All SAT Math Resources
Example Questions
Example Question #7 : How To Find The Solution To An Equation
If a = 1/3b and b = 4c, then in terms of c, a – b + c = ?
–5/3c
5/3c
–11/3c
c
–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 #71 : Equations / Inequalities
If x3 = 8, then x2(4/(3 – x))(2/(4 – x)) – (4/x2) = ?
0
15
16
35
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 #61 : How To Find The Solution To An Equation
Find the intersection of the following two equations:
3x + 4y = 6
15x - 4y = 3
(1, 0.5)
(18, 0)
(0.2, 0)
(3, 4)
(0.5, 1.125)
(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 #62 : 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?
3.33
None of the other answers
3.125
187.5
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 #63 : How To Find The Solution To An Equation
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?
411
548
375
500
475
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 #64 : How To Find The Solution To An Equation
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)(1 – 0.01c)
p(v – 0.01c)
(p – v)(0.01c)
(p – v)(1 – c)
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).
Example Question #41 : Equations / Inequalities
A theme park charges $10 for adults and $5 for kids. How many kids tickets were sold if a total of 548 tickets were sold for a total of $3750?
269
346
157
248
431
346
Let c = number of kids tickets sold. Then (548 – c) adult tickets were sold. The revenue from kids tickets is $5c, and the total revenue from adult tickets is $10(548 – c). Then,
5c + 10(548 – c) = 3750
5c + 5480 – 10c = 3750
5c = 1730
c = 346.
We can check to make sure that this number is correct:
$5 * 346 tickets + $10 * (548 – 346) tickets = $3750 total revenue
Example Question #42 : Equations / Inequalities
Two palm trees grow next to each other in Luke's backyard. One of the trees gets sick, so Luke cuts off the top 3 feet. The other tree, however, is healthy and grows 2 feet. How tall are the two trees if the healthy tree is now 4 feet taller than the sick tree, and together they are 28 feet tall?
8 and 20 feet
14 and 14 feet
12 and 16 feet
11 and 17 feet
cannot be determined
12 and 16 feet
Let s stand for the sick tree and h for the healthy tree. The beginning information about cutting the sick tree and the healthy tree growing is actually not needed to solve this problem! We know that the healthy tree is 4 feet taller than the sick tree, so h = s + 4.
We also know that the two trees are 28 feet tall together, so s + h = 28. Now we can solve for the two tree heights.
Plug h = s + 4 into the second equation: (s + 4) + s = 28. Simplify and solve for h: 2s = 24 so s = 12. Then solve for h: h = s + 4 = 12 + 4 = 16.
Example Question #43 : Equations / Inequalities
Solve for z:
3(z + 4)3 – 7 = 17
–2
8
–8
4
2
–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
Example Question #23 : How To Find The Solution To An Equation
Jen and Karen are travelling for the weekend. They both leave from Jen's house and meet at their destination 250 miles away. Jen drives 45mph the whole way. Karen drives 60mph but leaves a half hour after Jen. How long does it take for Karen to catch up with Jen?
She can't catch up.
For this type of problem, we use the formula:
When Karen catches up with Jen, their distances are equivalent. Thus,
We then make a variable for Jen's time, . Thus we know that Karen's time is (since we are working in hours).
Thus,
There's a logical shortcut you can use on "catching up" distance/rate problems. The difference between the faster (Karen at 60mph) and slower (Jen at 45mph) drivers is 15mph. Which means that every one hour, the faster driver, Karen, gains 15 miles on Jen. We know that Jen gets a 1/2 hour head start, which at 45mph means that she's 22.5 miles ahead when Karen gets started. So we can calculate the number of hours (H) of the 15mph of Karen's "catchup speed" (the difference between their speeds) it will take to make up the 22.5 mile gap:
15H = 22.5
So H = 1.5.
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