SAT Math Practice Test 2

This is our second free set of SAT Math practice questions. Working through practice questions is the best way to prepare for this section of the SAT.

Directions: Solve each problem and select the best of the answer choices given.

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Question 1
Franklin bought several kites at the store, and each kite cost 16 dollars. Richard purchased several different kites, and spent 20 dollars each.  If the ratio of the number of kites Franklin purchased to the number of kites Richard purchased is 3 to 2, what is the average cost of a kite purchased by Richard and Franklin?

A
16.8
B
17.2
C
17.6
D
17.8
E
18.0
Question 1 Explanation: 
The correct response is (C). For every 5 kites purchased by these men, 3 of them are Franklin’s and 2 of them are Richard’s. We can set up an equation to find the total money spent:

3(16) + 2(20) = Total Money Spent for every 5 kites
88 = Total Money Spent for every 5 kites

To find the average per kite cost, we can simply divide 88 by 5. 88/5 = 17.6. If you chose (A), it doesn’t logically make sense that the average price of the kites would be so close to the cheaper price, especially since 2 out of every 5 purchased will be $20. If you chose (B) or (D), these are close, but you probably make a small calculation error. If you chose (E), you simply took the average of the prices: 16 and 20. Remember that we’re given the ratio for a reason. It affects the average!
Question 2
Cecilia, Robbie, and Briony all bought stamps. The number of stamps Cecilia purchased was equal to a single digit. The number of stamps only one of them purchased was divisible by 3. The number of stamps one of them bought was an even number. Which of the following could represent the numbers of stamps each purchased?

A
3, 8, 24
B
7, 9, 17
C
6, 9, 12
D
5, 15, 18
E
9, 10, 13
Question 2 Explanation: 
The correct response is (E). It’s easier for this type of question to use process of elimination. Since all of the answer choices have at least one single-digit number in it, let’s look at the second requirement. If the number of stamps that ONLY ONE of them purchased was divisible by 3, we can eliminate answer choices that contain more than one multiple of 3: (A), (C), and (D).

The third requirement is that we have at least one even number. Between (B) and (E), only choice (E) contains an even number, 10.

If you chose (A), the phrase “only one” means we cannot have more than one multiple of 3. Both 3 and 24 are multiples of 3.

If you chose (B), one number must be even and all three of these numbers are odd.

If you chose (C), the phrase “only one” tells us there cannot be more than one multiple of 3. All three of these numbers are multiples of 3.

If you chose (D), remember that the phrase “only one” means there cannot be more than one multiple of 3 in an answer choice. Both 15 and 18 are multiples of 3.
Question 3
Circle A is inside Circle B, and the two circles share the same center O. If the circumference of B is four times the circumference of A, and the radius of Circle A is three, what is the difference between Circle B’s diameter and Circle A’s diameter?

A
6
B
9
C
12
D
18
E
24
Question 3 Explanation: 
The correct response is (D). Start by drawing the figure. If the radius of A is 3, then its diameter is 6. Its circumference is 2πr = 6π. B’s circumference is four times A’s circumference. B’s circumference = 24π = 2πr. B’s radius must be 12, and its diameter is 24.
SATmathQ3
The difference between the diameters is 24 – 6 = 18.

If you chose (A), this is the diameter of the smaller circle.

If you chose (B), this is the difference between the radii.

If you chose (C), this is the radius of the larger circle.

If you chose (E), this is the diameter of the larger circle.
Question 4
A delivery cart went from Candleford to Lark Rise and back at an average speed of 2/3 miles per hour. If the distance from Candleford to Lark Rise is 1 mile, and the trip back took half as much time as the trip there, what was the average speed of the delivery cart on the way to Lark Rise?

A
1/3
B
1/2
C
3/4
D
2/3
E
3/2
Question 4 Explanation: 
The correct response is (B). It’s important to first understand what this question is asking. “On the way to Lark Rise” means the way there. The question is asking the average speed for a portion of the total trip. To find it, we’ll need to know the distance for that part of the trip and the time spent on that part of the trip.

If the average speed of the entire journey was 2/3 miles per hour, then every 3 hours 2 miles were travelled. Since the total distance was 2 miles, the total time must have been 3 hours. If the way back took half as much time as the way there, then for every 3 hours, 2 hours was spent on the way there, and 1 hour was spent on the way back.

Average Speed = Distance/Time = 1 mile / 2 hours. The average speed for the way to Lark Rise was ½ mph.
Question 5
In the figure, the radius of circle B is three-fourths the radius of circle A. The distance from center A to the circumference of circle B is 3, and the distance from center B to the circumference of circle A is 2. What is the area of the smaller circle?

SATmathQ5

A
π
B
C
D
E
16π
Question 5 Explanation: 
The correct response is (D). To find the area of the smaller circle (Circle B), we’ll need to find the radius of circle B. The radius is 2 + x.

We’re told that the radius of B is three-fourths the radius of A, so we can set up a proportion:

2 + x / 3 + x = 3/4

Now cross-multiply to solve:
8 + 4x = 9 + 3x
4x = 1 + 3x
1x = 1
1 = x

If x = 1, the radius of circle B is 2 + 1 = 3. The area will be πr2 = 9π.

If you chose (A), this is much too small to be the area, since we know the area is the produce of pi and the square of the radius.

If you chose (B), remember that x + 2 is the radius of the smaller circle. Though the radius is 3, we have to square it to find the area.

If you chose (C), this is the circumference of the smaller circle, not the area.

If you chose (E), you found the area of the larger circle, not the smaller one.
Question 6
The midpoints of the sides of a square are connected to form a new inscribed square. How many times greater than the area of the inscribed square is the area of the original square?

A
1/2
B
2
C
8
D
8√2
E
16
Question 6 Explanation: 
SATmathQ6
The credited response is (B). This is a great question to pick numbers for, especially since the correct answer is a ratio. Let’s say the length of the original square is 4. The midpoint divides the sides into two halves, each with a value of 2. The isosceles triangles formed by the lines connecting the midpoints are 45-45-90 special right triangles. Therefore the hypotenuse of those triangles would be 2√2.

The area of the original square is 4x4 = 16.

The area of the inscribed square is 2√2 x 2√2 = 4x2 = 8. The original square’s area is two times greater than the inscribed square’s area.

If you chose (A), you probably misread the question. The ratio here is the original square’s area to the inscribed square’s area, not vice-versa.

If you chose (C), this is the area of the inscribed square.

If you chose (D), this is the perimeter of the inscribed square. Make sure you carefully re-read what the question is asking.

If you chose (E), this is the area of the original square.
Question 7
The diameter of a circle is increased by 80%. By what percent, approximately, is the area increased?

A
180%
B
225%
C
275%
D
325%
E
375%
Question 7 Explanation: 
The correct response is (B). This is a great question for plugging in your own values. Let’s say the value of the original diameter was 10. The original area would be equal to πr2 = (10)2π = 100π. If the area increased by 80%, the new area would be 18, and the new area would be equal to πr2 = (18)2π = 324π.

The increase is 324π - 100π = 224π. An increase of 224/100, or approximately 225%.

If you chose (A), be careful not to reuse values that are simply given in the question stem.

If you chose (C), this is close, but a little too large.

If you chose (D), this is the value of the new percent, not the percent increase.

If you chose (E), you may want to practice picking values for these types of questions.
Question 8
Shauna is throwing a bachelorette party for her best friend with x guests total. All of the guests plan to split the cost of renting a limo for y dollars. The day before, z guests cancel. Which of the following represents the percent increase in the amount each guest must pay towards the limo rental?

A
y/x
B
y (x – z)/z – x
C
100xy / x – z
D
100x / y (z – x)
E
100yz / y(x – z)
Question 8 Explanation: 
The correct response is (E). This is a great question to pick numbers. Let’s choose easy, manageable values. If x = 5 and y = 10, then each guest would originally have paid $2. Let’s say z = 3. Now with only 2 guests, each will pay $5. The percent change is always the change over the original amount: $5 - $2 / $2 = $3/$2 = 150%. Now we’ll plug in our picked numbers into the answer choices to see which one yields the same percent increase.

(A) 10/5 = 2

(B) 10 (5 – 3) / 3 – 5 = 20/-2 = -10

(C) 100(5)(10) / 5 – 3 = 5000 / 2 = 2500

(D) 100(5) / 10(3 – 5) = 500/-20 = -25

(E) 100(10)(3) / 10(5-3) = 3000/20 = 150 Correct!
Question 9
For all negative integers m other than -1, let ⏀m⏀ be defined as the product of all the negative odd integers greater than m. What is the value of ⏀-85⏀/⏀-84⏀?

A
-85
B
-84
C
-83
D
1
E
83
Question 9 Explanation: 
The correct response is (D). For a function question using symbols, we must carefully apply the described rule of the symbol to the given values. ⏀-85⏀ is defined as the product of ALL the negative odd integers GREATER than -85. So ⏀-85⏀= -83 x -81 x -79 x -77…-5 x -3 x -1.

Likewise, ⏀-84⏀is defined as -83 x -81 x -79 x -77…-5 x -3 x -1. When these numbers are divided, we can see that every value will cancel out. The answer is 1.
Question 10
On a number line with points LMNOP, the ratio of LM to MN is 1, and the ratio of NO to OP is 3/4. If the length of LP is 28 and the length of MO is 13, what is the ratio of LO to MP?

A
7/18
B
13/28
C
15/28
D
17/20
E
20/21
Question 10 Explanation: 
The correct response is (E). Let’s draw the number line:
SATmathQ10
I chose variables “x” and “y” to help express the ratios. Right now the ratio of LO to MP is (2x + 3y)/(x + 7y). If we could find the values of x and y we can determine the ratio.

Let’s set up a few equations based on the given information:
x + 3y = 13, or x = 13 – 3y
2x + 7y = 28

Using substitution: 2(13 – 3y) + 7y = 28
26 – 6y + 7y = 28
26 + y = 28
y = 2

Now we can find x: x + 3(2) = 13
x + 6 = 13
x = 7

Therefore the ratio is (2x + 3y)/(x + 7y) = 14 + 6 / 7 + 14 = 20/21.
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