Full Length PSAT Math Practice Test

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(x2)=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 ab? (a > b)
2 x2  11 x + 8=  3 x + 18
(A) 15
(B) 5
(C)  15
(D)  5
4- A line in the x yplane passes through origin and has a slope of 13. 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 x5, which of the following is equivalent to  11x  5 + 1x + 4?
(A) (x  5)(x + 4)(x  5) + (x + 4)
(B) (x + 4) + (x  5)(x + 4)(x  5)
(C) (x + 4)(x  5)(x + 4)  (x + 5)
(D) (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?
PSAT_Math12
(A) 56
(B) 530
(C) 630
(D) 1130
9- If a parabola with equation y=a x2 + 5 x + 10, where a is constant, passes through point (2,12), what is the value of a2?
(A)  2
(B) 2
(C)  4
(D) 4
10- What is the value of f(5) for the following function f?
f(x)=x2  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 yintercept represent?
(A) The yintercept represents the rate of grows of the plant which is 5 inches
(B) The yintercept represents the starting height of 6 inches
(C) The yintercept represents the rate of growth of plant which is 3 inches per year
(D) There is no yintercept
12- What is the solution of the following system of equations?
{ x2 + y4=1 5 y6 + 2 x=4
(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?
PSAT_Math1
(A) 3.8
(B) 4.8
(C) 7.2
(D) 24
14- If x0, what is the value of (10 (x) (y2))2(8 x y2)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?
y5=x  25 x + 10
(A) 50
(B) 50.0
(C) 50
16- What is the value of x in the following equation?
x2  9x + 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 25 full?

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

In the x yplane, 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=x2
(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 < x2 + 1 < x2 + 1
(B) x < x2 + 1 < x2 + 1
(C) x2 + 1 < x < x2 + 1
(D) x2 + 1 < x2 + 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?
PSAT_Math2

(A) 𝑎 < 𝑏 < 𝑐
(B) 𝑏 < 𝑎 < 𝑐
(C) 𝑎=𝑐
(D) 𝑏 < 𝑐=𝑎
27- What percent of cities are in the type of pollution A, C, and E respectively?
PSAT_Math3
(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?
PSAT_Math4
(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? 
PSAT_Math5
(A) 12
(B) 2
(C) 13
(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) 78
(B)  87
(C)  78
(D) 87
32- What's the maximum ratio of women to men in the four cities?
PSAT_Math6
(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?
PSAT_Math7
(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?
PSAT_Math8
(A) 120
(B) 128
(C) 132
(D) 160
35- In the rectangle below if y > 5 cm and the area of rectangle is 50 cm2 and the perimeter of the rectangle is 30 cm, what is the value of x and y respectively? 
PSAT_Math9
(A) 4,11
(B) 5,11
(C) 5,10
(D) 4,10
36- If a car has 80liter petrol and after one hour driving the car use 6liter petrol, how much petrol will remain after xhours 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)
PSAT_Math
(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?
PSAT_Math
(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?
{x + 4 y=105 x + 10 y=20
(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, cos⁡ (β) is equal to?
PSAT_Math11
(A) \frac{a}{b}
(B) \frac{𝑎}{\sqrt{𝑎^2 \ + \ 𝑏^2}}
(C) \frac{\sqrt{𝑎^2 \ + \ 𝑏^2}}{a \ b}
(D) \frac{b}{\sqrt{𝑎^2 \ + \ 𝑏^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{𝑎}{𝑏}=\frac{11}{10}
(B) \frac{𝑎}{𝑏}=\frac{21}{11}
(C) \frac{𝑎}{𝑏}=\frac{11}{21}
(D) \frac{𝑎}{𝑏}=\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 inches
To solve this problem, first recall the equation of a line:
y=m \ x \ + \ b
Where, m= slope and y=y-intercept
Remember 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 \triangleBAE and \triangleBCD 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 𝑎=𝑐
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, 𝑎=𝑐

 

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{𝑎}{\sqrt{𝑎^2 \ + \ 𝑏^2}}
Cos ⁡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{𝑎}{𝑏}=\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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