Linear / Rational / Variable Equations - GRE Quantitative Reasoning
Card 1 of 944
Quantity A: 
Quantity B: 
Quantity A:
Quantity B:
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Rather than manually finding common denominators and adding the fractions together, realize that
Since
Quantity A must be greater, and this can be seen without actually calculating its value.
Rather than manually finding common denominators and adding the fractions together, realize that
Since
Quantity A must be greater, and this can be seen without actually calculating its value.
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Column A: 
Column B: 
Column A:
Column B:
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Column B is greater for positive numbers.
The columns are equal for 0.
Column A is greater for negative numbers.
Because our answer changes depending on the value inserted, we cannot determine the relationship.
Column B is greater for positive numbers.
The columns are equal for 0.
Column A is greater for negative numbers.
Because our answer changes depending on the value inserted, we cannot determine the relationship.
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Find the solution to the following equation if x = 3:
y = (4x2 - 2)/(9 - x2)
Find the solution to the following equation if x = 3:
y = (4x2 - 2)/(9 - x2)
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Substituting 3 in for x, you will get 0 in the denominator of the fraction. It is not possible to have 0 be the denominator for a fraction so there is no possible solution to this equation.
Substituting 3 in for x, you will get 0 in the denominator of the fraction. It is not possible to have 0 be the denominator for a fraction so there is no possible solution to this equation.
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Quantity A: 
Quantity B: 
Quantity A:
Quantity B:
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We are given that y = 32. Plug this value of y into the second equation.
32 = x2 – 4
36 = x2
x = +/– 6.
Next find a value for Quantity A:
y/7 = 32/7
This number is less than +6, but more than –6. Thus, the relationship cannot be determined from the information given.
We are given that y = 32. Plug this value of y into the second equation.
32 = x2 – 4
36 = x2
x = +/– 6.
Next find a value for Quantity A:
y/7 = 32/7
This number is less than +6, but more than –6. Thus, the relationship cannot be determined from the information given.
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I. x = 0
II. x = –1
III. x = 1
I. x = 0
II. x = –1
III. x = 1
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A fraction is considered undefined when the denominator equals 0. Set the denominator equal to zero and solve for the variable.
A fraction is considered undefined when the denominator equals 0. Set the denominator equal to zero and solve for the variable.
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Solve:
Solve:
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First, distribute, making sure to watch for negatives.
Combine like terms.
Subtract 7x from both sides.
Add 18 on both sides and be careful adding integers.
First, distribute, making sure to watch for negatives.
Combine like terms.
Subtract 7x from both sides.
Add 18 on both sides and be careful adding integers.
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Solve:
Solve:
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First, distribute the 
to the terms inside the parentheses.
Add 6x to both sides.
This is false for any value of 
. Thus, there is no solution.
First, distribute the
Add 6x to both sides.
This is false for any value of
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Solve 
.
Solve
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By definition, the absolute value of an expression can never be less than 0. Therefore, there are no solutions to the above expression.
By definition, the absolute value of an expression can never be less than 0. Therefore, there are no solutions to the above expression.
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Quantity A: 
Quantity B: 11
Quantity A:
Quantity B: 11
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Expand 
out into 
.
Since 
, it can be seen that 
so Quantity B is greater.
Expand
Since
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Approximately, what was the percent growth of Beetleton's GDP from 2009 to 2010?
Approximately, what was the percent growth of Beetleton's GDP from 2009 to 2010?
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Percent growth is given as:
For Beetleton, this can be expressed as (in terms of billions of US dollars):
Percent growth is given as:
For Beetleton, this can be expressed as (in terms of billions of US dollars):
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The sum of two integers is 
. The larger integer is 
greater than the smaller integer. What is the positive difference between the two?
The sum of two integers is
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Let us write down what we are told in mathematical terms, designating the smaller integer as 
and the larger integer as 
.
The sum of the two integers is 
:
And the larger integer is 
% greater than the smaller integer:
Writing the first equation in terms of 
gives:
Which allows us to find 
:
Thus, the positive difference between the two is found as
Let us write down what we are told in mathematical terms, designating the smaller integer as
The sum of the two integers is
And the larger integer is
Writing the first equation in terms of
Which allows us to find
Thus, the positive difference between the two is found as
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Quantity A: 
Quantity B: 
Quantity A:
Quantity B:
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To solve this problem, expand each function described by Quantities A and B:
Quantity A: 
Quantity B: 
Now note that Quantities A and B only differ in that Quantity A is greater by 
.
Since we are told that 
is greater than 
and thus always positive, Quantity A must be greater than Quantity B for all possible values of 
.
To solve this problem, expand each function described by Quantities A and B:
Quantity A:
Quantity B:
Now note that Quantities A and B only differ in that Quantity A is greater by
Since we are told that
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In the equation below, 
, 
, and 
are non-zero numbers. What is the value of 
in terms of 
and 
?
In the equation below,
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The arithmetic mean of 
, 
, 
, and 
is 14.
Quantity A: 32
Quantity B: The arithmetic mean of 
and 
The arithmetic mean of
Quantity A: 32
Quantity B: The arithmetic mean of
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The definition of an arithmetic mean of a set of values is given as the sum of all the values divided by the total count of values:
Where 
represents the 
value in a set, and 
is the number of values in the set.
Quantity B can thus be defined as follows:
Which simplifies to:
or, simplifying:
We are told that the mean of 
, 
, 
, and 
is 14, which can be written as:
and then as
Plugging this value into our definition of Quantity B, we can find its numerical value:
So
The definition of an arithmetic mean of a set of values is given as the sum of all the values divided by the total count of values:
Where
Quantity B can thus be defined as follows:
Which simplifies to:
or, simplifying:
We are told that the mean of
and then as
Plugging this value into our definition of Quantity B, we can find its numerical value:
So
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Satoshi and Reginald are bottlecap collectors, and Satoshi has 46 more bottlecaps than Reginald. If they were both to receive 6 bottle caps each, Satoshi would have three times as many bottle caps as Reginald. How many bottlecaps does Satoshi have?
Satoshi and Reginald are bottlecap collectors, and Satoshi has 46 more bottlecaps than Reginald. If they were both to receive 6 bottle caps each, Satoshi would have three times as many bottle caps as Reginald. How many bottlecaps does Satoshi have?
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The mathematical approach to this problem would be to set up a system of equations; one which represents the current relationship of Satoshi's and Reginald's number of bottle caps:
And one which represents the relationship 6 bottle caps later, wherein Satoshi would have three times as many bottlecaps as Reginald:
Subtracting the first equation from the second equation provides:
Which simplifies to
Solving this yields the value for Reginald's number of bottle caps, which is 17, which can be plugged back into the very first equation to find Satoshi's number of bottle caps, which is 63.
There is, however, another way to find this solution, which is by first looking at the answer choices.
Since we know that Satoshi's number of bottlecaps plus six must be divisible by 3 (since it is three times Reginald's new number of caps), we can quickly eliminate the choices 86, 143, and 52. Furthermore, since Satoshi has 46 more bottlecaps than Reginald, it'd be impossible for him to have 21 bottlecaps. This leaves only the 63 option.
When taking standardized math tests, where time management is key, it can be helpful to look for quicker ways to find a solution, such as back-solving, rather than just solving the problem normally.
The mathematical approach to this problem would be to set up a system of equations; one which represents the current relationship of Satoshi's and Reginald's number of bottle caps:
And one which represents the relationship 6 bottle caps later, wherein Satoshi would have three times as many bottlecaps as Reginald:
Subtracting the first equation from the second equation provides:
Which simplifies to
Solving this yields the value for Reginald's number of bottle caps, which is 17, which can be plugged back into the very first equation to find Satoshi's number of bottle caps, which is 63.
There is, however, another way to find this solution, which is by first looking at the answer choices.
Since we know that Satoshi's number of bottlecaps plus six must be divisible by 3 (since it is three times Reginald's new number of caps), we can quickly eliminate the choices 86, 143, and 52. Furthermore, since Satoshi has 46 more bottlecaps than Reginald, it'd be impossible for him to have 21 bottlecaps. This leaves only the 63 option.
When taking standardized math tests, where time management is key, it can be helpful to look for quicker ways to find a solution, such as back-solving, rather than just solving the problem normally.
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Four less than three times a certain number is equivalent to five plus four times this same number. What is three less than three times this number?
Four less than three times a certain number is equivalent to five plus four times this same number. What is three less than three times this number?
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The key to solving this problem is deciphering the language and translating it into a numerical representation. The first part can be written as an equaltiy as follows:
Rearranging terms allows us to solve for this mystery number:
From there we can address the problem's question:
The key to solving this problem is deciphering the language and translating it into a numerical representation. The first part can be written as an equaltiy as follows:
Rearranging terms allows us to solve for this mystery number:
From there we can address the problem's question:
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