SAT Math No Calculator practice test #3 (DIFFICULT).

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Question 1 |

**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 |

Question 1 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 $πr$

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}= $9π$.Question 2 |

**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 |

Question 2 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 3 |

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

$\dfrac{y (x – z)}{z – x}$ | |

$\dfrac{100yz }{ y(x – z)}$ | |

$\dfrac{100x }{ y (z – x)}$ |

Question 3 Explanation:

The correct response is (C).

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 4 |

**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 |

Question 4 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 5 |

**If the product of x and y does not equal zero, which of the following could be true based on the figure above?**

I. (−x, y) lies above the x-axis and to the right of the y-axis.

II. (x, −y) lies below the x-axis and to the left of the y-axis.

III. (x, y) lies on the x-axis to the right of the y-axis.

I. (−x, y) lies above the x-axis and to the right of the y-axis.

II. (x, −y) lies below the x-axis and to the left of the y-axis.

III. (x, y) lies on the x-axis to the right of the y-axis.

I only | |

II only | |

I and II only | |

II and III |

Question 5 Explanation:

The correct answer is (C).

For choice I, if (−x, y) lies above the x-axis and to the right of the y-axis (in the first quadrant), then

For choice II, if (x, −y) is in the third quadrant, then

For choice III, if a point lies on the x-axis, then its value is 0, that would make the product of

Given the constraints of the problem, only the first two Roman numerals are possible answer choices.

For choice I, if (−x, y) lies above the x-axis and to the right of the y-axis (in the first quadrant), then

*x*must equal a negative number, since all x-values to the right of the y-axis are positive. The only way*-x*= positive is if*x*= negative. So,*y*must be positive since it is above the x-axis. This is possible.For choice II, if (x, −y) is in the third quadrant, then

*x*= negative, and*-y*= negative. This is only true if*x*= negative and*y*= positive. This is also possible.For choice III, if a point lies on the x-axis, then its value is 0, that would make the product of

*x*and*y*zero, which contradicts the given information.Given the constraints of the problem, only the first two Roman numerals are possible answer choices.

Question 6 |

**If a = $\dfrac{-1}{b}$ and c = $\dfrac{1}{d}$ for integers**

*a, b, c*and*d*, in which of the following ranges does the product*abcd*fall?

−4 < abcd < -2 | |

−2 < abcd < 0 | |

0 < abcd < 2 | |

2 < abcd < 4 |

Question 6 Explanation:

The correct answer is (B).

The expression

Only answer choice (B) correctly places

The expression

*abcd*can be rewritten as:*abcd*= $b(\dfrac{-1}{b})*(\dfrac{1}{d})d = (-1)(1) = -1$Only answer choice (B) correctly places

*abcd*in the range −2 < −1 < 0.Question 7 |

**The average of several test scores is 80. One make-up exam was given. Included with the other scores, the new average was 84. If the score on the make-up exam was 92, how many total exams were given?**

2 | |

3 | |

4 | |

5 |

Question 7 Explanation:

The correct answer is (B).

Recall that average is calculated by adding a group of numbers and then dividing by the count of those numbers. Plugging in what we’re given to the formula for average:

$80 = \dfrac{Sum}{x}$

$80x = Sum$

$84 = \dfrac{(Sum~+~Makeup~Score)}{(x+1)}$

Multiplying both sides by $x + 1$:

$84x + 84 = Sum~+~ Makeup~Score$

Let’s substitute 80x for Sum:

$84x + 84 = 80x + Makeup~Score$

$4x + 84 = Makeup~Score$

We are told that the makeup score was 92:

$4x + 84 = 92$

$4x = 8$

$x = 2$

The question asks for the

Recall that average is calculated by adding a group of numbers and then dividing by the count of those numbers. Plugging in what we’re given to the formula for average:

$80 = \dfrac{Sum}{x}$

$80x = Sum$

$84 = \dfrac{(Sum~+~Makeup~Score)}{(x+1)}$

Multiplying both sides by $x + 1$:

$84x + 84 = Sum~+~ Makeup~Score$

Let’s substitute 80x for Sum:

$84x + 84 = 80x + Makeup~Score$

$4x + 84 = Makeup~Score$

We are told that the makeup score was 92:

$4x + 84 = 92$

$4x = 8$

$x = 2$

The question asks for the

*total*number of exams, which is $x+1 = 3$Question 8 |

**If Eric was 22 years old**

*x*years ago and Shelley will be 24 years old in*y*years, what was the average of their ages 4 years ago?$\dfrac{x + y}{2}$ | |

$\dfrac{y + 38}{3}$ | |

$\dfrac{x - y + 28}{4}$ | |

$\dfrac{x - y + 38}{2}$ |

Question 8 Explanation:

The correct answer is (D).

The problem asks for the average of the ages of 2 people. Thus, the divisor will be 2, and answer choices (B) and (C) can be eliminated. That leaves us with just (A) and (D).

Let's start off by converting the problem into mathematical expressions.

If Eric was 22 years old

If Shelley will be 24 years old in

(If these expressions seem confusing to you, try plugging in some numbers to convince yourself that they hold true.)

This means, that 4 years ago, Eric was $x + 22 - 4 = x + 18$ years old.

Likewise, 4 years ago, Shelley was $24 - y - 4 = 20 - y$ years old.

Taking the average of these two ages, we get:

$\dfrac{x + 18 + 20 - y}{2} = \dfrac{x - y + 38}{2}$

The problem asks for the average of the ages of 2 people. Thus, the divisor will be 2, and answer choices (B) and (C) can be eliminated. That leaves us with just (A) and (D).

Let's start off by converting the problem into mathematical expressions.

If Eric was 22 years old

*x*years ago, then he is currently $22 + x$ years old.If Shelley will be 24 years old in

*y*years, then she is currently $24 - y$ years old.(If these expressions seem confusing to you, try plugging in some numbers to convince yourself that they hold true.)

This means, that 4 years ago, Eric was $x + 22 - 4 = x + 18$ years old.

Likewise, 4 years ago, Shelley was $24 - y - 4 = 20 - y$ years old.

Taking the average of these two ages, we get:

$\dfrac{x + 18 + 20 - y}{2} = \dfrac{x - y + 38}{2}$

Question 9 |

**Point A lies between**

*X*and*Y*. Point B lies on*YZ*and Point C lies on*XZ*.*BZ*is congruent to*CZ*, and*∠XYZ*= 90°.*XZ – CZ = XA*. What is the value of*∠ACB*?45° | |

60° | |

75° | |

90° |

Question 9 Explanation:

The correct answer is (A).

This is a Geometry question, so you’ll want to begin by drawing the figure on your own and filling in the given information. Let’s label

The problem tells us that

Because

We know that $∠X + ∠Y + ∠Z = 180°$. Simplifying, we get:

$∠X + 90° + ∠Z = 180°$

$∠X + ∠Z = 90°$

We also know that $∠X = 180 − 2a$ and $∠Z = 180 − 2b$

Substituting these values, we get: $(180 − 2b) + (180 − 2a) = 90$

Simplifying this second equation:

$360 − 2b − 2a = 90.$

Subtracting 360:

$−2b − 2a = −270 → 270 = 2a + 2b$

Let’s factor out the 2 from the variables:

$270 = 2(a + b)$

Now, let's go back to the very first equation. If we manipulate it, we can see that $a + b = 180 − z.$ Let’s substitute $180 − z$ for $a + b$:

$270 = 2(180 − z)$

$z = 45° = ∠ACB.$

This is a difficulty level 5 question.

This is a Geometry question, so you’ll want to begin by drawing the figure on your own and filling in the given information. Let’s label

*∠ACB*as*z*to keep track of it.

The problem tells us that

*BZ*and*CZ*are congruent. Thus, ∠*ZBC*and ∠*ZCB*(denoted by*b*for simplicity) are equal.The problem tells us that

*XZ – CZ = XA*. Looking at the triangle, we can also see that*XZ – CZ = XC*. Hence, sides*XC*and*XA*are equal, and so are the angles opposite them (denoted by*a*.)Because

*a, z,*and*b*are supplementary, $a + b + z = 180.$We know that $∠X + ∠Y + ∠Z = 180°$. Simplifying, we get:

$∠X + 90° + ∠Z = 180°$

$∠X + ∠Z = 90°$

We also know that $∠X = 180 − 2a$ and $∠Z = 180 − 2b$

Substituting these values, we get: $(180 − 2b) + (180 − 2a) = 90$

Simplifying this second equation:

$360 − 2b − 2a = 90.$

Subtracting 360:

$−2b − 2a = −270 → 270 = 2a + 2b$

Let’s factor out the 2 from the variables:

$270 = 2(a + b)$

Now, let's go back to the very first equation. If we manipulate it, we can see that $a + b = 180 − z.$ Let’s substitute $180 − z$ for $a + b$:

$270 = 2(180 − z)$

$z = 45° = ∠ACB.$

This is a difficulty level 5 question.

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