1 Choice D is correct
The correct answer is \(30\) The area of the floor is: \(5\) cm \(× \ 30\) cm \(= 150\) cm The number is tiles needed \(= 150\ ÷ \ 5=30\)

2 Choice B is correct
The correct answer is \( \ 1 \) Solving Systems of Equations by Elimination method. \(\cfrac{\begin{align} 4 \ x \ + \ 6 \ y \ = \ 10 \\ x \ + \ y \ = 3 \end{align}}{} \) Multiply the second equation by \( \ 4 \), then add it to the first equation. \(\cfrac{\begin{align} 4 \ x \ + \ 6 \ y \ = 10 \\  \ 4 \ ( x \ + \ y \ = 3) \end{align}}{} ⇒\) \(\cfrac{ \begin{align} 4 \ x \ + \ 6 \ y \ = 10 \\  \ 4 \ x \  \ 4 \ y \ = 12 \end{align} }{\begin{align} 2\ y \ =  \ 2 \\ ⇒ y \ =  \ 1 \end{align}} \)

3 Choice C is correct
The correct answer is \(0.69\) D To find the discount, multiply the number by (\(100\% \ –\) rate of discount). Therefore, for the first discount we get: (D) \((100\% \ – \ 30\%) =\) (D) \((0.60) = 0.60\) D For increase of \(10\%: \ (0.60\) D) \((100\% \ + \ 15\%) = (0.60\) D)\( (1.15) = 0.69\) D \(= 69\%\) of D

4 Choice A is correct
The correct answer is \(50\) cm\(^2\) The diagonal of the square is \(10\). Let \(x\) be the side. Use Pythagorean Theorem: \(a^2 \ + \ b^2=c^2\) \(x^2 \ + \ x^2 = 10^2 ⇒ 2 \ x^2=10^2 ⇒ 2 \ x^2 = 100 ⇒ x^2 = 50 ⇒x= \sqrt{50}\) The area of the square is: \(\sqrt{50} \ × \ \sqrt{50}=50 \)

5 Choice C is correct
The correct answer is \(0.0555\) \(1000\) times the number is \(55.5\). Let \(x\) be the number, then: \(1000 \ x=55.5\) \(x=\frac{55.5}{1000}=0.0555\)

6 Choice D is correct
The correct answer is \(9\) hours and \(36\) minutes Use distance formula: Distance \(=\) Rate \(×\) time \(⇒ 384 = 40 \ ×\) T, divide both sides by \(40\). \(\frac{384}{40} =\) T \(⇒\) T \(= 9.6\) hours. Change hours to minutes for the decimal part. \(0.6\) hours \(= 0.6 \ × \ 60=36\) minutes.

7 Choice A is correct
The correct answer is \(165\) \(x=35 \ + \ 130=165\)

8 Choice A is correct
The correct answer is \(88\) \(35\%\) of \(80\) equals to: \(0.35 \ × \ 80=28\) \(12\%\) of \(600\) equals to: \(0.15\ × \ 400=60\) \(35\%\) of \(80\) is added to \(15\%\) of \(400: \ 28 \ + \ 60=88\)

9 Choice E is correct
The correct answer is \(36\) The area of \(\triangle\)BED is \(25\), then: \(\frac{5 \ × \ AB}{2}=25→5 \ ×\) AB \(=50→\) AB \(=10\) The area of \(\triangle\)BDF is \(18\), then: \(\frac{3 \ × \ BC}{2}=12→3 \ ×\) BC \(=24→\) BC \(=8\) The perimeter of the rectangle is \(= 2 \ × \ (10 \ + \ 8)=36\)

10 Choice B is correct
The correct answer is \(2 \ y \ + \ x=π§\) \(y\) and \(z\) are colinear. \(x\) and \(3 \ y\) are colinear. Therefore, \(y \ + \ z= x \ + \ 3 \ y \), subtract \(x\) from both sides,then, \(z= 2 \ y \ + \ x\)

11 Choice B is correct
The correct answer is sin \(π΄ =\) cos \(B\) By definition, the sine of any acute angle is equal to the cosine of its complement. Since, angle A and B are complementary angles, therefore: sin \(A =\) cos \(B\)

12 Choice B is correct
The correct answer is \(1,000\) ml \(3\%\) of the volume of the solution is alcohol. Let \(x\) be the volume of the solution. Then: \(3\%\) of \(x=30\) ml \(⇒ 0.03 \ x=30 ⇒ x=30 \ ÷ \ 0.03=1,000\)

13 Choice D is correct
The correct answer is \(\frac{26}{27}\) If \(26\) balls are removed from the bag at random, there will be one ball in the bag. The probability of choosing a red ball is \(1\) out of \(27\). Therefore, the probability of not choosing a red ball is \(26\) out of \(27\) and the probability of having not a red ball after removing \(26\) balls is the same.

14 Choice C is correct
The correct answer is \(12\%\) The percent of girls playing tennis is: \( 60\% \ × \ 20\%=0.60 \ × \ 0.20=0.12=12\%\)

15 Choice C is correct
The correct answer is \((– \ 2,2)\) Plug in each pair of number in the equation: A. \((2,1)\): \(4 \ (2) \ + \ 6 \ (1)=15\) Nope! B. \((– \ 1,3)\): \(4 \ (– \ 1) \ + \ 6 \ (3)=14\) Nope! C. \((– \ 2,2)\): \(4 \ (– \ 2) \ + \ 6 \ (2)=4\) Bingo! D. \((2,2)\): \(4 \ (2) \ + \ 6 \ (2)=20\) Nope! E. \((2,8)\): \(4 \ (2) \ + \ 6 \ (8)=56\) Nope!

16 Choice C is correct
The correct answer is \(40\) ft Write a proportion and solve for \(x\). \(\frac{4}{3}=\frac{x}{30} ⇒ 3 \ x=4 \ × \ 30 ⇒ x=40\) ft

17 Choice B is correct
The correct answer is \(584\) km Add the first \(5\) numbers. \(30 \ + \ 62 \ + \ 47 \ + \ 70 \ + \ 50=259\) To find the distance traveled in the next \(5\) hours, multiply the average by number of hours. Distance \(=\) Average \(×\) Rate \(=65 \ × \ 5=325\) Add both numbers. \(259 \ + \ 325=584\)

18 Choice E is correct
The correct answer is \( \ 2\) Solve for \(x\). \(\frac{2 \ x}{12}=\frac{x \ + \ 1}{3}\) Multiply the second fraction by \(4\). \(\frac{2 \ x}{12}=\frac{4 \ (x \ +\ 1)}{3 \ × \ 4}\) Tow denominators are equal. Therefore, the numerators must be equal. \(2 \ x=4 \ x \ + \ 4\) \( \ 2 \ x = 4\) \(x=  \ 2 \)

19 Choice B is correct
The correct answer is \( \ 9 \ ≤ \ x \ ≤ \ 5\) \(x \ + \ 7  \ ≤ \ 2→\) \( \ 2 \ ≤ \ x \ + \ 7 \ ≤ \ 2→\) \( \ 2 \  \ 7 \ ≤ \ x \ + \ 7 \  \ 7 \ ≤ \ 2 \  \ 7→\) \(9 \ ≤ \ x \ ≤ \ 5\)

20 Choice E is correct
The correct answer is \(90\%\) The question is this: \(1.50\) is what percent of \(1.35\)? Use percent formula: part \(=\frac{percent}{100} \ ×\) whole \(1.50 = \frac{percent}{100} \ × \ 1.35 ⇒\) \(1.50=\frac{percent \ × \ 1.35}{100} ⇒\) \(150=\) percent \(× \ 1.35 ⇒\) percent \(=\frac{150}{1.35}= 90\)

21 Choice A is correct
The correct answer is \(169\) \(0.4 \ x=(0.25) \ × \ 32→x=20→(x \  \ 7)^2=(13)^2=169\)

22 Choice E is correct
The correct answer is \((12,5)\) When points are reflected over \(y\)axis, the value of \(y\) in the coordinates doesn’t change and the sign of \(x\) changes. Therefore, the coordinates of point B is \((12,5)\).

23 Choice A is correct
The correct answer is \( \frac{8}{10}\) tan \(θ=\frac{opposite}{adjacent}\) tan \(θ=\frac{8}{6}⇒\) we have the following right triangle. Then: \(c=\sqrt{8^2 \ + \ 6^2 }=\sqrt{64 \ + \ 36}=\sqrt{100}=10\) cos \(θ=\frac{adjacent}{hypotenuse}=\frac{8}{10}\)

24 Choice B is correct
The correct answer is \(\frac{1}{3}\) Set of number that are not composite between \(1\) and \(45\): A \(= \left\{1, 2, 3, 5, 7, 11, 13, 17, 19, 23 , 29 , 31 ,37 ,41 ,43 \right\}\) Probability \(= \frac{number \ of \ desired \ outomes}{number \ of \ total \ outcomes} =\frac{15}{45}=\frac{1}{3}\)

25 Choice D is correct
The correct answer is \(y= x\) The slop of line A is: \(m=\frac{y_{2} \  \ y_{1}}{x_{2} \  \ x_{1}}=\frac{7 \  \ 6}{5 \  \ 4}=1\) Parallel lines have the same slope and only choice D \((y=x)\) has slope of \(1\).

26 Choice E is correct
The correct answer is \(316\) \(y = 5 \ a \ b \ + \ b^4 \) Plug in the values of \(a\) and \(b\) in the equation: \(a=3\) and \(b=4 \) \(y = 5 \ (3) \ (4) \ + \ (4)^4 = 60 \ + \ (256) = 316\)

27 Choice A is correct
The correct answer is \(69\) The area of trapezoid is: \((\frac{30 \ + \ 12}{2}) \ × \ x=192→24 \ x=192→x=8\) \(y=\sqrt{12^2 \ + \ 5^2}=13\) Perimeter is: \(18 \ + \ 30 \ + \ 8 \ + \ 13=69\)

28 Choice D is correct
The correct answer is \(3 \ x^2 \ – \ 7 \ x \ + \ 13\) \((g \ – \ f)(x)=g(x) \ – \ f(x)=(3 \ x^2 \ + \ 5 \ – \ 4 \ x) \ – \ ( \ 8 \ + \ 3 \ x)\) \(3 \ x^2 \ + \ 5 \ – \ 4 \ x \ + \ 8 \ – \ 3 \ x = \ 3 \ x^2 \  \ 7 \ x \ + \ 13\)

29 Choice A is correct
The correct answer is \(x\) is multiplied by \(3\) Plug in \(z\times 3\) for \(z\) and simplify. \(x_{1}=\frac{8 \ y \ + \ \frac{r}{r \ + \ 1}}{\frac{6}{z \times 3 }}=\frac{8 \ y \ + \ \frac{r}{r \ + \ 1}}{\frac{1}{3}\times \frac{ 6}{z }}=\) \(\frac{8 \ y \ + \ \frac{r}{r \ + \ 1}}{\frac{6}{z}}={3} \ × \ \frac{8 \ y \ + \ \frac{r}{r \ + \ 1}}{\frac{6}{z}}=x \times 3\)

30 Choice E is correct
The correct answer is \(4.788\) The weight of \(10.5\) meters of this rope is: \(10.5 \ × \ 456\) g \(=4788\) g \(1\) kg \(=1000\) g, therefore, \(4788\) g \(÷ \ 1000=4.788\) kg

31 Choice B is correct
The correct answer is \(282\) \(g(x)= \ 3\), then \(f(g(x))= f( \ 3)=3 \ ( \ 3)^4 \  \ 2 \ ( \ 3)^3 \ + \ 5 \ ( \ 3)= 243 \ + \ 54 \  \ 15=282\)

32 Choice C is correct
The correct answer is \(20\) Check each option provided: A. \(4 \ \ \ \ \frac{7\ + \ 13 \ + \ 15 \ + \ 16 \ + \ 20}{5}=\frac{71}{5}=14.2\) B. \(13 \ \ \ \ \frac{4 \ + \ 7 \ + \ 15 \ + \ 16 \ + \ 20}{5}=\frac{62}{5}=12.4\) C. \(20 \ \ \ \ \frac{4 \ + \ 7 \ + \ 13 \ + \ 15 \ + \ 16}{5}=\frac{55}{5}=11\) D. \(16\ \ \frac{4 \ + \ 7 \ + \ 13 \ + \ 15 \ + \ 20}{5}=\frac{59}{5}=11.8\) E. \(7 \ \ \frac{4 \ + \ 13 \ + \ 15 \ + \ 16 \ + \ 20}{5}=\frac{68}{5}=13.6\)

33 Choice D is correct
The correct answer is \(130 \) miles Use the information provided in the question to draw the shape. Use Pythagorean Theorem: \(a^2 \ + \ b^2=c^2\) \(50^2 \ + \ 120^2=c^2 ⇒\) \(2500 \ + \ 14400= c^2⇒\) \(16,900=c^2⇒ c=130\)

34 Choice A is correct
The correct answer is \(\frac{3}{2}\) tangent \(\beta= \frac{1}{cotangent \ \beta}=\frac{3}{2}\)

35 Choice C is correct
The correct answer is \(72\) \(\frac{3}{5} \ × \ 120=72\)

36 Choice A is correct
The correct answer is \(\frac{200 \ x}{y}\) Let the number be A. Then: \(2 \ x=y\% \ × \) A Solve for A. \( 2 \ x=\frac{y}{100} \ ×\) A Multiply both sides by \(\frac{100}{y}\): \(2 \ x \ × \ \frac{100}{y}=\frac{y}{100} \ × \ \frac{100}{y} \ ×\) A A \(=\frac{200 \ x}{y}\)

37 Choice C is correct
The correct answer is \(125\) cm One liter \(= 1000\) cm\(^3→ 5\) liters \(= 5000\) cm\(^3\) \(5000=10 \ × \ 4 \ × \ h→h=\frac{5000}{40}=125\) cm

38 Choice A is correct
The correct answer is \(110 \ π\) in\(^2\) Surface Area of a cylinder \(= 2 \ π \ r \ (r \ + \ h)\), The radius of the cylinder is \(5 \ (10 \ ÷ \ 2)\) inches and its height is \(8\) inches. Therefore, Surface Area of a cylinder \(=2 \ π \ (5) \ (5 \ + \ 6)=110 \ π\) in\(^2\)

39 Choice C is correct
The correct answer is \(80\) feet The relationship among all sides of special right triangle \(30^\circ \  \ 60^\circ \  \ 90^\circ\) is provided in this triangle: In this triangle, the opposite side of \(60^\circ\) angle is half of the hypotenuse. Draw the shape of this question: The latter is the hypotenuse. Therefore, the latter is \(80\) ft.

40 Choice A is correct
The correct answer is II only I. \(a \ < \ 1→ \ 1 \ < \ a \ < \ 1\) Multiply all sides by \(b\). Since, \(b \ > \ 0→ \ b \ < \ b \ a \ < \ b\) (it is true!) II. Since, \( \ 1 \ < \ a \ < \ 1\),and \(a \ < \ 0→ \ a \ > \ a^2 \ > \ a\) (plug in \( \ \frac{1}{2}\), and check!) (It’s false) III. \( \ 1 \ < \ a \ < \ 1\), multiply all sdes by \(2\), then: \( \ 2 \ < \ 2 \ a \ < \ 2\) Subtract \(3\) from all sides. Then: \( \ 2 \  \ 3 \ < \ 2 \ a \  \ 3 \ < \ 2 \  \ 3→ \ 5 \ < \ 2 \ a \  \ 3 \ < \  \ 1\) (It is true!)
