All SAT Math Resources
Example Questions
Example Question #61 : Algebra
If 6h – 2g = 4g + 3h
In terms of g, h = ?
2g
g
3g
4g
5g
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.
Example Question #62 : Algebra
If 2x + y = 9 and y – z = 4 then 2x + z = ?
29
Cannot be determined
21
5
13
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.
Example Question #62 : Equations / Inequalities
11/(x – 7) + 4/(7 – x) = ?
7/(7 – x)
15/(7 – x)
15
(–7)/(7 – x)
15/(x – 7)
(–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 #2 : 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?
12
8
10
4
6
10
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 #63 : Algebra
If a = 1/3b and b = 4c, then in terms of c, a – b + c = ?
–5/3c
c
–11/3c
5/3c
–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.
Example Question #64 : Algebra
If x3 = 8, then x2(4/(3 – x))(2/(4 – x)) – (4/x2) = ?
16
15
35
0
22
15
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).
Example Question #65 : Algebra
Find the intersection of the following two equations:
3x + 4y = 6
15x - 4y = 3
(0.5, 1.125)
(1, 0.5)
(3, 4)
(0.2, 0)
(18, 0)
(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)
Example Question #15 : How To Find The Solution To An Equation
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?
187.5
None of the other answers
3.33
3.125
200
187.5
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.
Example Question #81 : Algebra
A given university has an average professor pay of $40,000 a year and an average administrator pay of $45,000 per year. If the ratio of professors to administrators is 4 to 3, and the total pay for professors and administrators in a year is $40,415,000, how many professors does the college have?
375
500
548
475
411
548
Set up a system of linear equations based on our data:
40,000P + 45,000A = 40,415,000
P/A = 4/3
To make things easiest, solve the second equation for A in terms of P:
A = (3/4) P
Replace this value into the first equation:
40,000P + 45,000 * (3/4)P = 40,415,000
Simplify:
40,000P + 33,750P = 40,415,000
73,750P = 40,415,000
P = 548 (The number of professors)
Example Question #81 : Algebra
Abby works at a car dealership and receives a commission "c" which is a percent of the profit the dealership makes from the sale, which is the difference between the price "p" of the car and the value "v" of the car. How much, in dollars, does the dealership earn per transaction?
p(v – 0.01c)
(p – v)(1 – 0.01c)
(p – v)(1 – c)
(p – v)(0.01c)
pv(0.01c)
(p – v)(1 – 0.01c)
To show that c is of the profit of the transaction, we must represent the profit as the difference between the price and the value of the car, or "(p – v)"
To show that Abby's commission in dollars is a percentage of the profit, we use 0.01 * c to convert the commission she earns to a percent.
To shift the earnings from Abby to the dealership (which is what the question requires of us), we must take 1 – 0.01c since this will accommodate for the remaining percentage. For example, it shifts 75% (0.75) to 25% (1 – 0.75 or 0.25).
Putting this all together, we get a final expression of:
(p – v)(1 – 0.01c) = dealership earnings
Check answer with arbitrary values: letting p = 300, v = 200, and c = 20, we get a value of 80 which makes sense as the $100 profit must be distributed evenly between Abby ($20) and the dealership ($80).