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I'm going to make a sticky thread where the very very basic

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I'm going to make a sticky thread where the very very basic  [#permalink]

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New post Updated on: 17 Aug 2005, 08:57
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I'm going to make a sticky thread where the very very basic mathematical principles are collected for everybody's reference. Please feel free to discuss and add to what I have here.
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Originally posted by HongHu on 03 Mar 2005, 12:15.
Last edited by HongHu on 17 Aug 2005, 08:57, edited 1 time in total.
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Numbers  [#permalink]

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New post Updated on: 09 Mar 2005, 15:58
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Even and Odd
Definition: Suppose k is an integer. If there exists an integer r such that k=2r+1, then k is an odd number. If there exists an integer r such that k=2r, then k is an even number.
Explanation: as long as an integer can be divided by 2, it is an even number.
Zero is an even number.

Positive and Negative
Definition: A positive number is a real number that is greater than zero. A negative number is a real number that is smaller than zero.
Zero is not positive, nor negative.

Originally posted by HongHu on 03 Mar 2005, 12:16.
Last edited by HongHu on 09 Mar 2005, 15:58, edited 2 times in total.
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Inequalities  [#permalink]

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New post Updated on: 14 Mar 2005, 08:21
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Basic rules for inequalities:
(in the example: a>b>0, c>d>0)

You need to flip signs when both side are multiplied by a negative number:
-a<-b, -c<-d

You need to flip signs when 1 is divided by both side:
1/a<1/b, 1/c<1/d

You can only add or multiply them when their signs are in the same direction:
a+c>b+d
ac>bd

You can only apply substractions and divisions when their signs are in the opposite directions:
a>b, d<c
a-d>b-c
a/d>b/c
(You can't say a/c>b/c. It is WRONG)

Deal with negative numbers:
-a<-b<0, -c<-d<0
Then
-a-c<-b-d<0
-a-(-d)<-b-(-c)
However the sign needs to be flipped one more time if you are doing multiplication or division (because you are multiplying/dividing a negative number):
(-a)*(-c)>(-b)*(-d)
(-a)/(-d)>(-b)/(-c)

For example:
If x<-4, y<-2, we know that xy>8, but we don't know how x/y compare to (-4)/(-2)=2 since you can only do division when their signs are in different directions
If x>-4 and y<-2 then x/y<2 but we don't know how xy is compared to 8 since we can only do multiplication when their signs are the same direction.

It is easier to do the derivation, though, if you first change them to postive. For example:
If x<-4, y<-2, then -x>4, -y>2, xy>8
If x<-4, y<2, then -x>4, y<2, -x/y>2, x/y<-2

Originally posted by HongHu on 03 Mar 2005, 12:17.
Last edited by HongHu on 14 Mar 2005, 08:21, edited 3 times in total.
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New post 04 Mar 2005, 05:32
very useful thread

do you, guys, know what the "line test method" is, which is discussed here

http://www.gmatclub.com/phpbb/viewtopic ... equalities

Did those guys mean just picking numbers or ..?
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Cancelling out "common terms" on both sides  [#permalink]

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New post Updated on: 26 Mar 2005, 22:08
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Cancelling out "common terms" on both sides of an equation

You need to be very careful when you do algebra derivations. One of the common mistake is to divide both side by "a common term". Remember you can only do this safely if the "common term" is a constant. However you CAN't do it if it contains a variable.

Example:

x(x-2)=x
You can't cancel out the x on both side and say x=3 is the solution. You must move the x on the right side to the left side.
x(x-2)-x=0
x(x-2-1)=0
The solutions are: x=0 and x=3
The reason why you can't divided both sides by x is that when x is zero, you can't divide anything by zero.

Equally important if not more, is that you CAN'T multiple or divide a "common term" that includes a variable from both side of an inequality. Not only it could be zero, but it could also be negative in which case you would need to flip the sign.

Example:

x^2>x
You CAN'T divided both sides by x and say x>1. What you have to do is to move the right side to the left:
x^2-x>0
x(x-1)>0
Solution would be either both x and x-1 are greater than zero, or both x and x-1 are smaller than zero. So your solution is: x>1 or x<0

Example:

x>1/x
Again you CAN'T multiply both sides by x because you don't know if x is positive or negative. What you have to do is to move the right side to the left:
x-1/x>0
(x^2-1)/x>0
If x>0 then x^2-1>0 =>x>1
If x<0 then x^2-1<0 =>x>-1
Therefore your solution is x>1 or 0>x>-1.
You could also break the original question to two branches from the beginning:
x>1/x
if x>0 then x^2>1 =>x>1
if x<0 then x^2<1 => x>-1
Therefore your solution is x>1 or 0>x>-1.

Originally posted by HongHu on 08 Mar 2005, 10:44.
Last edited by HongHu on 26 Mar 2005, 22:08, edited 3 times in total.
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New post 10 Mar 2005, 23:58
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I'll add on some more over a period of time, but here are for starters:

Divisibility rules: - Know them well

Integer is divisible by:
2 - Even integer
3 - Sum of digits are divisible by 3
4 - Integer is divisible by 2 twice or Last 2 digits are divisible by 4
5 - Last digit is 0 or 5
6 - Integer is divisbile by 2 AND 3
8 - Integer is divisible by 2 three times
9 - Sum of digits is divisible by 9
10 - Last digit is 0

Some other things to note:
- If 2 numbers have the same factor, then the sum or difference of the two numbers will have the same facor.

(e.g. 4 is a factor of 20, 4 is also a factor of 80, then 4 will be a factor of 60 (difference) and also 120 (sum))

- Remember to include '1' if you're asked to count the number of factors a number has
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New post 11 Mar 2005, 11:59
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and for 11 which i got on an OG problem recently

Start with the units digit, add every other digit and remember this number. Form a new number by adding the digits that remain. If the difference between these two numbers is divisible by 11, then the original number is divisible by 11.

Examples:
Is the number 824472 divisible by 11? Starting with the units digit, add every other number:2 + 4 + 2 = 8. Then add the remaining numbers: 7 + 4 + 8 = 19. Since the difference between these two sums is 11, which is divisible by 11, 824472 is divisible by 11.

Is the number 49137 divisible by 11? Starting with the units digit, add every other number:7 + 1 + 4 = 12. Then add the remaining numbers: 3 + 9 = 12. Since the difference between these two sums is 0, which is divisible by 11, 49137 is divisible by 11.
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New post 13 Mar 2005, 19:39
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adding more...

Adding on to the odd/even rule honghu wrote...

Adding/subtracting two odds or two evens --> even
Add/ Subtract an odd and an even --> od

Multiplication with at no even number --> odd
** Even number in a multiplication will always ensure an even product

Interesting properties:
- Adding n consecutive integers will yield a sum that is divisible by n (i.e. n will be a factor of the sum)

E.g. Adding 3,4,5,6 will give a sum with 4 as a factor
Adding 2,3,4 will give a sum with 3 as a factor

Basic principles dervied from adding, n, n+1, n+2 etc..

- Adding a consecutive set of odd integers will result in sum that is a multiple of the number of integers

E.g. Adding 2,4,6,8,10 --> sum will be multiple of 5 (sum=30)
Adding 1,3,5 --> sum will be multiple of 3 (sum=9)

- An even integer in a multiplication --> product divisible by 2
- 2 even integers in a multiplication --> product divisible by 4
etc
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New post 13 Mar 2005, 23:53
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Sometime back, someone asked me how to find the GCF and LCM. Thought it'll be useful to throw it in this thread so members preparing for gmat and need a quick refresh can go through this.

GCF -> Greatest Common Factor
-> Largest possible common factor between numbers

LCM -> Lowest Common Multiple
-> Largest possible common multiple between numbers

To find the GCF/LCM, you will need to do prime-factorization. This means reducing a number to its prime-factor form.

E.g. 1

GCF/LCM of 4,18

4 = 2*2
18= 2*3*3

To find the GCF, take the multiplication of the common factors (pick the lowest power of the common factors) In this case, GCF = 2.

To find the LCM, take the multiplication of all the factors (pick the higest power of the common factors). In this case, LCM=2*2*3*3=36

E.g. 2

GCF/LCM of 4,24

4 = 2*2
24 = 2*2*2*3

GCF = 2*2 = 4
LCM = 2*2*2*3 = 24
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New post Updated on: 17 Aug 2005, 07:28
Yes, that's a good one. I've been long thinking of doing one about the absolute value questions but have forgot about it. :oops:

Absolute values
The way to solve this kind of questions is to break the equation (inequality) into two parts, one is when the value is non negative, the other is when the value is negative.

For example:

|x-4|<9

You break it into two parts:
If x-4>=0, then x-4<9, solve for both you get x>=4, x<13. So your solution is 4<=x<13.
If x-4<0, then -(x-4)<9, ie x-4>-9. Solve for both you get x<4, x>-5. So your solution for this part is -5<x<4.

Combine the two solutions, you get -5<x<13 as your final solution.

Another example:
|x+4|>4
If x+4>=0, then x+4>4. Solve for both you get x>=-4, x>0. So your solution is x>0.
If x+4<0, then -(x+4)>4, ie. x+4<-4. Solve for both you get x<-4, x<-8. So your solution is x<-8.
You final solution is x>0 or x<-8.

The same strategy can apply to square questions.
For example: (x+4)^2>4
You could solve it this way:
x^2+8x+12>0
(x+2)(x+6)>0
x>-2 or x<-6
Or you can solve it this way:
If x+4>=0 then x+4>2. Solve for them you get x>-2.
If x+4<0 then x+4<-2. Solve for them you get x<-6.

|y|>|y+1|
if y>=0, y+1>=0, y>y+1, no solution.
if y<0, y+1<0, -y>-(y+1), solution is y<-1
if y>=0, y+1<0, y>-(y+1), no solution.
if y<0, y+1>=0, -y>y+1, solution is -1<=y<-1/2
So your final solution is y<-1/2
You could also solve this question by going the square route.
y^2>(y+1)^2
y^2>y^2+2y+1
2y+1<0
y<-1/2

Originally posted by HongHu on 15 Mar 2005, 11:59.
Last edited by HongHu on 17 Aug 2005, 07:28, edited 1 time in total.
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Working with Ratios  [#permalink]

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New post 30 Mar 2005, 09:00
Working with Ratios

Ratio questions are very easy to solve if you have mastered the way of thinking.

Basically if you have
a/b=c/d (or a:b=c:d)
then you can immediately derive a variaty of correlated ratios, such as:
a/(a+b)=c/(c+d)
a/(a-b)=c/(c-d)
(a+b)/(a-b)=(c+d)/(c-d)
(a+c)/(b+d)=c/d
(a-c)/(b-d)=c/d
etc

Basically, you can do all kinds of additions and subtractions.

Example:

a/b=3/5 (1)
2a-b=4 (2)
What is a?
From (1) we get a/(2a-b)=3/1, so a=3*4=12
Explanation: a is 3 share, b is 5 share. Two a is 6 share, 2a-b is one share. If one share is 4, then 3 share is 12.

Of course this question can be solved using the more traditional algebra approach:
b=5/3a
substitute in (2)
2a-5/3a=4
1/3a=4
a=12

You can see the two approaches are really the same in nature. However the first approach is very straight forward and does not involve calculation in fractions. Sometimes it can save you lots of time, especially when using this method with word problems such as mixture problems.

Mixture Problems

Example:

A fruit mixture is made up by 25% fruit A and 75% fruit B. Now if the amount of fruit A is doubled, what is their relative share in the new mixture?

A:B=25:75
2A:B=50:75=2:3
The new mixture total quantity is 2A+B
2A:(2A+B)=2:5
B:(2A+B)=3:5
Therefore the new shares are fruit A 40%, fruit B 60%.

Example 2:

In a picnic 60% people ate two hotdogs, 30% people ate one hamburger, and 10% people ate one hotdog. The total number of hotdog and hamburgers consumed is 80. How many hamburgers and hotdogs are consumed?

People:
T:H:O=6:3:1 (1)
Food:
2T+H+O=80
From (1)
T:H:O:(2T+H+O)=6:3:1:16
Therefore
T:(2T+H+O)=3:8=30:80 30 people ate two hotdogs
H:T=3:6=1:2=15:30 15 people ate one hamburger
O:T=1:6=5:30 5 people ate one hotdogs
Total people 50, total hotdogs 65, total hamburgers 15.
Verify, total hotdogs and hamburgers=65+15=80.
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Set theory  [#permalink]

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New post Updated on: 24 Nov 2005, 20:56
Formula:

Total = N(A) + N(B) + N(C) - N(A n B) - N(A n C) - N(C n B) + N(A n B n C)

If instead of numbers for (A n B) and (A n C) and (C n B), what is given is the total number of people who choose exactly two items, then the formula becomes:

Total = N(A) + N(B) + N(C) - (N(choose exactly two items)) - 2N(choose all three items)

Also,
Total = N(A) + N(B) + N(C) - (N(choose at least two items)) - N(choose all three items)
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Originally posted by HongHu on 14 Oct 2005, 07:17.
Last edited by HongHu on 24 Nov 2005, 20:56, edited 1 time in total.
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New post 19 Oct 2005, 17:32
Is there a useful formula or strategy of attacking remainder problems besides picking numbers? I only solve these mingaz by brute force..

e.g. question:

X is an integer, what is the remainder when it is divided by 10?
1) When divided by 5, the remainder is 2
2). When divided by 2, the remainder is 1
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New post 20 Oct 2005, 08:28
nero44 wrote:
Is there a useful formula or strategy of attacking remainder problems besides picking numbers? I only solve these mingaz by brute force..

e.g. question:

X is an integer, what is the remainder when it is divided by 10?
1) When divided by 5, the remainder is 2
2). When divided by 2, the remainder is 1


A useful strategy to help you thinking about reminder and multiple problems is to express your variable with an algebra expression.
In your example:
1) X=5m+2
Obiously if m is even then 5m is divided by 10 and the reminder of x/10 would be 2. If m is odd then the reminder would be 7. So insufficient.
2) X=2n+1
This says X is odd. Obviously all odd numbers divided by 10 yield many different reminders. So insufficient.
Combined, 2) would mean that m in 1) is odd and thus we would be able to determine the reminder.
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Re: Cancelling out "common terms" on both sides  [#permalink]

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New post 01 Nov 2005, 19:52
Guess inequality is one of my weak areas. :oops:

Could you further explain the following two examples?

Example:

x^2>x
You CAN'T divided both sides by x and say x>1. What you have to do is to move the right side to the left:
x^2-x>0
x(x-1)>0
Solution would be either both x and x-1 are greater than zero, or both x and x-1 are smaller than zero. So your solution is: x>1 or x<0

Example:

x>1/x
Again you CAN'T multiply both sides by x because you don't know if x is positive or negative. What you have to do is to move the right side to the left:
x-1/x>0
(x^2-1)/x>0
If x>0 then x^2-1>0 =>x>1
If x<0 then x^2-1<0 =>x>-1
Therefore your solution is x>1 or 0>x>-1.
You could also break the original question to two branches from the beginning:
x>1/x
if x>0 then x^2>1 =>x>1
if x<0 then x^2<1 => x>-1
Therefore your solution is x>1 or 0>x>-1.[/quote]
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New post 01 Nov 2005, 19:59
The second method seems easier...not following the first one.

Care to explain further?

|y|>|y+1|
if y>=0, y+1>=0, y>y+1, no solution.
if y<0, y+1<0, -y>-(y+1), solution is y<-1
if y>=0, y+1<0, y>-(y+1), no solution.
if y<0, y+1>=0, -y>y+1, solution is -1<=y<-1/2
So your final solution is y<-1/2
You could also solve this question by going the square route.
y^2>(y+1)^2
y^2>y^2+2y+1
2y+1<0
y<-1/2[/quote]
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New post 03 Nov 2005, 21:18
AMBA wrote:
The second method seems easier...not following the first one.

For your question about the first example, x^2>x, you could also adopt the second approach, that is to break the question into two parts.
If x>0, then you can divide both side by x and you'll get x>1.
If x<0, then obviously x^2 is always greater than x since x is negative and x^2 is positive.
So combine the two you get the correct answer: x>1 or x<0.
This approach is less mathematic, and more time consuming, but it may be more straight forward for some people. You need to evaluate the two methods yourself to see which approach you would like to take.

Quote:
Care to explain further?

|y|>|y+1|
if y>=0, y+1>=0, y>y+1, no solution.
if y<0, y+1<0, -y>-(y+1), solution is y<-1
if y>=0, y+1<0, y>-(y+1), no solution.
if y<0, y+1>=0, -y>y+1, solution is -1<=y<-1/2
So your final solution is y<-1/2
You could also solve this question by going the square route.
y^2>(y+1)^2
y^2>y^2+2y+1
2y+1<0
y<-1/2


Absolute value questions are definitely going to be on the test, and it is often easy to miss. Just remember that absolute values are always positive, but the things inside the absolute value sign may or may not be positive. This is why you need to break them into two or more parts, one being the inside is positive, another being the inside is negative.

If you are comfortable with the more algebra approach in the first example, then squaring absolute values may be a short cut for such questions sometimes. As you can see from the example, it took much less steps of reasoning. When I have time, I often try both ways, see if I get the same answer, then I'd know that I've got the correct answer.
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Re: Cancelling out "common terms" on both sides  [#permalink]

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New post 03 Nov 2005, 22:17
Thanks, HH...How about the following example. I was not clear on these 2 parts:

(x^2-1)/x>0
If x>0 then x^2-1>0 =>x>1
If x<0 then x^2-1<0 =>x>-1
Therefore your solution is x>1 or 0>x>-1.


AND

You could also break the original question to two branches from the beginning:
x>1/x
if x>0 then x^2>1 =>x>1
if x<0 then x^2<1 => x>-1
Therefore your solution is x>1 or 0>x>-1.


Example:

x>1/x
Again you CAN'T multiply both sides by x because you don't know if x is positive or negative. What you have to do is to move the right side to the left:
x-1/x>0
(x^2-1)/x>0
If x>0 then x^2-1>0 =>x>1
If x<0 then x^2-1<0 =>x>-1
Therefore your solution is x>1 or 0>x>-1.
You could also break the original question to two branches from the beginning:
x>1/x
if x>0 then x^2>1 =>x>1
if x<0 then x^2<1 => x>-1
Therefore your solution is x>1 or 0>x>-1.[/quote]
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Re: Cancelling out "common terms" on both sides  [#permalink]

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New post 03 Nov 2005, 22:25
AMBA wrote:
Thanks, HH...How about the following example. I was not clear on these 2 parts:

(x^2-1)/x>0

Here you want to multiple x to solve the equation. Because x is in the determinator we know that x can't be zero. However we don't know if x is positive or negative. And this is important because it will determine whether the sign of the inequality gets changed. Therefore you do the following breaking it into two parts: positive x or negative x.
Quote:

If x>0 then x^2-1>0 =>x>1
If x<0 then x^2-1<0 =>x>-1
Therefore your solution is x>1 or 0>x>-1.



The other way of solving it is to simplify the equation first. So (x^2-1)/x=x-1/x. Then you evaluate x-1/x>0, or x>1/x>0, by breaking it into two parts according to whether x is positive or negative.
Quote:
You could also break the original question to two branches from the beginning:
x>1/x
if x>0 then x^2>1 =>x>1
if x<0 then x^2<1 => x>-1
Therefore your solution is x>1 or 0>x>-1.



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New post 03 Nov 2005, 22:54
AMBA wrote:
Not quite sure how to derive the following 4 'if' scenario:

|y|>|y+1|

Here we have two absolute values. Therefore we have four scenarios.
First, if both of the insides are positive:
Quote:
if y>=0, y+1>=0, y>y+1, no solution.

y>y+1 would mean 0>1, which is impossible, that's why there is no solution.

Second, if both are negative, then |y|=-y and |y+1|=-(y+1), therefore you get:
Quote:
if y<0, y+1<0, -y>-(y+1), solution is y<-1

-y>-y-1 meaning 0>-1, always true. So the solution is all points that satisfy y<0 and y+1<0, for which we get y<-1.

Now you consider if one is negative and the other is positive and go through the same stipulation.
Quote:
if y>=0, y+1<0, y>-(y+1), no solution.
if y<0, y+1>=0, -y>y+1, solution is -1<=y<-1/2


Combine your answer in all the four cases: y<-1 and -1<=y<-1/2, you get your final solution:
Quote:
So your final solution is y<-1/2

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