #718 : Ged Math

Example Question #31 : Word Problems In Algebra

Example Question #720 : Ged Math

Example Question #31 : Word Problems In Algebra

Example Question #32 : Word Problems In Algebra

Example Question #33 : Word Problems In Algebra

All GED Math Resources

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