SAT Math No Calculator practice tests.

## Quiz #10

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Question 1 |

If x is 6 less than y and y is twice of z, then what is the value of x when z=2?

10 | |

8 | |

-2 | |

12 | |

-5 |

Question 1 Explanation:

The correct answer is (C).

Let's set up some equations to solve for $x$.

Translating from English to math, we get:

"x is 6 less than y" ⇒ $x = y - 6$

"y is twice of z" ⇒ $y = 2z$

Now plug in $z = 2$:

$y = 2z = 2(2) = 4$

$x = y - 6 = 4 - 6 = -2$

Therefore, $x$ is equal to $-2$.

Let's set up some equations to solve for $x$.

Translating from English to math, we get:

"x is 6 less than y" ⇒ $x = y - 6$

"y is twice of z" ⇒ $y = 2z$

Now plug in $z = 2$:

$y = 2z = 2(2) = 4$

$x = y - 6 = 4 - 6 = -2$

Therefore, $x$ is equal to $-2$.

Question 2 |

If $x + 3 = y$, then $2x + 6 = ?$

y | |

4y | |

3y | |

2y | |

cannot be determined |

Question 2 Explanation:

The correct answer is (D).

We can rewrite $x + 3 = y$ as $x = y - 3$

Plugging this value of $x$ into the 2

$2(y - 3) + 6$

Simplifying: $2y - 6 + 6 = 2y$

Therefore, $2y$ is the answer.

We can rewrite $x + 3 = y$ as $x = y - 3$

Plugging this value of $x$ into the 2

^{nd}equation, we get:$2(y - 3) + 6$

Simplifying: $2y - 6 + 6 = 2y$

Therefore, $2y$ is the answer.

Question 3 |

$3(x - 4) = 18$, what is the value of $x$?

14/3 | |

22/3 | |

6 | |

10 | |

22 |

Question 3 Explanation:

The correct answer is (C).

First, distribute:

$3(x - 4) = 18$

$3x - 12 = 18$

$3x = 30$

$x = 10$

First, distribute:

$3(x - 4) = 18$

$3x - 12 = 18$

$3x = 30$

$x = 10$

Question 4 |

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 person purchased?

3, 8, 24 | |

7, 9, 17 | |

6, 9, 12 | |

5, 15, 18 | |

9, 10, 13 |

Question 4 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.

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.

Once you are finished, click the button below. Any items you have not completed will be marked incorrect.

There are 4 questions to complete.

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## Quiz #38

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Question 1 |

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?

1/3 | |

1/2 | |

3/4 | |

2/3 | |

3/2 |

Question 1 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 = $\dfrac{Distance}{Time}$ $=$ $\dfrac{1 mile}{2 hours}.$

The average speed for the way to Lark Rise was $½$ $ mph.$

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 = $\dfrac{Distance}{Time}$ $=$ $\dfrac{1 mile}{2 hours}.$

The average speed for the way to Lark Rise was $½$ $ mph.$

Question 2 |

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?

π | |

3π | |

6π | |

9π | |

16π |

Question 2 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 of circle B 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:

$\dfrac{3}{4} * (3 + x) = 2 + x$

Solving, we get:

$\dfrac{9}{4} + \dfrac{3*x}{4} = 2 + x$

$\dfrac{1}{4} + \dfrac{3x}{4} = x$

$\dfrac{1}{4} = \dfrac{x}{4}$

$1 = x$

If $x = 1$, the radius of circle B $= 2 + 1 = 3$. The area will be $πr$

If you chose (A), this is much too small to be the area, since we know the area is $πr$

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.

To find the area of the smaller circle (Circle B), we’ll need to find the radius of circle B. The radius of circle B 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:

$\dfrac{3}{4} * (3 + x) = 2 + x$

Solving, we get:

$\dfrac{9}{4} + \dfrac{3*x}{4} = 2 + x$

$\dfrac{1}{4} + \dfrac{3x}{4} = x$

$\dfrac{1}{4} = \dfrac{x}{4}$

$1 = x$

If $x = 1$, the radius of circle B $= 2 + 1 = 3$. The area will be $πr$

^{2}= $9π$.If you chose (A), this is much too small to be the area, since we know the area is $πr$

^{2}and $r > 2.$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 3 |

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?

1/2 | |

2 | |

8 | |

8√2 | |

16 |

Question 3 Explanation:

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 $4 * 4 = 16.$

The area of the inscribed square is $2√2 * 2√2 = 4 * 2 = 8.$ The original square’s area is $2x$ 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 4 |

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?$\dfrac{y}{x}$ | |

$\dfrac{y (x – z)}{z – x}$ | |

$\dfrac{100xy}{x – z}$ | |

$\dfrac{100x }{ y (z – x)}$ | |

$\dfrac{100yz }{ y(x – z)}$ |

Question 4 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 $\dfrac{10}{5}$ = $\$2$.

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: $\dfrac{5 - 2}{2}$ = $\dfrac{3}{2}$ = $150\%$.

Now we’ll plug in our picked numbers into the answer choices to see which one yields the same percent increase.

(A) $\dfrac{10}{5}$ = $2$

(B) $\dfrac{10 (5 – 3) }{ 3 – 5 }$= $\dfrac{20}{-2}$ = $-10$

(C) $\dfrac{100(5)(10) }{ 5 – 3}$ = $\dfrac{5000 }{ 2 }$= $2500$

(D) $\dfrac{100(5) }{ 10(3 – 5)}$ = $\dfrac{500}{-20}$= $-25$

(E) $\dfrac{100(10)(3) }{ 10(5-3)}$ = $\dfrac{3000}{20}$ = $150$ ⇒ Correct!

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 $\dfrac{10}{5}$ = $\$2$.

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: $\dfrac{5 - 2}{2}$ = $\dfrac{3}{2}$ = $150\%$.

Now we’ll plug in our picked numbers into the answer choices to see which one yields the same percent increase.

(A) $\dfrac{10}{5}$ = $2$

(B) $\dfrac{10 (5 – 3) }{ 3 – 5 }$= $\dfrac{20}{-2}$ = $-10$

(C) $\dfrac{100(5)(10) }{ 5 – 3}$ = $\dfrac{5000 }{ 2 }$= $2500$

(D) $\dfrac{100(5) }{ 10(3 – 5)}$ = $\dfrac{500}{-20}$= $-25$

(E) $\dfrac{100(10)(3) }{ 10(5-3)}$ = $\dfrac{3000}{20}$ = $150$ ⇒ Correct!

Question 5 |

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 $\dfrac{⏀(-85)⏀}{⏀(-84)⏀}$?-85 | |

-84 | |

-83 | |

1 | |

83 |

Question 5 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 * -81 * -79 * -77…-5 * -3 * -1$

Likewise, ⏀-84⏀ $ = -83 * -81 * -79 * -77…-5 * -3 * -1$

When these numbers are divided, we can see that every value will cancel out. The answer is $1$.

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 * -81 * -79 * -77…-5 * -3 * -1$

Likewise, ⏀-84⏀ $ = -83 * -81 * -79 * -77…-5 * -3 * -1$

When these numbers are divided, we can see that every value will cancel out. The answer is $1$.

Question 6 |

For their school uniform, each student can choose from 4 types of tops and 3 types of bottoms. How many combinations of tops and bottoms are there?

7 | |

12 | |

1 | |

10 | |

24 |

Question 6 Explanation:

The correct answer is (B).

This is a simple probability question. For every top, there are 3 types of bottoms. We know that there are 4 types of tops.

Thus, there are $3 * 4 = 12$ combinations of tops and bottoms.

This is a simple probability question. For every top, there are 3 types of bottoms. We know that there are 4 types of tops.

Thus, there are $3 * 4 = 12$ combinations of tops and bottoms.

Once you are finished, click the button below. Any items you have not completed will be marked incorrect.

There are 6 questions to complete.

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## Quiz #43

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*Quiz #43*. You scored %%SCORE%% out of %%TOTAL%%. Your performance has been rated as %%RATING%%
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Question 1 |

The table above represents a relationship between and

*$x$*and*$y$*. Which of the following linear equations describes the relationship?y = x + 1 | |

y = x + 3 | |

y = 3x | |

y = 2x | |

y = 3x - 1 |

Question 1 Explanation:

The correct answer is (E).

This is the only option that satisfies all the values of $x$ and $y$ in the table.

This is the only option that satisfies all the values of $x$ and $y$ in the table.

Question 2 |

If the average (arithmetic mean) of $2x$ and $4x$ is $18$, what is the value of $x$?

2 | |

3 | |

6 | |

9 | |

10 |

Question 2 Explanation:

The correct answer is (C). Use the formula for average:

$Average = \dfrac{Sum~of~all~elements}{Number~of~elements}$

$= \dfrac{(2x + 4x)}{2} = 18$

$\dfrac{(6x)}{2} = 18$

$6x = 18 * 2$

$6x = 36$

$x = 6$

$Average = \dfrac{Sum~of~all~elements}{Number~of~elements}$

$= \dfrac{(2x + 4x)}{2} = 18$

$\dfrac{(6x)}{2} = 18$

$6x = 18 * 2$

$6x = 36$

$x = 6$

Question 3 |

If

*a*,*b*,*c*are integers, let*a*W(*b*,*c*) be defined to be true only if*b*≤*a*≤*c*. If -3W(*b*,0) is true, which of the following could be a possible value of*b*?I. $-3$

II. $5$

III. $-5$

I only | |

II only | |

I and II | |

I and III | |

All are correct |

Question 3 Explanation:

If $-3W(b,0)$ is true, then $b ≤ -3 ≤ 0$.

Thus, $b = -3$ or $b = -5.$

Thus, $b = -3$ or $b = -5.$

Question 4 |

Trains A, B and C passed through a station at different speeds. Train A’s speed was 2 times Train B’s speed and Train C’s speed was 3 times Train A’s speed. What was Train C’s speed in miles per hour, if the Train B’s speed was 9 miles per hour?

54 | |

9 | |

27 | |

18 | |

45 |

Question 4 Explanation:

The correct answer is (A).

We can describe the situation using equations:

"Train A’s speed was 2 times Train B’s speed" ⇒ $A = 2B$

"Train C’s speed was 3 times Train A’s speed" ⇒ $C = 3A$

We know the speed of Train B is $\dfrac{9 miles}{hour}$

Plugging the value of $B$ into $A$, we get:

$A = 2 * 9 = 18$

Plugging the value of $A$ into $C$, we get:

$C = 3A = 3*18 = 54$

We can describe the situation using equations:

"Train A’s speed was 2 times Train B’s speed" ⇒ $A = 2B$

"Train C’s speed was 3 times Train A’s speed" ⇒ $C = 3A$

We know the speed of Train B is $\dfrac{9 miles}{hour}$

Plugging the value of $B$ into $A$, we get:

$A = 2 * 9 = 18$

Plugging the value of $A$ into $C$, we get:

$C = 3A = 3*18 = 54$

Question 5 |

The graph of

*y*=*f*(*x*) is shown above. If*f*(3) =*k*, which of the following could be the value of*f*(*k*)?3 | |

3.5 | |

4 | |

4.5 | |

2 |

Question 5 Explanation:

The correct answer is (E).

We know that

Now, plugging in the value of $k = 4$, we get:

We know that

*f*(3) =*k*. Using the graph, we can see that*f*(3) = 4.Now, plugging in the value of $k = 4$, we get:

*f*(*k*) =*f*(*4*) = 2*f*(4) = 2
There are 5 questions to complete.

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