1 Choice B is correct
The correct answer is \(0.01\) \(0.65\) equals \(65\) M. Then: \(0.65=65\) M \(→\) M \(=\frac{0.65}{65}=0.01\)

2 Choice C is correct
The correct answer is \(7\) hours Use distance formula: Distance \(=\) Rate \(\times\) time \(⇒\) \(490 = 70 \ \times\) T, divide both sides by \(70\). \(\frac{490}{ 70} =\) T \(⇒\) T \(= 7\) hours.

3 Choice C is correct
The correct answer is \($54\) Use simple interest formula: I \(=\) prt (I \(=\) interest, p \(=\) principal, r \(=\) rate, t \(=\) time) t is for one year. For \(3\) months, t is \(\frac{1}{4}\) or \(0.25\) I \(=(5,400) \ (0.04) \ (0.25)=$54\)

4 Choice A is correct
The correct answer is \(93.5\) D To find the discount, multiply the number by (\(100\% \ –\) rate of discount). Therefore, for the first discount we get: (D) (\(100\% \ – \ 15\%\)) = (D) \((0.85) = 0.85\) D For increase of \(10\%: \ (0.85\) D) \((100\% \ + \ 10\%) =\) \((0.85\) D) \((1.10) = 93.5\) D \(=93.5\) \(93.5\%\) of D

5 Choice E is correct
The correct answer is \(20\) average \(= \frac{sum \ of \ terms}{number \ of \ terms} ⇒\) \(16 = \frac{11 \ + \ 14 \ + \ 19 \ + \ x}{4} ⇒\) \(64 = 44 \ + \ x ⇒\) \(x = 20\)

6 Choice B is correct
The correct answer is \(6\) Three times a certain number, increased by \(18\), is equal to \(36\). Write an equation and solve. \(3 \ x \ + \ 18=36→\) \(3 \ x=36 \  \ 18=18→\) \(x=\frac{18}{3}=6\)

7 Choice C is correct
The correct answer is \(N \ + \ 2\) Alex has \(N\) toy cars. John has \(5\) more cars than Alex. Therefore, Alex has \(N \ + \ 5\) toy cars. John gives Alex \(3\) cars. Now, Alex has \((N \ + \ 5 \ – \ 3) \ N \ + \ 2\) toy cars.

8 Choice B is correct
The correct answer is \(520\) km Add the first \(5\) numbers. \( 45 \ + \ 50 \ + \ 55 \ + \ 35 \ + \ 60 = 245\) To find the distance traveled in the next \(5\) hours, multiply the average by number of hours. Distance \(=\) Average \(\times\) Rate \(= 55 \times 5 = 275\) Add both numbers. \(245 \ + \ 275 =520 \) km

9 Choice A is correct
The correct answer is \(27\) \(\frac{x \ + \ 3}{6}=5→\) \(x \ + \ 3=5 \ × \ 6=30→\) \(x=30 \  \ 3=27\)

10 Choice D is correct
The correct answer is \(108\) \(20\) percent of a number is \(120\). Therefore, the number is \(600\). \(0.20 \ x=120→\) \(x=\frac{120}{0.20}=600\) \(18\) percent of \(600\) is \(108\). \(0.18 \times 600=108\)

11 Choice B is correct
The correct answer is \(200\%\) Write the equation and solve for B: \(0.60\) A \(=0.30\) B , divide both sides by \(0.30\), then you will have \(\frac{0.60}{0.30}\) A \(=\) B, therefore: B \(= 2\) A, and B is \(2\) times of A or it’s \(200\%\) of A.

12 Choice A is correct
The correct answer is \(300\) The ratio of boy to girls is \(3:5\). Therefore, there are \(3\) boys out of \(8\) students. To find the answer, first divide the total number of students by \(8\), then multiply the result by \(3\). \(800 \ ÷ \ 8 = 100 ⇒\) \(100 \ × \ 3 = 300\)

13 Choice C is correct
The correct answer is \($0.318\) One pound of cheese costs \($0.86\). One pound \(=16\) ounces \(16\) ounces of cheese costs \($0.86\). Then, \(1\) ounce of chees costs \((0.86 \ ÷ \ 16) \ $0.053\). \(6\) ounces of cheese costs \((6 \ × \ $0.053) \ $0.318\).

14 Choice C is correct
The correct answer is \(13.5\%\) The question is this: \(450.40\) is what percent of \(520\)? Use percent formula: part \(=\frac{ percent}{100} \ ×\) whole \(450.20 = \frac{percent}{100} \ × \ 520 ⇒\) \(450.20 = \frac{percent \ ×\ 520}{100} ⇒\) \(45020 =\) percent \(× \ 520 ⇒\) percent \(= \frac{45020}{520} = 86.5\) \(450.20\) is \(86.5\%\) of \(520\). Therefore, the discount is: \(100\% \ – \ 86.5\% = 13.5\%\)

15 Choice A is correct
The correct answer is \(30\) Let \(x\) be the number. Write the equation and solve for \(x\). \(\frac{2}{3} \times 18= \frac{2}{5}\). \(x ⇒ \frac{2 \ × \ 18}{3}= \frac{2 \ x}{5}\) use cross multiplication to solve for \(x\). \(5 \ × \ 36= 2 \ x \ × \ 3 ⇒\) \(180=6 \ x ⇒ x=30\)

16 Choice D is correct
The correct answer is \(\frac{19}{20}\) If \(15\) balls are removed from the bag at random, there will be five balls in the bag. The probability of choosing a brown ball is \(1\) out of \(20\). Therefore, the probability of not choosing a brown ball is \(19\) out of \(20\) and the probability of having not a brown ball after removing \(19\) balls is the same.

17 Choice B is correct
The correct answer is \(75\%\) The failing rate is \(15\) out of \(60 = \frac{15}{40}\) Change the fraction to percent: \(\frac{15}{60} \ × \ 100\%=25\%\) \(25\) percent of students failed. Therefore, \(75\) percent of students passed the exam.

18 Choice A is correct
The correct answer is \(8\) Let \(x\) be the number. Write the equation and solve for \(x\). \((40 \ – \ x) \ ÷ \ x = 4\), Multiply both sides by \(x\). \((40\ – \ x) = 4 \ x\), then add \(x\) both sides. \(40 = 5 \ x\), now divide both sides by \(5\). \(x = 8\)

19 Choice D is correct
The correct answer is \(0\) \(N \ × \ \frac{2}{5} \ × \ 3=0\), then \(N\) must be \(0\).

20 Choice E is correct
The correct answer is \(18\%\) The percent of girls playing tennis is: \(60\% \ × \ 30\%=\) \(0.60 \ × \ 0.30=0.18=18\%\)

21 Choice A is correct
The correct answer is \(9\) \(4 \ x \ +\ 7=→43\) \(4 \ x=43 \  \ 7=36→\) \(x=\frac{36}{4}=9\)

22 Choice E is correct
The correct answer is \(20\) Let \(x\) be the number. Write the equation and solve for \(x\). \(30\%\) of \(x=6⇒\) \(0.30 \ x=6 ⇒\) \(x=6 \ ÷ \ 0.30=20\)

23 Choice A is correct
The correct answer is \(I \ > \ 3000 \ x \ + \ 32000\) Let \(x\) be the number of years. Therefore, \($3,000\) per year equals \(3000 \ x\). starting from \($32,000\) annual salary means you should add that amount to \(3000 \ x\). Income more than that is: \(I \ > \ 3000 \ x \ + \ 32000\)

24 Choice C is correct
The correct answer is \(25\%\) \($\ 14\) is what percent of \($\ 56\)? \(14 \ ÷ \ 56 = 0.25 = 25\%\)

25 Choice A is correct
The correct answer is \(13\) \(\frac{z}{3} =3→\) \(z=3 \ ×\ 3=9\) \(z \ + \ 4=9 \ + \ 4=13\)

25 Choice A is correct
The correct answer is \(13\) \(\frac{z}{3} =3→\) \(z=3 \ ×\ 3=9\) \(z \ + \ 4=9 \ + \ 4=13\)

26 Choice C is correct
The correct answer is \(143\) \(\frac{x \  \ 5}{6} \ + \ 5=28→\) \(\frac{x \  \ 5}{6}=28 \  \ 5=23→\) \(x \  \ 5=23 \ × \ 6=138→\) \(x=138 \ + \ 5=143\)

27 Choice A is correct
The correct answer is \(2.5\) The width of a rectangle is \(4 \ x\) and its length is \(6 \ x\). Then, the perimeter of the rectangle is \( 20 \ x\). Perimeter of a rectangle \(= 2\) (width \(+\) length) \(=2 \ (4 \ x \ + \ 6 \ x)=20 \ x\) The perimeter of the rectangle is \(50\). Then: \(20 \ x=50→x=2.5\)

28 Choice C is correct
The correct answer is \(x \ + \ 20\) John has \(x\) dollars and he receives \($120\). Therefore, he has \(x \ + \ 120\). He then buys a bicycle that costs \($100\). Now, he has: \(x \ + \ 120 \  \ 100= x \ + \ 20\)

29 Choice C is correct
The correct answer is \($\ 500\) Use simple interest formula: I \(=\) prt (I \(=\) interest, p \(=\) principal, r \(=\) rate, t \(=\) time) I \(= (4,000) \ (0.025) \ (5)=500\)

30 Choice C is correct
The correct answer is \(27\%\) David needs an \(45\%\) average to pass for five exams. Therefore, the sum of \(4\) exams must be at least \(4 \ ×\ 65 = 260\) The sum of \(4\) exams is: \(62 \ + \ 70 \ + \ 75 \ + \ 80 = 287\). The minimum score David can earn on his fifth and final test to pass is: \(287 \ – \ 260 = 27\)

31 Choice A is correct
The correct answer is \(8\) hours The distance between Alex and Jack is \(6\) miles. Alex running at \(4.5\) miles per hour and Jack is running at the speed of \(6\) miles per hour. Therefore, every hour the distance is \(1.5\) miles less. \(12 \ ÷\ 1.5 = 8\)

32 Choice D is correct
The correct answer is \(18\) The formula for the area of the circle is: A \(=π \ r^2 \) The area of the circle is \(81 \ π\). Therefore: A \(=π \ r^2 ⇒\) \(81 \ π = π \ r^2\) Divide both sides by \(π\): \(81 = r^2 ⇒ r=9\) Diameter of a circle is \(2 \times\) radius. Then: Diameter \(= 2 \ ×\ 9 = 18\)

33 Choice C is correct
The correct answer is \(15\) \(3 \ ÷ \ \frac{1}{5} = 15\)

34 Choice B is correct
The correct answer is \(1\) \((8 \  \ 6)\ × \ 4=8 \ + \ \Box\) Then: \(2 \ × \ 4=8 \ + \ \Box, \ 8 =8 \ +\ \Box\), then: \(\Box = 1\)

35 Choice A is correct
The correct answer is \(10 \ x \ + \ 6 \ y\) There are \(y\) tables that can each seat \(6\) people and there are \(x\) tables that can each seat \(10\) people. Therefore, \(6 \ y \ + \ 10 \ x\) people can be seated in the classroom.

36 Choice A is correct
The correct answer is \(7\) 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) \(= 2 \ (2 \ x \ + \ x)=42 ⇒\) \(6 \ x=42 ⇒\) \(x=7\) Length of the rectangle is \(7\) meters.

37 Choice D is correct
The correct answer is \(76.40\) average (mean) \(= \frac{sum \ of \ terms}{number \ of \ terms} ⇒\) \(77 = \frac{sum \ of\ terms}{42} ⇒\) sum \(= 77 \ × \ 42 = 3234\) The difference of \(92\) and \(68\) is \(24\). Therefore, \(24\) should be subtracted from the sum. \(3234 \ – \ 25 = 3209\) mean \(= \frac{sum \ of \ terms}{number \ of \ terms} ⇒\) mean \(= \frac{3209}{42} = 76.40\)

38 Choice A is correct
The correct answer is \((250) \ (0.80) \ (0.80)\) To find the discount, multiply the number by (\(100\% \ –\) rate of discount). Therefore, for the first discount we get: \((250) (100\% \ – \ 10\%) = (250) \ (0.80) = 200\) For the next \(10\%\) discount: \((250) \ (0.80) \ (0.80)\)

39 Choice C is correct
The correct answer is \(40\) average \(= \frac{sum \ of \ terms}{number \ of \ terms} ⇒\) (average of \(8\) numbers) \(16 = \frac{sum \ of \ numbers }{8} ⇒\) sum of \(8\) numbers is: \(16 \ × \ 8 = 128\) (average of \(6\) numbers) \(8 = \frac{sum \ of \ numbers}{6} ⇒\) sum of \(6\) numbers is \(8 \ × \ 6 = 48\) sum of \(8\) numbers \(–\) sum of \(6\) numbers \(=\) sum of \(2\) numbers \(128 \ – \ 48 = 80\) average of \(2\) numbers \(= \frac{80}{2} = 40\)

40 Choice E is correct
The correct answer is \(\frac{1}{5}\) The deck contains \(13\) Hearts. Then, the probability of choosing a Hearts is \(\frac{13}{65}=\frac{1}{5}\)

41 Choice A is correct
The correct answer is \(\frac{1}{10}\) \(4,500\) out of \(45,000\) equals to \(\frac{4500}{45000} = \frac{45}{450} = \frac{1}{10}\)

42 Choice D is correct
The correct answer is \(20\) \(x \ + \ 3=6→\) \(x=6 \  \ 3=3\) \(2 \ y \ \ 2=10→\) \(2 \ y=8→\) \(y=4\) \(x \ y \ +\ 8=3 \ × \ 4 \ + \ 8=20\)

43 Choice A is correct
The correct answer is \(\frac{1}{4}\) From the options provided, only \(\frac{1}{4}\) is greater than \(\frac{1}{6}\).

44 Choice C is correct
The correct answer is \(19\) \(x=7\), then: \(2 \ x \ + \ 5=2 \ ×\ 7 \ +\ 5=14 \ + \ 5=19\)

45 Choice E 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: \(0.9 \ ×\ 50 = 45\)

46 Choice A is correct
The correct answer is \(121\) cm\(^2\) The perimeter of the trapezoid is \(24\). Therefore, the missing side (height) is: \(= 49 \ – \ 10 \ – \ 16 \ – \ 12 = 11\) Area of a trapezoid: A \(= \frac{1}{2} \ h \ (b1 \ + \ b2) =\) \(\frac{1}{2} \ (11) \ (10 \ + \ 12) = 121\) cm \(^2\)

47 Choice E is correct
The correct answer is \(26\%\) the population is increased by \(10\%\) and \(15\%\). \(10\%\) increase changes the population to \(110\%\) of original population. For the second increase, multiply the result by \(115\%\). \((1.10) \ × \ (1.15) = 1.26 = 126\%\) \(26\) percent of the population is increased after two years.

48 Choice C is correct
The correct answer is \(50\) Plug in \(104\) for F and then solve for C. C \(= \frac{4}{8}\) (F \(– \ 30) ⇒\) C \(= \frac{4}{8} \ (110 \ –\ 30) ⇒\) C \(= \frac{5}{8} \ (80) = 50\)

49 Choice A is correct
The correct answer is \(384\) cm\(^3\) If the length of the box is \(24\), then the width of the box is one third of it, \(8\), and the height of the box is \(2\) (one fourth of the width). The volume of the box is: Volume of a box \(=\) (length) \(×\) (wdth) \(×\) (height) \(= (24) \ × \ (8) \ × \ (2) = 384\) cm\(^3\)

50 Choice B is correct
The correct answer is \(15\) \(2 \ x \ + \ 10=40→\) \(2\ x=40 \  \ 10=30→\) \(x=\frac{30}{2}=15\)
