A Mathophobe is afraid of math. You don't need to be. Memorize a
few simple rules and your times table, and you can solve most of the math
problems you will find on job qualification exams. Besides, you can figure
out which deal is a better price when confronted with (intentionally) confusing
pricing structures. It's not hard, but you do need to be careful.
The important thing to remember about mathematics is that, if you work carefully and don't make any mistakes, you will get the right answer, no matter how you do it. I will try to show you the fast and easy way to solve multiple-choice questions (the kind you get in those tests), but if any of the tricks I tell you about is too confusing, use a different rule or formula.
A few things you need to already know coming in: your times table (up to 9x9) and how to do long multiplication and division, and of course how to add up sums. I can't help you with these, you should have learned them in grade school. If you are rusty, get a practice book or use one of the online tutorials. If you need help, find somebody willing to help you. You really need these skills -- maybe less so if you know how to use the calculator in your phone, but they are good skills to have.
Then I will show you how to leverage these skills to ace the test. Don't worry about the names I gave my shortcut rules, I just picked a random name. Mostly it's the ideas themselves you want to have at your fingertips.
I have grouped my remarks into these topics. You probably should read through them in order once, then when working through your practice tests questions, you can jump to them as needed:
Vocabulary and Symbols
Useful Numbers to Know
Multiple-Choice Tests (How to Do Them)
Word Problems
Tables
Letters As Numbers
Adding Integers (Whole Numbers)
Integer Multiplication
Integer Division
Decimals
Fractions
Factoring
Circles
Integer -- The word "integer" is not well known outside mathematical circles, but the test will sometimes use it. It just means whole numbers (no fractional part, no decimals). It can be positive or negative.
Per -- Many problems give values as "something per some-other" like "miles per hour" or "hours per day" or "dollars per unit," or whatever. If you think of "per" as meaning "divided by" you will get most of your math right. So "miles per hour" means the total number of miles divided by the hours it took you to drive that far, and "hours per day" means the total number of hours you worked that week divided by the number of days you worked.
Percent -- You already know what percent is, but for mathematical purposes, just think of it as "divided by 100." The word is actually made up of "per" (meaning divided by) and "cent" (from the Latin word for 100), so it literally means "divided by 100." Originally the symbol "%" was written with two zeros under the bar, so it visually showed the same thing. When you read "25% of something," it means 25 times that something, divided by 100. 25% of your $800 paycheck is 25x800 = 20000, divided by 100, = $200.
Of -- Some fraction or some percent (percent is a fraction) of a quantity means "multiplied times" that quantity. "25% of a dozen" of "1/4 of 12" (same thing) means 0.25 x 12 = 1/4 x 12/1 = 12/4 = 3. See Multiplying Fractions (Rule F5) below.
Sum, Difference -- These are the technical names for the result of adding (sum) and subtracting (difference). When you see the word "sum" you know that it is the result of addition, and when you see the word "difference" you know that it is the result of subtracting. Usually a difference is assumed to be positive, so the subtraction is flipped over if necessary to subtract the smaller number from the larger. Some test questions might assume you know that, so you should.
Product, Multiplier, Multiplicand -- These are the technical names for the parts of a multiplication, where you multiply the multiplier times the multiplicand to get the product. You will probably see "product" a lot more than "multiplier" and "multiplicand" almost never, so at least you should know that the "product" is the result after multiplying.
Divisor, Dividend, Quotient -- These are the technical names for the parts of a division, where the divisor is divided into the dividend to get the quotient as an answer. You should know all three terms, they are often used when discussing a division. It might help you to remember that a "dividend" in business is the profit of a corporation that is divided by the number of stocks so each stockholder gets a fair share of the profits; the word is often erroneously used to refer to what the stockholder receives, which is actually the quotient. Don't be confused.
Fraction, Numerator, Denominator -- Technically, a fraction is a division, nothing more, but we usually use the word to refer to the way we write the number before we actually divide it, the numerator over a horizontal bar (sometimes shown in linear text as a slant "/") over the denominator. The whole thing, two numbers with a horizontal line between them, is called a fraction. The numerator is the dividend, and the denominator is the divisor, different words for the same things. You should know that.
Prime -- A number is prime if there is no number (other than one and itself) that divides it exactly. It's a good idea to memorize the first half-dozen or so prime numbers: 2, 3, 5, 7, 11, 13, 17. Larger primes will turn up, but not very often. You can figure out all the prime numbers less than 100 by knowing your times table: they are all the number not in the table anywhere except in the 1-row or column. Rule D2 tells you how to figure out if a number is divisible by small primes. Any number less than 100 not divisible by some number less than ten is prime.
X, x, y -- Any letter by itself stands for a number that you don't know the value yet. A letter is a number. Usually they will use lower-case italics for values, the most common is x and then if they need a second one, y. Any letter by itself followed by an equal "=" and then some more letters and numbers and symbols, the letter on the left is what you are expected to determine the value, or sometimes think about without figuring out its value:
-- Pi (pronounced "pie") is a Greek letter used in place of the number
which is the ratio between a circle's circumference and its diameter, 3.14159265...
There is no way to write the number exactly, so we use the Greek letter
pi. For approximations you may be encouraged to use 3.14. When doing round-number
calculations, 3 is good enough.
+ -- Plus means add.
- -- Minus means subtract, or used in front of a number (or letter representing a number), that the number is negative (less than zero, a debt rather than a credit)
-- Times means multiply. We all learned to use a plain "x" for multiply,
but that is sometimes confused with the letter "x" so linear computer text
usually uses the asterisk (star) for multiply. Mathematicians sometimes
use a raised dot, but you probably won't see that in your tests. More often,
when a number and a letter (representing another number) are to be multiplied,
they are just put next to each other, "3x" which you pronounce
"three ex" like you might say "three cars" but it means that you multiply
whatever number the letter x stands for times three.
-- Divide. We learned to use the two dots with a line between them -- it
sort of is a visual representation of a fraction, which is another way
to write "divide" (see Fractions below) -- to
mean divided by, but computer linear text does not normally have that symbol,
so the computer geek chose the forward slant "/" (kind of like a fraction
tipped over to the left) to mean divide. You will probably see it more
often than the divide symbol you learned in school.
--
Square Root. The number or value inside (to the right of the checkmark,
under the horizontal bar) represents the square (times itself) of the number
the whole square root formula represents. You probably won't see much of
this, it's a way of working backwards from your times table. 3*3 = 9 so
the square root of nine is three. If you know an area of a square, you
can figure out what its side is. I have not seen any test problems like
this, but it's here for completeness.
x2 -- A raised number 2 means "squared," the number times itself x * x. A raised 3 means "cubed," three times x * x * x, but you won't see very much of that except maybe for measurements of volume (length times width times height) which might be measured in cubic yards "y3".
A negative exponent means the number is in the denominator of a fraction,
that is x-1 means 1/x
and x-1 means 1/(x*
x)
which is one divided by x squared. Fractional exponents are
roots, so that a number with an exponent 1/2 is the square root of
that number, and 1/3 is the cube root. Any test that expects you to know
about fractional exponents is beyond the scope of this help.
I may add more words and symbols here, as I come across them.
The first half-dozen or so prime numbers, 2, 3, 5, 7, 11, 13, 17, are good to know when you are reducing fractions.
Besides 2, 3, 4, 5, 6, 9 and 12, a few other numbers like 15 and 1.5 are good numbers to recognize as (Rule M1) easy multiplies (and sometimes divides).
50%, 25%, 20%, 10%, 1%, 33% -- these are 1/2, 1/4, 1/5, 1/10, 1/100, and (slightly rounded) 1/3. Sometimes it's easier to divide by 2, 3, 4, 5, 10, or 100 than to worry about multiplying. If you are good at memorizing, it might also be useful to remember that 1/8 is 12.5%, 1/6 is approximately 17% or 16.7%, 1/7 is near 14%, and 1/9 is about 11%. If you know your times table, you can look for products that are very close to 100, the way 7x14 = 98, or 6x17 = 102. Sometimes knowing that 66% or 67% are about 2/3 is helpful. 75% is exactly 3/4, and 40% is 2/5.
1.4 (approximately) -- is the square root of 2, it's the ratio of the length of the diagonal to the side of a square, and sometimes turns up in odd places.
3,4,5 -- (Pythagoras) In problems involving the diagonal of a rectangle, or a right triangle (one square corner), test problems tend to favor this particular set of numbers because they come out even. Most likely they will be some multiple, the same number times each. The two short (square) sides are 3x and 4x, and the diagonal is 5x, where "x" is some whole number multiplied to hide the fact that they are using Pythagorean numbers.
2, 4, 8, 16, 32, 64, 128, 256... are the doublings of one. When
you recognize one of them, they factor quickly, just count off their position
in the list, and that's how many twos to write down. Computer jocks like
these numbers more than test-makers, so maybe you won't see them so often.
Suppose you have a question like this:
What is 1234 x 5678?
| A. 69,043 | B. 6,912 | C. 7,006,652 | D. 12,345,678 | E. 7,006,526 |
The first thing you see is that two of these answers are very close to each other. That means that one of them is probably the right answer. You can quickly verify that assumption using round numbers: Here you have the product of a number near 1000 times another near 5000 (in round numbers). Always do your first calculation in small round numbers, it saves a lot of time.
Count the total number of zeros to the right of the two rounded numbers, in this case 3+3=6; that is the number of zeros to add back onto your round number estimate. Remove those zeros and do the simple math, 1 x 5 = 5. then put the zeros back on, giving you 5,000,000. There are only two of the choices anywhere near that, which confirms that the correct answer is one of them. The two answers are both greater than our guess, but we rounded both values down, so that is to be expected. Look for results about the same number of digits, more than half your guess and less than double.
Now look at the last digit of each of the original two numbers we are supposed to multiply, and multiply just those last digits, 4 x 8 = 32. Whenever you are multiplying any two whole numbers the last digit of the answer is always the last digit of the product of the two last digits. That's always true for any whole-number arithmetic except divide (divide is always harder). So we look in our two best choices for a number that ends in '2'. There is only one, which is the correct answer. Choose it and move on.
See how easy that was? We can do a lot of that, different tricks, not
always, but we can save enough time so we can afford to spend more doing
the hard ones. This is just an example, I'll give you the specific things
to look for when we get to them, called Rules below.
Bill and Mary went into business together making widgets. They bought the parts for each widget for $5 and they paid Bill's little sister $2 to put them together. Mary's brother sold them door to door for $10, but he wanted $2 each for his efforts. The store they bought the parts from saw how many they were buying, so he gave them a $1 rebate on each. How much profit did Bill and Mary earn on each widget?The first thing to look for is what they are asking you to calculate. That's usually a question near the end, like "What?" or "How much?" or it can be only implied, like "Figure out..." It doesn't have to be at the end, they could start out like "What is the profit on..." and then give the details you need to figure it out.
In this problem, they want you to figure out the profit. "Profit" is one of those words you need to understand, it means the sum of all the costs subtracted from the total money you get for it. Sometimes they give you the total costs and/or income items, sometimes they are on a per-unit or per-batch basis. You need to notice how they figured each number, then divide that number by the number of units it applies to. In this problem, everything is given "per widget" so there is nothing to multiply or divide.
They sold each widget for $10, plus they got a $1 rebate on their materials.
That's a total of $11 income. They paid out $5 for materials, $2 for labor,
and another $2 commission on sales, for a total $9 costs. The profit is
the difference, $11-$9 = $2. Most of the problems you see like this will
have complications you need to work through. Use round numbers for your
first cut. If that gets you close to only one answer, you are done.
| Name | S | M | T | W | T | F | S |
| Anne | -- | 4 | 5 | 3 | 4 | -- | -- |
| Bill | -- | 8 | 9 | 10 | 5 | 7 | 4 |
| Fred | 6 | -- | -- | 4 | 4 | 2 | 6 |
| Mary | 2 | 4 | 6 | -- | -- | 4 | 5 |
| Zoe | -- | 8 | 7 | -- | 8 | 9 | 6 |
They might ask you for the number of hours Fred worked that week, which means you should ignore everything except the one line labelled "Fred". Or they might ask for the number of overtime hours they need to pay for people working on Sunday, which means you need to add up only the hours in that column.
Or they might tell you that they need to pay overtime to any employee
who works more than 40 hours in a week, on the hours over 40, which means
that you need to scan down the list for people who worked more than (or
perhaps close to) 40, then add up only those hours and subtract off 40.
In this list, Anne worked only four days about a half-day each, there's
no way that adds up to more than 40, so I don't need to add her hours up
at all. Bill has some long days, he might need me to add up the hours.
If the table had big complicated numbers -- say hours and minutes -- I
would round off to hours (or maybe even single-digit numbers, which this
table already is) to see if it gets close, before doing the careful sum.
Fred and Mary both worked five days, all less than 8 hours, so they didn't
do any overtime, no summation needed. Zoe, maybe, maybe not, I'd have to
add up the numbers and see. Look for what they are asking, then see how
much you can do with round numbers. We only had two employees working marginally
more (or in the case of Zoe, less) than 40 hours, so if the multiple choices
they offer include some very large numbers, you can discount them immediately.
Like suppose I'm a used car dealer, and I can buy five cars off the market, but the guy doesn't know how much he will charge me yet, I want to figure out how much I need to sell them for to make a profit, so I do my calculations with "X" for the price of each car, then when I know, I just substitute the correct value for "X" and finish the calculation. I might want to figure it for different commisions for my floor sales team, so I can use "Y" for the commission, then try different numbers in the same calculation. Let's see how that works. I don't even know the sales price yet (that's what I'm trying to figure out) -- let's call that "S" -- but let's suppose my overhead can be figured at $500 for the lot, and I want a $200 profit at the end of the day. So the formula for each car might look like this:
S = X+YS+500/5 + 200/5Notice that the commission is paid out on the sale price (which we don't know yet), so I just multiplied the commission rate Y times the sale price S -- put two letters, or a number and a letter next to each other, and that means they are multiplied, so the total I need to pay my supplier is 5X (think: "five eXes" like you might think "five cars"). Algebra is the mathematical discipline for figuring out what numbers work for "S" when it appears more than once in the formula, or in an odd place. Many of the tests you need to take don't really do much algebra, so I'll leave that for the next lesson.
You do need to know about the letters. Letters are just numbers that we don't know yet. There are rules for doing math using letters, so that the answer comes out correctly. Next time, not today.
Did you ever get a puzzle where the guy says, "Think of a number, any
number." Then he tells you to multiply it times three, then divide by your
original number, or something like that, then he tells you your result
is three. How did he do that? He thinks of your original number as X. Multiply
it times 3 and you have 3X. He still doesn't know what it is, but you can
divide 3X by X and the result is always 3, no matter what X was (except
zero: you can't divide by zero). X is just a number we don't know the value
yet.
If there are several numbers in the list that are the same value, just count them, and multiply one of them by how many are that value. Then I can add in the other values. If there are pairs of numbers the same, add one pair, then multiply it times the number of pairs. Sometimes the pairs don't exactly match, but each pair adds up to the same value, so count them as pairs anyway. If you are not used to thinking of numbers in pairs, this may not be so useful. Don't waste a lot of time on tricks that don't work for you.
Often a list will be a lot of numbers very close to some nominal middle
-- like that table above, where Bill's and Zoe's
hours were (with a couple exceptions) just one or two hours above or below
one eight-hour day. So I run through the list thinking, "Right on (=8),
+1 over, +2 over (now +3), -3 down (that cancels the +3, leaving me right
on again), then -1 down at the end of five days (which should be 40 in
a 40-hour week), so the sum so far is -1 under 40 (=39), but I'm only interested
in the hours over 40. When I add in that last +4 I end up with a net +3
hours of overtime, with much less hard work. Zoe worked out the same way,
-1, then cancelled by +1, leaving -2 at the end of the fifth day (no overtime,
no big numbers to add up).
Do the same for the next size (one digit smaller), but don't add them, just count those 5 or higher, and add that count to your base rounded sum. Discard any numbers more than two digits shorter, and count the remaining numbers (all two digits shorter than the longest number). They probably won't give you so many, but if there are more than fifty or so, round the count up or down to the nearest multiple of 100, then throw the last two digits away and add the hundreds (as ones) to your rounded sum. Less than fifty, throw them all away.
The basic principle is, the most significant digits are most significant,
so we add them. The less significant digits (one position over) don't matter
so much, so we effectively round them up or down and count those that make
it over into the first position, assuming that the ones we didn't count
merged with the ones we did count. Smaller numbers don't matter at all
unless there are a lot of them. Let's see how this works. Here's a list
of 20 random numbers to add up, in the first column:
| 9398 | 9 | |||||
| 1187 | 1 | o | ||||
| 161 | o | |||||
| 308 | ||||||
| 7520 | 8 | |||||
| 6665 | 7 | o | ||||
| 82 | ||||||
| 21 | o | |||||
| 565 | + | o | ||||
| 3636 | 4 | |||||
| 15 | o | |||||
| 47 | o | |||||
| 66 | ||||||
| 466 | ||||||
| 254 | ||||||
| 17 | o | |||||
| 7665 | 8 | o | ||||
| 31 | o | |||||
| 8557 | 9 | o | ||||
| _844 | __ | + | ||||
| 47505 | 46 | 2 | 11 |
There are seven 4-digit numbers, which I rounded to the nearest (4-digit) multiple of 1000, then worked with only the first digit, shown in the second column. These will add up to my base rounded sum 46 at the bottom. There are seven 2-digit numbers, which I simply ignored. Of the remaining six 3-digit numbers, I marked only the two over 500, so I can add 2 (for the two marked numbers) to my base sum, then put the three zeros back on to give an estimated (round) sum of 48,000, which is pretty close for only eight 1-digit adds in 20 numbers. If there were subtractions mixed in, I would add up the positives and negatives separately (Rule A5).
In the last column I marked and counted the odd numbers, a total of eleven, which is odd, so the correct sum will be an odd number (Rule A2) near 48,000, and that is what to look for in the multiple choices they give. Notice that our estimate could be off by a thousand or more (the second digit, but probably not the first unless the second digit is 9 or 0), so if they offer two odd answers that close, you need to work harder on this one.
The next approximation, which is your final approach, just add up the
last one or two digits (however many digits you need to distinguish their
answers), ignoring all the carries out those two (or one) digits. Adding
just the last one or two digits will be exact in those last one or two
digits (Rule A3), and it's faster than adding all four digits. You have
already discarded all the offered answers not close to your rounded base
sum, you only need to distinguish the few that are close.
The twos row is extremely useful, you want to be very fast at doubling a number, or cutting it in half. Practice this through the day, as you drive or walk around, look for license plate numbers or house numbers or phone numbers on billboards, and practice doubling and halving them. Why is this useful? You do not need to memorize the fours, fives or eights rows of the times table, two doubles and you have times four, three doubles and you have times eight. You could even skimp on the three and six rows, because doubling a number then adding that number back on is times three, double again for times six.
Half a number is as important as doubling, because that gets you times 5: Just multiply times ten (insert a zero on the right), then take half of the result. You can do this in your head in the other order, take half the number, then if the original was odd, insert a 5 on the right, otherwise insert a 0.
Multiplying by 25 is easy, because 25 is a quarter of 100. A quarter is half of half. An eighth is half three times, but you won't see that so often as you see quarters. So multiply by 25 is times 100 (shift over two, inserting two zeros) then half twice. When we get to decimals below, 0.25 is exactly one quarter, just half twice, no times 100 needed.
Multiplying by 15 is easy, just times ten (shift it over and insert a zero), then add half of that result. Multiplying by 1.5 (one and one half) is the same, just skip the times ten part.
Practice doubling and halving on numbers you see on the street. You
will ace large parts of your test by using round numbers and then applying
Rule
M1 where you can.
Sometimes multiplying the leftmost single digits from the two numbers
is all you need to arrive at a reasonable approximation to the desired
result. The actual product will generally be bigger than this first approximation,
but not by much. If you round instead of snipping off, then your approximation
will be much closer, but might overshoot.
Or you can recognize that 25 is 1/4 of 100 (and 1/4 is half twice, Rule
M1), then (Rule M6 below) we have one quarter
of 100x800, knock of four zeros leaving 1x8=8, then take a quarter of 8
(half twice) ->4->2, than add the four zeros back on, a lot less effort.
Some digits are magical when they are multiplied together, they give you a result that ends in the same digit. There are four with this property: 0,1,5,6. If both numbers end with the same one of these, then the product will end in the same digit.
If that doesn't help, multiply the rightmost single digits from the
two numbers, and the last digit of your product will always be the last
digit of the correct whole product. Multiply two digits and it will be
correct for two digits, and so on, all the way up to the full multiply.
This works for division too, but only if you choose a round number that
is an exact multiple of the divisor (see my explanation in the Factoring
example).
[Interesting factoid: Did you know that long division was the straw
that broke the back of Roman numerals? Can you imagine how to do division
at all in Roman numerals? Sailors knew for a long time before Columbus
that the earth is round -- Columbus underestimated the circumference, and
thought he would be arriving in India when he actually arrived in the Caribbean
islands -- but the invention of the sextant required mariners to do long
division in order to figure out where they were from the stars, and Arabic
numbers were far easier to divide than Roman. But that's another story.]
Even numbers divide exactly by 2.
Any number that ends in 0 or 5 is divisible by 5.
Add up the digits of any number (if the sum of the digits is more than 9, add up the digits of the sum) and if the result is 9, the number is exactly divisible by 9. If the sum is divisible by 3 (3, 6, or 9) it is divisible by 3. For example, the sum of the digits in 1234567890 is 45, add 4+5 again to get 9. So this big long number is exactly divisible by three and nine, also by two and five.
Any even number divisible by 3 is also divisible by 6.
If the last two digits are divisible by 4, the whole number is. Sometimes that's not so easy to figure, but if the second (tens) digit is even, then the last (units) digit must be divisible by 4 (0, 4, or 8). If the second digit is odd, then the units digit must be even but not divisible by 4 (2 or 6). For example: 384, the tens digit is even and the last digit is 4, so 384 is divisible by 4. You can cactually do the divide by halving twice. Another example, 7469 is odd, not divisible by 4. Take 895636, the tens is odd and the units is even but not divisible by 4, so the whole number is. In 56894 the tens is odd and the units is divisible by 4, so the whole number is not. This is pretty subtle, probably more work than it's worth your time preparing for the test. But you never know.
There is no easy way to know if a number is exactly divisible by 7.
I suggest you commit the 7-row of the times table to memory, so if you
see a number like 42 or 63 you know immediately that it is an exact multiple
of seven.
Example: 7200 
150 = 720x10 / 15x10 = 720 / 15
The sum of the digits in 720 is 9, and in 15 is 6, so both are divisible by 3, so
720 / 15 = 240x3 / 5x3 = 240 / 5Any number ending in zero is divisible by 5 (lop off the zero and double it), so
240 / 5 = 48x5 / 5 = 48 / 1 = 48No long division is necessary.
| 3.59 | 3.89 | |||
| ------ | : | ------ | => | 3.59 x 27 : 3.89 x 23 |
| 23 | 27 |
It's still a messy multiply, but arguably easier than two divisions -- especially after rounding to 3.6 and 3.9 (or 4, which is a couple doubles ;-) The decimals make it harder (see rules below for dealing with decimals), but since we are only comparing, we can move the decimal over the same number of places on each side without affecting the compare, 36x27 to 39x23 instead of 3.6 and 3.9.
Let's do the multiplications slowly. 36 and 27 are both exact multiples of 3, so I have 3x12x3x9. Rule M6 lets me collect the 3x3 =9 and then 9x9 =81, which is a fast multiply (times 80 is three doubles 12->24->48->96, shift it over for times 10 ->960, then add the original 12 -> 972) or even faster the other way around: times 12 is times ten (shift over and insert 0 ->810) then add double ->162 +810 = 972.
39 rounds quickly to 40, which is two doubles and a shift: 23->46->92
-> 920. This is less than 972, but still close, so let's apply Rule
M5 and treat 39 as 40-1, that is, subtract 23 from 920 = 897, and now
we know for sure that Soggy Flakes is the better price. I will explain
this method of resolving a pair of divisions like this in more detail when
we get to fractions, below.
We use decimals all the time for money, dollars and cents, and for calculations involving percentages. Some of the problems you will run into also use the same kind of numbers for other values, so in this section we will treat them in a general way, with special emphasis on money and percentages.
The way we speak of money is the best way to understand decimal in general. We write $12.34 but we say "twelve dollars and 34 cents." There is no decimal point in what we say, just whole numbers of dollars and whole numbers of cents. That's exactly how you want to think about decimals, just whole numbers of cents (or whatever).
Remember what we did in Rule M3, lopping off the extra zeros on the right, then adding them back on at the end? We're going to do exactly the same thing with the decimal point, only we count the other direction. Moving the decimal point to the right is the opposite of chopping off zeros; moving the point to the left is like cutting off zeros -- mathematically it's exactly the same -- so we can keep a single count, zeros - decimal places, and then fix it up at the end by going the other way.
Let's see how that works: $12.34 is twelve dollars and 34 cents, or you can think of it a one thousand two hundred thirty-four whole-number (integer) cents, 1,234. it's easy to work in whole numbers, solve the problem, then move the decimal point back as many places as you took off. It's that simple.
Percentages often hide the fact that you have a simple fraction, like 25% is really just one quarter, or 20% is just one fifth. This won't always be helpful, but I saw one test question that asked you to calculate 20% of something that dividing by five would be much easier (the other number was an exact multiple of 5).
These are not shortcuts, you do these things to get exact answers. If
you are starting with round numbers, then apply the decimal rules, then
the answer will still be rounded.
I saw one test question that asked you to compare four numbers all starting with "0." (that is, they were all less than one), and pick the least. I mentally moved the decimal point to the right one place, which made three numbers greater than 1, and one of them still with zero in front, the easy winner.
I saw one test question where rounding to a single digit (27.8 -> 30
and 9.6 -> 10) gave a result near two of the offered answers, so I tried
a two-digit round (27.8 -> 28) and got exactly one of them. It also helps
to remember which direction you rounded. For adding and multiplying, if
you rounded both numbers up, then your round-number answer will be higher
than the true result; if you rounded both down, then your rounded answer
will be lower. For subtracting and dividing, rounding the divisor or the
number being subtracted the opposite direction has the same effect, but
that's harder to remember.
Make sure you have the same number of digits to the right of the decimal
point in every number to be added or subtracted before you take the last-digit
shortcut. For example, when adding 2.75 + 0.003 + 0.158, one of the numbers
has only two digits to the right of its decimal, so you need to insert
a zero. Then you can add up the last digits 0+3+8 = 11, and know that the
correct answer will have "1" as the third digit to the right of the decimal
point. Combined with the round-numbers rule (Rule D6), this will almost
always get you to the answer much faster than adding up everything.
If some numbers have extra zeros and some have decimals, and you need to multiply them all together, count the zeros against the decimals (or the decimals against the zeros, whichever is less), then if you had more zeros than decimals, add the excess zeros back on, or if more decimal places, move the decimal point in that many places. For example 2000 x 1.23 has three zeros and two decimal places, a net one excess zero to be added onto the product 2 x 123 = 246, final answer 2460.
On the other hand, 400 x 1.23 x 4.5 has one more decimal place (2+1 decimal places, less 2 zeros) than zeros, so you do the math on 4 x 123 x 45, which is 4x45=2x90=180, 180 x 123 is (Rule M5) 200x123 - 20x123 (so I only need to do one multiply, a quick double and shift ;-) 24600-2460 = 22140, then move the decimal place back in one place to 2214.0 (or lop off that lone zero to the right of the decimal point, =2214).
You can still do the last-digit shortcut, so long as you are careful
to watch where it falls with respect to the decimal point, but you need
to count off any zeros before you do this, because zero times anything
is always zero, which is not very helpful. For example, when we want to
multiply 400 x 1.23 x 4.5, there are total three digits to the right of
the decimal point, less two zeros to the left, for a net one decimal in
the answer from 4 x 123 x 45; when we figure 4 x 3 x 5, the 4 x 5 part
gives us a zero in the last digit, cancelling the final decimal point,
so we know the answer is a whole number, but not much else. Sometimes it
works that way.
A "Proper Fraction" is less than one in value, that is, the denominator is greater than the numerator. A proper fraction written next to a whole number is presumed to be added to it as part of a single value. In linear computer text we use a diagonal line instead of a horizontal line, so the numerator comes out on the same line next to the whole number; people usually insert a hyphen "-" there to separate them, but the operation is plus, not minus. It's really hard to do arithmetic on these mixed fractions, so you need to know how to convert from one form to the other and back.
Here are some important things to remember:
You can make new fractions out of whole numbers by dividing by one: 3 = 3/1.
It's the same addition or subtraction, except there you need to multiply the denominators also (but only once: you will have the same denominator everywhere when you are done).
For example, we got a result of multiplying two fractions above, =6/12. To reduce this, we factor the top and bottom, 6 = 2x3, and 12 = 2x2x3, then cancel the threes and one of the twos, leaving 1/2, which is the same result we got by cancelling before we multiplied. It cannot be different, unless we made a mistake.
Test results also normally expect you to convert your improper fractions
to mixed form. When we added two fractions above and got 7/6, we need to
"normalize" it (convert it to normal form) by dividing the denominator
into the numerator to find the integer part, then leaving the remainder
as a fraction: seven divided by six is one, with a remainder 1/6, written
as 1
.
You should be able to recognize either form in the multiple-choice test
answers listed.
Going the other way, just do the divide. 7/20 expressed as a decimal,
first count the zeros, one in the denominator means that the result will
have one more digit to the right of the decimal than we get by dividing
7/2 which is 7x5 (move the decimal over) = 3.5, then move the decimal over
again = 0.35. I would check my work by doing the math in round numbers,
where 7 is near ten, and ten divided by twenty is about one half, so the
answer should be something less than one but greater than one tenth.
For example, to factor 36, we see first that it is even, so we rewrite
it as 2x18. It's still even, so repeat: 2x2x9. Now we see that nine is
not yet prime, so we can rewrite it one more time, 2x2x3x3. These are all
the prime factors of 36.
For example, if you are factoring 182, it's even,
so start with 2x91. Unless you know your times table really well (I don't)
you might not know whether 91 is there or not. But 91 is less than 100,
so if it has any factors at all, at least one of them will be less than
ten. It's not even, the digits do not sum to a multiple of 3, it doesn't
end in 5, so the next (and last) prime factor to try is 7, which does divide
it exactly -- who does long division? Not me! I see that 91 is greater
than 70 (=7x10), so (Rule M5) I subtract off 70=10x7,
leaving 21, which I immediately recognize as 3x7. The final quotient I
get by adding up the partial quotients 10+3=13, therefore 7x13=91. The
prime factors of 182 are 2x7x13.
I think of the joke about the farm kid, the first in his family to go
to college. He came home for Christmas vacation and his father asked what
he learned. "Math, physics, English, Latin," he replied. "Say something
in math," his father said. The kid thought for a moment, the said "Pi-R-square."
The father scolded him, "You learned nothing at all. Pie are not square,
pie are round. Cornbread are square."
The Khan Academy has gotten
very good reviews for the quality of their material. Look for the link
"Math by Subject" then search the topics there for what you need help with.
They have both practice questions and also basic math instruction for topics
you cannot remember from school. It's a good place to start. They won't
tell you how to ace the multiple-choice questions (come here for that ;-)
but they do tell you how to solve the problems in the approved way. You
should know both, because sometimes the shortcuts don't work.
Formulas
Algebra
Working draft 2017 June 13a
