## Full Length PSAT Math Practice Test

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PSAT Math
Practice Test 1

Section 1

(No Calculator)

17 questions

Total time for this section: 25 Minutes

You MAY NOT use a calculator on this Section.

1- John works for an electric company. He receives a monthly salary of $$4,500$$ plus $$5\%$$ of all his monthly sales as bonus. If $$x$$ is the number of all John’s sales per month, which of the following represents John’s monthly revenue in dollars?
(A) $$0.05 \ x$$
(B) $$0.95 \ x \ − \ 4,500$$
(C) $$0.05 \ x \ + \ 4,500$$
(D) $$0.95 \ x \ + \ 4,500$$
2- If $$f(x^2)=3 \ x \ + \ 4$$, for all positive value of $$x$$, what is the value of $$f(121)$$?
(A) $$367$$
(B) $$37$$
(C) $$29$$
(D) $$- \ 29$$
3- If a and b are solutions of the following equation, which of the following is the ratio $$\frac{a}{b}$$? $$(a\ > \ b)$$
$$2 \ x^2 \ - \ 11 \ x \ + \ 8= \ - \ 3 \ x \ + \ 18$$
(A) $$\frac{1}{5}$$
(B) $$5$$
(C) $$- \ \frac{1}{5}$$
(D) $$- \ 5$$
4- A line in the $$x \ y-$$plane passes through origin and has a slope of $$\frac{1}{3}$$. Which of the following points lies on the line?
(A) $$(2,1)$$
(B) $$(4,1)$$
(C) $$(9,3)$$
(D) $$(6,3)$$
5- Which of the following is the solution of the following inequality?
$$2 \ x \ + \ 4 \ > \ 11 \ x \ - \ 12.5 \ - \ 3.5 \ x$$
(A) $$x \ < \ 3$$
(B) $$x \ > \ 3$$
(C) $$x \ ≤ \ 4$$
(D) $$x \ ≥ \ 4$$
6- If $$a, b$$ and $$c$$ are positive integers and $$3 \ a=4 \ b=5\ c$$, then the value of $$a \ + \ 2 \ b \ + \ 15 \ c$$ is how many times the value of $$a$$?
(A) $$11.5$$
(B) $$12$$
(C) $$12.5$$
(D) $$15$$
7- If $$x≠ \ - \ 4$$ and $$x≠5$$, which of the following is equivalent to  $$\frac{1}{\frac{1}{x \ - \ 5} \ + \ \frac{1}{x \ + \ 4}}$$?
(A) $$\frac{(x \ − \ 5)(x \ + \ 4)}{(x \ − \ 5) \ + \ (x \ + \ 4)}$$
(B) $$\frac{(x \ + \ 4) \ + \ (x \ − \ 5)}{(x \ + \ 4)(x \ − \ 5)}$$
(C) $$\frac{(x \ + \ 4)(x \ − \ 5)}{(x \ + \ 4) \ − \ (x \ + \ 5)}$$
(D) $$\frac{(x \ + \ 4) \ + \ (x \ − \ 5)}{(x \ + \ 4) \ − \ (x \ − \ 5)}$$
8- The table above shows the distribution of age and gender for $$30$$ employees in a company. If one employee is selected at random, what is the probability that the employee selected be either a female under age $$45$$ or a male age $$45$$ or older?
(A) $$\frac{5}{6}$$
(B) $$\frac{5}{30}$$
(C) $$\frac{6}{30}$$
(D) $$\frac{11}{30}$$
9- If a parabola with equation $$y=a \ x^2 \ + \ 5 \ x \ + \ 10$$, where a is constant, passes through point $$(2, 12)$$, what is the value of $$a^2$$?
(A) $$− \ 2$$
(B) $$2$$
(C) $$- \ 4$$
(D) $$4$$
10- What is the value of $$f(5)$$ for the following function $$f$$?
$$f(x)=x^2 \ - \ 3 \ x$$
(A) $$5$$
(B) $$10$$
(C) $$15$$
(D) $$20$$
11- John buys a pepper plant that is $$6$$ inches tall. With regular watering the plant grows $$4$$ inches a year. Writing John’s plant’s height as a function of time, what does the $$y-$$intercept represent?
(A) The $$y−$$intercept represents the rate of grows of the plant which is $$5$$ inches
(B) The $$y−$$intercept represents the starting height of $$6$$ inches
(C) The $$y−$$intercept represents the rate of growth of plant which is $$3$$ inches per year
(D) There is no $$y−$$intercept
12- What is the solution of the following system of equations?
$$\begin{cases}\frac{ - \ x}{2} \ + \ \frac{y}{4} = 1 \\ \frac{ - \ 5 \ y}{6} \ + \ 2 \ x =4 \end{cases}$$
(A) $$x=48, y=22$$
(B) $$x=50, y=20$$
(C) $$x=20, y=50$$
(D) $$x=22, y=48$$
13- What is the length of AB in the following figure if AE $$=4$$, CD $$=6$$ and AC $$=12$$?
(A) $$3.8$$
(B) $$4.8$$
(C) $$7.2$$
(D) $$24$$
14- If $$x≠0$$, what is the value of $$\frac{(10 \ (x) \ (y^2))^2}{(8 \ x \ y^2)^2}$$?
(A) 25/16
(B) 25/16
(C) 1.56
(D) 1.56
(E) 25:16
(F) 25 : 16
(G) 25:16
(H) 1.562
(I) 1.5625
15- In the following equation, what is the value of $$y \ - \ 3 \ x$$?
$$\frac{y}{5}=x \ - \ \frac{2}{5} \ x \ + \ 10$$
(A) 50
(B) 50.0
(C) 50
16- What is the value of $$x$$ in the following equation?
$$\frac{x^2 \ - \ 9}{x \ + \ 3} \ + \ 2 \ (x \ + \ 4)=15$$
(A) 10/3
(B) 10/3
(C) 3.33
(D) 3.333
(E) 10:3
(F) 10 : 3
17- The length of a rectangle is $$3$$ meters greater than $$4$$ times its width. The perimeter of the rectangle is $$36$$ meters.  What is the area of the rectangle in meters?
(A) 45
(B) 45
(C) 45.0

PSAT Math
Practice Test 1

Section 2

(Calculator)

31 questions

Total time for this section: 45 Minutes

You may use a scientific calculator on this Section.

18-

If $$8 \ + \ 2 \ x$$ is $$16$$ more than $$20$$, what is the value of $$6 \ x$$?

(A) $$40$$
(B) $$55$$
(C) $$62$$
(D) $$84$$
19-

If a gas tank can hold $$25$$ gallons, how many gallons does it contain when it is $$\frac{2}{5}$$ full?

(A) $$50$$
(B) $$125$$
(C) $$62.5$$
(D) $$10$$
20-

In the $$x \ y -$$plane, the point $$(4,3)$$ and $$(3,2)$$ are on line A. Which of the following equations of lines is parallel to line A?

(A) $$y=3 \ x$$
(B) $$y=\frac{x}{2}$$
(C) $$y=2 \ x$$
(D) $$y= x$$
21-

If $$y=n \ x \ + \ 2$$, where $$n$$ is a constant, and when $$x=6$$, $$y=14$$, what is the value of $$y$$ when $$x=10$$?

(A) $$10$$
(B) $$12$$
(C) $$18$$
(D) $$22$$
22-

A football team won exactly $$80\%$$ of the games it played during last session. Which of the following could be the total number of games the team played last season?

(A) $$49$$
(B) $$35$$
(C) $$12$$
(D) $$32$$
23-

The capacity of a red box is $$20\%$$ bigger than the capacity of a blue box. If the red box can hold $$30$$ equal sized books, how many of the same books can the blue box hold?

(A) $$9$$
(B) $$15$$
(C) $$21$$
(D) $$25$$
24-

The sum of six different negative integers is $$- \ 70$$. If the smallest of these integers is $$- \ 15$$, what is the largest possible value of one of the other five integers?

(A) $$− \ 14$$
(B) $$− \ 10$$
(C) $$− \ 5$$
(D) $$− \ 1$$
25-

If $$x$$ is greater than $$0$$ and less than $$1$$, which of the following is true?

(A) $$x \ < \ \sqrt{x^2 \ + \ 1} \ < \ \sqrt{x^2} \ + \ 1$$
(B) $$x \ < \ \sqrt{x^2} \ + \ 1\ < \ \sqrt{x^2 \ + \ 1}$$
(C) $$\sqrt{x^2 \ + \ 1} \ < \ x \ < \ \sqrt{x^2} \ + \ 1$$
(D) $$\sqrt{x^2} \ + \ 1 \ < \ \sqrt{x^2 \ + \ 1} \ < \ x$$
26-

If $$a$$ is the mean (average) of the number of cities in each pollution type category, $$b$$ is the mode, and $$c$$ is the median of the number of cities in each pollution type category, then which of the following must be true?

(A) $$π \ < \ π \ < \ π$$
(B) $$π \ < \ π \ < \ π$$
(C) $$π=π$$
(D) $$π \ < \ π=π$$
27- What percent of cities are in the type of pollution A, C, and E respectively?
(A) $$60\%, 40\%, 90\%$$
(B) $$30\%, 40\%, 90\%$$
(C) $$40\%, 60\%, 90\%$$
(D) $$40\%, 60\%, 90\%$$
28- How many cities should be added to type of pollutions B until the ratio of cities in type of pollution B to cities in type of pollution E will be $$0.625$$?
(A) $$2$$
(B) $$3$$
(C) $$4$$
(D) $$5$$
29- The ratio of boys and girls in a class is $$4:7$$. If there are $$44$$ students in the class, how many more boys should be enrolled to make the ratio $$1:1$$?
(A) $$8$$
(B) $$10$$
(C) $$12$$
(D) $$16$$

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30- In the following right triangle, if the sides AB and BC become twice longer, what will be the ratio of the perimeter of the triangle to its area?
(A) $$\frac{1}{2}$$
(B) $$2$$
(C) $$\frac{1}{3}$$
(D) $$3$$
31- What is the ratio of the minimum value to the maximum value of the following function?
$$- \ 2 \ ≤ \ x \ ≤ \ 3$$
$$f(x)= \ - \ 3 \ x \ + \ 1$$
(A) $$\frac{7}{8}$$
(B) $$- \ \frac{8}{7}$$
(C) $$- \ \frac{7}{8}$$
(D) $$\frac{8}{7}$$
32- What's the maximum ratio of women to men in the four cities?
(A) $$0.98$$
(B) $$0.97$$
(C) $$0.96$$
(D) $$0.95$$
33- What's the ratio of percentage of men in city A to percentage of women in city C?
(A) $$0.9$$
(B) $$0.95$$
(C) $$1$$
(D) $$1.05$$
34- How many women should be added to city D until the ratio of women to men will be $$1.2$$?
(A) $$120$$
(B) $$128$$
(C) $$132$$
(D) $$160$$
35- In the rectangle below if $$y \ > \ 5$$ cm and the area of rectangle is $$50$$ cm$$^2$$ and the perimeter of the rectangle is $$30$$ cm, what is the value of $$x$$ and $$y$$ respectively?
(A) $$4, 11$$
(B) $$5, 11$$
(C) $$5, 10$$
(D) $$4, 10$$
36- If a car has $$80-$$liter petrol and after one hour driving the car use $$6-$$liter petrol, how much petrol will remain after $$x-$$hours driving?
(A) $$6 \ x \ − \ 80$$
(B) $$80 \ + \ 6 \ x$$
(C) $$80 \ - \ 6 \ x$$
(D) $$80 \ - \ x$$
37- In the triangle below, if the measure of angle A is $$37$$ degrees, then what is the value of $$y$$? (figure is NOT drawn to scale)
(A) $$62$$
(B) $$70$$
(C) $$78$$
(D) $$86$$
38- The following graph shows the mark of six students in mathematics. What is the mean (average) of the marks?
(A) $$15$$
(B) $$14.5$$
(C) $$14$$
(D) $$13.5$$
39- If $$f(x)=3 \ x \ + \ 4 \ (x \ + \ 1) \ + \ 2$$ then $$f(4 \ x)=$$?
(A) $$28 \ x \ + \ 6$$
(B) $$16 \ x \ − \ 6$$
(C) $$25 \ x \ + \ 4$$
(D) $$12 \ x \ + \ 3$$
40- Which of the following values for $$x$$ and $$y$$ satisfy the following system of equations?
$$\begin{cases}x \ + \ 4 \ y =10\\ 5 \ x \ + \ 10 \ y =20\end{cases}$$
(A) $$x=3, y=2$$
(B) $$x=2, y =− \ 3$$
(C) $$x=− \ 2, y=3$$
(D) $$x=3, y=− \ 2$$
41- Given the right triangle ABC bellow, cosβ‘ $$(β)$$ is equal to?
(A) $$\frac{a}{b}$$
(B) $$\frac{π}{\sqrt{π^2 \ + \ π^2}}$$
(C) $$\frac{\sqrt{π^2 \ + \ π^2}}{a \ b}$$
(D) $$\frac{b}{\sqrt{π^2 \ + \ π^2}}$$
42- Solve the following inequality.
$$|\frac{x}{2} \ - \ 2 \ x \ + \ 10| \ < \ 5$$
(A) $$− \ \frac{10}{3} \ < \ x \ < \ 10$$
(B) $$− \ 10 \ < \ x \ < \ \frac{10}{3}$$
(C) $$\frac{10}{3} \ < \ x \ < \ 10$$
(D) $$- \ 10 \ < \ x \ < \ - \ \frac{10}{3}$$
43- If $$x$$ is directly proportional to the square of $$y$$, and $$y=2$$ when $$x=12$$, then when $$x=75, \ y=$$?
(A) $$\frac{1}{5}$$
(B) $$1$$
(C) $$5$$
(D) $$12$$
44- If  $$\frac{a \ - \ b}{b}=\frac{10}{11}$$, then which of the following must be true?
(A) $$\frac{π}{π}=\frac{11}{10}$$
(B) $$\frac{π}{π}=\frac{21}{11}$$
(C) $$\frac{π}{π}=\frac{11}{21}$$
(D) $$\frac{π}{π}=\frac{21}{10}$$
45- $$f(x)=a \ x^2 \ + \ b \ x \ + \ c$$ is a quadratic function where $$a, b$$ and $$c$$ are constant. The value of $$x$$ of the point of intersection of this quadratic function and linear function $$g(x)=2 \ x \ - \ 3$$ is $$2$$. The vertex of $$f(x)$$ is at $$(- \ 2, 5)$$. What is the product of $$a, b$$ and $$c$$?
(A) 1
(B) 1.0
(C) 1
46- A ladder leans against a wall forming a $$60^\circ$$ angle between the ground and the ladder. If the bottom of the ladder is $$45$$ feet away from the wall, how many feet is the ladder?
(A) 90
(B) 90.0
(C) 90
47- The volume of cube A is $$\frac{1}{3}$$ of its surface area. What is the length of an edge of cube A?
(A) 2
(B) 2
(C) 2.0
48- If $$3 \ x \ + \ 6 \ y=\frac{- \ 3 \ y^2 \ + \ 15}{x}$$, what is the value of $$(x \ + \ y)^2$$? $$(x≠0)$$
(A) 5
(B) 5.0
(C) 5
 1- Choice C is correct The correct answer is $$0.05 \ x \ + \ 4,500$$$$x$$ is the number of all John’s sales per month and $$5\%$$ of it is: $$5\% \ × \ x=0.05 \ x$$John’s monthly revenue: $$0.05 \ x \ + \ 4,500$$ 2- Choice B is correct The correct answer is $$37$$$$x^2=121→x=11$$ (positive value) Or $$x= - \ 11$$ (negative value)Since $$x$$ is positive, then: $$f(121)=f(11^2 )=3 \ (11) \ + \ 4=33 \ + \ 4=37$$ 3- Choice D is correct The correct answer is $$- \ 5$$$$2 \ x^2 \ - \ 11 \ x \ + \ 8= \ - \ 3 \ x \ + \ 18→$$$$2 \ x^2 \ - \ 11 \ x \ + \ 3 \ x \ + \ 8 \ - \ 18=0→$$$$2 \ x^2 \ - \ 8 \ x \ - \ 10=0→$$$$2 \ (x^2 \ - \ 4 \ x \ - \ 5)=0→$$ Divide both sides by $$2$$.Then:$$x^2 \ - \ 4 \ x \ - \ 5=0$$, Find the factors of the quadratic equation. $$→(x \ - \ 5) \ (x \ + \ 1)=0→x=5$$ or $$x= \ - \ 1$$$$a \ > \ b$$, then: $$a=5$$ and $$b= \ - \ 1$$$$\frac{a}{b}=\frac{5}{- \ 1}= \ - \ 5$$ 4- Choice C is correct The correct answer is $$(9,3)$$First, find the equation of the line.All lines through the origin are of the form $$y=m \ x$$, so the equation is $$y=\frac{1}{3} \ x$$.Of the given choices, only choice C $$(9,3)$$, satisfies this equation: $$y=\frac{1}{3} \ x→3=\frac{1}{3} \ (9)=3$$ 5- Choice A is correct The correct answer is $$x \ < \ 3$$$$2 \ x \ + \ 4 \ > \ 11 \ x \ - \ 12.5 \ - \ 3.5 \ x→$$ Combine like terms:$$2 \ x \ + \ 4 \ > \ 7.5 \ x \ - \ 12.5→$$ Subtract $$2 \ x$$ from both sides: $$4>5.5x-12.5$$Add $$12.5$$ both sides of the inequality.$$16.5 \ > \ 5.5 \ x$$, Divide both sides by $$5.5$$.$$\frac{16.5}{5.5} \ > \ x→x \ < \ 3$$ 6- Choice A is correct The correct answer is $$11.5$$$$3 \ a=4 \ b→b=\frac{3 \ a}{4}$$ and $$3 \ a=5 \ c→c=\frac{3 \ a}{5}$$$$a \ + \ 2 \ b \ + \ 15 \ c=a \ + \ (2 \ × \ \frac{3 \ a}{4}) \ + \ (15 \ × \ \frac{3 \ a}{5})=a \ + \ 1.5 \ a \ + \ 9 \ a=11.5 \ a$$The value of $$a \ + \ 2 \ b \ + \ 15 \ c$$ is $$11.5$$ times the value of $$a$$. 7- Choice A is correct The correct answer is $$\frac{(x \ − \ 5)(x \ + \ 4)}{(x \ − \ 5) \ + \ (x \ + \ 4)}$$To rewrite $$\frac{1}{\frac{1}{x \ - \ 5} \ + \ \frac{1}{x \ + \ 4)}}$$, first simplify $$\frac{1}{x \ - \ 5} \ + \ \frac{1}{x \ + \ 4}$$. $$\frac{1}{x \ - \ 5} \ + \ \frac{1}{x \ + \ 4}=$$$$\frac{1(x \ + \ 4)}{(x \ - \ 5) \ (x \ + \ 4)} \ + \ \frac{1 \ (x \ - \ 5)}{(x \ + \ 4) \ (x \ - \ 5)}=$$$$\frac{(x \ + \ 4) \ + \ (x \ - \ 5)}{(x \ + \ 4) \ (x \ - \ 5)}$$Then:$$\frac{1}{\frac{1}{x \ - \ 5} \ + \ \frac{1}{x \ + \ 4}}=$$$$\frac{1}{\frac{(x \ + \ 4) \ + \ (x \ - \ 5)}{(x \ + \ 4) \ (x \ - \ 5)}}=$$$$\frac{(x \ - \ 5) \ (x \ + \ 4)}{(x \ - \ 5) \ + \ (x \ + \ 4)}$$. (Remember, $$\frac{1}{\frac{1}{x}}=x$$)This result is equivalent to the expression in choice A. 8- Choice D is correct The correct answer is $$\frac{11}{30}$$Of the $$30$$ employees, there are $$5$$ females under age $$45$$ and $$6$$ males age $$45$$ or older.Therefore, the probability that the person selected will be either a female under age $$45$$ or a male age $$45$$ or older is: $$\frac{5}{30} \ + \ \frac{6}{30}=\frac{11}{30}$$ 9- Choice D is correct The correct answer is $$4$$Plug in the values of x and y of the point $$(2, 12)$$ in the equation of the parabola. Then:$$12=a \ (2)^2 \ + \ 5 \ (2) \ + \ 10→$$$$12=4 \ a \ + \ 10 \ + \ 10→$$$$12=4 \ a \ + \ 20$$$$→4 \ a=12 \ - \ 20=- \ 8→$$$$a=\frac{- \ 8}{4}=- \ 2→$$$$a^2=(- \ 2)^2=4$$ 10- Choice B is correct The correct answer is $$10$$The input value is $$5$$. Then: $$x=5$$$$f(x)=x^2 \ - \ 3 \ x→$$$$f(5)=5^2 \ - \ 3 \ (5)=25 \ - \ 15=10$$ 11- Choice B is correct The correct answer is The $$y−$$intercept represents the starting height of $$6$$ inchesTo solve this problem, first recall the equation of a line:$$y=m \ x \ + \ b$$Where, $$m=$$ slope and $$y=y-$$interceptRemember that slope is the rate of change that occurs in a function and that the $$y-$$intercept is the $$y$$ value corresponding to $$x=0$$.Since the height of John’s plant is $$6$$ inches tall when he gets it. Time (or $$x$$) is zero.The plant grows $$4$$ inches per year.Therefore, the rate of change of the plant’s height is $$4$$.The $$y-$$intercept represents the starting height of the plant which is $$6$$ inches. 12- Choice D is correct The correct answer is $$x=22, y=48$$$$\begin{cases}\frac{ - \ x}{2} \ + \ \frac{y}{4} = 1 \\ \frac{ - \ 5 \ y}{6} \ + \ 2 \ x =4 \end{cases} \rightarrow$$ Multiply the top equation by $$4$$. Then,$$\begin{cases}- \ 2 \ x \ + \ y =4\\\frac{- \ 5 \ y}{6} \ + \ 2 \ x =4\end{cases} \rightarrow$$ Add two equations.$$\frac{1}{6} \ y=8→y=48$$, plug in the value of $$y$$ into the first equation $$→x=22$$ 13- Choice B is correct The correct answer is $$4.8$$Two triangles $$\triangle$$BAE and $$\triangle$$BCD are similar. Then: $$\frac{AE}{CD}=\frac{AB}{BC}→\frac{4}{6}=\frac{x}{12}→$$$$48 \ - \ 4 \ x=6 \ x→$$$$10 \ x=48→x=4.8$$ 14- Choice I is correct The correct answer is $$\frac{25}{16}$$First, simplify the numerator and the denominator. $$\frac{(10 \ (x) \ (y^2)^2}{(8 \ x \ y^2 )^2} =\frac{100 \ x^2 \ y^4}{64 \ x^2 \ y^4 )}$$Remove $$x^2 \ y^4$$ from both numerator and denominator. $$\frac{100 \ x^2 \ y^4}{64 \ x^2 \ y^4 )}=\frac{100}{64}=\frac{25}{16}$$ 15- Choice C is correct The correct answer is $$50$$$$\frac{y}{5}=x \ - \ \frac{2}{5} \ x \ + \ 10$$, Multiply both sides of the equation by $$5$$. Then:$$5 \ × \ \frac{y}{5}=5 \ × \ (x \ - \ \frac{2}{5} \ x \ + \ 10)→$$$$y=5 \ x \ - \ 2 \ x \ + \ 50→y=3 \ x \ + \ 50$$Now, subtract $$3 \ x$$ from both sides of the equation. Then:$$y \ - \ 3 \ x=50$$ 16- Choice F is correct The correct answer is $$\frac{10}{3}$$First, factorize the numerator and simplify.$$\frac{x^2 \ - \ 9}{x \ + \ 3} \ + \ 2 \ (x \ + \ 4)=15→$$$$\frac{(x \ - \ 3) \ (x \ + \ 3)}{(x \ + \ 3)} \ + \ 2 \ x \ + \ 8=15$$Divide both sides of the fraction by $$(x \ + \ 3)$$. Then:$$x \ - \ 3 \ + \ 2 \ x \ + \ 8=15→3 \ x \ + \ 5=15$$Subtract $$5$$ from both sides of the equation. Then:$$→3 \ x=15 \ - \ 5=10→x=\frac{10}{3}$$ 17- Choice C is correct The correct answer is $$\frac{10}{3}$$Let $$L$$ be the length of the rectangular and $$W$$ be the with of the rectangular. Then,$$L=4 \ W \ + \ 3$$The perimeter of the rectangle is $$36$$ meters. Therefore:$$2 \ L \ + \ 2 \ W=36$$$$L \ + \ W=18$$Replace the value of $$L$$ from the first equation into the second equation and solve for $$W$$:$$(4 \ W \ + \ 3) \ + \ W=18→5 \ W \ + \ 3=18→5 \ W=15→W=3$$The width of the rectangle is $$3$$ meters and its length is:$$L=4 \ W \ + \ 3=4 \ (3) \ + \ 3=15$$The area of the rectangle is: length $$×$$ width $$= 3 \ × \ 15 = 45$$ 17- Choice C is correct The correct answer is $$\frac{10}{3}$$Let $$L$$ be the length of the rectangular and $$W$$ be the with of the rectangular. Then,$$L=4 \ W \ + \ 3$$The perimeter of the rectangle is $$36$$ meters. Therefore:$$2 \ L \ + \ 2 \ W=36$$$$L \ + \ W=18$$Replace the value of $$L$$ from the first equation into the second equation and solve for $$W$$:$$(4 \ W \ + \ 3) \ + \ W=18→5 \ W \ + \ 3=18→5 \ W=15→W=3$$The width of the rectangle is $$3$$ meters and its length is:$$L=4 \ W \ + \ 3=4 \ (3) \ + \ 3=15$$The area of the rectangle is: length $$×$$ width $$= 3 \ × \ 15 = 45$$ 18- Choice D is correct The correct answer is $$84$$The description $$8 \ + \ 2 \ x$$ is $$16$$ more than $$20$$ can be written as the equation $$8 \ + \ 2 \ x=16 \ + \ 20$$, which is equivalent to $$8 \ + \ 2 \ x=36$$.Subtracting $$8$$ from each side of $$8 \ + \ 2 \ x=36$$ gives $$2 \ x=28$$.Since $$6 \ x$$ is $$3$$ times $$2 \ x$$, multiplying both sides of $$2 \ x=28$$ by $$3$$ gives $$6 \ x=84$$ 19- Choice D is correct The correct answer is $$10$$$$\frac{2}{5} \ × \ 25=\frac{50}{5}=10$$ 20- Choice D is correct The correct answer is $$y= x$$The slop of line A is: $$m=\frac{y_{2} \ - \ y_{1}}{x_{2} \ - \ x_{1}}=\frac{3 \ - \ 2}{4 \ - \ 3}=1$$Parallel lines have the same slope and only choice D $$(y=x)$$ has slope of $$1$$. 21- Choice D is correct The correct answer is $$22$$Substituting $$6$$ for $$x$$ and $$14$$ for $$y$$ in $$y = n \ x \ + \ 2$$ gives $$14=(n) \ (6) \ + \ 2$$,which gives $$n=2$$. Hence, $$y=2 \ x \ + \ 2$$.Therefore, when $$x = 10$$, the value of $$y$$ is:$$y=(2) \ (10) \ + \ 2 = 22$$ 22- Choice B is correct The correct answer is $$35$$Choices A, C and D are incorrect because $$80\%$$ of each of the numbers is a non-whole number.A. $$49$$,      $$80\%$$ of $$49 = 0.80 \ × \ 49=39.2$$B. $$35$$,      $$80\%$$ of $$35=0.80 \ × \ 35=28$$C. $$12$$,      $$80\%$$ of $$12=0.80 \ × \ 12=9.6$$D. $$32$$,      $$80\%$$ of $$32=0.80 \ × \ 32=25.6$$Only choice B gives a whole number. 23- Choice D is correct The correct answer is $$25$$The capacity of a red box is $$20\%$$ bigger than the capacity of a blue box and it can hold $$30$$ books.Therefore, we want to find a number that $$20\%$$ bigger than that number is $$30$$. Let $$x$$ be that number. Then:$$1.20 \ × \ x=30$$, Divide both sides of the equation by $$1.2$$. Then:$$x=\frac{30}{1.20}=25$$ 24- Choice C is correct The correct answer is $$− \ 5$$The smallest number is $$- \ 15$$.To find the largest possible value of one of the other five integers, we need to choose the smallest possible integers for four of them.Let $$x$$ be the largest number. Then:$$- \ 70=(- \ 15) \ + \ (- \ 14) \ + \ (- \ 13) \ + \ (- \ 12) \ +(- \ 11) \ + \ x→- \ 70=- \ 65 \ + \ x$$$$→x=- \ 70 \ + \ 65=- \ 5$$ 25- Choice A is correct The correct answer is $$x \ < \ \sqrt{x^2 \ + \ 1} \ < \ \sqrt{x^2} \ + \ 1$$Let $$x$$ be equal to $$0.5$$, then: $$x=0.5$$$$\sqrt{x^2 \ + \ 1}=\sqrt{0.5^2 \ + \ 1}=\sqrt{1.25}≈1.12$$$$\sqrt{x^2 } \ + \ 1=\sqrt{0.5^2} \ + \ 1=0.5 \ + \ 1=1.5$$Then, option A is correct.$$x \ < \ \sqrt{x^2 \ + \ 1} \ < \ \sqrt{x^2} \ + \ 1$$ 26- Choice C is correct The correct answer is $$π=π$$Let’s find the mean (average), mode and median of the number of cities for each type of pollution. Number of cities for each type of pollution: $$6, 3, 4, 9, 8$$ average (mean) $$= \frac{sum \ of \ terms}{number \ of \ terms}= \frac{6 \ + \ 3 \ + \ 4 \ + \ 9 \ + \ 8}{5}=\frac{30}{5}=6$$Median is the number in the middle. To find median, first list numbers in order from smallest to largest. $$3, 4, 6, 8, 9$$ Median of the data is $$6$$. Mode is the number which appears most often in a set of numbers.Therefore, there is no mode in the set of numbers. Median $$=$$ Mean, then, $$π=π$$ 27- Choice A is correct The correct answer is $$60\%, 40\%, 90\%$$Percent of cities in the type of pollution A: $$\frac{6}{10} \ × \ 100=60\%$$Percent of cities in the type of pollution C: $$\frac{4}{10} \ × \ 100=40\%$$Percent of cities in the type of pollution E: $$\frac{9}{10} \ × \ 100=90\%$$ 28- Choice A is correct The correct answer is $$2$$Let $$x$$ be the number of cities need to be added to type of pollutions B. Then:$$\frac{x \ + \ 3}{8}=0.625→x \ + \ 3=8 \ × \ 0.625→x \ + \ 3=5→x=2$$ 29- Choice C is correct The correct answer is $$12$$The ratio of boy 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 $$1:1$$ 30- Choice A is correct The correct answer is $$\frac{1}{2}$$AB $$=5$$ And BC $$=12$$AC $$=\sqrt{12^2 \ + \ 5^2 }=\sqrt{144 \ + \ 25}=\sqrt{169}=13$$Perimeter $$=5 \ + \ 12 \ + \ 13=30$$Area $$=\frac{5 \ × \ 12}{2}=5 \ × \ 6=30$$In this case, the ratio of the perimeter of the triangle to its area is: $$\frac{30}{30}=1$$If the sides AB and BC become twice longer, then:AB $$=24$$ And AC $$=10$$BC $$=\sqrt{24^2 \ + \ 10^2}=\sqrt{576 \ + \ 100}=\sqrt{676}=26$$Perimeter $$=26 \ + \ 24 \ + \ 10=60$$Area $$=\frac{10 \ × \ 24}{2}=10 \ × \ 12=120$$In this case the ratio of the perimeter of the triangle to its area is: $$\frac{60}{120}=\frac{1}{2}$$ 31- Choice B is correct The correct answer is $$- \ \frac{8}{7}$$Since $$f(x)$$ is linear function with a negative slop, then when $$x=- \ 2, \ f(x)$$ is maximum and when $$x=3, \ f(x)$$ is minimum.Then the ratio of the minimum value to the maximum value of the function is: $$\frac{f(3)}{f(- \ 2)}=\frac{- \ 3 \ (3) \ + \ 1}{- \ 3 \ (- \ 2) \ + \ 1}=\frac{- \ 8}{7}=- \ \frac{8}{7}$$ 32- Choice B is correct The correct answer is $$0.97$$Ratio of women to men in city A: $$\frac{570}{600}=0.95$$ Ratio of women to men in city B: $$\frac{291}{300}=0.97$$ Ratio of women to men in city C: $$\frac{665}{700}=0.95$$ Ratio of women to men in city D: $$\frac{528}{550}=0.96$$ Choice B provides the maximum ratio of women to men in the four cities. 33- Choice D is correct The correct answer is $$1.05$$Percentage of men in city A $$= \frac{600}{1170} \ × \ 100=51.28\%$$ Percentage of women in city C $$= \frac{665}{1365} \ × \ 100=48.72\%$$ Percentage of men in city A to percentage of women in city C $$= \frac{51.28}{48.72}=1.05$$ 34- Choice C is correct The correct answer is $$132$$Let the number of women should be added to city D be $$x$$, then:$$\frac{528 \ + \ x}{550}=1.2→$$$$528 \ + \ x=550 \ × \ 1.2=660→$$$$x=132$$ 35- Choice C is correct The correct answer is $$5, 10$$The perimeter of the rectangle is: $$2 \ x \ + \ 2 \ y=30→$$$$x \ + \ y=15→$$$$x=15 \ - \ y$$ The area of the rectangle is: $$x \ × \ y=50→$$$$(15 \ - \ y) \ (y)=50→$$$$y^2 \ - \ 15 \ y \ + \ 50=0$$ Solve the quadratic equation by factoring method.$$(y \ - \ 5) \ (y \ - \ 10)=0→y=5$$ (Unacceptable, because $$y$$ must be greater than $$5$$) or $$y=10$$If $$y=10 →x \ × \ y=50→x \ × \ 10=50→x=5$$ 36- Choice C is correct The correct answer is $$80 \ - \ 6 \ x$$The amount of petrol consumed after $$x$$ hours is: $$6 \ × \ x=6 \ x$$Petrol remaining after $$x$$ hours driving: $$80 \ - \ 6 \ x$$ 37- Choice D is correct The correct answer is $$86$$In the figure angle A is labeled $$(3 \ x \ - \ 2)$$ and it measures $$37$$.Thus, $$3 \ x \ - \ 2=37$$ and $$3 \ x=39$$ or $$x=13$$. That means that angle B, which is labeled $$(5 \ x)$$, must measure $$5 \ × \ 13=65$$.Since the three angles of a triangle must add up to $$180, \ 37 \ + \ 65 \ + \ y \ - \ 8=180$$, then:$$y \ + \ 94=108→y=180 \ - \ 94=86$$ 38- Choice B is correct The correct answer is $$14.5$$average (mean) $$= \frac{sum \ of \ terms}{number \ of \ terms}=$$$$\frac{9 \ + \ 12 \ + \ 15 \ + \ 16 \ + \ 19 \ + \ 16 \ + \ 14.5}{7}=14.5$$ 39- Choice A is correct The correct answer is $$28 \ x \ + \ 6$$If $$f(x) = 3 \ x \ + \ 4 \ (x \ + \ 1) \ + \ 2$$, then find $$f(4 \ x)$$ by substituting $$4 \ x$$ for every $$x$$ in the function.This gives:$$f(4 \ x) = 3 \ (4 \ x) \ + \ 4 \ (4 \ x \ + \ 1) \ + \ 2$$, It simplifies to:$$f(4 \ x)= 3 \ (4 \ x) \ + \ 4 \ (4 \ x \ + \ 1) \ + \ 2=12 \ x \ + \ 16 \ x \ + \ 4 \ + \ 2=28 \ x \ + \ 6$$ 40- Choice C is correct The correct answer is $$x=− \ 2, y=3$$$$\begin{cases}x \ + \ 4 \ y =10\\ 5 \ x \ + \ 10 \ y =20\end{cases}\rightarrow$$ Multiply the top equation by $$- \ 5$$ then,$$\begin{cases}- \ 5 \ x \ - \ 20 \ y =- \ 50\\ 5 \ x \ + \ 10 \ y =20\end{cases}\rightarrow$$ Add two equations $$- \ 10 \ y=- \ 30→y=3$$, plug in the value of $$y$$ into the first equation $$x \ + \ 4 \ y=10→x \ + \ 4 \ (3)=10→x \ + \ 12=10$$Subtract $$12$$ from both sides of the equation. Then:$$x \ + \ 12=10→x=- \ 2$$ 41- Choice B is correct The correct answer is $$\frac{π}{\sqrt{π^2 \ + \ π^2}}$$Cos β‘$$β=\frac{Adjacent \ side}{hypotenuse}$$To find the hypotenuse, we need to use Pythagorean theorem. $$a^2 \ + \ b^2=c^2→c=\sqrt{a^2 \ + \ b^2 }$$cos $$(β)=\frac{a}{c}=\frac{a}{\sqrt{a^2 \ + \ b^2}}$$ 42- Choice C is correct The correct answer is $$\frac{10}{3} \ < \ x \ < \ 10$$$$|\frac{x}{2} \ - \ 2 \ x \ + \ 10| \ < \ 5→|- \ \frac{3}{2} \ x \ + \ 10| \ < \ 5→- \ 5< \ - \ \frac{3}{2} \ x \ + \ 10 \ < \ 5$$Subtract $$10$$ from all sides of the inequality.$$→- \ 5 \ - \ 10 \ < \ - \ \frac{3}{2} \ x \ + \ 10 \ - \ 10 \ < \ 5 \ - \ 10→- \ 15 \ < \ - \ \frac{3}{2} \ x \ < \ - \ 5$$Multiply all sides by $$2$$.$$→2 \ × \ (- \ 15) \ < \ 2 \ × \ (- \ \frac{3 \ x}{2}) \ < \ 2 \ × \ (- \ 5)→- \ 30 \ < \ - \ 3 \ x \ < \ - \ 10$$Divide all sides by $$- \ 3$$.(Remember that when you divide all sides of an inequality by a negative number, the inequality sing will be swapped. $$<$$ becomes $$>$$)$$→\frac{- \ 30}{- \ 3} \ > \ \frac{- \ 3 \ x}{- \ 3} \ > \ \frac{- \ 10}{- \ 3}$$$$→10 \ > \ x \ > \ \frac{10}{3}→\frac{10}{3} \ < \ x \ < \ 10$$ 43- Choice C is correct The correct answer is $$5$$$$x$$ is directly proportional to the square of $$y$$. Then:$$x=c \ y^2$$$$12=c \ (2)^2→12=4 \ c→c=\frac{12}{4}=3$$The relationship between $$x$$ and $$y$$ is:$$x=3 \ y^2$$$$x=75$$$$75=3 \ y^2→y^2=\frac{75}{3}=25→y=5$$ 44- Choice B is correct The correct answer is $$\frac{π}{π}=\frac{21}{11}$$The equation $$\frac{a \ - \ b}{b}=\frac{10}{11}$$ can be rewritten as $$\frac{a}{b} \ - \ \frac{b}{b}=\frac{10}{11}$$, from which it follows that $$\frac{a}{b \ - \ 1}=\frac{10}{11}$$, or $$\frac{a}{b}=\frac{10}{11} \ + \ 1=\frac{21}{11}$$. 45- Choice C is correct The correct answer is $$1$$The intersection of two functions is the point with $$2$$ for $$x$$.Then:$$f(2)=g(2)$$ and $$g(2)=(2 \ × \ (2)) \ - \ 3=4 \ - \ 3=1$$Then, $$f(2)=1→a \ (2)^2 \ + \ b \ (2) \ + \ c=1→4 \ a \ + \ 2 \ b \ + \ c=1$$ (i)The value of $$x$$ in the vertex of the parabola is: $$x=- \ \frac{b}{2 \ a}→- \ 2=- \ \frac{b}{2 \ a}→b=4 \ a$$ (ii)In the point $$(- \ 2, 5)$$, the value of the $$f(x)$$ is $$5$$.$$f(- \ 2)=5→a \ (- \ 2)^2 \ + \ b \ (- \ 2) \ + \ c=5→4 \ a \ - \ 2 \ b \ + \ c=5$$ (iii)Using the first two equations:$$\begin{cases}4 \ a \ + \ 2 \ b \ + \ c=1\\4 \ a \ - \ 2 \ b \ + \ c=5\end{cases}$$Equation $$1$$ minus equation $$2$$ is:(i) $$-$$ (iii) $$→4 \ b=- \ 4→b=- \ 1$$ (iv)Plug in the value of b in the second equation:$$b=4 \ a →a=\frac{b}{4}=- \ \frac{1}{4}$$Plug in the values of a and be in the first equation. Then:$$→4(- \ \frac{1}{4}) \ + \ 2 \ (- \ 1) \ + \ c=1→- \ 1 \ - \ 2 \ + \ c=1→c=1 \ + \ 3→c=4$$The product of $$a, \ b$$ and $$c=(- \ \frac{1}{4}) \ × \ (- \ 1) \ × \ 4=1$$ 46- Choice C is correct The correct answer is $$90$$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 $$30^\circ$$ angle is half of the hypotenuse. Draw the shape of this question. The latter is the hypotenuse. Therefore, the latter is $$90$$ feet. 47- Choice C is correct The correct answer is $$2$$Let $$x$$ be the length of an edge of cube, then the volume of a cube is: $$V=x^3$$ The surface area of cube is: $$SA=6 \ x^2$$The volume of cube $$A$$ is $$\frac{1}{3}$$ of its surface area. Then:$$x^3=\frac{6 \ x^2}{3}→x^3=2 \ x^2$$, divide both side of the equation by $$x^2$$. Then:$$\frac{x^3}{x^2} =\frac{3 \ x^2}{x^2} →x=2$$ 48- Choice C is correct The correct answer is $$5$$$$3 \ x \ + \ 6 \ y=\frac{- \ 3 \ y^2 \ + \ 15}{x}$$, Multiply both sides by $$x$$.$$x \ × \ (3 \ x \ + \ 6 \ y)=x \ × \ (\frac{- \ 3 \ y^2 \ + \ 15}{x})→3 \ x^2 \ + \ 6 \ x \ y=- \ 3 \ y^2 \ + \ 15$$$$→3 \ x^2 \ + \ 6 \ x \ y \ + \ 3 \ y^2=15→3 \ × \ (x^2 \ + \ 2 \ x \ y \ + \ y^2 )=15→x^2 \ + \ 2 \ x \ y \ + \ y^2=\frac{15}{3}$$$$x^2 \ + \ 2 \ x \ y \ + \ y^2=(x \ + \ y)^2$$, Then:$$→(x \ + \ y)^2=5$$

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