All GRE Math Resources
Example Question #51 : Linear / Rational / Variable Equations
What is the value of (5 + x)(10 – y) when x = 3 and y = –3?
This is a simple plug-in and PEMDAS problem. First, plug in x = 3 and y = –3 into the x and y. You should follow the orders of operation and compute what is within the parentheses first and then find the product. This gives 8 * 13 = 104. The answer is 104.
Example Question #1 : How To Find The Solution To An Equation
If x = 4, and y = 3x + 5, then 2y – 1 equals
Start by plugging in x = 4 to solve for y: y = 3 * 4 + 5 = 17. Then 2 * 17 – 1 = 33
Example Question #52 : Linear / Rational / Variable Equations
Sarah’s current age is three times Ron’s age two years ago. Sarah is currently 14 years older than Ron. What is the sum of Sarah and Ron’s current age?
The best way to solve this problem is to turn the two statements into equations calling Sarah’s age S and Ron’s age R. So, S = 3(R – 2) and S = 14 + R. Now substitute the value for S in the second equation for the value of S in the first equation to get 14 + R = 3(R – 2) and solve for R. So R equals 10 so S equals 24 and the sum of 10 and 24 is 34.
Example Question #53 : Linear / Rational / Variable Equations
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.
Example Question #5 : How To Find The Solution To An Equation
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.
Example Question #63 : Gre Quantitative Reasoning
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.
Example Question #6 : How To Find The Solution To An Equation
If 2x + y = 9 and y – z = 4 then 2x + z = ?
Cannot be determined
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.
Example Question #3 : How To Find The Solution To An Equation
11/(x – 7) + 4/(7 – x) = ?
(–7)/(7 – x)
15/(x – 7)
7/(7 – x)
15/(7 – x)
(–7)/(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.
Example Question #4 : How To Find The Solution To An Equation
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.
Example Question #72 : How To Find The Solution To An Equation
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.