All GED Math Resources
Example Questions
Example Question #11 : Word Problems In Algebra
The French club wants to make and sell some pizzas for a fundraiser. It will cost $250 to rent the equipment to make the pizzas and $2 worth of ingredients to make each pizza. The pizzas will be sold for $4.50 apiece.
How many pizzas must be made and sold for the French club to make a profit of at least $500?
Let be the number of pizzas made and sold. Each pizza will require $2 worth of ingredients, so the ingredients in total will cost . Add this to the cost to rent the equipment and the cost will be .
The pizzas will cost $4.50 each, so the money raised will be .
The profit will be the difference between the revenue and the cost:
The French club wants a profit of at least $500, so we set up and solve the inequality:
At least 300 pizzas must be made and sold.
Example Question #11 : Word Problems In Algebra
Jeff, the barista at Moonbucks Coffee, is having a problem. He needs to make fifty pounds of Premium Blend coffee by mixing together some Kona beans, which cost $24 per pound, with some Ethiopian Delight beans, which cost $10 per pound. The Premium Blend coffee will cost $14.20 per pound. Also, the coffee will sell for the same price mixed as it would separately.
Let be the number of pounds of Kona beans and be the number of pounds of Ethiopian Delight beans. Which of the following systems of equations could you set up to solve this problem?
The number of pounds of coffee beans totals 50, so one of the equations would be
.
The total price of the Kona beans is the unit price, $24 per pound, multiplied by the quantity, pounds. This is dollars. Similarly, the total price of the Ethiopian delight beans is dollars, and the price of the mixture is dollars. Add the prices of the Kona and Ethiopian Delight beans to get the price of the mixture:
These are the equations of the system.
Example Question #691 : Ged Math
Leslie has some nickels, some dimes, and some quarters. She has twice as many dimes as she has nickels, and she has four more quarters than she has dimes. If she has dimes, how much does she have, in terms of , in nickels, dimes, and quarters?
dollars
dollars
dollars
dollars
dollars
Since Leslie has twice as many dimes as nickels, the number of nickels she has is half the number of dimes, or half of . This means she has nickels. Also, since she has four more quarters than dimes, she has quarters.
She has
in nickels,
in dimes, and
in quarters.
In total, the number of dollars Leslie has is
Leslie has dollars.
Example Question #141 : Algebra
John had 48 candies. He ate 2, and said he gave almost 25% of the remaining to charity. What is the most reasonable number of candies he had left?
After John has ate 2 candies out of his total of 48 candies, he has 46 candies left.
25% of 46 is equivalent to:
However, candies are counted in units, which means if John could have only given 11 candies to charity if he said he has given almost 25% away. The trick answer is 34 because 11.5 candies cannot be rounded to 12 candies.
Hence:
Example Question #142 : Algebra
Suppose a customer paid $500 for a new phone. The store had applied a 20% discount, and the tax after the discount was 8.25%. What was the approximate price of the phone before the applied discount and tax rate?
$573.37
$592.37
$580.37
$583.37
$577.37
$577.37
Let x be the cost of the phone before applying the 20% discount. After applying the discount, the value will be equal to some amount y before applying tax.
After amount y has been taxed the 8.25%, the new value will be the price of the phone, which is $500. The equation representing this relationship is:
We have a system of equations. Substitute y in terms of x into the 2nd equation. Solving the value of x will give the original value of the phone.
Therefore, the price of the phone is approximately $577.37.
Example Question #11 : Word Problems In Algebra
Mary spent $48 for shoes. This was $14 less than twice what she spent for a shirt. How much was the shirt?
Every word problem has an unknown number. In this problem, the unknown number or value is the price of the shirt. The variable will represent the unknown number.
In this problem, will represent how much Mary spent for the shirt. The problem states that $48 was $14 less than 2 times what she spent on for the shirt. An equation will be written in order to solve this problem.
Then solve for .
The first step is to remove the subtraction of $14 by using the inverse operation, which would be addition. Remember, because it is an equation, whatever operation you choose to perform on one side of the equation must also be done to the other side of the equation. Therefore, you will add $14 to both sides of this equation.
In order to solve for , you will need to divide both sides by the coefficient, (the number next to the variable), which is 2.
Example Question #12 : Word Problems In Algebra
Pat makes fifteen birdhouses every month, while John makes twenty. If the two men work together to make birdhouses, how long will it take them to make of them? (Presume that they make the houses separately and do not interfere with each other's work.)
months
months
Cannot be computed from the data given
months
months
months
This kind of question is really just a basic rate problem. What it is asking is "How many months at rate does it take to make birdhouses?" The rate is just the combination of the two men's rates:
Thus, you can set up the equation:
Thus,
Example Question #13 : Word Problems In Algebra
A tank leaks at a rate of ml per second. How many hours will it take for the tank to drain, given that it is liters? Round to the nearest hour.
To solve this equation, first get everything into the same units. Recall that there are ml for every one liter. Thus, liters are equal to ml. Now, what you basically have is a rate equation. Recall that the total work is represented as follows:
For your data, the total "work" is the ml. Start by calculating the time as a decimal:
This gives you:
However, remember that this is in seconds. You need to convert back to hours. You do this by dividing by , giving you . Rounding up, this gives you hours.
Example Question #12 : Word Problems In Algebra
A writer make dollars of profit per book sold. The printing run of a lot of books cost dollars. How many books must the author sell before the book becomes profitable?
This question could be set up like an equation as follows:
The variable indicates the number of books that would have to be sold. This comes out to:
Now, don't be tricked! You can't sell partial books. Thus, you will need to sell books in order to turn a profit. If you only sell , you will not have sold enough to make a profit.
Example Question #13 : Word Problems In Algebra
An author writes a book that sells for dollars. He has published at a cost of dollars per book. How many books must he sell before his profit is at least dollars per book?
books
books
books
books
books
books
This question is a bit hard. You need to think it out step by step. First, you could write an equation like this:
This represents the idea of trying to calculate when the profit per book will be three dollars. Now, we know that profit is equal to:
Thus, you can rewrite your equation:
Now, the original cost is calculated by multiplying by . This is the same as . The sales amount is just , where indicates the total number of books sold. This will also be the total for in your equation. Thus, you can write out the following equation:
Now, just solve for :
However, you will need to sell one more book than . (That would be just a little too insufficient.) Thus, the answer is .