Algebra&word Problems Tutorial

Algebra & word problems tutorial 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' and 'Exponents, ratios and 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. 2a + 3

= 11

a =?

solution at the bottom of this page

Solve this problem by finding the value for a. 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 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? A $9,600 B $8,000 C $6,400 D $4,000 E $3,200 If you feel brave you can have a go at it now. My advice would be to follow these guidelines. •

Summarize: Write equations, which contain the information given to you in the question.

•

o

Use sensible variable names for example the first letter of the thing that the variable represents.

o

Identify answer i.e. write down exactly what you are looking for.

Solve: find the solution for the equstions you have written down.

If you have had a go at the question then check your answer. If not then we will answer 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 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% × S ) + ( 5% × B ) = 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 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. ( 10% × S ) + ( 5% × B ) = 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% × B = 800 We know that 25% = = 800 ×B

, (see fractions)

Multiply both sides by 4

B = 800 × 4 B = 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 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? A $9,600 B $8,000 C $6,400 D $4,000 E $3,200 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 we 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 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 60% B 52% C 50% D 40% E 5% If you would like to try the question now then follow these guidelines. •

•

Summarize: o

Organize the information you are given into a table.

o

Identify answer i.e. mark on the table exactly what you are looking for.

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. If not then we will answer 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 $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 non-members?' In the question we have three different types of tickets, 'club-members', 'children' and 'non-members' and three different types of information are given or asked for, '% of tickets sold', 'price of ticket' and '% of total income'. Therefore we would sketch a table, which is 3 × 3. % tickets sold price

% total income

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 $11 each and 10% are sold to children who pay $5 each.' % tickets sold price club-members

50%

$11

children

10%

$5

% total income

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?'' % tickets sold price club-members

50%

$11

children

10%

$5

non-members

% total income

$15

?

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.

% tickets sold price club-members

50%

$11

children

10%

$5

non-members

% total income

$15

?

We will add a totals row because we are working with percentages and an income column so that we can later work out the percentages for the income. We can fill in 100% for the totals of the percentages. % tickets sold price club-members

50%

$11

children

10%

$5

non-members total

100%

% total income

$15

?

-

100%

income

Then it is a matter of filling in as many cells as we can calculate until we have enough information to find the answer. In this case we know the % tickets sold will sum to 100% so the percentage sold to club members will be %non-members = 100% - ( %club-members + %children ) = 100% - ( 50% + 10% ) = 40% % tickets sold

price

% total income

club-members

50%

$11

children

10%

$5

non-members

40%

$15

?

total

100%

-

100%

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 = number of tickets sold × price of ticket from club-members = 50 × $11 = $550 from children = 10 × $5 = $50 from non-members = 40 × $15 = $600 total income = $550 + $50 + $600 = $1200 % tickets sold

price

% total income

income

club-members

50%

$11

$550

children

10%

$5

$50

non-members

40%

$15

?

$600

total

100%

-

100%

$1,200

Now that we have the total income and the income from non-members we can find the percentage we need. non-members / total = 600/1200 =

= 50%

% tickets sold price

% total income

income

club-members

50%

$11

$550

children

10%

$5

$50

non-members

40%

$15

? = 50%

$600

total

100%

-

100%

$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 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 60% B 52% C 50% D 40% E 5% 50% was contributed by non-members so the answer is C.

Quadratic equations Quadratic equations are of the form ax2 + bx + c = 0 where a, b and c are real numbers and, more specifically, in the GRE they will be integers. For example x2 + 5x - 6 = 0 To solve an equation like this you will have to find the values of 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. If you have little patience for following the math and you would just like a quick method for solving quadratic equations then skip ahead.

Factorizing quadratic equations The easiest way to solve a quadratic equation is by factorization. x2 + 5x - 6 = (x + g) (x + h)

Where we g and h are unknowns

2

= x + gx + hx + gh = x2 + (g + h)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. x2 + 5x - 6 = (x - 1) (x + 6) = 0 You can check the factorization by multiplying out (x - 1) (x + 6), if you wish. This will now give us the solution for the product of these two factors to be 0, one of the two factors must be 0, i.e. x - 1 = 0 or x + 6 = 0 Therefore 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 x2 + bx + c = 0. 1. Find the factors of c 2. Decide which of these pairs of factors sum to give b 3. Reverse the signs of the two numbers you have found and these will be the solutions for the equation An example So now an example for you to try. Solve x2 - 7x + 10 = 0 Explanation: 1. Find the factors of 10.

1 and 10 -1 and -10 2 and 5 -2 and -5 2. The pair which sum to -7 are -2 and -5. 3. Reversing the signs, we find 2 and 5 which are the solutions to this quadratic equation. So the solution 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 evaluate them all. Pick a number which will make your calculations easy for example 0. If x = 0 then,

=

=3

A: x - 3 = 0 - 3 = -3 B: x - 2 = 0 - 2 = -2 C: x + 1 = 0 + 1 = 1 D: x + 2 = 0 + 2 = 2 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: o

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.

Solve: o

Keep working even if you are not sure exactly what to do.

o

When using equations, keep eliminating variables until you have only the one you need to find.

o

With tables, keep calculating more elements in the table until you arrive at the answer you need.

...and algebra. •

•

Quadratic equations: x2 + bx + c = 0. o

Find the factors 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 solutions for the equation

Substitution: o

Pick an nice easy number to substitute the variable with.

o

Evaluate all the expressions and answer choices.