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SAT Math : Algebra

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Example Questions

Example Question #21 : How To Find The Solution To An Equation

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?

Possible Answers:

248

431

346

269

157

Correct answer:

346

Explanation:

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

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Example Question #71 : Algebra

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?

Possible Answers:

11 and 17 feet

12 and 16 feet

cannot be determined

14 and 14 feet

8 and 20 feet

Correct answer:

12 and 16 feet

Explanation:

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.

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Example Question #22 : How To Find The Solution To An Equation

Solve for z:

3(z + 4)³ – 7 = 17

Possible Answers:

–8

–2

Correct answer:

–2

Explanation:

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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Example Question #72 : Algebra

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?

Possible Answers:

$\dpi{100} \small 1 \ hour$

$\dpi{100} \small 1.5 \ hours$

$\dpi{100} \small 3 \ hours$

She can't catch up.

$\dpi{100} \small 2 \ hours$

Correct answer:

$\dpi{100} \small 1.5 \ hours$

Explanation:

For this type of problem, we use the formula:

$\dpi{100} \small distance = rate\times time$

When Karen catches up with Jen, their distances are equivalent. Thus,

$\dpi{100} \small rate (Jen) \times time (Jen)=rate(Karen)\times time(Karen)$

We then make a variable for Jen's time, $\dpi{100} \small t$ . Thus we know that Karen's time is $\dpi{100} \small t-.5$ (since we are working in hours).

Thus,

45(t+0.5)=60t

45t+22.5=60t

22.5=15t

$t=1.5\ hours$

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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Example Question #24 : How To Find The Solution To An Equation

Bill and Bob are working to build toys. Bill can build $\dpi{100} \small k$ toys in 6 hours. Bob can build $\dpi{100} \small k$ toys in 3 hours. How long would it take Bob and Bill to build $\dpi{100} \small 4k$ toys working together?

Possible Answers:

$\dpi{100} \small 2\ hours$

$\dpi{100} \small 4\ hours$

$\dpi{100} \small 8\ hours$

$\dpi{100} \small 12\ hours$

$\dpi{100} \small 9\ hours$

Correct answer:

$\dpi{100} \small 8\ hours$

Explanation:

Bill builds $\dpi{100} \small \frac{k}{6}$ toys an hour. Bob builds $\dpi{100} \small \frac{k}{3}$ toys an hour. Together, their rate of building is $\dpi{100} \small \frac{k}{6}+\frac{k}{3}=\frac{k}{2}$ . Together they can build $\dpi{100} \small k$ toys in 2 hours. They would be able to build $\dpi{100} \small 4k$ toys in 8 hours.

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Example Question #25 : How To Find The Solution To An Equation

A hybrid car gets 40 miles per gallon. Gasoline costs $3.52 per gallon. What is the cost of the gasoline needed for the car to travel 120 miles?

Possible Answers:

$\dpi{100} \small \$ 9.54$

$\dpi{100} \small \$ 10.56$

$\dpi{100} \small \$ 12.53$

$\dpi{100} \small \$ 14.08$

$\dpi{100} \small \$ 10.36$

Correct answer:

$\dpi{100} \small \$ 10.56$

Explanation:

The car will be using $\dpi{100} \small \frac{120\ miles}{40\ mpg}=3\ gallons$ of gas during this trip. Thus, the total cost would be $\dpi{100} \small 3\times \$ 3.52=\$ 10.56$ .

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Example Question #73 : Algebra

Jon invested part of $16,000 at 3% and the rest at 5% for a total return of $680.

Quantity A: The amount Jon invested at 5% interest

Quantity B: The amount Jon invested at 3% interest

Possible Answers:

Quantity A is greater

The relationship cannot be determined from the information given

Quantity B is greater

The two quantities are equal

Correct answer:

Quantity A is greater

Explanation:

First, let $\dpi{100} \small x$ represent the invested amount at 3% and set up an equation like this:

$\dpi{100} \small .03x+.05(16,000-x)=680$

Solve for $\dpi{100} \small x$ , and you'll find that Jon invested $6,000 at 3% and $10,000 at 5%.

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Example Question #74 : Algebra

Audrey, Penelope and Clementine are all sisters. Penelope is 8 years older than Clementine and 2 years younger than Audrey. If the sum of Penelope and Clementine's age is Audrey's age, how old is Clementine's age?

Possible Answers:

Correct answer:

Explanation:

Let = Audrey's age, = Penelope's age, and = Clementine's age.

Since A=P+C , then C=A-P .

Furthermore, P=C+8 , and A=P+2 .

Through substitution, C = [(C + 8) + 2] - (C + 8), C = 2 .

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Example Question #75 : Algebra

If $\dpi{100} \small 3x+y =13$ and $\dpi{100} \small x-2y=-12$ , what is the value of $\dpi{100} \small x$ ?

Possible Answers:

$\dpi{100} \small 2$

$\dpi{100} \small 1$

$\dpi{100} \small \frac{1}{3}$

$\dpi{100} \small \frac{2}{3}$

$\dpi{100} \small 3$

Correct answer:

$\dpi{100} \small 2$

Explanation:

We could use the substitution or elimination method to solve the system of equations. Here we will use the elimination method.

To solve for $\dpi{100} \small x$ , combine the equations in a way that makes the $\dpi{100} \small y$ terms drop out. The first equation has $\dpi{100} \small y$ and the second $\dpi{100} \small -2y$ , so multiplying the first equation times 2 then adding the equations will eliminate the $\dpi{100} \small y$ terms.

Multiplying the first equation times 2: $\dpi{100} \small 2(3x+y = 13)\rightarrow 6x+2y = 26$

Adding this result to the second equation: $\dpi{100} \small 6x+x + 2y - 2y = 26-12\rightarrow 7x=14$

Isolate $\dpi{100} \small x$ by dividing both sides by 7:

$\dpi{100} \small \frac{7x}{7}=\frac{14}{7}$

$\dpi{100} \small x=2$

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Example Question #32 : How To Find The Solution To An Equation

If $\frac{1}{4}x-\frac{1}{6}y=\frac{1}{6}$ and $\frac{y}{z}=\frac{1}{2}$ , then what is the value of 3x-z ?

Possible Answers:

Correct answer:

Explanation:

Since the expression we want just involves z and x, but not y, we start by solving $\frac{y}{z}=\frac{1}{2}$ for y $(y=\frac{1}{2}z)$ .

Then we can plug that expression in for y in the first equation.

$\frac{1}{4}x-\frac{1}{6}y=\frac{1}{6}$

$\frac{1}{4}x-\frac{1}{6}(\frac{1}{2}z)=\frac{1}{6}$

$\frac{1}{4}x-\frac{1}{12}z=\frac{1}{6}$

Multiply everything by 12 to get rid of fractions.

$\small 3x - z = 2$

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SAT Math : Algebra

Study concepts, example questions & explanations for SAT Math

All SAT Math Resources

Example Questions

Example Question #21 : How To Find The Solution To An Equation

Example Question #71 : Algebra

Example Question #22 : How To Find The Solution To An Equation

Example Question #72 : Algebra

Example Question #24 : How To Find The Solution To An Equation

Example Question #25 : How To Find The Solution To An Equation

Example Question #73 : Algebra

Example Question #74 : Algebra

Example Question #75 : Algebra

Example Question #32 : How To Find The Solution To An Equation

All SAT Math Resources

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