Algebra - GRE Quantitative Reasoning
Card 0 of 3896
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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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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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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Find the intersection of the following two equations:
3x + 4y = 6
15x - 4y = 3
Find the intersection of the following two equations:
3x + 4y = 6
15x - 4y = 3
The point of intersection for two lines is the same as the values of x and y that mutually solve each equation. Although you could solve for one variable and replace it in the other equation, use elementary row operations to add the two equations since you have a 4y and -4y:
3x + 4y = 6
15x - 4y = 3
18x = 9; x = 0.5
You can now plug x into the first equation:
3 * 0.5 + 4y = 6; 1.5 +4y = 6; 4y = 4.5; y = 1.125
Therefore, our point of intersection is (0.5, 1.125)
The point of intersection for two lines is the same as the values of x and y that mutually solve each equation. Although you could solve for one variable and replace it in the other equation, use elementary row operations to add the two equations since you have a 4y and -4y:
3x + 4y = 6
15x - 4y = 3
18x = 9; x = 0.5
You can now plug x into the first equation:
3 * 0.5 + 4y = 6; 1.5 +4y = 6; 4y = 4.5; y = 1.125
Therefore, our point of intersection is (0.5, 1.125)
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John has $50 for soda and he must buy both diet and regular sodas. His total order must have at exactly two times as many cans of diet soda as cans of regular soda. What is the greatest number of cans of diet soda John can buy if regular soda is $0.50 per can and diet soda is $0.75 per can?
John has $50 for soda and he must buy both diet and regular sodas. His total order must have at exactly two times as many cans of diet soda as cans of regular soda. What is the greatest number of cans of diet soda John can buy if regular soda is $0.50 per can and diet soda is $0.75 per can?
From our data, we can come up with the following two equations:
0.50R + 0.75D = 50
2R = D
Replace the D value in the second equation into the first one:
0.5R + 0.75 * 2R = 50
0.5R + 1.5R = 50; 2R = 50; R = 25
However, note that the question asks for the number of diet cans, so this will have to be doubled to 50.
From our data, we can come up with the following two equations:
0.50R + 0.75D = 50
2R = D
Replace the D value in the second equation into the first one:
0.5R + 0.75 * 2R = 50
0.5R + 1.5R = 50; 2R = 50; R = 25
However, note that the question asks for the number of diet cans, so this will have to be doubled to 50.
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Two cars start 25 mile apart and drive away from each other in opposite directions at speeds of 50 and 70 miles per hour. In approximately how many minutes will they be 400 miles apart?
Two cars start 25 mile apart and drive away from each other in opposite directions at speeds of 50 and 70 miles per hour. In approximately how many minutes will they be 400 miles apart?
The cars have a distance from each other of 25 + 120t miles, where t is the number of hours, 25 is their initial distance and 120 is 50 + 70, or their combined speeds. Solve this equation for 400:
25 + 120t = 400; 120t = 375; t = 3.125
However, the question asked for minutes, so we must multiply this by 60:
3.125 * 60 = 187.5 minutes.
The cars have a distance from each other of 25 + 120t miles, where t is the number of hours, 25 is their initial distance and 120 is 50 + 70, or their combined speeds. Solve this equation for 400:
25 + 120t = 400; 120t = 375; t = 3.125
However, the question asked for minutes, so we must multiply this by 60:
3.125 * 60 = 187.5 minutes.
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√(3x) = 9
What is x?
√(3x) = 9
What is x?
To solve, remove the radical by squaring both sides
(√3x) 2 = 92
3x = 81
x = 81/3 = 27
To solve, remove the radical by squaring both sides
(√3x) 2 = 92
3x = 81
x = 81/3 = 27
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√(8y) + 18 = 4
What is y?
√(8y) + 18 = 4
What is y?
First, simplify the equation:
√(8y) + 18 = 4
√(8y) = -14
Then square both sides
(√8y) 2 = -142
8y = 196
y = 196/8 = 24.5
First, simplify the equation:
√(8y) + 18 = 4
√(8y) = -14
Then square both sides
(√8y) 2 = -142
8y = 196
y = 196/8 = 24.5
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Quantitative Comparison
3_x_ + 4_y_ = 5
x – y = 6
Quantity A: x
Quantity B: y
Quantitative Comparison
3_x_ + 4_y_ = 5
x – y = 6
Quantity A: x
Quantity B: y
First let's solve for y using the second equation, x – y = 6.
x = 6 + y. Then plug this in to the other equation.
3 (6 + y) + 4_y_ = 5
18 + 3_y_ + 4_y_ = 5
18 + 7_y_ = 5
7_y_ = –13
y = –13/7. Now plug this value back into x = 6 + y.
x = 6 – 13/7 = 29/7. x is positive and y is negative, so clearly x is larger, so Quantity A is bigger.
First let's solve for y using the second equation, x – y = 6.
x = 6 + y. Then plug this in to the other equation.
3 (6 + y) + 4_y_ = 5
18 + 3_y_ + 4_y_ = 5
18 + 7_y_ = 5
7_y_ = –13
y = –13/7. Now plug this value back into x = 6 + y.
x = 6 – 13/7 = 29/7. x is positive and y is negative, so clearly x is larger, so Quantity A is bigger.
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For what value of x does 4(3_x_ – 2) = 12?
For what value of x does 4(3_x_ – 2) = 12?
We have to solve the equation 4(3_x_ – 2) = 12. First, we can distribute the left side.
4(3_x_) – 4(2) = 12
12_x_ – 8 = 12
Then we add 8 to both sides.
12_x_ = 20
Divide both sides by 12.
x = 20/12
Simplify 20/12 by dividing the numerator and denominator by 4.
x = 20/12 = 5/3
The answer is 5/3.
We have to solve the equation 4(3_x_ – 2) = 12. First, we can distribute the left side.
4(3_x_) – 4(2) = 12
12_x_ – 8 = 12
Then we add 8 to both sides.
12_x_ = 20
Divide both sides by 12.
x = 20/12
Simplify 20/12 by dividing the numerator and denominator by 4.
x = 20/12 = 5/3
The answer is 5/3.
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Solve for z:
3(z + 4)3 – 7 = 17
Solve for z:
3(z + 4)3 – 7 = 17
1. Add 7 to both sides
3(z + 4)3 – 7 + 7= 17 + 7
3(z + 4)3 = 24
2. Divide both sides by 3
(z + 4)3 = 8
3. Take the cube root of both sides
z + 4 = 2
4. Subtract 4 from both sides
z = –2
1. Add 7 to both sides
3(z + 4)3 – 7 + 7= 17 + 7
3(z + 4)3 = 24
2. Divide both sides by 3
(z + 4)3 = 8
3. Take the cube root of both sides
z + 4 = 2
4. Subtract 4 from both sides
z = –2
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