1- Choice D is correct
The correct answer is \(–\ 122\) Use PEMDAS (order of operation): \([6\ ×\ (–\ 24)\ +\ 8]\ –\ (–\ 4)\ +\ [4\ ×\ 5]\ ÷\ 2 = [–\ 144\ +\ 8]\ –\ (–\ 4)\ +\ [20]\ ÷\ 2 =\) \([–\ 144\ +\ 8]\ –\ (–\ 4)\ +\ 10 =\) \([–\ 136]\ –\ (–\ 4) \ +\ 10 = [–\ 136]\ +\ 4\ +\ 10 = \ –\ 122\)
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2- Choice D is correct
The correct answer is \(4\ x^2\ +\ 2\ x\ y \ - \ 2 \ y^2\) Use FOIL method. \((2\ x\ +\ 2\ y)\ (2\ x\ -\ y) = 4\ x^2\ -\ 2\ x\ y\ +\ 4\ x\ y\ -\ 2\ y^2=4\ x^2\ +\ 2\ x\ y\ -\ 2\ y^2\)
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3- Choice D is correct
The correct answer is \(91\) To solve absolute values equations, write two equations. \(x\ -\ 10\) could be positive \(3\), or negative \(3\). Therefore, \(x\ -\ 10=3 ⇒ x=13\) \(x\ -\ 10=-\ 3 ⇒ x=7\) Find the product of solutions: \(7\ ×\ 13 = 91\)
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4- Choice B is correct
The correct answer is \(-\frac{1}{2}\) The equation of a line in slope intercept form is: \(y=m\ x\ +\ b\) Solve for \(y\). \(4\ x\ -\ 2\ y=12 ⇒ -\ 2\ y=12\ -\ 4\ x ⇒ y=(12\ -\ 4\ x)\ ÷\ (-\ 2)\) ⇒ \(y=2\ x\ -\ 6\) The slope of this line is \(2\). The product of the slopes of two perpendicular lines is \(-\ 1\). Therefore, the slope of a line that is perpendicular to this line is: \(m_1\ ×\ m_2 = -\ 1 ⇒ 2\ ×\ m_2 = -\ 1 ⇒ m_2 = \ -\frac{1}{2}\)
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5- Choice C is correct
The correct answer is \(36\) Plug in the value of\(x\) and \(y\). \(x=3\) and \(y=-\ 2\) \(5\ (x\ -\ 2\ y)\ +\ (2\ -\ x)^2= 5\ (3\ -\ 2\ (-\ 2))\ +\ (2\ -\ 3)^2=5\ (3\ +\ 4)\ +\ (-\ 1)^2 =36\)
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6- Choice C is correct
The correct answer is \(10\) Let \(x\) be the number. Write the equation and solve for \(x\). \(40\%\) of \(x=4⇒ 0.40\ x=4 ⇒ x=4\ ÷\ 0.40=10\)
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7- Choice A is correct
The correct answer is \(3\) \(A\) is \(4\) times of \(B\), then: \(A\) = \(4\ B\) ⇒ \((A = 12)\ 12 = 4\ ×\ B ⇒ B = 12\ ÷\ 4 = 3\)
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8- Choice C is correct
The correct answer is \(6\) hours The distance between Jason and Joe is \(9\) miles. Jason running at \(5.5\) miles per hour and Joe is running at the speed of \(7\) miles per hour. Therefore, every hour the distance is \(1.5\) miles less. \(9\ ÷\ 1.5 = 6\)
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9- Choice D is correct
The correct answer is \(80\%\) The failing rate is \(11\) out of \(55 = \frac{11}{55}\) Change the fraction to percent: \(\frac{11}{55} ×\ 100\%=20\%\) \(20\) percent of students failed. Therefore, \(80\) percent of students passed the exam.
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10- Choice D is correct
The correct answer is \(729\) \(3^6 = 3\ ×\ 3\ ×\ 3\ ×\ 3\ ×\ 3\ ×\ 3 = 729\)
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11- Choice D is correct
Solve for \(x\). \(-\ 2\ ≤\ 2\ x\ -\ 4\ <\ 8\) ⇒ (add \(4\) all sides) \(-\ 2\ +\ 4\ ≤\ 2\ x\ -\ 4\ +\ 4\ <\ 8\ +\ 4\) ⇒ \(2\ ≤\ 2\ x\ <\ 12\) ⇒ (divide all sides by \(2\)) \(1\ ≤\ x\ <\ 6\) \(x\) is between \(1\) and \(6\)
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12- Choice D is correct
The correct answer is \(28\) Let \(x\) be the width of the rectangle. Use Pythagorean Theorem: \(a^2\ +\ b^2 = c^2\) \(x^2\ +\ 8^2 = 10^2 ⇒ x^2\ +\ 64 = 100 ⇒ x^2 = 100\ –\ 64 = 36 ⇒ x = 6\) Perimeter of the rectangle \(= 2\) (length \(+\) width) \(= 2\ (8\ +\ 6) = 2\ (14) = 28\)
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13- Choice B is correct
The correct answer is \(70\) cm\(^2\) The perimeter of the trapezoid is \(36\) cm. Therefore, the missing side (height) is \(= 36\ –\ 8\ –\ 12\ –\ 6 = 10\) Area of a trapezoid: A = \(\frac{1}{2}\ h\ (b_1\ +\ b_2) = \frac{1}{2}\ (10)\ (6\ +\ 8) = 70\)
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14- Choice B is correct
The correct answer is \(\frac{1}{4}\) The probability of choosing a Hearts is \(\frac{13}{52} =\frac{1}{4}\)
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15- Choice D is correct
The correct answer is \(\frac{2}{3} ,\ 67\%,\ 0.68,\ \frac{4}{5}\) \(\frac{2}{3} = 0.666…\) \(0.68\) \(67\% = 0.67\) \(\frac{4}{5} = 0.80\) Therefore \(\frac{2}{3}\ <\ 67\%\ <\ 0.68\ <\ \frac{4}{5}\)
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16- Choice C is correct
The correct answer is \(87.5\) average (mean) \(=\frac{sum\ of\ terms}{number\ of\ terms} ⇒ 88 =\frac{sum\ of\ terms}{50} ⇒ sum = 88\ ×\ 50 = 4400\) The difference of \(94\) and \(69\) is \(25\). Therefore, \(25\) should be subtracted from the sum. \(4400\ –\ 25 = 4375\) mean \(=\frac{sum\ of\ terms}{number\ of\ terms}\) ⇒ mean =\( \frac{4375}{50} = 87.5\)
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17- Choice D is correct
The correct answer is \(\frac{5}{36}\) To get a sum of \(6\) for two dice, we should get \(3\) and \(3\), or \(2\) and \(4\), or \(4\) and \(2\), or \(1\) and \(5\), or \(5\) and \(1\). Therefore, there are \(5\) options. Since, we have \(6\ ×\ 6 = 36\) total options, the probability of getting a sum of \(6\) is \(5\) out of \(36\) or \(\frac{5}{36}\).
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18- Choice D is correct
The correct answer is \(8\) Use formula of rectangle prism volume. V \(=\) (length) (width) (height) ⇒ \(2000 = (25)\ (10)\) (height) ⇒ height \(= 2000\ ÷\ 250 = 8\)
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19- Choice C is correct
The correct answer is \(60\) To find the number of possible outfit combinations, multiply the number of options for each factor: \(3\ ×\ 5\ ×\ 4 = 60\)
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20- Choice D is correct
The correct answer is \(24\) \(4\ ÷\ \frac{1}{6} = 24\)
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21- Choice B is correct
The correct answer is \(32\) The diagonal of the square is \(8\). Let \(x\) be the side. Use Pythagorean Theorem: \(a^2\ +\ b^2 = c^2\) \(x^2\ +\ x^2 = 8^2 ⇒ 2\ x^2 = 8^2 ⇒ 2\ x^2 = 64 ⇒x^2 = 32 ⇒\ x= \sqrt{32}\) The area of the square is: \(\sqrt{32}\ ×\ \sqrt{32} = 32\)
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22- Choice B is correct
The correct answer is \(\frac{1}{4}\) Probability \(=\frac{number \ of\ desired\ outcomes}{number \ of\ total\ outcomes} = \frac{18}{12\ +\ 18\ +\ 18\ +\ 24} = \frac{18}{72} = \frac{1}{4}\)
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23- Choice D is correct
The correct answer is \(30\) Find the difference of each pairs of numbers: \(2,\ 3,\ 5,\ 8,\ 12,\ 17,\ 23,\) ___\(,\ 38\) The difference of \(2\) and \(3\) is \(1,\ 3\) and \(5\) is \(2,\ 5\) and \(8\) is \(3,\ 8\) and \(12\) is \(4,\ 12\) and \(17\) is \(5,\ 17\) and \(23\) is \(6,\ 23\) and next number should be \(7\). The number is \(23\ +\ 7 = 30\)
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24- Choice A is correct
The correct answer is \(10\) meters The width of the rectangle is twice its length. Let \(x\) be the length. Then, width\(=2\ x\) Perimeter of the rectangle is \(2\) (width \(+\) length) \(= 2\ (2\ x\ +\ x)=60 ⇒ 6\ x=60 ⇒ x=10\) The length of the rectangle is \(10\) meters.
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25- Choice D is correct
The correct answer is \(16\) average \(=\frac{sum\ of\ terms}{number\ of\ terms}\) ⇒ (average of \(6\) numbers) \(12 = \frac{sum\ of\ numbers}{6}\) ⇒sum of \(6\) numbers is \(12\ ×\ 6 = 72\) (average of \(4\) numbers) \(10 =\frac{sum\ of\ numbers}{4}\) ⇒sum of \(4\) numbers is \(10\ ×\ 4 = 40\) sum of \(6\) numbers \(–\) sum of \(4\) numbers \(=\) sum of \(2\) numbers \(72\ –\ 40 = 32\) average of \(2\) numbers \(=\frac{32}{2} = 16\)
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26- Choice C is correct
The correct answer is \(-\ 2\) Solving Systems of Equations by Elimination Multiply the first equation by \((–2)\), then add it to the second equation. \(\cfrac{\begin{align}-\ 2\ (2\ x\ +\ 5\ y= 11) \\ \ 4\ x\ -\ 2\ y=-\ 14 \end{align}}{} \) \(⇒-\ 4\ x\ -\ 10\ y= -\ 22\\4\ x\ -\ 2\ y=-\ 14 ⇒ -\ 12\ y= -\ 36 ⇒ y= 3\) Plug in the value of \(y\) into one of the equations and solve for \(x\). \(2\ x\ +\ 5\ (3)= 11 ⇒ 2\ x\ +\ 15= 11 ⇒ 2\ x= -\ 4 ⇒ x= -\ 2\)
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27- Choice C is correct
The correct answer is \(27\) Solve for the sum of five numbers. average \(=\frac{sum\ of\ terms}{number\ of\ terms} ⇒ 24 = \frac{sum\ of\ 5\ numbers}{5}\) ⇒ sum of \(5\) numbers \(= 24\ ×\ 5 = 120\) The sum of \(5\) numbers is \(120\). If a sixth number \(42\) is added, then the sum of \(6\) numbers is \(120\ +\ 42 = 162\) average \(=\frac{sum\ of\ terms}{number\ of\ terms} =\frac{162}{6} = 27\)
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28- Choice C is correct
The correct answer is \(12\) The ratio of boys to girls is \(4:7\). Therefore, there are \(4\) boys out of \(11\) students. To find the answer, first, divide the total number of students by \(11\), then multiply the result by \(4\). \(44\ ÷\ 11 = 4 ⇒ 4\ ×\ 4 = 16\) There are \(16\) boys and \(28\ (44\ –\ 16)\) girls. So, \(12\) more boys should be enrolled to make the ratio \(1:1\)
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29- Choice A is correct
The correct answer is \(\frac{1}{22}\) \(2,500\) out of \(55,000\) equals to \(\frac{2500}{55000} =\frac{25}{550} =\frac{1}{22}\)
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30- Choice A is correct
The correct answer is \(120\ x\ +\ 14,000\ ≤\ 20,000\) Let \(x\) be the number of new shoes the team can purchase. Therefore, the team can purchase \(120\ x\). The team had \($20,000\) and spent \($14000\). Now the team can spend on new shoes \($6000\) at most. Now, write the inequality: \(120\ x\ +\ 14,000\ ≤\ 20,000\)
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31- Choice D is correct
The correct answer is \(60\) Jason needs an \(75\%\) average to pass for five exams. Therefore, the sum of \(5\) exams must be at lease \(5\ ×\ 75 = 375\) The sum of \(4\) exams is: \(68\ +\ 72\ +\ 85\ +\ 90 = 315\). The minimum score Jason can earn on his fifth and final test to pass is: \(375\ –\ 315 = 60\)
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32- Choice D is correct
The correct answer is \(\frac{1}{4}\) Isolate and solve for \(x\). \(\frac{2}{3}\ x\ +\ \frac{1}{6}= \frac{1}{3} ⇒ \frac{2}{3}\ x= \frac{1}{3}\ -\frac{1}{6} =\frac{1}{6} ⇒ \frac{2}{3}\ x=\frac{1}{6}\) Multiply both sides by the reciprocal of the coefficient of \(x\). \((\frac{3}{2})\ \frac{2}{3}\ x= \frac{1}{6}\ (\frac{3}{2}) ⇒ x= \frac{3}{12}=\frac{1}{4}\)
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33- Choice B is correct
The correct answer is \($840\) Use simple interest formula: \(I=\)prt \((I =\) interest, \(p =\) principal, \(r =\) rate, \(t =\) time) \(I=(12000)\ (0.035)\ (2)=840\)
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34- Choice D is correct
The correct answer is \(48\ x^8\ y^6\) Simplify. \(6\ x^2\ y^3\ (2\ x^2\ y)^3= 6\ x^2\ y^3\ (8\ x^6\ y^3 ) = 48\ x^8\ y^6\)
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35- Choice C is correct
The correct answer is \(66\ π\ i n^2\) Surface Area of a cylinder \(= 2\ π\ r\ (r\ +\ h)\), The radius of the cylinder is \(3\ (6\ ÷\ 2)\) inches and its height is \(8\) inches. Therefore, Surface Area of a cylinder \(= 2\ π\ (3)\ (3\ +\ 8) = 66\ π\)
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36- Choice C is correct
The correct answer is \(\frac{125}{512}\) The square of a number is \(\frac{25}{64}\), then the number is the square root of \(\frac{25}{64}\) \(\sqrt{\frac{25}{64}}=\frac{5}{8}\) The cube of the number is: \((\frac{5}{8})^3 =\frac{125}{512}\)
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37- Choice B is correct
The correct answer is \(28\) Write the numbers in order: \(2,\ 19,\ 27,\ 28,\ 35,\ 44,\ 67\) The Median is the number in the middle. So, the median is \(28\).
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38- Choice D is correct
The correct answer is \(170\) Use the information provided in the question to draw the shape. Use Pythagorean Theorem: \(a^2\ +\ b^2 = c^2\) \(80^2\ +\ 150^2 = c^2 ⇒ 6400\ +\ 22500 = c^2 ⇒ 28900 = c^2 ⇒ c = 170\)
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39- Choice B is correct
The correct answer is \(40\) Plug in \(104\) for F and then solve for \(C\). \(C = \frac{5}{9}\ (F\ –\ 32) ⇒ C = \frac{5}{9}\ (104\ –\ 32) ⇒ C = \frac{5}{9}\ (72) = 40\)
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40- Choice A is correct
The correct answer is \(45\) First, find the number. Let \(x\) be the number. Write the equation and solve for \(x\). \(150\%\) of a number is \(75\), then: \(1.5\ ×\ x=75 ⇒ x=75\ ÷\ 1.5=50\) \(90\%\) of \(50\) is: \(0.9\ × \ 50 = 45\)
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41- Choice D is correct
The correct answer is \(2\) Solve for \(y\). \(4\ x\ -\ 2\ y=6 ⇒ -\ 2\ y=6\ -\ 4\ x ⇒ y=2\ x\ -\ 3\) The slope of the line is \(2\).
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42- Choice D is correct
The correct answer is \(120\) cm\(^3\) Volume of a box = length \(×\) width \(×\) height \(= 4\ ×\ 5\ ×\ 6 = 120\)
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43- Choice A is correct
The correct answer is \(4\ x^4\ +\ 4\ x^3\ -\ 12\ x^2\) \((6\ x^3\ -\ 8\ x^2\ +\ 2\ x^4 )\ -\ (4\ x^2\ -\ 2\ x^4\ +\ 2\ x^3 ) ⇒ (6\ x^3\ -\ 8\ x^2\ +\ 2\ x^4 )\ -\ 4\ x^2\ +\ 2\ x^4\ -\ 2\ x^3 ⇒ 4\ x^4\ +\ 4\ x^3\ -\ 12\ x^2\)
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44- Choice C is correct
The correct answer is \(38\%\) the population is increased by \(15\%\) and \(20\%\). \(15\%\) increase changes the population to \(115\%\) of original population. For the second increase, multiply the result by \(120\%\). \((1.15)\ ×\ (1.20) = 1.38 = 138\%\) \(38\) percent of the population is increased after two years.
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45- Choice C is correct
The correct answer is \(60,000\) Three times of \(24,000\) is \(72,000\). One-sixth of them canceled their tickets. One-sixth of \(72,000\) equals \(12,000\ (\frac{1}{6}\ ×\ 72000 = 12000)\). \(60,000\ (72000\ –\ 12000 = 60000)\) fans are attending this week
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46- Choice D is correct
The correct answer is \(97.6\) The area of the square is \(595.36\). Therefore, the side of the square is the square root of the area. \(\sqrt{595.36}=24.4\) Four times the side of the square is the perimeter: \(4\ × \ 24.4 = 97.6\)
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