This practice test covers the second section of PSAT Math. For these questions the use of a calculator is allowed.

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Question 1 |

### Joe eats at a restaurant once every night. His bill before a tax of 8.25% is $25.00. He also leaves a tip of 15% on his meal before tax.

### Approximately how much money does Joe spend at this restaurant every month, considering a month has 31 days?

## $964.33 | |

## $964.39 | |

## $964.75 | |

## $964.78 |

Question 1 Explanation:

The correct answer is (D). The cost of one night of food is:

$25.00 × 1.15 × 1.0825 = 31.12

Multiply by by 31 to find his monthly expenditure:

31.12 × 31 = 964.78

Round in the last step, not the intermediate steps, to ensure accuracy.

$25.00 × 1.15 × 1.0825 = 31.12

Multiply by by 31 to find his monthly expenditure:

31.12 × 31 = 964.78

Round in the last step, not the intermediate steps, to ensure accuracy.

Question 2 |

### The scatterplot below shows the increase in households with 3 cars from 2000 to 2010 in a small community.

### Based on the line of best fit shown, which of the following values is closest to the average yearly increase in the number of households with 3 cars?

## 5 | |

## 7 | |

## 11 | |

## 15 |

Question 2 Explanation:

The correct answer is (B). Use the scatterplot to estimate these values:

2001 = 125 households

2008 = 175 households

Average $= \frac{175 − 125}{7}$ $= \frac{50}{7} ≅ 7$

2001 = 125 households

2008 = 175 households

Average $= \frac{175 − 125}{7}$ $= \frac{50}{7} ≅ 7$

Question 3 |

### If *k* is a positive constant not equal to 1, which of the following could be a graph of *y* + *x* = *k*(*x* − *y*) in the *xy*-plane?

Question 3 Explanation:

The correct answer is (B).

$y + x = k(x – y)$

$y + x = kx – ky$

$y + ky = kx – x$

$y(1 + k) = x(k – 1)$

$y = \dfrac{k – 1}{k + 1}x$

So, when $x = 0$, $y = 0$. This means the line has to grow through the origin. Only answer choice B shows a line through the origin.

$y + x = k(x – y)$

$y + x = kx – ky$

$y + ky = kx – x$

$y(1 + k) = x(k – 1)$

$y = \dfrac{k – 1}{k + 1}x$

So, when $x = 0$, $y = 0$. This means the line has to grow through the origin. Only answer choice B shows a line through the origin.

Question 4 |

### The height of an airplane is increasing at a linear rate. Ten minutes after takeoff, the plane is 1,225 feet above sea level. Twenty-five minutes after takeoff, the plane is 15,570 feet above sea level.

Which of the following is the value of the constant rate at which the plane’s height is increasing per minute?$956.334$ | |

$965.334$ | |

$8338.334$ | |

$14,345$ |

Question 4 Explanation:

The correct answer is (A). The two points of the plane are (10, 1225) and (25, 15570). The rate at which the height is increasing per minute is:

$\dfrac{15570 - 1125}{25 - 10} = 956.334$ $\text{feet/minute}$

$\dfrac{15570 - 1125}{25 - 10} = 956.334$ $\text{feet/minute}$

Question 5 |

### One Gigabyte is 1024 Megabytes. A DVD can hold 4.7 Gigabytes of data. A floppy disc can hold 1.44 Megabytes of data. If the data from *x* floppy discs can be fit into one DVD, what is the value of *x*?

$3342$ | |

$3343$ | |

$3555$ | |

$3558$ |

Question 5 Explanation:

The correct answer is (A).

A DVD can hold $1024 \cdot 4.7$ megabytes, which is $4812.8$. $\frac{4812.8}{1.44} = 3342.222$. Since only full discs can be fit, the largest whole number less than $3342.222$ is $3342$.

A DVD can hold $1024 \cdot 4.7$ megabytes, which is $4812.8$. $\frac{4812.8}{1.44} = 3342.222$. Since only full discs can be fit, the largest whole number less than $3342.222$ is $3342$.

Question 6 |

*Questions 6 and 7 refer to the table below:*

### In a historical town of Alexandria, the number of voters increased each year as shown in the table below:

### Which age group had the greatest increase in voters between 594 AD and 596 AD?

18–30 | |

$31–45$ | |

$46-51$ | |

$52-65$ |

Question 6 Explanation:

The correct answer is (A). The number of voters from 18 to 30 nearly doubled between 594 and 596, a higher increase than the other three age groups.

Question 7 |

### If the population percentage of 46–65 year-olds in 597 AD is 10% less than the population percentage of this group in 596 AD, then approximately what percent would that be?

$25\%$ | |

$35\%$ | |

$40\%$ | |

$45\%$ |

Question 7 Explanation:

The correct answer is (A). The voter percentage of 46–65 year-olds in 596 AD is:

$\dfrac{11690 + 2000}{39490} = 34.67\%$

The voter percentage of this group in 597 AD is 10% less. So, it is:

$34.67 - 10 = 24.67\% ≅ 25\%$

$\dfrac{11690 + 2000}{39490} = 34.67\%$

The voter percentage of this group in 597 AD is 10% less. So, it is:

$34.67 - 10 = 24.67\% ≅ 25\%$

Question 8 |

### The graphs of a system of equations—a parabola, a straight line, and a circle—are shown in the figure below.

### How many solutions does the system have?

$0$ | |

$1$ | |

$2$ | |

$3$ |

Question 8 Explanation:

The correct answer is (C). Solutions to the system are points where all three equations intersect. This occurs at two coordinates only.

Question 9 |

$x^2 + y^2 = 185$ $x = –6y$

**If $(x, y)$ is a solution to the system of equations shown above, then $y^2 =$**$3$ | |

$4$ | |

$5$ | |

$6$ |

Question 9 Explanation:

The correct answer is (C).

$x^2 + y^2 = 185$

$x = –6y$

What is the value of $y^2$?

$(–6y)^2 + y^2 = 185$

$36y^2 + y^2 = 185$

$37y^2 = 185$

$y^2 = \dfrac{185}{37} = 5$

$x^2 + y^2 = 185$

$x = –6y$

What is the value of $y^2$?

$(–6y)^2 + y^2 = 185$

$36y^2 + y^2 = 185$

$37y^2 = 185$

$y^2 = \dfrac{185}{37} = 5$

Question 10 |

**If $\dfrac{9x^2}{3x + 1} = \dfrac{1}{3x + 1} + B$, what is $B$ in terms of $x$?**

$3x + 1$ | |

$3x – 1$ | |

$2x + 1$ | |

$2x – 1$ |

Question 10 Explanation:

The correct answer is (B).

$\dfrac{9x^2}{3x + 1} = \dfrac{1}{3x + 1} + B$. Since you have to find $B$ in terms of $x$, isolate $B$ and simplify.

$B = \dfrac{9x^2}{3x + 1} – \dfrac{1}{3x + 1}$ $= \dfrac{9x^2 – 1}{3x + 1}$

Simplify numerator using $a^2 – b^2 = (a + b)(a – b)$

$= \dfrac{(3x)^2 – 1^2}{3x + 1}$ $= \dfrac{(3x + 1)(3x – 1)}{3x + 1} = 3x – 1$

$\dfrac{9x^2}{3x + 1} = \dfrac{1}{3x + 1} + B$. Since you have to find $B$ in terms of $x$, isolate $B$ and simplify.

$B = \dfrac{9x^2}{3x + 1} – \dfrac{1}{3x + 1}$ $= \dfrac{9x^2 – 1}{3x + 1}$

Simplify numerator using $a^2 – b^2 = (a + b)(a – b)$

$= \dfrac{(3x)^2 – 1^2}{3x + 1}$ $= \dfrac{(3x + 1)(3x – 1)}{3x + 1} = 3x – 1$

Question 11 |

**The change in enthalpy of a reaction is represented by the formula $q = mcΔt$, $q$ representing the heat of the system, $m$ representing the mass of the substance being heated, $c$ representing the substance’s specific heat, and $t$ representing temperature. Which of the following expressions could be used to solve for $m$?**

$\dfrac{q}{m} = cΔt$ | |

$m = \dfrac{q}{cΔt}$ | |

$Δt = \dfrac{q}{mc}$ | |

$q = mc$ |

Question 11 Explanation:

The correct answer is (B).

In order to solve this equation for the value of $m$, we need to isolate $m$. This done by moving all the other variables to one side through division.

$\dfrac{q}{cΔt} = \dfrac{mcΔt}{cΔt}$

$m = \dfrac{q}{cΔt}$

In order to solve this equation for the value of $m$, we need to isolate $m$. This done by moving all the other variables to one side through division.

$\dfrac{q}{cΔt} = \dfrac{mcΔt}{cΔt}$

$m = \dfrac{q}{cΔt}$

Question 12 |

**Which of the following $x$ values does NOT satisfy the inequality $–|3x + 5| < |2x + 2|$?**

$5$ | |

$10$ | |

$15$ | |

All real values satisfy this inequality |

Question 12 Explanation:

The correct answer is (D).

The left side of this equation will always be negative, as the negative sign is outside of the absolute value. This means that the left side is always less than the right side, regardless of the $x$-value. This means that all real values satisfy this inequality.

The left side of this equation will always be negative, as the negative sign is outside of the absolute value. This means that the left side is always less than the right side, regardless of the $x$-value. This means that all real values satisfy this inequality.

Question 13 |

**The minimum temperature readings in a city for $365$ days are shown in the box plot above. The range and interquartile range for this data are**

range $= 60$, IQR $= 30$ | |

range $= 60$, IQR $= 40$ | |

range $= 60$, IQR $= 60$ | |

range $= 60$, IQR $= 20$ |

Question 13 Explanation:

The correct answer is (D).

The range is $max – min = 70 – 10 = 60$. The interquartile range is $50 – 30 = 20$.

The range is $max – min = 70 – 10 = 60$. The interquartile range is $50 – 30 = 20$.

Question 14 |

**The license plate for a special vehicle will consist of three letters, picked from A to Z. Letters can be repeated. How many license plates can be given out for this special vehicle?**

$15,600$ | |

$17,576$ | |

$18,200$ | |

$18,576$ |

Question 14 Explanation:

The correct answer is (B).

Three letters are picked from A to Z. So, that is $26$ letters to pick from. Problem says that the letters can be repeated – so you can have AAA, AAB, or ABA, etc . The first letter can be picked in $26$ ways, the second letter in $26$ ways, and the third letter in $26$ ways. So, totally $26 \cdot 26 \cdot 26 = 17,576$ $3$-letter plates can be given out for this vehicle.

Three letters are picked from A to Z. So, that is $26$ letters to pick from. Problem says that the letters can be repeated – so you can have AAA, AAB, or ABA, etc . The first letter can be picked in $26$ ways, the second letter in $26$ ways, and the third letter in $26$ ways. So, totally $26 \cdot 26 \cdot 26 = 17,576$ $3$-letter plates can be given out for this vehicle.

Question 15 |

**When $3x^2 + 4x + 1$ is divided by $3x + 1$, the remainder is**

$x + 1$ | |

$0$ | |

$2x + 1$ | |

$2x – 1$ |

Question 15 Explanation:

The correct answer is (A).

Doing the long division:

Doing the long division:

Question 16 |

**A circus clown walks down a straight rope from a point $9$ feet above the ground as shown. At the same time, a trapeze artist swings down a parabolic path shown. After crossing the trapeze artist for the first time at $9$ feet, approximately how far does the clown walk down the rope before meeting the trapeze artist again?**

$3$ | |

$5$ | |

$6$ | |

$7$ |

Question 16 Explanation:

The correct answer is (D).

The first point where they meet is $(1, 9)$. The second point of meeting is $(6, 4)$. The distance that the clown walked down the rope is

$\sqrt{(4 – 9)^2 + (6 – 1)^2} = \sqrt{25 + 25} = 7.07$ ft.

This is approximately equal to $7$ feet. Note that the question asks “approximately how far”, thereby giving you a clue to round your answer. Use the distance formula and your calculator to find the square root.

The first point where they meet is $(1, 9)$. The second point of meeting is $(6, 4)$. The distance that the clown walked down the rope is

$\sqrt{(4 – 9)^2 + (6 – 1)^2} = \sqrt{25 + 25} = 7.07$ ft.

This is approximately equal to $7$ feet. Note that the question asks “approximately how far”, thereby giving you a clue to round your answer. Use the distance formula and your calculator to find the square root.

Question 17 |

**Use the following information for questions 17 and 18.****A materials engineer is testing three properties $P$, $Q$, and $R$ of a new alloy. All of them have the same units of measurement. Scatterplots showing the relation between $P$ and $Q$ and the relation between $Q$ and $R$ are shown above. Which of the following models best describe the relation between $P$ and $R$?**

$P = 7 – 2R$ | |

$P = 2R – 7$ | |

$P = R + 5$ | |

$P = R – 5$ |

Question 17 Explanation:

The correct answer is (B).

First, find the relationship between $P$ and $Q$ by listing the scatterplot values for $P$ and $Q$.

[insert chart here]

The relation can be expressed as $P = 2Q + 3$.

Now find the relation between $Q$ and $R$.

[insert chart here]

The relation between $Q$ and $R$ is $Q = R – 5$.

The question asks for the relationship between $P$ and $R$. You have $P = 2Q + 3$ and $Q = R – 5$.

So, $P = 2(R – 5) + 3 = 2R – 10 + 3 = 2R – 7$.

First, find the relationship between $P$ and $Q$ by listing the scatterplot values for $P$ and $Q$.

[insert chart here]

The relation can be expressed as $P = 2Q + 3$.

Now find the relation between $Q$ and $R$.

[insert chart here]

The relation between $Q$ and $R$ is $Q = R – 5$.

The question asks for the relationship between $P$ and $R$. You have $P = 2Q + 3$ and $Q = R – 5$.

So, $P = 2(R – 5) + 3 = 2R – 10 + 3 = 2R – 7$.

Question 18 |

**If the property $Q = 6$ units, then what is the ratio of $R$ to $P$?**

$\dfrac{7}{15}$ | |

$\dfrac{11}{15}$ | |

$\dfrac{13}{15}$ | |

$\dfrac{17}{15}$ |

Question 18 Explanation:

The correct answer is (B).

When $Q = 6$, $R = Q + 5 = 6 + 5 = 11$

$P = 2Q + 3 = 2 \cdot 6 + 3 = 15$

The ratio of $R$ to $P = \frac{11}{15}$.

When $Q = 6$, $R = Q + 5 = 6 + 5 = 11$

$P = 2Q + 3 = 2 \cdot 6 + 3 = 15$

The ratio of $R$ to $P = \frac{11}{15}$.

Question 19 |

**The ratio of poor people to rich people in a town in the year $1990$ was $6$ to $1$. The same ratio in the year $2000$ was $10$ to $3$. What is the percent increase of rich people in the town from $1990$ to $2000$?**

$2.67$ | |

$6.69$ | |

$8.79$ | |

$9.23$ |

Question 19 Explanation:

The correct answer is (C).

In $1990, \frac{poor}{rich} = \frac{6}{1}$. So, the percentage of rich people $= \frac{1}{7} \cdot 100 = 14.29%$

In $2000, \frac{poor}{rich} = \frac{10}{3}$. So, the percentage of rich people $= \frac{3}{13} \cdot 100 = 23.08%$

So, the percentage increase is $23.08 – 14.29 = 8.79%$

In $1990, \frac{poor}{rich} = \frac{6}{1}$. So, the percentage of rich people $= \frac{1}{7} \cdot 100 = 14.29%$

In $2000, \frac{poor}{rich} = \frac{10}{3}$. So, the percentage of rich people $= \frac{3}{13} \cdot 100 = 23.08%$

So, the percentage increase is $23.08 – 14.29 = 8.79%$

Question 20 |

**Based on the diagram above, if chord $CD$ has a length of $10$, what is the length of $AB$?**

$10$ | |

$\dfrac{10\sqrt{3}}{3}$ | |

$5\sqrt{3}$ | |

$\dfrac{10\sqrt{2}}{3}$ |

Question 20 Explanation:

The correct answer is (B).

You can use the $30-60-90$ triangles created when triangle $ACD$ is divided into two. This gives $CD = 5$, $AE = \frac{5\sqrt{3}}{3}$, and $AC = \frac{10\sqrt{3}}{3}$. $AB$ is a radius of the circle and has the same value as $AC$.

You can use the $30-60-90$ triangles created when triangle $ACD$ is divided into two. This gives $CD = 5$, $AE = \frac{5\sqrt{3}}{3}$, and $AC = \frac{10\sqrt{3}}{3}$. $AB$ is a radius of the circle and has the same value as $AC$.

Question 21 |

**If a circle with its origin at $(0, 0)$ passes through the point $(–4, 0)$, then which one of the following points is on the circumference of the circle?**

$(–4, 1)$ | |

$(0, 5)$ | |

$(–1, \sqrt{15})$ | |

$(2, \sqrt{14})$ |

Question 21 Explanation:

The correct answer is (C).

Since the circle has its origin at $(0, 0)$ and passes through $(–4, 0)$, its radius has a length of $4$ units. So, any other point on the circle must satisfy the equation $x^2 + y^2 = r^2 = 4^2 = 16$. Of the choices given, only choice (C) $(–1, \sqrt{15})$ satisfies the equation. $(–1)^2 + (\sqrt{15})^2 = 1 + 15 = 16$.

Since the circle has its origin at $(0, 0)$ and passes through $(–4, 0)$, its radius has a length of $4$ units. So, any other point on the circle must satisfy the equation $x^2 + y^2 = r^2 = 4^2 = 16$. Of the choices given, only choice (C) $(–1, \sqrt{15})$ satisfies the equation. $(–1)^2 + (\sqrt{15})^2 = 1 + 15 = 16$.

Question 22 |

**A stone is thrown from the ground along a parabolic path defined by the function $y = –x^2 + 4x$. The stone hits the tip of a vertical pole that is $4$ feet tall at the vertex of its path. How far is the pole from the point where the stone was thrown?**

$1$ | |

$2$ | |

$4$ | |

$16$ |

Question 22 Explanation:

The correct answer is (B).

The stone takes a parabolic path which opens downward, since the coefficient of $x^2$ is $–1$. The path of the stone intersects the tip of the pole that is $4$ feet tall. This can be represented as $y = 4$. Together, you have a linear quadratic system.

$y = –x^2 + 4x$

$y = 4$

Equate them and solve for $x$.

$–x^2 + 4x = 4$

$x^2 – 4x = –4$

$x^2 – 4x + 4 = 0$

$x^2 – 2x – 2x + 4 = 0$

$x(x – 2) – 2(x – 2) = 0$

$(x – 2)(x – 2) = 0$

$(x – 2) = 0$

$x = 2$, $y = 4$

The stone touches the pole at $x = 2$.

The stone takes a parabolic path which opens downward, since the coefficient of $x^2$ is $–1$. The path of the stone intersects the tip of the pole that is $4$ feet tall. This can be represented as $y = 4$. Together, you have a linear quadratic system.

$y = –x^2 + 4x$

$y = 4$

Equate them and solve for $x$.

$–x^2 + 4x = 4$

$x^2 – 4x = –4$

$x^2 – 4x + 4 = 0$

$x^2 – 2x – 2x + 4 = 0$

$x(x – 2) – 2(x – 2) = 0$

$(x – 2)(x – 2) = 0$

$(x – 2) = 0$

$x = 2$, $y = 4$

The stone touches the pole at $x = 2$.

Question 23 |

**Two lab tests were conducted on $12$ guinea pigs. Each of them was placed in a sound proof cage. In Lab Test-$1$ a low decibel sound was sent into the cage through a small hole in the cage. The time it took for each guinea pig to recognize the sound was noted in seconds. In Lab Test-$2$ a high decibel sound was sent into the cage and the time it took for each guinea pig to recognize the sound was noted. The results are shown in the dot plots below. Which of the following inferences can be made?**

Lab Test-$1$ has a higher standard deviation than Lab Test-$2$ | |

Lab Test-$2$ has a higher standard deviation than Lab Test-$1$ | |

Both tests have the same standard deviation | |

More students should take the tests in order to compare the standard deviations |

Question 23 Explanation:

The correct answer is (A).

You are asked to compare the standard deviation of two sets of data. In Test-$1$, several data values are at $60$ and $61$ and a few values are at $55$. So the average is likely to be between less than $60$. This means that the standard deviation will be a higher number.

In Test-$2$, the mean is likely to be about $59$, since the values are distributed evenly around $59$. Since the data is evenly spread, the standard deviation is most likely less than the standard deviation of test-$1$. Alternatively, plugging in the values from the dot-plots into the calculator shows that Test-$1$ has a standard deviation of $2.69$, and test $2$ has a standard deviation of $2.58$.

You are asked to compare the standard deviation of two sets of data. In Test-$1$, several data values are at $60$ and $61$ and a few values are at $55$. So the average is likely to be between less than $60$. This means that the standard deviation will be a higher number.

In Test-$2$, the mean is likely to be about $59$, since the values are distributed evenly around $59$. Since the data is evenly spread, the standard deviation is most likely less than the standard deviation of test-$1$. Alternatively, plugging in the values from the dot-plots into the calculator shows that Test-$1$ has a standard deviation of $2.69$, and test $2$ has a standard deviation of $2.58$.

Question 24 |

**In a housing community, $15$ people play Tennis, $20$ people play Soccer, and $25$ people play Baseball. $5$ People play Tennis and Soccer but not baseball. $4$ people play Tennis and Baseball but not Soccer. $6$ people play Soccer and Baseball but not Tennis. There are $35$ people in the housing community.**

How many people play all three games?

How many people play all three games?

$3$ | |

$4$ | |

$5$ | |

$9$ |

Question 24 Explanation:

The correct answer is (C).

Let $x$ be the number of people who play all the games. Place the numbers from the problem in the Venn diagram.

Those who play tennis only is $15 – 5 – 4 – x = 6 – x$. Those who play soccer only is $20 – 5 – 6 – x = 9 – x$. Those who play baseball only is $25 – 4 – x – 6 = 15 – x$. There are totally $35$ people. Set up an equation.

$6 – x + 5 + x + 4 + 9 – x + 6 + 15 – x = 35$

$45 – 2x = 35$

$2x = 10$

$x = 5$. So, $5$ people play all the games.

Let $x$ be the number of people who play all the games. Place the numbers from the problem in the Venn diagram.

Those who play tennis only is $15 – 5 – 4 – x = 6 – x$. Those who play soccer only is $20 – 5 – 6 – x = 9 – x$. Those who play baseball only is $25 – 4 – x – 6 = 15 – x$. There are totally $35$ people. Set up an equation.

$6 – x + 5 + x + 4 + 9 – x + 6 + 15 – x = 35$

$45 – 2x = 35$

$2x = 10$

$x = 5$. So, $5$ people play all the games.

Question 25 |

**When an object is in motion, its velocity $v$ is related to its kinetic energy $E$, and its mass $m$, by the formula $v = \sqrt{\frac{2E}{m}}$. If the velocity of an object of mass $m2$ is twice the velocity of an object of mass $m1$, and the kinetic energy of both objects are the same, then what is the ratio of $\frac{m1}{m2}$?**

$1$ | |

$2$ | |

$4$ | |

$8$ |

Question 25 Explanation:

The correct answer is (C).

Though strange looking, this problem is easy to solve. There are $2$ objects, of mass $m1$ and $m2$.

Let $v1$ and $v2$ be their velocities. Their kinetic energies are the same.

$v1 = \sqrt{\dfrac{2E}{m1}} \cdot v2 = \sqrt{\dfrac{2E}{m2}}$

$v2 = 2v1$ - given fact

So, $\sqrt{\dfrac{2E}{m2}} = 2 \sqrt{\dfrac{2E}{m1}}$. Square both sides.

$\dfrac{2E}{m2} = 4 \dfrac{2E}{m1}$. Don’t forget to square the $2$.

Therefore, $\dfrac{m1}{m2} = 4$.

Though strange looking, this problem is easy to solve. There are $2$ objects, of mass $m1$ and $m2$.

Let $v1$ and $v2$ be their velocities. Their kinetic energies are the same.

$v1 = \sqrt{\dfrac{2E}{m1}} \cdot v2 = \sqrt{\dfrac{2E}{m2}}$

$v2 = 2v1$ - given fact

So, $\sqrt{\dfrac{2E}{m2}} = 2 \sqrt{\dfrac{2E}{m1}}$. Square both sides.

$\dfrac{2E}{m2} = 4 \dfrac{2E}{m1}$. Don’t forget to square the $2$.

Therefore, $\dfrac{m1}{m2} = 4$.

Question 26 |

**In the figure above, a semicircle of radius $r$ has a chord $(CD) ̅$ that is parallel to diameter $(AB) ̅$ and is $\frac{1}{4}$ of $(AB) ̅$. What is distance between chords $(AB) ̅$ and $(CD) ̅$ in terms of $r$?**

$\sqrt{15}r$ | |

$\dfrac{\sqrt{15}}{4}r$ | |

$\dfrac{\sqrt{15}}{2}r$ | |

$\sqrt{4}r$ |

Question 26 Explanation:

The correct answer is (B).

Let the center of the semicircle be the point $O$. Draw a line $OC$ and $OP$. So, $OA = r OC = r$. $OP$ is the distance between the chords. $OPC$ is a right-angled triangle. $OC^2 = OP^2 + PC^2$.

$r^2 = OP^2 + \left( \dfrac{r}{4}\right)^2$

$OP^2 = r^2 – \dfrac{r^2}{16} = \dfrac{15}{16} r^2$

$OP = \dfrac{\sqrt{15}}{4}r$

Let the center of the semicircle be the point $O$. Draw a line $OC$ and $OP$. So, $OA = r OC = r$. $OP$ is the distance between the chords. $OPC$ is a right-angled triangle. $OC^2 = OP^2 + PC^2$.

$r^2 = OP^2 + \left( \dfrac{r}{4}\right)^2$

$OP^2 = r^2 – \dfrac{r^2}{16} = \dfrac{15}{16} r^2$

$OP = \dfrac{\sqrt{15}}{4}r$

Question 27 |

**If $\cos x = –p$ where the radian measure of angle $x$ is such that $π < x < \frac{3π}{2}$ and $\cos y = p$, then $y$ can be which one of the following?**

$π – x$ | |

$x + \dfrac{π}{2}$ | |

$x + 2π$ | |

$x – 2π$ |

Question 27 Explanation:

The correct answer is (A).

Note that the angles are in radians. The angle $x$ is such that $π < x < \frac{3π}{2}$. So, the angle is in the third quadrant. It is given that $\cos x = –p$. Mark a point P1 at an angle of $x$ radians. Note that the reference angle of point P1 is $x – π$.

Since $\cos y = p$, and cosine is positive in 1st and 4th quadrants, the point in the unit circle for which $\cos y = p$ is in the 1st quadrant (P3) or in the 4th quadrant (P2). If it is in the 1st quadrant, the angle of the point P3 is $x – π$. This could be the answer but is not in the answer choice. If the point is in the 4th quadrant (P2), then its reference angle is $(x – π)$, but since the angle is measured counterclockwise, the angle of the point P2 is $–(x – π) = π – x$.

So, $\cos (π – x) = p$. Since it is given that $\cos y = p$, it follows that $y = π – x$. This is given in the answer choice.

Note that the angles are in radians. The angle $x$ is such that $π < x < \frac{3π}{2}$. So, the angle is in the third quadrant. It is given that $\cos x = –p$. Mark a point P1 at an angle of $x$ radians. Note that the reference angle of point P1 is $x – π$.

Since $\cos y = p$, and cosine is positive in 1st and 4th quadrants, the point in the unit circle for which $\cos y = p$ is in the 1st quadrant (P3) or in the 4th quadrant (P2). If it is in the 1st quadrant, the angle of the point P3 is $x – π$. This could be the answer but is not in the answer choice. If the point is in the 4th quadrant (P2), then its reference angle is $(x – π)$, but since the angle is measured counterclockwise, the angle of the point P2 is $–(x – π) = π – x$.

So, $\cos (π – x) = p$. Since it is given that $\cos y = p$, it follows that $y = π – x$. This is given in the answer choice.

Question 28 |

**The product of $(5 + 5i) \cdot (5 − 6i)$ can be expressed in the form $a + bi$. What is the sum $a + b$?**

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Question 28 Explanation:

The correct answer is $50$.

$(5 + 5i) \cdot (5 − 6i) = 25 − 30i + 25i + 30 = 55 − 5i$

The sum of $a$ and $b = 55 + (−5) = 50$

$(5 + 5i) \cdot (5 − 6i) = 25 − 30i + 25i + 30 = 55 − 5i$

The sum of $a$ and $b = 55 + (−5) = 50$

Question 29 |

**A cylinder with a radius of $5$ cm is carved out of a hexagonal prism with a height of $10$ cm and a side length of $10$ cm. If the density of this cylinder is $3.2$ grams/cubic cm, what is the mass of the cylinder, to the nearest whole gram? (figure not drawn to scale)**

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Question 29 Explanation:

The correct answer is $5801$.

The volume of the cylinder is $π \cdot 5^2 \cdot 10 = 250π$. The area of the base of the hexagonal prism is $3 \cdot 100 \cdot \dfrac{\sqrt{3}}{2}$ $= 150\sqrt{3}$. So, the volume of the hexagonal prism $= A \cdot h = 150\sqrt{3} \cdot 10 = 1500\sqrt{3}$. The difference is $1500\sqrt{3} − 250π = 1812.68 cm^3$. The density is $3.2$ grams/cubic cm.

So, the mass of the prism is $1812.68 \cdot 3.2 = 5800.58 ≅ 5801$ rounded up.

The volume of the cylinder is $π \cdot 5^2 \cdot 10 = 250π$. The area of the base of the hexagonal prism is $3 \cdot 100 \cdot \dfrac{\sqrt{3}}{2}$ $= 150\sqrt{3}$. So, the volume of the hexagonal prism $= A \cdot h = 150\sqrt{3} \cdot 10 = 1500\sqrt{3}$. The difference is $1500\sqrt{3} − 250π = 1812.68 cm^3$. The density is $3.2$ grams/cubic cm.

So, the mass of the prism is $1812.68 \cdot 3.2 = 5800.58 ≅ 5801$ rounded up.

Question 30 |

**The New Express Bank issues its Credit Cards worldwide. When a customer makes a purchase using these Credit Cards in a foreign currency, the bank converts the purchase price at the daily foreign rate and then charges a $5\%$ fee on the converted cost.**Marla lives in the USA, and is currently on vacation in Shanghai, China. For a purchase that she made in Shanghai for $75$ Yuans, the bank posted a charge of $\$12.42$ to her account, including the $5\%$ fee. What Foreign Exchange Rate in (Chinese Yuan per US dollar) was used to calculate Marla’s purchase? Round your answer to the nearest hundredths.

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Question 30 Explanation:

The correct answer is $6.34$.

Marla purchased something for $75$ Yuans. The bank charged her $\$12.42$ including a $5\%$ fee. What exchange rate, Chinese Yuan per $1$ USD did the bank charge?

The bank converts the purchase price at the current exchange rate and then charges $4\%$. So let $1$ USD $= x$ Chinese Yuans

So, $75$ Yuans $= \dfrac{75 Yuans}{x \frac{Yuans}{Dollar}} = \dfrac{75}{x}$ USD

After adding $5\%$ fee to this purchase cost, the converted cost with fee is $\frac{75}{x} + 0.05 \cdot \frac{75}{x} = \frac{75}{x} (1.05)$. This is given to be $\$12.42$.

So $\dfrac{75}{x} (1.05) = 12.42$

$x = \dfrac{(75 \cdot 1.05)}{12.42} = 6.340$

Since the question asks you to round to the nearest hundredths, the exchange rate is $1$ USD $= 6.34$ Chinese Yuans.

Marla purchased something for $75$ Yuans. The bank charged her $\$12.42$ including a $5\%$ fee. What exchange rate, Chinese Yuan per $1$ USD did the bank charge?

The bank converts the purchase price at the current exchange rate and then charges $4\%$. So let $1$ USD $= x$ Chinese Yuans

So, $75$ Yuans $= \dfrac{75 Yuans}{x \frac{Yuans}{Dollar}} = \dfrac{75}{x}$ USD

After adding $5\%$ fee to this purchase cost, the converted cost with fee is $\frac{75}{x} + 0.05 \cdot \frac{75}{x} = \frac{75}{x} (1.05)$. This is given to be $\$12.42$.

So $\dfrac{75}{x} (1.05) = 12.42$

$x = \dfrac{(75 \cdot 1.05)}{12.42} = 6.340$

Since the question asks you to round to the nearest hundredths, the exchange rate is $1$ USD $= 6.34$ Chinese Yuans.

Question 31 |

**A bank in China sells a debit card worth $2000$ Yuans. Charges to the debit card incur no additional fee on top of the purchases. Marla wants to know the minimum amount of purchase she can make using the Debit Card so that it is cheaper than the Credit card. Round your answer to the nearest whole number of Yuans.**

**Write down your answer and then click below to see if you are correct.**

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Question 31 Explanation:

The correct answer is 1905.

Thinking and dealing with one currency will help avoid errors in calculations. If she wants to purchase something for USD $x$ using the Credit Card, she will be charged USD $1.05x$ including the fee. If she buys the Debit Card for $2000$ Yuans, she would pay $(2000 Yuans)/(\frac{6.34 Yuans}{USD}) =$ USD $315.46$. So, she can purchase up to USD $315.46$ using this debit card and her cost will be USD $315.46$ since she has to prepay this amount upfront. That’s how debit cards work. Make a table like shown below to see how this works.

[insert table here]

A few values are shown – say for USD $100$ in purchase cost, the cost using credit card is USD $105$ and this is cheaper than the cost or purchasing this item using debit card which is $333.33$. So, for some purchase that has cost of USD $x$, the cost using credit card is $1.05x$ and this is going to be more than the cost of paying it with a debit card, which is $333.33$ USD. So, you can setup and equation to find this USD $x$.

$1.05x > = 315.46$.

$x > \dfrac{315.46}{1.05} = 300.44$ USD.

So, for a purchase price up to USD $300.44$, the cost using credit card is cheaper than the cost using a debit credit card. After USD $300.44$, the debit card will be cheaper than the credit card. Convert this amount to Yuans – $300.44 \cdot 6.34 = 1904.78$ and round it to a whole number – the answer is $1905$.

Thinking and dealing with one currency will help avoid errors in calculations. If she wants to purchase something for USD $x$ using the Credit Card, she will be charged USD $1.05x$ including the fee. If she buys the Debit Card for $2000$ Yuans, she would pay $(2000 Yuans)/(\frac{6.34 Yuans}{USD}) =$ USD $315.46$. So, she can purchase up to USD $315.46$ using this debit card and her cost will be USD $315.46$ since she has to prepay this amount upfront. That’s how debit cards work. Make a table like shown below to see how this works.

[insert table here]

A few values are shown – say for USD $100$ in purchase cost, the cost using credit card is USD $105$ and this is cheaper than the cost or purchasing this item using debit card which is $333.33$. So, for some purchase that has cost of USD $x$, the cost using credit card is $1.05x$ and this is going to be more than the cost of paying it with a debit card, which is $333.33$ USD. So, you can setup and equation to find this USD $x$.

$1.05x > = 315.46$.

$x > \dfrac{315.46}{1.05} = 300.44$ USD.

So, for a purchase price up to USD $300.44$, the cost using credit card is cheaper than the cost using a debit credit card. After USD $300.44$, the debit card will be cheaper than the credit card. Convert this amount to Yuans – $300.44 \cdot 6.34 = 1904.78$ and round it to a whole number – the answer is $1905$.

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