Integers are numbers like -10, 0, 2, 5, 189.  You might know them as whole numbers or numbers that do not have decimals.  Rational numbers, in contrast, are numbers that can be expressed as a/b where a and b are integers.  This would include numbers such as ½, ¼ -4 (which you can think of as -4/1), 0 and so on. Most problem solving GMAT questions that deal with operations on rational numbers essentially require you to understand defined operations, to translate word problems into equations and to be able to do conversions. Question Defined Operations There are certain established operations.  We all know that + represents addition and – represents subtraction and / represents division and ² represents squaring a number and so on.  Aptitude tests often come up with their own symbols to represent a particular operation.  Let’s examine the operation represented by ∆ below. The question reads: let x ∆ y = x²/_y for all positive values of x and y. This means that ∆ wants you to square the first number and divide that by the second number.  Thus 6 ∆ 9 = 6²/9 = 36/9 = 4 Word Problems Word problems may look tricky but if you work through each bit of information at a time, you might find that they are not as difficult as you imagined. Let’s try an algebraic word problem: Allyn has t tattoos, which is half as many as Krystal and 4 times as many as Joshua. How many tattoos do the 3 of them have together? Look at the first bit of information.  Allyn has t tattoos.  It is half as many as Krystal’s, meaning Krystal has 2t tattoos.  Allyn has 4 times as many as Joshua, meaning Joshua has t/4 tattoos.   Thus, they have t + 2t + t/4 = 13t/4 tattoos in total. Here’s a much longer, wordier problem: In the first leg of a dog-sledding competition, the teams raced across a 120-mile route. If the Swiss team took 10 hours to finish the first leg, and the average speed of the Canadian team was 25 percent greater than the average speed of the Swiss team, how many hours did it take the Canadian team to finish the first leg of the competition? What do you know?

• The Swiss team took 10 hours to complete 120 miles.  So their average speed is 120miles / 10hours = 12mph
• The Canadian team’s average speed was 25% greater.  25% greater than 12mph is 16mph.
• The Canadian team traveled 120 miles at 16mph meaning they took 120miles/16mph = 7.5 hours.

You can do the same for this question involving percentages: A barrel contains 40 kilograms of syrup that is 35% sugar by weight. If 10 kilograms of sugar are added to the barrel, the resulting syrup will be what percent sugar by weight? What do you know?

• 35% of a 40kg of syrup is sugar.  So 12kg of the syrup is sugar.
• 10kg of sugar is added.  There is now 22kg of sugar.  Overall, the syrup weighs 50kg now.
• So the sugar percentage is now 22/50 %

Given the choices

1. 17%
2. 28%
3. 37%
4. 48%
5. 60%

you should be able to figure the answer without even calculating the precise answer of 22/50 simply by estimating.  22/50% is close to 25/50% which is 50%.  Since 22/50 is less than 25/50, the answer should be choice D: 48%. Conversions One of the simplest operations on numbers is to know how to convert between units.  Sometimes the GMAT will make up units such as in the following question. If 100 “fnords” is equal to 1 “norton”, then how much bigger is 53 “nortons” than 23 “nortons” in “fnords”? If you break this question down like you did the word problems, you would realize that it is essentially asking: how many fnords are there in 30 nortons. Since 1 norton = 100 fnords 30 nortons = 3000 fnords.

Once you’ve figured out whether to use permutation or combination, there is actually very little work to be done if you know the formulas for permutation and combination or if you know where the function is hidden on your calculator. As I mentioned before, permutation is used when the order matters and combination when you just want to choose, but not order, the items.  Let’s dive straight into the formulas then go through a few examples. The number of ways of permuting r objects out of n objects is given by n P r =n!/(n-r)!   where n! = n * (n-1) * (n-2) * … * (2) * (1) The number of ways of combining r objects out of n objects is given by n C r = n!/r!(n-r)! Note that 0! is defined as 1. Example 1 How many ways are there of arranging 4 letters out of the following F R I E N D? FRIEND has 6 different letters, and since the order of letters matter, I know I have to use the permutation formula.  Applying the formula directly, the answer is given by 6 P 4 = 6!/(6-4)! – 6!/2! – 360 You can double-check this by using the fundamental counting principle that we covered in Part I.  Drawing a tree diagram will also work, but it might get a little messy as the tree gets bigger and bigger. Example 2 If there are 3 entrees and 5 desserts, how many ways are there of choosing 1 entrée and 2 desserts?  Note, that the question does not say anything about the order in which the entrees and desserts are eaten, so we know to use the combination principle.  Because we are choosing entrees and desserts separately, we have to apply the combination formula twice. First, to choose 1 entrée out of 3, we apply the formula 3 C 1 = 3!/1!(3-1)!  – 3!/1!2! – 3 Next, to choose 2 desserts out of 5, we apply the formula 5 C 2 = 5!/2!(5-2)! – 5!/2!3! – 10 Then because for each entrée, there are 10 possible 2-dessert combinations and there are 3 ways of choosing 1 entrée, to get the total number of possibilities, we take 3*10 = 30. Example 3 What if the question throws you a curveball and asks you to permute something that has a repeated item.  For example, how many ways are there of arranging the letters A G H A S T? The ‘A’ is repeated twice so first we pretend that the letters are distinct and find the number of possibilities.  Then divide that value by 2! because ‘A’ is repeated twice. AGHAST has 6 letters, and if we permute all the letters, we get 6! Because A is repeated, we divide 6! by 2! to get the answer 360 Supposing I asked you to permute only 4 out of 6 of the letters from AGHAST, then you would do 6 P 4 as per normal, and divide that answer by 2! since A is repeated.  The answer should be 180. Please visit the Grockit forum or leave a comment here to discuss further.

Once you’ve figured out whether to use permutation or combination, there is actually very little work to be done if you know the formulas for permutation and combination or if you know where the function is hidden on your calculator. As I mentioned before, permutation is used when the order matters and combination when you just want to choose, but not order, the items.  Let’s dive straight into the formulas then go through a few examples. The number of ways of permuting r objects out of n objects is given by n P r =  n!/(n-r)! where n! = n * (n-1) * (n-2) * … * (2) * (1) The number of ways of combining r objects out of n objects is given by n C r = n!/r!(n-r)! Note that 0! is defined as 1. Example 1 How many ways are there of arranging 4 letters out of the following F R I E N D? FRIEND has 6 different letters, and since the order of letters matter, I know I have to use the permutation formula.  Applying the formula directly, the answer is given by 6 P 4 = 6!/(6-4)! -= 6!/2! =360 You can double-check this by using the fundamental counting principle that we covered in Part I.  Drawing a tree diagram will also work, but it might get a little messy as the tree gets bigger and bigger. Example 2 If there are 3 entrees and 5 desserts, how many ways are there of choosing 1 entrée and 2 desserts?  Note, that the question does not say anything about the order in which the entrees and desserts are eaten, so we know to use the combination principle.  Because we are choosing entrees and desserts separately, we have to apply the combination formula twice. First, to choose 1 entrée out of 3, we apply the formula 3 C 1 = 3!/1!(3-1)! = 3!/1!2! = 3 Next, to choose 2 desserts out of 5, we apply the formula 5 C 2 = 5!/2!(5-2)! = 5!/2!3! = 10 Then because for each entrée, there are 10 possible 2-dessert combinations and there are 3 ways of choosing 1 entrée, to get the total number of possibilities, we take 3*10 = 30. Example 3 What if the question throws you a curveball and asks you to permute something that has a repeated item.  For example, how many ways are there of arranging the letters A G H A S T? The ‘A’ is repeated twice so first we pretend that the letters are distinct and find the number of possibilities.  Then divide that value by 2! because ‘A’ is repeated twice. AGHAST has 6 letters, and if we permute all the letters, we get 6! Because A is repeated, we divide 6! by 2! to get the answer 360 Supposing I asked you to permute only 4 out of 6 of the letters from AGHAST, then you would do 6 P 4 as per normal, and divide that answer by 2! since A is repeated.  The answer should be 180. Check out Grockit for more GMAT permutations and combinations practice!

Once you’ve figured out whether to use permutation or combination, there is actually very little work to be done if you know the formulas for permutation and combination or if you know where the function is hidden on your calculator. As I mentioned before, permutation is used when the order matters and combination when you just want to choose, but not order, the items.  Let’s dive straight into the formulas then go through a few examples. The number of ways of permuting r objects out of n objects is given by n P r =  n!/(n-r)! where n! = n * (n-1) * (n-2) * … * (2) * (1) The number of ways of combining r objects out of n objects is given by n C r = n!/r!(n-r)! Note that 0! is defined as 1. Example 1 How many ways are there of arranging 4 letters out of the following F R I E N D? FRIEND has 6 different letters, and since the order of letters matter, I know I have to use the permutation formula.  Applying the formula directly, the answer is given by 6 P 4 = 6!/(6-4)! -= 6!/2! =360 You can double-check this by using the fundamental counting principle that we covered in Part I.  Drawing a tree diagram will also work, but it might get a little messy as the tree gets bigger and bigger. Example 2 If there are 3 entrees and 5 desserts, how many ways are there of choosing 1 entrée and 2 desserts?  Note, that the question does not say anything about the order in which the entrees and desserts are eaten, so we know to use the combination principle.  Because we are choosing entrees and desserts separately, we have to apply the combination formula twice. First, to choose 1 entrée out of 3, we apply the formula 3 C 1 = 3!/1!(3-1)! = 3!/1!2! = 3 Next, to choose 2 desserts out of 5, we apply the formula 5 C 2 = 5!/2!(5-2)! = 5!/2!3! = 10 Then because for each entrée, there are 10 possible 2-dessert combinations and there are 3 ways of choosing 1 entrée, to get the total number of possibilities, we take 3*10 = 30. Example 3 What if the question throws you a curveball and asks you to permute something that has a repeated item.  For example, how many ways are there of arranging the letters A G H A S T? The ‘A’ is repeated twice so first we pretend that the letters are distinct and find the number of possibilities.  Then divide that value by 2! because ‘A’ is repeated twice. AGHAST has 6 letters, and if we permute all the letters, we get 6! Because A is repeated, we divide 6! by 2! to get the answer 360 Supposing I asked you to permute only 4 out of 6 of the letters from AGHAST, then you would do 6 P 4 as per normal, and divide that answer by 2! since A is repeated.  The answer should be 180. Check out Grockit for more GMAT permutations and combinations practice!

A linear equation is any equation where the highest power of the unknown, which I shall call x, is 1.  To illustrate more clearly with a few examples:

x+1 = 4; 10x = 3; x = 18 – 4x are three examples of linear equations

x² + 2 = 2x and x³ = 8 are not linear equations because there are x’s that are raised to a higher power than 1.

A linear equation with 1 variable is the simplest type to solve.  There is 1 equation and 1 unknown, which means that the unknown can always be determined.  To solve such an equation, you need to rearrange the equation to have like terms on either side of the equal sign.  Put another way, you are trying to isolate x (or whatever the variable is called) on one side of the equation.

For example, if 2x = 234, to isolate x, we have to divide the entire equation by 2.  Doing this, we get x = 117.

If there are x’s and numbers on either side of the equal sign, we add and subtract values to isolate x on one side.  Suppose 2x – 17 = 18 – 3x

The first thing we could do is to add 17 to both sides to get: 2x – 17 + 17 = 18 – 3x + 17

This reduces to 2x = 35 – 3x

Now, we need to have all the x’s on one side so we add 3x to both sides to get: 2x + 3x = 35 – 3x + 3x

This reduces to 5x = 35

Dividing by 5 on both sides, we get x = 7

What I just went through was a fairly simple algebraic equation.  The questions on the GMAT will look more complicated but you are essentially doing the same thing: manipulating both sides of the equation in the same way to isolate x.  Let’s try a practice problem from Grockit.

To tackle this question, we multiply both sides by 2+3/x to get 3 = 2+3/x. .  (This is also known as cross multiplying where in general if a/b-c/d,

then ad = bc

To simplify 3 = 2+3/x,

we multiply the entire equation by x to get 3x = 2x + 3.  This leaves you with a much simpler equation that you already know how to solve.

What’s a little trickier than manipulating algebraic equations is translating a word problem into an algebraic equation.  Here’s another practice problem:

Jack and his brother are sharing a monster piece of licorice that is 28 inches long. Since Jack is older, his share is 8 inches longer than his brother’s. How long, in inches, is Jack’s brother’s piece?

The way to solve this problem is to let something be x.  Here’s what happens if we let Jack’s piece be x inches.

Jack’s piece = x inches

Jack’s brother’s piece = x – 8 inches

Total length of licorice = Jack’s piece + Jack’s brother’s piece = 28 = x + (x-8)

This means that x = 18 inches.  But remember that the question wants the length of Jack’s brother’s piece, which we have defined as x – 8.  So the correct answer is 10 inches.

Here’s what happens if we let Jack’s brother’s piece be x inches.

Jack’s brother’s piece = x inches

Jack’s piece = x + 8 inches

Total length of licorice = 28 = x + (x+8) and we determine that x = 10.  In this case, since we have already defined Jack’s brother’s piece to be x, there is no further step we need to take.

In general, here are a few things to keep in mind.

• if there is only one unknown, you only need one equation to determine the value of the unknown
• in dealing with algebraic equations, remember that anything you do to one side (be it adding, subtracting, multiplying or dividing) you need to do to this other side too.
• in dealing with word problems, define something to be x and see if you can define other things in terms of x only.  (For example, in the question about licorice, you would not want to let Jack’s piece be x inches and his brother’s be y inches)  Don’t introduce unnecessary variables if it can be expressed in terms of an existing variable.