All PSAT Math Resources
Example Questions
Example Question #85 : How To Find The Solution To An Equation
Pets Plus makes bird houses. Their monthly fixed expenses are $750. The cost for each bird house is $15. The bird houses sell for $40.
What is the monthly break-even point at Pets Plus?
Let the number of bird houses sold each month.
The break-even point is where the revenue is the same as the costs:
Solve for :
Therefore, Pets Plus must sell 30 bird houses to break-even.
Example Question #91 : How To Find The Solution To An Equation
Pets Plus makes bird houses. Their monthly fixed expenses are $750. The cost for each bird house is $15. The bird houses sell for $40.
If Pets Plus sells 50 bird houses, what is the profit?
Let = the number of birdhouses sold each month.
Substituting in 50 for gives an answer of 500, so the profit on 50 birdhouses is $500.
Example Question #52 : Algebra
George is three times older than Joey. The sum of their ages is 16. What is the product of their ages?
Let = Joey's age and = George's age.
Then the equation to solve becomes .
Therefore, Joey is 4 years old and George is 12 years old, so the product of their ages is 48.
Example Question #53 : Algebra
Three consecutive even numbers add to 42. What is the middle number?
Let = 1st even number, = 2nd even number, and = 3rd even number.
Then the equation to solve becomes .
Thus , so the middle number is 14.
Example Question #51 : Equations / Inequalities
Consider the following equation:
Which of the following must be true?
The quantity inside the absolute value brackets must equal either or , depending on whether the quantity inside the brackets is positive or negative. We therefore have two seperate equations:
To solve the first equation, add 9 to both sides:
Subtract from both sides:
This is the first solution. Now let's look at the second equation. The distributive law gives us:
Add 9 to both sides:
Add to both sides:
Divide both sides by 3:
Therefore, is either 4 or 6.
Statement does NOT have to be true because can also equal 4.
Statement must be true because both 4 and 6 are positive .
Finally, statement always holds because 4 and 6 are both even.
Example Question #141 : Equations / Inequalities
What property of arithmetic is demonstrated here?
Commutative
Transitive
Symmetric
Identity
Reflexive
Reflexive
The statement expresses the idea that any number is equal to itself. This is the reflexive property of equality.
Example Question #5 : How To Find The Solution To A Quadratic Equation
If f(x) = -x2 + 6x - 5, then which could be the value of a if f(a) = f(1.5)?
We need to input 1.5 into our function, then we need to input "a" into our function and set these results equal.
f(a) = f(1.5)
f(a) = -(1.5)2 +6(1.5) -5
f(a) = -2.25 + 9 - 5
f(a) = 1.75
-a2 + 6a -5 = 1.75
Multiply both sides by 4, so that we can work with only whole numbers coefficients.
-4a2 + 24a - 20 = 7
Subtract 7 from both sides.
-4a2 + 24a - 27 = 0
Multiply both sides by negative one, just to make more positive coefficients, which are usually easier to work with.
4a2 - 24a + 27 = 0
In order to factor this, we need to mutiply the outer coefficients, which gives us 4(27) = 108. We need to think of two numbers that multiply to give us 108, but add to give us -24. These two numbers are -6 and -18. Now we rewrite the equation as:
4a2 - 6a -18a + 27 = 0
We can now group the first two terms and the last two terms, and then we can factor.
(4a2 - 6a )+(-18a + 27) = 0
2a(2a-3) + -9(2a - 3) = 0
(2a-9)(2a-3) = 0
This means that 2a - 9 =0, or 2a - 3 = 0.
2a - 9 = 0
2a = 9
a = 9/2 = 4.5
2a - 3 = 0
a = 3/2 = 1.5
So a can be either 1.5 or 4.5.
The only answer choice available that could be a is 4.5.
Example Question #162 : Gre Quantitative Reasoning
Solve for x: 2(x + 1)2 – 5 = 27
3 or 4
3 or –5
–2 or 5
–3 or 2
–2 or 4
3 or –5
Quadratic equations generally have two answers. We add 5 to both sides and then divide by 2 to get the quadratic expression on one side of the equation: (x + 1)2 = 16. By taking the square root of both sides we get x + 1 = –4 or x + 1 = 4. Then we subtract 1 from both sides to get x = –5 or x = 3.
Example Question #1 : How To Find The Solution To A Quadratic Equation
Two consecutive positive multiples of three have a product of 54. What is the sum of the two numbers?
12
3
15
9
6
15
Define the variables to be x = first multiple of three and x + 3 = the next consecutive multiple of 3.
Knowing the product of these two numbers is 54 we get the equation x(x + 3) = 54. To solve this quadratic equation we need to multiply it out and set it to zero then factor it. So x2 + 3x – 54 = 0 becomes (x + 9)(x – 6) = 0. Solving for x we get x = –9 or x = 6 and only the positive number is correct. So the two numbers are 6 and 9 and their sum is 15.
Example Question #8 : How To Find The Solution To A Quadratic Equation
Solve 3x2 + 10x = –3
x = –1/3 or –3
x = –1/9 or –9
x = –2/3 or –2
x = –4/3 or –1
x = –1/6 or –6
x = –1/3 or –3
Generally, quadratic equations have two answers.
First, the equations must be put in standard form: 3x2 + 10x + 3 = 0
Second, try to factor the quadratic; however, if that is not possible use the quadratic formula.
Third, check the answer by plugging the answers back into the original equation.
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