2. THE MATH TEST
36 questions in 54 minutes: that’s 1 ½ minutes
per question.
Each problem is stand alone.
About 25% of the questions are arithmetic and
algebra problems.
About 15%, although not true “word problems,”
are more than arithmetic, using tables or
graphs (mostly the coordinate system).
About 60% are true “word problems”. So need
to practice word problems along with your
basic arithmetic and algebra skills.
All questions are multiple choice.
Preparation:
o Review basic math skills.
o Take practice tests.
3. MULTIPLE CHOICE STRATEGIES USUALLY
IRRELEVANT
Because a math problem has a number answer
and so the choices are all numbers!
So you have to do the problem, get the answer
and make the choice.
Sometimes you can still use the procedure of
going through the choices one by one and
eliminating incorrect answers until you find the
correct one.
o This usually happens in a linear equation
problem:
Find the y-intercept and eliminate those
that don’t have it.
Then use slope to choose between the
remaining possibilities
4. o Very rarely a question might have word answers: One example
is “What is the result of dividing a positive integer X by a
positive number less than 1?” (two answers can be eliminated)
A number greater than X
A number less than X
A negative number (may be eliminated: +/+ = +)
An irrational number (eliminate: only a result of square
root)
5. BASIC MATH SKILLS NEEDED
Operations with fractions
Interchange between fractions,
decimals and percents (lots of
these)
Operations with signed numbers
(integers)
Find greatest common factor (GCF)
Find least common multiple (LCM)
Order of operations (PEMDAS)
6. Arrange numbers from small to large (a list with positive numbers, negative
numbers, fractions and decimals, even square roots).
Place value and rounding (for MOST problems: you pick the closest answer!)
Some basic geometric formulas.
A few basic conversion facts
o METRIC
o English
Probably at least one scientific notation problem
7. BASIC ALGEBRA SKILLS NEEDED
Change simple word statements into symbolic
expressions/equations.
Evaluate an algebraic expression given value(s) to
substitute.
Solve basic equations/inequalities.
o Combine like terms.
o Pull out GCF or distribute to remove parentheses
o Do the same thing to both sides to isolate the
variable:
Add/subtract
Multiply/divide
Multiply fractional coefficients by reciprocal.
Cross multiply in the case of a proportion.
Be able to FOIL two binomials.
8. BASIC GRAPHING SKILLS NEEDED
Linear equations: y = mx + b
Identify the equation from the graph using
intercept and slope.
Identify the equation from a set of (x, y)
coordinates:
o Look for a (0, y) and there’s your
intercept.
o Or find intercept by extending the
pattern of ordered pairs.
o Identify the slope by noting how much y
changes for a given change in x. (+
slope: “positive covariation”, - slope:
“negative covariation”)
Know difference between independent (x)
and dependent (y) variables.
9. BASIC STATISTICS SKILLS NEEDED
Find the mean (average) of a set of numbers.
Find the median (middle number) of a set of numbers.
Find the mode (most common number) of a set of numbers.
Perhaps identify a graphed data set as symmetrical, left skewed
or right skewed, unimodal, bimodal, uniform.
Will not be any standard deviations, z-scores, etc.
10. WORD PROBLEMS: BASICS
What is a word problem?
o Most math problems give the “set-up”: you just
do the mechanics of evaluating (expressions) or
solving (equations).
o A word problem makes you:
READ
THINK
Come up with the expression or equation
(the HARD part).
Then of course do the straightforward
mechanics of evaluating or solving (the
EASY part).
It doesn’t tell you what to do. Once you accept
responsibility of thinking it out for yourself:
AMAZING, IT IS POSSIBLE, THESE CAN BE
DONE!
11. HOW TO DO A WORD PROBLEM
Read the problem.
What is being asked for? (usually one or two answers).
o Assign an algebraic expression to each answer being asked for
using the given relationship between them (e.g. “a number” = “x” and “5
more than the number” = “x + 5”).
o Write an equation with those algebraic expressions in it
Often it’s just add the expressions and set them equal to a value
actually given in the problem (think: how would anyone calculate that
given value? – the equation is that calculation).
$10000 in two accounts, one 2.5%, the other 5%, total interest = $300
Interest in 2.5% account + interest in 5% account = total interest
.025x + .05(10000 – x) = 300
12. Now solve the equation for
x: this is just mechanical,
like any problem that gives
you the set-up, only you
came up with this set-up!
o“x” = something and that
is one of the answers.
oIn the above example,
subtract that from 10,000
to get the other answer
(“10,000 - x”).
13. NOTE: SOME “WORD” PROBLEMS JUST REQUIRE SIMPLE ARITHMETIC
No need to set up an equation.
o The unknown is already “by itself” and you are being given all the numbers
to calculate it!
Imagine yourself in the problem:
o What would you naturally do in such a situation?
o Add? Subtract? Multiply? Divide?
o One step at a time!
Cannot do it mechanically! You have to think and imagine what one would
naturally do in such a situation.
Plumber charges $50 to show up and $30/hour; what is the cost of a
5 hour job?
Flat rate + hourly cost x number of hours = cost of job
50 + 30 x 5 = 200
14. BASIC SET-UP SKILLS THATCAN BE
USED FOR WORD PROBLEMS
Ratio and proportion (LOTS of these).
o Set up and solve a proportion
o Find the parts of a whole given a ratio
of parts and the total (MANY of
these).
Conversions (LOTS of these)
o Unit cancellation
o Or by proportion if you prefer
15. Percentage problems (LOTS of these)
o Given whole and percent, find the
part
o Given part and whole, find the
percent
o Given part and percent, find the
whole.
o Find percent increase or decrease.
Rate problems:
o Distance = rate x time (d = rt)
o Or r = d/t, or t = d/r
But some word problems don’t easily fit
one of these categories.
17. MATH TEST EXAMPLES OF QUESTIONS
OUR APPROACH IN THIS WORKSHOP
• Now we will work through some word problems: some pretty much
fit one of the categories, some not exactly.
• Try the problem on your own.
• Click to see answer.
• Subsequent slides explain the process.
• Timing to imitate test would be 1.5 minutes per question.
18. 1. A cell phone on sale at 30%
off costs $210. What was the
original price of the phone.
a. $240
b. $273
c. $300
d. $320
ANSWER:
C: $300:
19. Note: There are many ways to solve any problem,
but setting up a simple equation almost always
works. Although it involves percents, this problem
is not the classic “find the percent decrease”: here
they gave you that and they want the original price.
FIRST: assign algebraic expressions to the
answers wanted – in this case only one answer is
wanted and so we assign: original price = x.
SECOND: write the equation –you must THINK:
how would one calculate the sale price (the dollar
amount given in the problem)?
20. For this you must understand that a sales discount is a percentage of
the original price which is subtracted from the original price to get the
sale price.
You also must know that to take a percentage of a number is to
multiply the percent in decimal form by the amount (known or
unknown). In this case the amount is unknown (the wanted answer),
assigned to be “x”.
How would you calculate the given sale price? To get the dollar
amount of the discount you would multiply the given discount percent
(expressed as a decimal) times the unknown (“x”) original price. That
dollar discount would then be subtracted from the same unknown
(“x”) original price:
x - .30x = 210
21. Now that you have the set-up problem, the rest is mechanical:
1x - .30x = 210 (“x” = “1x”)
.7x = 210 (combine like terms)
x = 300 (divide each side by .7)
Notice that once we get through with a word problem we will have “x = some
number”. THIS IS WHEN WE LOOK BACK TO SEE WHAT WE ASSIGNED TO
THE LETTER X! In this case it was the unknown original price, which we now
know, and can choose from the list of choices.
22. Now something tougher, tougher in more ways than one:
2. You are taking a test and you are allowed to work a class
period of 45 minutes. 20 problems are multiple choice and 30
of the problems are true/false. If they have equal value, how
much time would you estimate for each type of problem if you
believe you are twice as fast at true-false problems?
a. 90 seconds per m/c; 45 seconds for t/f
b. 60 seconds per m/c; 30 seconds for t/f
c. 70 seconds per m/c; 35 seconds for t/f
d. 80 seconds per m/c; 40 seconds for t/f
ANSWER:
C: 70 seconds per m/c; 35 seconds per t/f
23. Note: this could certainly be done
by trial and error – in fact doing it
that way would save us some other
trouble at the end, but we want to
practice our problem set up. Also
the “trouble” at the end will be a
good warning that we always have
to pick the answer that is closest to
correct. This happens a lot in TEAS
6 (in this unusual case we will
actually have to check our answer
to be sure it is the best answer!).
24. FIRST: again, assign algebraic expressions to the
answers wanted – in this case two answers are
wanted, and so we must assign two different
algebraic expressions based on the relationship
between them that is given.
The wanted answers are the time it takes to do each
kind of problem.
The relationship is: you can do t/f questions twice as
fast as m/c questions. One of the answers is always
assigned the expression “x”. Let’s assign the
multiple choice time to be “x”. Now, we can do t/f
twice as fast, which means the t/f time will be half of
the m/c time, so the t/f time will be the algebraic
expression “.5x”. Which to call “x”? – will work out
either way.
25. SECOND: again, now we write the
equation; as so often is the case,
here our two expressions will add up
to a number given in the problem.
What is given? The total time of 45
minutes is given. But the answers
are in seconds. So let’s just do this:
45 min times 60 sec in each minute =
2700 seconds. Now the time to do
the 30 t/f and the 20 m/c must add to
2700 seconds, so the equation is:
30(.5x) + 20x = 2700
There’s that set-up problem!
26. Now that we have that set-up problem, we
just do the mechanics to solve it:
15x +20x = 2700 (30 times .5 = 15)
35x = 2700 (combine like terms)
x = 77.14285714 (divide both sides by 35,
calculator result)
So now: WHAT WAS “X”? Looking back we
see that we called “x” the time to do a m/c
problem. But t/f we called “.5x” meaning
half that time. Dividing our “x” answer in half
(just the answer above “÷2” on your
calculator), we get the time to do t/f =
38.57142857
27. If we round these answers to the nearest 5
seconds (to match the possible answers) we
would get 80 sec for the m/c and 40 sec for the
t/f. That is answer D. BUT WATCH OUT:
rounding both of them up just might run us over
our time limit! CHECKING the 80 and 40 we get:
30 t/f times 40 sec each + 20 m/c times 80 sec
each = 2800 sec but we have only 2700 sec so
that would run us over. Looking at answer C we
see 70 sec per m/c and 35 sec for t/f which
would be rounding way down, but we have no
choice. We must choose answer C which would
give us as much time as we could have without
running over the 2700 sec (in fact using answer
C we get: 30 times 35 + 20 times 70 = 2450
sec). B would be less time, and A, again, over.
28. Finally, a problem involving rate:
3. Your interview is scheduled for 8:00 in the morning and you need to
allow 20 minutes for your trip to the interview. You oversleep and leave
10 minutes late. How fast will you have to travel to get there on time?
a. half as fast
b. twice as fast
c. three times as fast
d. four times as fast
ANSWER:
B: twice as fast
Note: this could actually be done be intuition: “if I have to get there in half the
time I would have to go twice as fast.” But we want to understand the set-up.
29. FIRST: as usual, identify what is being
asked for and assign algebraic
expression(s). In this problem the word
answers are all about the rate = how fast.
But there are two rates which we need to
compare and the word relationship
between them is what we want to find.
We could call the rate we travel to get to
our destination in 20 minutes “x”. But then
what do we call the rate to get there in 10
minutes? Usually a relationship is given
and therefore the other expression would
have “x” in it, but here we are trying to find
the relationship, so we will have to call the
rate for 10 minutes “y”.
30. SECOND: Use the rate formula d = rt. As
in many rate problems a key is to note
which of these three parts (distance, rate
or time) is being held constant. It is the
same distance to the interview, so
distance (d) is held constant. That means
that each rate multiplied by time must be
equal. Our one rate is x (rate for 20
minutes) and the other is y (rate for 10
minutes). So if one rt = the other rt we
have (parentheses indicate multiplication):
(x)(20) = (y)(10) That’s the set up!
31. Now the mechanics:
We called “y” the rate to get there in 10 minutes, so let’s solve for y:
y = (x)(20/10) (divide both sides by 10 to isolate y)
y = (x)(2) (20 divided by 10 is 2)
y = 2x (the usual way to write “2 times x”)
And there’s our answer. The rate to get there in 10 minutes (“y”) is
equal to twice the rate to get there in 20 minutes (x). So we have to
travel twice as fast.
32. WRAP UP
We listed the skills you need for the TEAS
6 math test:
o Basic arithmetic
o Basic algebra
o Basic graphing (= basic linear
equations)
o Basic statistics
AND word problem skills for problems
involving writing an equation:
o What is being asked for?
o Assign algebraic expression(s).
o Write an equation.
33. AND word problem skills for
problems involving only
arithmetic:
o Imagine yourself in the
problem.
o The natural decision on the
operation will come (add,
subtract, multiply, or divide).
Then just do the arithmetic.
We listed the common word
problem set-up skills:
ratio/proportion, percentage, and
rate problems, and conversions
(by unit cancellation or
proportion).
34. REVIEW OF WORD PROBLEMS USING AN
EQUATION
We worked through three example word problems
one a pretty typical percentage problem, one
problem asking for two answers, and one using the
classic rate formula d = rt (but in an unusual way
involving comparing two unknown rates).
In each problem we saw how you could:
o FIRST: identify what was being asked for and
assign algebraic expression(s) to those
unknown answer(s).
o SECOND: using the given number(s), write an
equation with the assigned expressions in it.
o Solve that equation, then look back to see what
answer(s) were assigned to the variable(s);
calculate the other answer if necessary from its
algebraic expression having the same variable.
35. OTHER WORKSHOPS FOR TEAS 6 MATH IN
PREPARATION
We listed lots of skills you need and we
practiced a few word problems.
Where can you get more on all those skills?
Math books, You Tube, Khan Academy to
name a few possibilities
We are hoping to make additional TEAS 6
math test workshops:
o Basic Math (arithmetic) Skills
o Basic Algebra/graphing Skills
o Word Problem (and statistics) Skills
TEAS 6 Math Test CAN be conquered by
o Brushing up your skills
o PRACTICE tests
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