Algebra & word problems tutorial by Joel Chippindale Chippindale,, 1st May, 1999 Algebra is a common question topic in the GRE and if you include word problems which are really just algebra problems in disguise then even more of GRE quantitative questions are algebra problems. For many of these word problems the most difficult part of the question is understanding exactly what they are telling you and what you need to find out. This tutorial will assume that you have already worked through our 'Fractions ' Fractions'' and 'Exponents, ratios and percents' percents ' tutorials. It will also assume that you a fair knowledge of algebra. For example you should be able to solve the following equation to find a. 2 a + 3 = 11 a =?
solution at the bottom of this page
Solve this problem by finding the value for a. If you have difficulty with it then send an email to
[email protected] to
[email protected] and I will consider writing a more basic algebra tutorial. This tutorial will begin by giving you two methods for making sense of and solving word problems. Then we will revise the solution of quadratic equations and finally look at substitution techniques which will help when you do not know how to find the solution to a question. Solution: a = 4
Word problems with equations We will begin with an example of a word problem and then look at how to use equations to solve it .
Carl has twice as much money mone y invested in stocks as in bonds. Stocks earn 10% interest per year and bonds 5% per p er year. If Carl earned a total of $800 dollars from his stocks and bonds last year how much money did he have invested in stocks?
A $ 9, 6 0 0 B $ 8, 0 0 0 C $ 6, 4 0 0 D $ 4, 0 0 0 E $ 3, 2 0 0
If you feel brave you can have a go at it now. My advice would be to follow these guidelines. •
•
Summarize: Summarize: Write equations, which contain the information given to you in the question. Use sensible variable names for example the fir st letter of the thing that the o variable represents. Identify answer i.e. write down exactly what you are looking for. for. o Solve: Solve: find the solution for the equstions you have written down.
If you have had a go at the question then check your answer. answer. If not then we will answer a nswer the question together.
Summarize using equations In the question you are given a great deal of information and you need to be able to summarize it in a more manageable form. Often it is a good idea to translate the question into equations. It is important to use variable names that will make sense to you when you are translating these questions into equations. 'Carl has twice as much money invested in stocks as in bonds. Stocks earn 10% interest per year and bonds 5% per year. If Carl earned a total of $800 dollars from his stocks and bonds last year how much money did he have invested in stocks?' We will use 'S' to represent stocks and 'B' to represent bonds. Using the first letters of each word makes it easy to remember which is which and avoids any confusion that might arise from using more traditional variable names such as 'x' and 'y'. 'Carl has twice as much money invested in stocks as in bonds.' Translates to: S = 2B Note: many people get confused with the phrase 'twice as much' and write 2S = B. This is a very common mistake and must be avoided. If you find that you get confused writing the equation try replacing the variables with numbers and then read the sentence again to see if it makes sense. For example in this case if S = 2B, then if B = 1, S = 2. This makes sense because stocks are '2' which is twice as much as bonds which are '1'. 'Stocks earn 10% interest per year and bonds 5% per year. If Carl earned a total of $800 dollars from his stocks and bonds last year...' Stocks earned 10% of S and bonds earned 5% of B and this t his totaled $800 so, ( 10% × S ) + ( 5% × B ) = 800
It is also important to write down what you are trying to find. It is all to easy to do the correct working and get to a related or intermediate answer which you find in the list of answers A-E and to choose it in your haste to finish the question. '...how much money did he have invested in stocks?' You are trying to find the amount in stocks so write down S=? To summarize we have: S = 2B ( 10% 10% × S ) + ( 5% 5% × B ) = 800 800 S =? Two equations with two unknowns so we can solve them.
Solving equations We have already done much of the work in solving this problem by changing it from the word problem 'Carl has twice as much money invested in stocks as in bonds. Stocks earn 10% interest per year and bonds 5% per year. If Carl earned a total of $800 dollars from his stocks and bonds last year how much money did he have invested in stocks?' to the algebraic problem S = 2B ( 10% × S ) + ( 5% × B ) = 800 td> S =? To solve the system of equations you want to reduce the problem from two variables in two equations to one variable in one equation. Usually the easiest way to do this is by substitution i.e. replacing one of the variables by the other. other. ( 10% 10% × S ) + ( 5% 5% × B ) = 800 800 ( 10% × 2B ) + ( 5% × B ) = 800 ( 20% × B ) + ( 5% × B ) = 800
We know that S = 2B so we can replace the S in the second equation with 2B. multiply out 10% × 2B 20% of B and 5% of B are 25% of B
25% 25% × B = 800 800 We know that 25% = = 800 ×B B = 80 800 × 4
Multiply both sides by 4
, (see fractions fractions))
B = 32 3200 Careful at this point not to assume that you have finished. You have found the amount of money invested in bonds, no you need to use the equation S = 2B and calculate the amount invested in stocks. S = 2B S = 2 × 3200 S = 6400 Returning to the question.
Carl has twice as much money mone y invested in stocks as in bonds. Stocks earn 10% interest per year and bonds 5% per p er year. If Carl earned a total of $800 dollars from his stocks and bonds last year how much money did he have invested in stocks?
A $ 9, 6 0 0 B $ 8, 0 0 0 C $ 6, 4 0 0 D $ 4, 0 0 0 E $ 3, 2 0 0
The amount invested in stocks was $6,400 and the answer is C.
Word problems with a table Another example of a word problem that w e will use a different technique to solve. We will summarize the information given in the form of a table.
At a football game 50% of the seats are sold to season ticket holders who pay $11 each and 10% are sold to children c hildren who pay $5 each. All the remaining tickets are sold to non-members at $15 each. What proportion of the total gate receipts for the game is contributed by non-members?
A 6 0 % B 5 2 % C 5 0 % D 4 0 % E 5 %
If you would like to try the question now then follow these guidelines. •
Summarize: Summarize: Organize the information you are given into a table. o Identify answer i.e. mark on the table exactly what you are looking for. for. o
•
Solve: Solve: Keep calculating more elements in the table until you arrive at the answer you need.
If you have had a go at the question then check your answer. answer. If not then we will answer a nswer the question together.
Summarize using a table Yet again the question contains a great deal of information. This time we can put all the information into a table and this will make the question simple to solve. 'At a football game 50% of the seats are sold to club-members who pay pa y $11 each and 10% are sold to children who pay $5 each. All the remaining tickets are sold to non-members at $15 each. What proportion of the total gate receipts for the game is contributed by nonmembers?' In the question we have three different types of tickets, 'club-members', 'children' a nd 'nonmembers' and three different types of information are given or asked for, for, '% of tickets sold', 'price of ticket' and '% of total i ncome'. Therefore we would sketch a table, which is 3 × 3. % tick ticket ets s sol sold d
pric price e
% tota totall inc incom ome e
club-members children non-members
And begin to fill in the information. 'At a football game 50% of the seats are sold to club-members who pay pa y $11 each and 10% are sold to children who pay $5 each.' % tick ticket ets s sol sold d
pric price e
club-members
5 0%
$11
children
1 0%
$5
% tota totall inc incom ome e
non-members
'All the remaining tickets are sold to non-members at $15 each. What proportion of the total gate receipts for the game is contributed by non-members?'' % tick ticket ets s sol sold d
pric price e
club-members
5 0%
$11
children
1 0%
$5
non-members
$ 15
% tota totall inc incom ome e
?
Where '?' represents what we need to find to answer the question.
We have the summary now let's answer the question.
Solve using a table We reduced the question to the following table. % tick ticket ets s sol sold d
pric price e
club-members
5 0%
$11
children
1 0%
$5
% tota totall inc incom ome e
$ 15
non-members
?
We will add a totals row because we ar e working with percentages and a n income column so that we can later work out the percentages for the income. We can fill in 100% for the tot als of the percentages. % tick ticket ets s sold sold pric price e club-members
5 0%
$1 1
children
1 0%
$5
non-members 1 0 0%
total
% tota totall in income ome
$15
?
-
1 0 0%
inco incom me
Then it is a matter of filling in as many cells as we can calculate until we have enough information to find the answer. answer. In this case we know the % tickets sold will sum to 100% so the percentage sold to club members will be %non-member %non-members s = 100% - ( %club-members %club-members + %children %children ) = 100% - ( 50% + 10% ) = 40% % tickets sold
price
% total income
club-members
5 0%
$11
children
1 0%
$5
non-members
4 0%
$15
?
total
1 0 0%
-
1 0 0%
income
Now we will work out the amount of income from each group. The income from each group will be the number of tickets sold multiplied by the price of each ticket. Since we do not
know the total number of tickets sold we can assume that there were 100 tickets because this will make the mathematics easier income income = number of tickets tickets sold sold × price of ticket ticket from club-mem club-members bers = 50 × $11 $11 = $550 $550 from from childre children n = 10 × $5 = $50 $50 from non-memb non-members ers = 40 × $15 $15 = $600 $600 total income income = $550 + $50 + $600 $600 = $1200 $1200 % tickets sold
price
% total income
income
club-members
5 0%
$11
$ 55 0
children
1 0%
$5
$50
non-members
4 0%
$ 15
?
$ 60 0
total
1 0 0%
-
1 0 0%
$ 1 , 2 00
Now that we have the total income and the income fr om non-members we can find the percentage we need.
non-members / total = 600/1200 =
= 50%
% tick ticket ets s sold sold pric price e
% tota totall in income ome
inco incom me
club-members
5 0%
$11
$ 5 50
children
1 0%
$5
$ 50
non-members
4 0%
$1 5
? = 50%
$6 0 0
total
1 0 0%
-
1 0 0%
$1,200
Returning to the question.
At a football game 50% of the seats are sold to season ticket holders who pay $11 each and 10% are sold to children c hildren who pay $5 each. All the remaining tickets are sold to non-members at $15 each. What proportion of the total gate receipts for the game is contributed by non-members?
A 6 0 % B 5 2 % C 5 0 % D 4 0 % E 5 %
50% was contributed by non-members so the answer is C.
Quadratic equations
Quadratic equations are of the form ax 2 + bx + bx + c = c = 0 where a, b and c are c are real numbers and, more specifically, in the GRE they will be integers. For example x 2 + 5 x x - 6 = 0 To solve an equation like this you will have to find the values of x for x for which the equation holds true. Normally there will be two such values. We will begin by looking at the factorization of quadratic equations in detail because you need to know it in the GRE and it can also be used to solve quadratics. Then we will see how to solve them more quickly quic kly.. If you have littl e patience for following the math and you would just like a quick method for solving quadratic equations then skip ahead. ahead. Factorizing quadratic equations The easiest way to solve a quadratic equation is by factorization. x 2 + 5 x x - 6 = ( x + x + g) ( x + x + h)
Where we g and h are unknowns
= x 2 + gx + gx + hx + hx + gh = x 2 + (g (g + h) x + x + gh Therefore, to factorize, we need to find values of g and h so that. g + h = 5 and gh = -6 What are the factors of -6? 1 and -6 -1 and 6 2 and -3 -2 and 3 Which of these pairs sum to 5? -1 + 6 = 5 Therefore g and h must be -1 and 6. x 2 + 5 x x - 6 = ( x x - 1) ( x + x + 6) = 0 You can check the factorization by multiplying out ( x x - 1) ( x + x + 6), if you wish. This wil l now give us the solution for the product of these two factors to be 0, one of the two tw o factors must be 0, i.e. x x - 1 = 0 or x or x + +6=0 Therefore x = x = 1 or -6
Notice that the signs have changed from g and h, which were -1 and 6, to the solutions for x , which are 1 and -6. A quick method We can condense this method of solving a quadratic equation into 3 steps if the equation is in the form x form x 2 + bx + bx + c = c = 0. of c 1. Find the factors of c 2. Decide which of these pairs of factors sum to give b 3. Reverse Reverse the signs signs of the two numbers numbers you have have found found and these these will be the solutions solutions for the equation An example So now an example for you to try. Solve x Solve x 2 - 7 x + x + 10 = 0 Explanation: 1. Find Find the the fact factor ors s of 10. 10. 1 and 10 -1 and -10 2 and 5 -2 and -5 2. The The pair pair whi which ch sum sum to to -7 are are -2 and -5. 3. Reversing Reversing the the signs, we find 2 and and 5 which are are the solutions solutions to this quadrati quadratic c equation. So the solution is x is x = = 2 or 5
Solving by substitution If you do not know how to do an algebra question then you can often use the fact that the GRE is a multiple choice test to your advantage. You can substitute the value of your choice for the variable in the algebraic expression and then evaluate the answer choices to see which one is equal to the original expression. Example
is equivalent to which of the following expressions? A x
3 B x
2 C x
+ 1 D x
+ 2 E x
+ 3
If you followed 'solving quadratic equations' carefully you should be able to see how to answer this question using factorization but let us assume, for now, that you do not know.
Substitute the number of your choice into the equations to replace x and x and evaluate them all. Pick a number which will make your calculations easy for example 0. If x = 0 then,
=
=3
A: x A: x -- 3 = 0 - 3 = -3 -3 B: x B: x -- 2 = 0 - 2 = -2 -2 C: x C: x + + 1 = 0+ 1= 1 D: x D: x + + 2 = 0+ 2= 2 E: x E: x + + 3 = 0+ 3= 3 Both the expression and answer E have a value of 3, therefore E is the correct answer. It is left as an exercise for you to solve this question using factorization.
A summary of word problems and algebra In this tutorial we have looked at techniques for solving word problems. •
•
Summarize: Summarize: Organize the information you are given into equations or a table. o When using equations, use sensible variable names. o Identify exactly which value you have to find. o Solve: Solve: o o
o
Keep working even if you are not sure exactly what to do. When using equations, keep eliminating variables until you have only the one you need to find. With tables, keep calculating more elements in the table until you arrive at the answer you need.
...and algebra. •
•
bx + c = c = 0. Quadratic equations: equations: x 2 + bx + Find the factors of c of c o Decide which of these pairs of factors sum to give b o Reverse the signs of the two numbers you have found and these will be the o solutions for the equation Substitution: Substitution: Pick an nice easy number to substitute the variable with. o Evaluate all the expressions and answer choices. o
Exponents, Ratios & Percents Tutorial by Joel Chippindale, Chippindale, 10th October, 1999 This tutorial covers not only exponentials but also ratios and percentages. These topics have been brought together into a single tutorial because they are all important basic skills which you will need in the GRE and none of them is individually large enough to justify a tutorial of its own. If you can handle fractions (see previous math tutorial) and the topics in this tutorial you should be able to do nearly half of a ll the GRE math questions.
Common Exponentials We will begin with an example of an exponential. 5 3 = 5 × 5 × 5 = 125 '5 cubed' '5 to the power of 3' '5 raised to the 3 rd power' 5 3 is an exponential. It is a shorthand way of writing '5 times itself 3 times'. 5 is the base, i.e. the number which is multiplied by its elf. elf. 3 is the exponent or power, power, i.e. the number times that the base is multiplied by itself. In a similar way, 27 = 2 × 2 × 2 × 2 × 2 × 2 × 2 2 times itself 7 times. ...and, y5 = y × y × y × y × y y times itself 5 times.
Squares and Cubes Squares, cubes and their respective inverses, square roots and cube roots, are the exponents that come up most regularly in the GRE. It is therefore useful to memorize them, so as to have them at your fingertips. Note: I have left out the cubes of 7 and above, not because they do not exist, but simply because you do not need to know them.
Number ( x )
Square ( x 2 )
Cube ( x 3 )
1
1
1
2
4
8
3
9
27
4
16
64
5
25
125
6
36
216
7
49
-
8
64
-
9
81
-
10
100
-
11
121
-
12
144
-
13
169
-
14
196
-
15
225
-
16
256
-
If you can remember these then it will s ave you time and lessen your chances of making mistakes in the GRE.
Square and Cube Roots The square root of a number is the number which when multiplied by itself gives you the original number. For example, = 3 because 32 = 9 = 5 because 5 2 = 25 Similarly the cube root of a number is that which multiplied by itself 3 times produces the original number. For example,
= 2 because 23 = 8 = 4 because 4 3 = 64 Note: 64 is the smallest number which is both a square (8 2) and a cube (43).
Properties Of Exponents To The Power Of Zero What does it mean to raise a number to the power of zero? For example What is 50 ? By definition we say that any number to the power of zero is equal to one. For example, 50 = 1 2250 = 1 a0 = 1
To The Power Of A Negative Number What does it mean to raise a number to the power of a negative number? For example What is 5(-2)? By definition we say that it is equal to the inverse (or reciprocal) of the positive power. power.
5(-2) =
=
4(-3) =
=
x(-n) =
Fractional Exponents The only fractional exponents you will come across in the GRE are 1/2 and 1/3. These correspond to square root and cube root respectively. respectively. For example,
161/2 =
= ±4
271/3 =
=3
Multiplication, Division & Powers Of Exponents Often when a mathematical expression contains more than one exponent it is possible to simplify it.
Multiplication Of Exponents If you multiply two exponents with the same base then you simply have to add two exponents. for example, 52 × 54 = ( 5 × 5 ) × ( 5 × 5 × 5 × 5 ) = 5(2+4) = 56 5 times itself twice times 5 times itself 4 times is 5 times itself 6 times. More generally zn × zm = z
(n+m)
Division Of Exponents You can do a very similar operation to simplify the division of exponents that have the same base. base. This time, instead of adding the two exponents, you subtract them. For example,
7 5 ÷ 73 =
= 7(5-3) = 72
...and more generally, generally, zn ÷ zm = z
(n-m)
Exponents Of Exponents You can also simplify an exponent of an exponent, this time you multiply the exponents. For example, 2 to the power of 3 all to the power of 4, (23)4 = 2(3×4) = 212 (zn)m = z(n×m) = znm
Different Bases You cannot use the rules of multiplication and division with exponents which have different bases. For example, 3 4 × 52 The first exponent has the base 3 a nd the other 5, so you cannot simplify the expression. However, However, if one of the bases is a power of the other, other, you can transform them into an expression where they have a common base. For example, 2 5 × 82 In this case, 23 = 8 Therefore you can replace the 8 in the original expression with 2 3 = = = = =
25 × (23)2 25 × 2(3×2) 25 × 2 6 2(5+6) 211
That's all you need to know about exponents for now.
Ratios Ratio's can be expressed in three different ways, i .e. the following ratios are all the same:
1:2
1 to 2
An Example of Ratios The ratio of foreign to local students is 1 to 4 in The University of Macondo. If there are 1000 local students, how many foreign students are there? Use the rule of three to answer thi s question. 1 : 4 ? : 1000 Therefore, ? = 1 × 1000 ÷ 4 = 250 There are 250 foreign students at The University of Macondo.
Note: the order of the numbers in a ratio is very important. In the previous example if the ratio had been 4 to 1 then there would have been 4000 foreign students.
Triple Ratios The numbers of dogs, goats and hens on a farm are in the ratio 2:3:10 respectively. If there are 60 hens then how many dogs are there? When faced with a triple ratio you should always pick out the ratio of the two things you are interested in. In this case you know how many hens there are and you want to know how many dogs there are so pick out the ratio that relates hens to dogs. The ratio of dogs to hens is 2:10 or expressed in its simplest form 1:5. You should be able to see that this means there are 1/5 the number of dogs as hens and therefore if there are 60 hens there w ill be 12 dogs.
Percentages Percentages are a real favourite of the GRE question writers. You must be comfortable with them if you are to do well in the tes t. Percent means per hundred and therefore can easily be represented as a fraction with denominator 100. We already saw how the most common percentages could be expressed as both fractions and decimals in the Fractions tutorial. If you did not do the Fractions tutorial or do not remember the conversions then remind yourself here here..
Example of a Percentage Question What is 20% of 360 ? To calculate the percentage of a number you simply multiply the percentage by that number 20% × 360
...and since you did the Fractions tutorials you know that 20% =
× 360 = 72 So 20% of 360 is 72. Note: a neat trick for dividing di viding by 5 is to first multiply by 2 and then divide by 10. In this example 360 multiplied by 2 is 720 divided by 10 is 72. Easy. Easy.
Sales and Discounts Sales and discounts are a common topic of GRE questions. For example,
A chair normally priced at $48, is discounted by 25% in a sale. What is the chair's sale price? In these questions you could work out the discount and then take the di scount away from the original price but you can save time by going straight to the sale price. In this case a 25% discount will mean the sale price is 75% of the original. 75% × 48
...and you know that 75% =
× 48 = 36 The sale price of the chair is $36.
Percent Change There are many questions that will a sk you to calculate the percentage increase or decrease from one number to another. another. In these cases you will have to calculate the difference and then work that out as a percent of the original number. number. We will begin with an simple example, What is the percent increase fr om 50 to 75 ? First calculate the difference 75 - 50 = 25 Then work out the percentage of the original
=
= 50%
The percentage increase from 50 to 75 is 50%. It is important to note that the reverse, the percentage decrease from 75 to 50 is not the same. The absolute difference is sti ll 25 but the base number is now 75.
=
= 33
%
The percentage decrease from 75 to 50 is 33
%.
Now it is time to put your new skills into action with some practice questions.
Fractions Tutorial by Joel Chippindale Chippindale,, 15th September, 1999 We are going to start tutorials on the quantitative section with the very basics. In this tutorial you will dive straight into arithmetic and the unsung heroes of the GRE, fractions. We will begin with a look at how useful fractions are, follow up with reminders of how to do all the operations (addition, subtraction, multiplication and division) and round everything off with a review of improper and mixed fractions.
Why Do You Need Fractions? Now we all remember fractions from school, denominators, numerators, improper, improper, mixed etc. You probably also remember hating them, so why would you want to learn about them for a GRE test? If you didn't already know here is the bad news. You are not allowed to use a calculator in the GRE. That's right, NO CALCULATORS, CALCULATORS, just plain old paper and pencil. And this is where fractions come into their own. Fractions are quicker and more accurate to manipulate by hand than decimals. To prove the point lets do an example.
Example What is 64 × 0.125 ?
Try to repeat the calculation yourself. Although you probably know how to do it, you will w ill find it takes a long time and time is an asset that you cannot afford to waste in the GRE. Also, there are many calculations to do and each one of those introduces the possibility of making a mistake. Add a zero here, forget a decimal point there, forget to carry the 2 somewhere else.
An easier way Those of you who are real math whizzes wi ll have spotted that 0.125 is a fraction in disguise.
0.125 =
We will use this fact to rephrase the question:
What is 64 ×
64 ×
=
?
=8
A piece of cake! cake! ...if you already knew that 0.125 =
.
How Do You Spot A Fraction? We have already seen that 0.125 is equal to 1/8. How did we know that? The easiest way to have these fractions at your fingertips is to memorize them.
Fractions Percentages and Decimals This table shows the most commonly used fractions (in the GRE) and their decimal a nd percent equivalents.
Fraction
Decimal
Percentage
0.5
50%
0.333...
0.666...
33 %
66 %
You will notice that one third is written 0.333... , that is to say the 3 repeats an infinite number of times. This is one of the other advantages of fractions, some numbers cannot be written down in decimal form.
0.25
25%
0.75
75%
Why didn't we include
?
Because we have already included it in the table.
=
0.2
20%
0.4
40%
Followed by 60% and 80%.
0.1666...
0.8333...
You should be able to see why
,
and
16 %
83 %
have not been included.
0.125
12.5%
0.375
37.5%
0.625
62.5%
0.875
87.5%
0.111...
11%
0.222...
A clear pattern emerges from
,
22%
...
0.1
10%
0.05
5%
0.01
1%
Do you think you can remember all that? Have one last look over the fractions in the table and then see if you can do the exercise on the next page.
An Exercise Using Fractions Yes, it is time for you to do a little work. Here's tha t table again but this time it is full of gaps. Print it out and fill in the blank spaces with the appropriate fraction, decimal or percentage.
Fraction Decimal Percenta ge
Fraction Decimal Percenta ge
50%
0.333...
75%
0.125
Did you just skip this exercise? Do have a go at filling in the spaces before you go onto the next part of the tutorial. It will help you to remember all these fractions. You can compare your answers with the previous page. page.
Multiplication & Division of Fractions Now that we know how to find a fraction, even if it is disguised as a decimal or as a percentage, it is time to remind ourselves how to use them.
Multiplication Multiplication is the easiest of the operations to perform on fractions. Simply multiply the two numerators (top numbers) to find the resulting numerator, numerator, then multiply the two tw o denominators (bottom numbers) to find the new denominator. denominator.
Example of Multiplication ×
=
=
Multiply numerators 3 and 2, and denominators 5 and 7, to give 6/35.
Division Division is multiplication by the inverse of the divisor (the second fraction). First find the inverse (also called the reciprocal) which with a fraction involves turning it upside down, i.e. swapping the top and bottom numbers. Then multiply as before
Example of Division ÷
=
×
=
=
Invert 2/5 to give 5/2 and then multiply numerators and divisors as before.
Shortcuts The key to doing well in the math section of the GMAT is to do the minimum amount of work possible, so if there is a shortcut, take it. Always cancel any common factors from the numerators and denominators before you begin to do any multiplication.
Example of Cancellation ×
=
=
In this case you can cancel the common factor of 3 from the 3 and the 6 before multiplying the two fractions.
Addition & Subtraction of Fractions Addition Addition is easy when the denominators are the same. You add the numerators and keep the denominator the same.
Example of Addition (1) +
=
=
Both fractions are expressed in fifths and so the denominator is 5 in both fractions. In this case, simply add the two numerators 1 and 3 to give you 4. The denominator stays the same, so the sum is 4/5. Sadly, Sadly, it is much more common to find that the denominators are different. In these cases you have to reorganize the fractions so that they have t he same denominator. denominator. This is called a common denominator. If the two denominators have no common factors or are variables, then you simply multiply them together to find the common c ommon denominator. denominator. You then have to multiply the numerators by the same value that you multiplied the denominator by so that the fractions still have the same value. Now you have two fractions with the same denominator you can add them up as before.
Example of Addition (2) + The denominators are 3 and 4, which are different. Therefore we need to find a common denominator. denominator. Since the denominators 3 and 4 have no common factors, multiply them together 3 × 4 = 12. Then multiply the numerators to give us the same fractions but expressed in twelfths i.e. with the same denominator.
=
×
=
=
×
=
Now we can add them.
+
=
+
=
If the two denominators do have a common factor then it is usually best to find the lowest common denominator you can, which will be smaller than that given by multiplying the two denominators together.
Example of Addition (3) + First find a common denominator. denominator. We could multiply the denominators 8 and 12 together to give us a common denominator of 96. But, since they have a common factor of 4 we can work with the lowest common denominator which is 24 and will make our calculations a little simpler. Then multiply the numerators.
=
=
×
=
×
=
Now we can add them.
+
=
+
=
Subtraction Subtraction is performed in exactly the same way as addition but you subtract the numerators rather than add them.
Improper & Mixed Fractions A fraction in which the numerator is greater than the denominator is called an improper fraction. For example:
In this case the numerator, numerator, 5, is greater than the denominator, denominator, 2. This will be true of any fraction which is greater than 1. If we divide the numerator by the denominator we can split the improper fraction into its whole number and fraction parts: 5 ÷ 2 = 2 with remainder 1 Therefore 5/2 is 2 wholes and 1 half left over or two and a half.
=2 Two and a half is a mixed fraction. I.e. a whole number added to a fraction. It is very unusual to see an improper fraction listed in the answers to a GMAT question, so you will often need to convert your answer into a mixed fraction.
Example What is
written as a mixed fraction?
60 ÷ 7 = 8 with remainder 4 Therefore 60/7 is 8 wholes with 4 sevenths left over, over, or eight and four sevenths.
=8 In this tutorial we have discussed why we need to use fractions, revised the operations +, -, × and ÷ with fractions and finally looked at the different forms that fractions may take.
Geometry Tutorial by Joel Chippindale, Chippindale, 12th November, 1999 Geometry is a common topic in Quantitative GRE questions. The questions are generally (although not always) a little easier than the majority of the other questions and a large part of doing well on these types of questions will be how good you are at remembering everything that is mentioned in this tutorial. This tutorial will cover everything you need to know about geometry for the GRE, from angles and lines through triangles, quadrilaterals and circles to Cartesian geometry and 3D geometry. geometry. None of theses concepts is difficult and you will pleased to learn that trigonometry is NOT required in the GRE.
Geometry Tips There is really only one very important tip that I want to give you for geometry, draw your own diagrams. diagrams. This is especially important with Data Sufficiency questions since the diagrams provided by ETS in these questions are often deliberately misleading. When drawing diagrams the following guidelines are useful. •
To paraphrase ETS, "In the diagrams in the GRE everything that looks like a straight line should be considered a straight line and where lines appear to touch they do but the sca le of the drawings may not be accurate, i.e. just because t wo lines look the same length does not mean they are the same length, just because one angle looks larger than another does not mean that it is. Lines which are parallel or perpendicular or the same length will be marked as such on the diagram or described in the accompanying text."
•
•
If you know the lengths of lines or the size of angles try to make the diagram more or less to scale. This should reduce the number of mistakes you make since you can see what is going on more clearly. Unless you know lines or angles are definitely equal then draw them as different as you can. This will stop you making wrong assumptions w hen you are answering the question.
If you are answering questions from a book (or from a printed version of Test Tutor Tutor questions) then you should still draw your own diagram and avoid just adding notes to a n existing diagram because in the real test all diagrams will be on a computer screen.
Lines and Angles Angles: Angles: All angles in the GRE are measured in degrees and there are 360 o in a full circle.
Crossing Lines: Lines: When 2 lines cross they form 2 pairs of angles. The angles opposite each other are equal and angles next to each other add up to a straight line, i.e. 180 o. Therefore, in the diagram below, all angles marked x marked x are are equal and all angles marked y are y are o equal, finally x finally x + + y = y = 180 .
Parallel lines: lines: Parallel lines, on a plane, are those which never cross. They are always the same distance apart however far you extend them. The symbol < is often used to show that two lines are parallel. If a line crosses 2 parallel lines then it will form 2 sets of 4 equal angles, see the diagram below. All angles marked x marked x are are equal and all angles marked y are y are equal.
Perpendicular Lines: Lines: Lines which meet at right angles, i.e. 90 o, are perpendicular lines.
You should be able to see that all the other angles in the diagram a bove are also right angles.
Triangles You will probably see more questions involving triangles than any other part of geometry in the GRE so pay attention to these next two pages.
Basic Triangle Facts Angles: Angles: The sum of the angles in a triangle is always 180o. Area: Area: The area of a triangle is 1/2 × base × height, where the height is the perpendicular distance from the base to the top of the triangle.
Area =
×b×h
Triangle Inequality: Inequality: The triangle inequality says that no one side of a triangle can be longer than or equal to the sum of the other two. The reason for this is fairly obvious when you think about it because if one side is l onger than or equal to the sum of the other two then you cannot join the two shorter sides to form a triangle.
Therefore if we have the following triangle.
We can write down the following inequalities. r
Types of Triangles •
•
Isosceles: Isosceles: An isosceles triangle is a triangle where two of the sides are equal, and two of the angles are equal.
Equilateral: Equilateral: An equilateral triangle is a triangle where all three sides are equal, and all three angles are equal. You should should be able to see that since the angles sum to o o 180 , each angle must be 60 .
•
Right Angle: Angle: A right-angled triangle is a triangle with one right angle in it. I leave it as an exercise for you to work out why a triangle cannot have two right angles. We will see why right-angled triangles are so importa nt next.
The Pythagorus Theorem
If you have a right angled triangle with sides of length a, b and c then a2 + b2 = c2. For this formula to work side c must be the hypotenuse which is the longest side. You can also identify the hypotenuse because it is opposite the right angle. This is one of the most important formulas that you will learn for the GRE, it is also one of the formulas that people make the most mistakes with s o be careful. When we do some examples with Pythagorus we will see some of the most common mistakes made.
A Note on Square Roots You will need to know about simplifying square roots if you are to use Pythagorus successfully in the GRE. Basically if you are taking the square root of a number which has a factor which is a perfect square (like 4, 9, 16, 25....) then you can take the factor out of the square root. For example, if you have a right angled triangle with short sides 4 and 2 what is the length of the hypotenuse? To solve this you will use Pythagorus. Set c to c to be the length of the hypotenuse.
c 2 = 42 + 22 c 2 = 16 + 4 c 2 = 20
c = c =
You will never find amongst the answers in a GRE question because 4 is a factor of 20 and is also a perfect square and so we can simplify this square root.
c = c =
c = c =
c = c =
×
c = c = 2 ..and this is the answer that you would find in a GRE question. Whenever you have a root in your answer you need to make sure that it is in its simplest form which means taking any factors which are perfect squares out of the r oot.
Common Triangles Although we have, in theory, covered everything you need to know about tr iangles it would probably be worth familiarizing yourself with some of the triangles that turn up on a regular basis in the GRE.
3-4-5 Triangle: Triangle: This is a right-angled triangle with sides of length 3, 4 and 5. You can check that it is a r ight-angled using Pythagorus i.e. does 3 2 + 42 = 52 ?.
This triangle is used in the GRE because the lengths of its sides are integer values (which makes it easier to work with w ithout the aid of a calculator. calculator. This triangle will often turn up in disguise, for example all the triangles below are basically the 3-4-5 tr iangle, they have just been scaled up or down.
The 30-40-50 triangle is 10 times as big. The 12-16-20 triangle is 4 times as big. The 1.5-22.5 triangle is half the size.
5-12-13 Triangle: Triangle: This is a right-angled triangle with sides of length 5, 12 and 13. You can check that it is a r ight-angled using Pythagorus i.e. does 5 2 + 122 = 132 ?.
This is much less common in the GRE than the 3-4-5 triangle but is the only other distinct right-angled triangle which has small integer sides and for this reason it also appears in the GRE.
Isosceles Right Triangle: Triangle: This right-angled triangle with two sides the same length is a GRE favourite because you can work out all the angles in it. Since the two unknown angles are equal they must be 45 o.
Also if we assume that the short sides have length 1, what is the l ength of the hypotenuse? 12 + 12 = x 2 2 = x 2
Set the hypotenuse to be x be x and and get practicing your Pythagorus.
= x
So length of the hypotenuse would the diagram below).
, which gives us the ratios between the sides (see
30o - 60o - 90o Triangle: Triangle: This is the triangle formed when you cut an equilateral triangle in half (see below) and is a GRE favourite because you know all the angles in it.
We can also calculate the ratios between the lengths of the s ides for this triangle. Assume the length of side of the equilateral triangle is 2. Then the hypotenuse of the 30 o - 60o - 90o triangle is 2 (since it is one of the sides of the equilateral triangle). The base is 1 (since it is half of one of the sides of the equilateral triangle). And we will use Pythagorus to find the length of the final side. 12 + = 22 x 2
Set the height to be x be x and and we already know the base is 1 and the hypotenuse is 2.
1 + x + x 2 = 4 x 2 = 3 x = root 3
Which gives us sides of lengths 1,
and 2 as seen in the diagram below.
Most Quadrilaterals... The most common quadrilaterals that you will see in the GRE are the parallelogram, the rhombus, the rectangle and the square. Parallelogram A parallelogram is the quadrilateral formed by 2 pairs of parallel sides (thus the name).
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2 pairs of parallel sides. 2 pairs of equal sides. Diagonals bisect each other i.e. cut each other exactly in half.
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Opposite angles are equal.
• •
Rhombus A rhombus is a parallelogram in which all the sides are equal.
2 pairs of parallel sides. 4 equal sides. Diagonals bisect each other i.e. cut each other exactly in half. Diagonals are a re perpendicular. •
•
•
•
•
Opposite angles are equal.
Rectangle A rectangle is a parallelogram where all the angles are right angles (90o).
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2 pairs of parallel sides. 2 pairs of equal sides. Diagonals bisect each other i.e. cut each other exactly in half. Diagonals are equal.
•
All angles are 90o.
• • •
Square A square is a special s pecial kind of parallelogram (and rhombus and rectangle) where all the sides are equal and all the angles are right angles.
2 pairs of parallel sides. 4 equal sides. Diagonals bisect each other i.e. cut each other exactly in half. Diagonals are a re perpendicular. Diagonals are equal. •
•
•
•
•
•
All angles are equal.
Area The area of all the previous quadrilaterals, parallelogram, rhombus, rectangle and square is calculated in exactly the same way; Area = base × height. Area = b × h Note: Note : In all cases height is measured perpendicularly from the base so in the cases of a rhombus and a parallelogram it is not the same as the length of the side. See diagram below.
...And The Trapezium The last quadrilateral you will meet in the GR E is the trapezium. A trapezium is a quadrilateral with 1 pair of parallel sides. All you need to know about a trapezium is how to ca lculate its area; Area = average of the two parallel sides × perpendicular distance between them.
Area =
( b1 + b2 ) × h
Circles First some 'circle' 'circ le' vocabulary. vocabulary.
Circumference The edge of a circle. Diameter A line which joins two points on the circumference of the circle passing through the center of the circle. Radius A line which joins the center of the circle to the circumference. Chord A line which joins two points on the circumference of the circle. Tangent A line which touches the circumference at only one point. A tangent is always perpendicular to a radius or diameter which touches the circumference in the same place.
Area The area of a circle is PI × the radius squared. Area =
r2
You will not usually need to know the value of PI but just in case it does come up a test you should be aware that it is a little bit more than 3. A good decimal approximation is 3.1 or 3.14 and the fraction 22/7 also gives a good ap proximation.
Circumference The length of the circumference of a circle is 2 × PI × the radius. Circ Circum umfe fere renc nce e =2
r
GRE students commonly mix up these formulas, if you find you are using the wrong formula try to remember that area comes in square units (meters squared, m 2, or centimeters squared, cm2) and so it is the area formula which has radius squared in it.
Sectors and Arcs A sector is like a slice of pizza, it is just a part of circle cut out by two radii. An arc is part of the circumference of the circle. If you are asked to calculate the area of a sector (or the length of an arc) you should work out the area of the whole circle (or length of the whole circumference) and then take the appropriate fraction of that, remembering that there are 360o in a complete circle.
For example. If we were to work out the area of the sector with radius 6 and an angle of 60 o which is shown in the diagram above then we would firs t work out the area of the entire circle and then multiply by the fraction of the circle that the sector covers (60 o out of 360 o).
Sector Area = Area of Circle ×
=
62 ×
=6
Cuboids and Cylinders There are only two 3D forms that you have to deal with in the GRE and they are the c uboid (or 'box') and the cylinder (or 'tube'). For these forms you should know how to calculate their volumes and their surface areas.
Cuboid
Volume The volume of a cuboid is easy to calculate just multiply length × width × height. Volume = lwh
Surface Area A cuboid is made up of 6 surfaces the front and back, the top and bottom and the 2 ends. Simply calculate the area of each of these and add them up.
Surface Area = lh + lh + wh + wh + lw + lw + lw = lw = 2(lh 2(lh + wh + lw )
Cylinder
Volume The volume of a cylinder is the area of its base (a circle radius r) × height. Volume =
r 2h
Surface Area To calculate the surface area of a cylinder we need to think about what surfaces make up a cylinder. cylinder. It has a top and a bottom which are both circles of radius r and r and the side can be formed from a rectangle of height h and width 2 r so r so that it stretches all the way round the circumference of the circles at the top and bottom. To find the surface area of the whole we work out the area of each piece and an d add them together t ogether..
Surface Area =
r 2 +
r 2 + 2
rh = 2
r (r + r + h)
Coordinate Geometry Every point in coordinate geometry is specified by two coordinates, an x coordinate x coordinate which determines the horizontal position of the point and a y coordinate y coordinate which determines the vertical position of the point. The x The x -axis -axis is measured from left to r ight (i.e. left is negative and right is positive) and the y -axis -axis is measured from bottom to top. For example the point ( -2, 4 ) is shown in the diagram below.
Equations of Lines A line can be described by a linear equation in x in x and and y the y the solutions of which form the points on a line. The standard form of the equation of a line is: y = y = m x + x + c Where m is the gradient (or slope) of the line and c is the y intercept, y intercept, where the line crosses the y -axis. -axis. For Example,
You can see that the y intercept y intercept is -3 and that the gradient is 2 (i.e. the line rises by 2 each time you move 1 space left).
Geometry Reference Equations You must learn the following equations to be able to answer geometry questions successfully. successfully. Remember you cannot take notes into the GRE.
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Area of a Triangle =
bh
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Pythagorus Theorem: Theorem: a2 + b2 = c2
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Area of a Trapezium =
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Area of a Circle =
r2
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Circumference = 2
r
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Volume of a Cuboid = lwh
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Surface Area of a Cuboid = 2(lh 2(lh + wh + lw )
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Volume of a Cylinder =
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Surface Area of a Cylinder = 2
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Equation of a Line: Line: y = y = m x + x + c
( b1 + b2 ) × h
r 2h r (r + r + h)
Vocabulary It is useful to know the following vocabulary. •
Bisect: Bisect: Cut exactly in half.
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Circumference: Circumference: The perimeter of a circle.
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Gradient: Gradient: The rate at which a line l ine goes up (or down).
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Parallel: Parallel: Lines which lie in a plane and never meet.
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Perimeter: Perimeter: The length of the border of a shape.
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Perpendicular: Perpendicular: At right angles.
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Slope: Slope: The rate at which a line goes up (or down). See also 'Gradient'.
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Surface Area: Area: The area of the faces of a shape.
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Trisect: Trisect: Cut exactly in three.
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y intersect y intersect:: Point at which a line crosses the y-axis.
