Here are a few notes I wrote for a student of mine who’s taking the SSAT Upper Level on Saturday, December 9. They might also help you study for the SAT, ACT, SSAT, ISEE, or any math test you may be taking.

Slope-Intercept form of an equation:

The slope-intercept form of an equation is y = mx+b, where m is the slope, and b is the y-intercept (the value of y when x is zero). If you are presented with a formula such as 3y + 5x = 5, you have to rearrange the formula so you get y = mx+b. In this case, it would be y = (-5/3)x + 5/3.

If a line is perpendicular to another line, the slope of the perpendicular line is the negative reciprocal of the first line. For example, if the first line has a slope of 2, the perpendicular line has a slope of –½ . Just flip the fraction (or put one over the slope if the slope is an integer), and then multiply it by -1. So the slope of a line perpendicular to a line with slope 3 is -1/3 and the slope of a line perpendicular to a line with slope 33/44 is -44/33. Notice that b, the y-intercept, doesn’t matter when you’re finding the slope of a perpendicular line – all b would affect is WHERE the perpendicular line intersects the first line, not the angle at which it intersects the line.

Imaginary and Complex Numbers:

The imaginary number (or imaginary unit), i, is the square root of negative 1. A “complex number” is a number of the form ax+bi, where a, x, and b are all real numbers, and I, of course, is imaginary. If you have two complex numbers, you simply use FOIL to multiply them, the same way you would multiply any other binomials.

See, for example, this page from the publishers of the “SSAT & ISEE for Dummies” book we’ve been using:

http://www.dummies.com/education/math/algebra/how-to-multiply-binomials-using-the-foil-method/

For an example of how to do the same with complex numbers, see the top of this page:

http://www.montereyinstitute.org/courses/DevelopmentalMath/COURSE_TEXT2_RESOURCE/U16_L4_T2_text_final.html

Remember – the imaginary unit, i, is the square root of -1, so anything times i2 is NEGATIVE that number. For example, 3i * 5i = 15i2, which is -15, and -3i * 5i = -15i2, which is +15. It’s the reverse of when you do the last terms in a normal FOIL.
 
Set Notation :

As we saw in the problem we did tonight, A with a bar over it like this : Ā means the complement of Set A, which means everything in the “universal set” or “universe” of all things in all the sets we are considering that are not in Set A. So if the universal set, U, contains the numbers 1 through 10, inclusive, and A is the subset of U containing 2,4, 6, and 8, Ā is 1,3,5,7,9, and 10. This could also be represented using brackets to hold the set – that is Ā = {1,3,5,7,9, 10}.

You should also know the terms “union” and “intersection”, and their symbols, U and ᴒ (this is the closest I could find to the symbol, which looks like an upside-down capital U).  So {1,2,3} U (4,5,6} stands for the union (joining) of those sets into one set {1,2,3,4,5,6}. The union of two or more sets just makes a bigger set containing all its members.

The intersection of two or more sets includes only the members common to all the sets involved. So {1,2,3} ᴒ {3,5,7} would be the intersection of those two sets. Since they only have one common member, that set is {3}. The intersection of the set of all prime numbers, all even numbers, and all numbers less than 10, would also have only one member, the number 2. Questions could be asked that way, in words rather than set notation, perhaps using sets with infinite numbers of members, such as prime numbers, even numbers, numbers less than 10, etc.

Volume and Area Problems:

Important Area Formulas:

The area of a rectangle is equal to the length times the width.
The area of a square is the side length times itself. That’s why a number multiplied by itself, or taken to the second power, is called that number “squared.”

The area of a circle is , where r is the radius of the circle, and  is a constant that is approximately 3.14.

For more information on area, see the following page:

http://www.mathsisfun.com/area.html

Important Volume Formulas:

You should know the volume of a cube, cylinder, cone,  rectangular solid (i.e., a box), and a sphere. Bonus points if you know the formula for a pyramid and a tetrahedron. Here’s a page with those formulas and more.

http://onlinemschool.com/math/formula/volume/

Surface Area:

You should know that the surface area of a cube is 6(e^2), where e is the edge length. Be prepared to find the surface area of a cube when given the volume – e.g., if the volume of a cube is 27, the edge length is the “cube root” (remember, the 3rd power of any number is that number “cubed,” get it?), which is 3. Since there are 6 square sides, or “faces,” the surface area is 6*(3^2) , which is 6*9 = 54.

You may also need to find the surface area of a cylinder, which is shaped like a can. So it has two circles on the top, and a rectangle twisted around the middle. So find the area of the circle on the top and bottom (they’re the same), using the radius or diameter or circumference of the circle (depends on what they give you), then multiply the circumference of the circle times the height of the cylinder. Then add the areas of the top, the bottom, and the side.

See, for example:

https://www.google.com/search?q=surface+area+of+a+right+circular+cylinder&pws=0&gl=us&gws_rd=cr

and

http://www.learnalberta.ca/content/memg/Division03/Cylinder/CylinderSA/index.html
 
 
Rate and Distance/Work Problems:

The relationship between the speed at which you travel and the distance travel is D=RT, where D is the total distance traveled, R is the rate (speed) at which you travel, and T is the time you traveled. So if you traveled 50 miles per hour for two hours, the distance you traveled is 50 miles/hour * 2 hours = 100 miles. For some more complex distance -rate-time problems, see:

https://www.gcflearnfree.org/algebra-topics/distance-word-problems/1/

http://www.purplemath.com/modules/distance.htm

http://www.analyzemath.com/math_problems/rate_time_dist_problems.html
 
Work Problems:
These are similar to distance problems, in that the formula for work done is W = RT, where W is the amount of work done, R is the rate at which the work is done, and T is the time. So we’re just substituting work for distance in the distance formula.

HOWEVER, work problems on most tests I’ve seen are usually different from distance problems because two machines or people can work on the same job, shortening the time required to do the job. So you’re likely to run across a test problem such as “Printer A can print a huge print job in 3 hours, and Printer B can do the same job in 5 hours. How long would it take Printers A and B, working together at the same rate, to do the job?” This problem requires multiple applications of the work formula to solve.

First, we solve for RA, the rate at which Printer A can do the job by itself. RA = 1 job/3 hours, when we divide R by T. RB = 1 job/5 hours, using the same logic. Therefore, when both printers work together, we get 1 job = (RA+RB) TA+B, where TA+B  is the time it takes Printers A and B, working together at the same individual rates, to do the job.  So TA+B  = 1/ (RA+RB), which is 1/(1/3 +1/5) .

So we have to find a common denominator for 1/3 and 1/5.The best common denominator to use is the lowest common multiple of both denominators. If you don’t know how to find the lowest common multiple of two numbers, see here: http://www.math.com/school/subject1/lessons/S1U3L3DP.html
Since there are no common factors of 3 and 5, we just multiply 3 and 5. We take 1/3 * 5/5 to get 5/15 for RA, and 1/5 *3/3 = 3/15 for RB. Adding 5/15 and 3/15, we get 8/15 jobs per hour for the rate. Since TA+B = 1/(RA+RB), the time is 15/8 hours, or 1.875 hours.
 
Hope this helps!
 
Sample SSAT Test:
 
http://ssatprep.com/free-practice/