Algebra - GRE Quantitative Reasoning
Card 0 of 3896
Simplify:
(2_x_ + 4)/(x + 2)
Simplify:
(2_x_ + 4)/(x + 2)
(2_x_ + 4)/(x + 2)
To simplify you must first factor the top polynomial to 2(x + 2). You may then eliminate the identical (x + 2) from the top and bottom leaving 2.
(2_x_ + 4)/(x + 2)
To simplify you must first factor the top polynomial to 2(x + 2). You may then eliminate the identical (x + 2) from the top and bottom leaving 2.
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A function f(x) = –1 for all values of x. Another function g(x) = 3_x_ for all values of x. What is g(f(x)) when x = 4?
A function f(x) = –1 for all values of x. Another function g(x) = 3_x_ for all values of x. What is g(f(x)) when x = 4?
We work from the inside out, so we start with the function f(x). f(4) = –1. Then we plug that value into g(x), so g(f(x)) = 3 * (–1) = –3.
We work from the inside out, so we start with the function f(x). f(4) = –1. Then we plug that value into g(x), so g(f(x)) = 3 * (–1) = –3.
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What is f(–3) if f(x) = _x_2 + 5?
What is f(–3) if f(x) = _x_2 + 5?
f(–3) = (–3)2 + 5 = 9 + 5 = 14
f(–3) = (–3)2 + 5 = 9 + 5 = 14
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For all values of x, f(x) = 7_x_2 – 3, and for all values of y, g(y) = 2_y_ + 9. What is g(f(x))?
For all values of x, f(x) = 7_x_2 – 3, and for all values of y, g(y) = 2_y_ + 9. What is g(f(x))?
The inner function f(x) is like our y-value that we plug into g(y).
g(f(x)) = 2(7_x_2 – 3) + 9 = 14_x_2 – 6 + 9 = 14_x_2 + 3.
The inner function f(x) is like our y-value that we plug into g(y).
g(f(x)) = 2(7_x_2 – 3) + 9 = 14_x_2 – 6 + 9 = 14_x_2 + 3.
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A store sells potatoes for \$0.24 and tomatoes for \$0.76. Fred bought 12 individual vegetables. If he paid \$6.52 total, how many potatoes did Fred buy?
A store sells potatoes for \$0.24 and tomatoes for \$0.76. Fred bought 12 individual vegetables. If he paid \$6.52 total, how many potatoes did Fred buy?
Set up an equation to represent the total cost in cents: 24P + 76T = 652. In order to reduce the number of variables from 2 to 1, let the # tomatoes = 12 – # of potatoes. This makes the equation 24P + 76(12 – P) = 652.
Solving for P will give the answer.
Set up an equation to represent the total cost in cents: 24P + 76T = 652. In order to reduce the number of variables from 2 to 1, let the # tomatoes = 12 – # of potatoes. This makes the equation 24P + 76(12 – P) = 652.
Solving for P will give the answer.
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Kim is twice as old as Claire. Nick is 3 years older than Claire. Kim is 6 years older than Emily. Their ages combined equal 81. How old is Nick?
Kim is twice as old as Claire. Nick is 3 years older than Claire. Kim is 6 years older than Emily. Their ages combined equal 81. How old is Nick?
The goal in this problem is to have only one variable. Variable “x” can designate Claire’s age.
Then Nick is x + 3, Kim is 2x, and Emily is 2x – 6; therefore x + x + 3 + 2x + 2x – 6 = 81
Solving for x gives Claire’s age, which can be used to find Nick’s age.
The goal in this problem is to have only one variable. Variable “x” can designate Claire’s age.
Then Nick is x + 3, Kim is 2x, and Emily is 2x – 6; therefore x + x + 3 + 2x + 2x – 6 = 81
Solving for x gives Claire’s age, which can be used to find Nick’s age.
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If 6h – 2g = 4g + 3h
In terms of g, h = ?
If 6h – 2g = 4g + 3h
In terms of g, h = ?
If we solve the equation for b, we add 2g to, and subtract 3h from, both sides, leaving 3h = 6g. Solving for h we find that h = 2g.
If we solve the equation for b, we add 2g to, and subtract 3h from, both sides, leaving 3h = 6g. Solving for h we find that h = 2g.
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If 2x + y = 9 and y – z = 4 then 2x + z = ?
If 2x + y = 9 and y – z = 4 then 2x + z = ?
If we solve the first equation for 2x we find that 2x = 9 – y. If we solve the second equation for z we find z = –4 + y. Adding these two manipulated equations together we see (2x) + (y) = (9 – y)+(–4 + y).
The y’s cancel leaving us with an answer of 5.
If we solve the first equation for 2x we find that 2x = 9 – y. If we solve the second equation for z we find z = –4 + y. Adding these two manipulated equations together we see (2x) + (y) = (9 – y)+(–4 + y).
The y’s cancel leaving us with an answer of 5.
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11/(x – 7) + 4/(7 – x) = ?
11/(x – 7) + 4/(7 – x) = ?
We must find a common denominator and here they changed the first fraction by removing a negative from the numerator and denominator, leaving –11/(7 – x). We add the numerators and keep the same denominator to find the answer.
We must find a common denominator and here they changed the first fraction by removing a negative from the numerator and denominator, leaving –11/(7 – x). We add the numerators and keep the same denominator to find the answer.
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25_x_2 – 36_y_2 can be factored into:
25_x_2 – 36_y_2 can be factored into:
This is the difference of squares. You must know this formula for the GRE!
_a_2 – _b_2 = (a – b)(a + b)
Here a = 5_x_ and b = 6_y_, so the difference of squares formula gives us (5_x_ – 6_y_)(5_x_ + 6_y_).
This is the difference of squares. You must know this formula for the GRE!
_a_2 – _b_2 = (a – b)(a + b)
Here a = 5_x_ and b = 6_y_, so the difference of squares formula gives us (5_x_ – 6_y_)(5_x_ + 6_y_).
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Simplify (4x)/(x2 – 4) * (x + 2)/(x2 – 2x)
Simplify (4x)/(x2 – 4) * (x + 2)/(x2 – 2x)
Factor first. The numerators will not factor, but the first denominator factors to (x – 2)(x + 2) and the second denomintaor factors to x(x – 2). Multiplying fractions does not require common denominators, so now look for common factors to divide out. There is a factor of x and a factor of (x + 2) that both divide out, leaving 4 in the numerator and two factors of (x – 2) in the denominator.
Factor first. The numerators will not factor, but the first denominator factors to (x – 2)(x + 2) and the second denomintaor factors to x(x – 2). Multiplying fractions does not require common denominators, so now look for common factors to divide out. There is a factor of x and a factor of (x + 2) that both divide out, leaving 4 in the numerator and two factors of (x – 2) in the denominator.
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what is 6/8 X 20/3
what is 6/8 X 20/3
6/8 X 20/3 first step is to reduce 6/8 -> 3/4 (Divide top and bottom by 2)
3/4 X 20/3 (cross-cancel the threes and the 20 reduces to 5 and the 4 reduces to 1)
1/1 X 5/1 = 5
6/8 X 20/3 first step is to reduce 6/8 -> 3/4 (Divide top and bottom by 2)
3/4 X 20/3 (cross-cancel the threes and the 20 reduces to 5 and the 4 reduces to 1)
1/1 X 5/1 = 5
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Quantitative Comparison
Quantity A: 
Quantity B: 
Quantitative Comparison
Quantity A:
Quantity B:
If x = 0, (x – 5)2 = 25 and (x + 5)2 = 25, so the quantities are equal.
If x = 1, (x – 5)2 = (–4)2 = 16 and (x + 5)2 = 62 = 36, so B is greater.
This is a contradiction, so the answer cannot be determined.
If x = 0, (x – 5)2 = 25 and (x + 5)2 = 25, so the quantities are equal.
If x = 1, (x – 5)2 = (–4)2 = 16 and (x + 5)2 = 62 = 36, so B is greater.
This is a contradiction, so the answer cannot be determined.
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Jack has 14 coins consisting of nickels and dimes that total \$0.90. How many nickels does Jack have?
Jack has 14 coins consisting of nickels and dimes that total \$0.90. How many nickels does Jack have?
In order to solve this question we must first set up two equations. We know the number of nickels and the number of dimes equals 14 (n + d = 14). We also know the value of nickels and dimes.
For the second equation we simply multiply the number of nickels we have by their value, added to the number of dimes we have by their value to get the total (0.05n + 0.10d = 0.90).
Solve the first equation for n giving us n = 14 – d. We can then substitute 14 – d into the second equation wherever there is an “n”. Giving us 0.05 (14 – d) + 0.10d = 0.90.
When we solve the equation we find the number of dimes is d = 4; therefore the remaining 10 coins must be nickels.
In order to solve this question we must first set up two equations. We know the number of nickels and the number of dimes equals 14 (n + d = 14). We also know the value of nickels and dimes.
For the second equation we simply multiply the number of nickels we have by their value, added to the number of dimes we have by their value to get the total (0.05n + 0.10d = 0.90).
Solve the first equation for n giving us n = 14 – d. We can then substitute 14 – d into the second equation wherever there is an “n”. Giving us 0.05 (14 – d) + 0.10d = 0.90.
When we solve the equation we find the number of dimes is d = 4; therefore the remaining 10 coins must be nickels.
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If a = 1/3b and b = 4c, then in terms of c, a – b + c = ?
If a = 1/3b and b = 4c, then in terms of c, a – b + c = ?
To begin we must find how a and c relate to each other. Using the second equation we know that we can plug in 4c everywhere there is a b in the first equation, giving us a = 4/3c.
Now we can plug into the last equation. We plug in 4/3c for a, 4c for b, and leave c as it is. We must find a common denominator (4/3c – 12/3c + 3/3c) and add the numerators to find that our equation equals –5/3c.
To begin we must find how a and c relate to each other. Using the second equation we know that we can plug in 4c everywhere there is a b in the first equation, giving us a = 4/3c.
Now we can plug into the last equation. We plug in 4/3c for a, 4c for b, and leave c as it is. We must find a common denominator (4/3c – 12/3c + 3/3c) and add the numerators to find that our equation equals –5/3c.
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If x3 = 8, then x2(4/(3 – x))(2/(4 – x)) – (4/x2) = ?
If x3 = 8, then x2(4/(3 – x))(2/(4 – x)) – (4/x2) = ?
There is really no need to alter this equation using algebra. Simply find that x = 2 and plug in. We see that 4(4)(1) – (1)=15. Remember to use correct order of operations here (parentheses, exponents, multiplication, division, addition, subtraction).
There is really no need to alter this equation using algebra. Simply find that x = 2 and plug in. We see that 4(4)(1) – (1)=15. Remember to use correct order of operations here (parentheses, exponents, multiplication, division, addition, subtraction).
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x2 + 5x – 24 = 0
y2 – 9y + 20 = 0
Quantity A
x
Quantity B
y
x2 + 5x – 24 = 0
y2 – 9y + 20 = 0
Quantity A
x
Quantity B
y
If x2 + 5x – 24 = 0,
(x – 3)(x + 8) = 0 or x = –8 or +3.
y2 – 9y + 20 = 0, then
(y – 5)(y – 4) = 0, or y = +4 or +5.
y is always greater than x.
If x2 + 5x – 24 = 0,
(x – 3)(x + 8) = 0 or x = –8 or +3.
y2 – 9y + 20 = 0, then
(y – 5)(y – 4) = 0, or y = +4 or +5.
y is always greater than x.
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One of the roots of the equation x2 + kx - 12 = 0 is 3, and k is a constant.
Quantity A: The value of k
Quantity B: -1
One of the roots of the equation x2 + kx - 12 = 0 is 3, and k is a constant.
Quantity A: The value of k
Quantity B: -1
We can factor the equation x2 + kx - 12 = 0, knowing that we will have (x - 3) as one of the parentheses since the root is equal to 3.
x - 3 = 0
x = 3
We also know that the other root will be -4, because we multiply the 4 and -3 in (x + 4)(x - 3) to get our constant, -12.
This means that kx is equal to 4x - 3x = x. Therefore k = 1, and quantity A > quantity B.
Quantity A is greater.
We can factor the equation x2 + kx - 12 = 0, knowing that we will have (x - 3) as one of the parentheses since the root is equal to 3.
x - 3 = 0
x = 3
We also know that the other root will be -4, because we multiply the 4 and -3 in (x + 4)(x - 3) to get our constant, -12.
This means that kx is equal to 4x - 3x = x. Therefore k = 1, and quantity A > quantity B.
Quantity A is greater.
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y = x2 - 10
y = 15
Quantity A: y/3
Quantity B: x
y = x2 - 10
y = 15
Quantity A: y/3
Quantity B: x
We know that Quantity A = y / 3 = 15 /3 = 5.
If we plug in 15 for y, we can solve for x, for Quantity B.
y = x2 - 10
y = 15
15 = x2 - 10 (Add 10 to both sides.)
25 = x2
x = 5 or -5
Since 5 is equal to 5 but is greater than -5, we cannot determine the relationship between Quantities A and B.
We know that Quantity A = y / 3 = 15 /3 = 5.
If we plug in 15 for y, we can solve for x, for Quantity B.
y = x2 - 10
y = 15
15 = x2 - 10 (Add 10 to both sides.)
25 = x2
x = 5 or -5
Since 5 is equal to 5 but is greater than -5, we cannot determine the relationship between Quantities A and B.
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If b – 3 = a, then (a – b)2 =
If b – 3 = a, then (a – b)2 =
The quantity can be regrouped to be –3 = a – b. Thus, (a – b)2 = (–3)2 = 9.
The quantity can be regrouped to be –3 = a – b. Thus, (a – b)2 = (–3)2 = 9.
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