MATH Evaluation 6_EDGE []

Republic of the Philippines

EDGE ECE REVIEW SPECIALIST BOARD OF ELECTRONICS ENGINEERING ELECTRONICS ENGINEER LICENSURE EXAMINATION

MATHEMATICS INSTRUCTION: Do not write anything on this questionnaire. Select the correct answer for each of the following questions. Mark only one answer for each for each item by shading the box corresponding to the letter of your choice on the answer sheet provided. 1. An angle greater than π/2 radians and less than π radians. a. Acute Angle b. Obtuse Angle c. Straight Angle d. Reflex Angle 2. At 12:00 noon ship B is 100 miles east of ship A. If ship B sails west at 10 mi/hr and ship A sails south at 20 mi/hr, when will the ships be closest to each other? a. 1:00 PM b. 2:00 PM c. 3:00 PM d. 4:00 PM z= x 2 +y 2 z=distance between two ships since ship B is traveling 10 mi/h, after t hours x=100-10t. similarly, y=20t. Thus, z= (100-10t)2 +(20t)2 = 10,000-2000t+100t 2 +400t 2 = 500t 2 -2000t+10,000 2

z =500t 2 -2000t+10,000 2z

dz =0=1000t-2000 dt 0=1000t-2000

t=2 therefore at 2:00PM, they will be closest to each other.

The graph intersects the x axis when y=0. y  x  x2 0  x  x2 0  x (1  x ) x 0 V

x=1

   y 2dx =  ( x  x 2 )2 dx =  ( x 2  2 x 3 

x 4 )dx 1

 x3 2x 4 x5  =     3 4 5   0   2 1  =      0   4 5   

5. Insert three geometric means between 2 and 162. a. 6, 18, 36 b. 4, 18, 54 c. 6, 20, 54 d. 6, 18, 54 2, a2, a3, a4, 162 a  2 n  5 an  162 an  arn1 162  2r 5 1 r 4  81

3. The arithmetic mean of 80 numbers is 55. If two numbers namely 274 and 850 are removed what is the arithmetic mean of the remaining numbers? a. 42 b. 28 c. 30 d. 32 S  55  4400 80 4400  274  850 A.M.   42 78 4. Find the volume of the solid of revolution obtained by revolving the region bounded by y=x-x2 and the x axis about the x-axis. a. π/10 b. π/20 c. π/30 d. π/40

r 3 a2  2  3   6 2 a3  2  3   18 3 a4  2  3   54

6. What is the equation of the circle with center at the origin and a radius of 5? a. x2 + y2 = 1 b. x2 + y2 = 25 c. x2 + y2 = 10 d. x2 + y2 = 5 S tan dard equation of a circle with center at the origin : x2  y2  r 2 With r  5, the equation is : x2  y2   5 

2

x 2  y 2  25

7. A quadrilateral ABCD is inscribed in a circle, if AB = 90 cm, CD = 70 cm, AD = 50 cm, AC = 97.29 cm and BD = 101.76 cm. respectively. Find the distance BC. a. 72 b. 68 c. 74 d. 77

Using Ptolemy's Theorem: (AB)(CD)  (AD)(BC)  (AC)(BD) 90(70)  50(BC)  97.29(101.76) BC  72

8. Given the area in the first quadrant bounded by x2 = 8y, the line y – 2 = 0 and the y-axis. What is the volume generated when this area is revolved about the line y – 2 = 0? a. 53.31 cu. units b. 45.87 cu. units c. 26.81 cu. units d. 33.98 cu. Units

dθ =2π (Since 1 revolution is 2π radians, dt dθ the value of =2π rad/sec.) dt dx Find: when θ=0 (The beam is closest to the police car dt when θ=0.) Given:

tan θ =

x 30

x=30tanθ

dx dθ =30sec 2θ dt dt Substituting θ=0 and

dθ dx =2π, we compute : dt dt

dx =30(1)2 (2π) = 60π ft/sec dt

11. Find the area of the region bounded by y=x2-5x+6, the xaxis, and the vertical lines x=0 and x=4. a. 14/3 b. 1/6 c. 17/3 d. 5/6 Determine where the graph crosses the x-axis: x2  8  2 

y=x 2 -5x+6

x  4

0=x 2 -5x+6 0=(x-2)(x-3) x=2 x=3 we integrate separately on the three intervals [0,2], [2,3], and [3,4].

2 4 y   2  5 5 2nd Proposition of Pappus: V  2y A

A

 2 3

 4  2  

6 3

2

 x 3 5x 2  I1=  02 (x 2 -5x+6)dx=    6x  3 2  0 14  8 20  =   12    0   3 3 2 

 4  16  V  2     5  3  V  26.81 m3

3

9. A statue 2 meters high stands on a column that is 3 meters high. An observer in level with the top of the statue observed that the column and the statue subtend the same angle. How far is the observer from the statue? a. 5 2 meters

b. 20 meters

c. 20 meters

d. 2 5 meters

 x 3 5x 2  I2 =  32 (x 2 -5x+6)dx=    6x  2 3 2 45    8 20  = 9   18      12  2   3 2  9 14 1 =   2 3 6 4

 x 3 5x 2  I3 =  34 (x 2 -5x+6)dx=    6x  3 2  3 45  64    =  40  24    9   18  2  3    16 9 5 =   3 2 6 14 1 5 17 Total area = I1+I2 +I3     3 6 6 3

12. Transform the fraction by rationalizing the denominator 4x tan2 

2 tan 

1  tan  2 4 2  5 x    2x 2 x x 4 2 1   x2 x

5

4x 2 x2  4

2

 5x 2  20  4x 2

x  20 2

x2 5 m

10. A police car is 30 ft away from a long straight wall. Its beacon, rotating 1 revolution per second, shines a beam of light on the wall. How fast is the beam moving when it is closest to the police car? a. 20π b. 30π c. 40π d. 60π

x - x2 - 4

  a. x  x  x2  4   

c. 4x  x2  4

b. x x 2  4 d. x  x 2 4

4x x+ x2 -4 . x- x2 4 x+ x2 -4 4x2 +4x x2 -4 x2 - x2 -4

 

  4x  x+ x2 -4    =x  x+ x2 -4    4  

13. Simplify:

 3 - j2 1+ j4   2 - j3  4 + j1

21 220 j 221 221 221 20 j c. 22 21 a.

21 220 j 220 221 210 21 j d. 221 221 b.

 3 - j2 1+ j4   11  j10 x 11  j10  21  j220  2 - j3  4 + j1 11  j10 11  j10 121  100 

21 220 j 221 221

14. Ryan has 800 ft of fencing. He wishes to form a rectangular enclosure and then divide it into three sections by running two lengths of fence parallel to one side. What should the dimensions of the enclosure be in order to maximize the enclosure area? a. 100 ft, 150 ft b. 50 ft, 100 ft c. 150ft, 300 ft d. 100 ft, 200 ft

m

y  y1 x  x1

y 1 x 1 2x  2  y  1 2x  y  3  0 2

17. Find the limit: sin2x/sin3x as x approaches to 0. a. 1/3 b. 3/4 c. 2/3 d. 0

The area of the enclosure is still the product of its length and width A=xy The 800 ft of fencing is to be divided into six pieces, 4 of length x, and 2 of length y. 4x+2y=800 2y=800-4x

 sin2x  Lim   x 0  sin3x  Re writing the quotient in the form : sin x Lim 1 x 0 x  sin2x   2 2x  2 1 2  sin2x  Lim     Lim  sin3x   3 1 3 x 0  sin3x  x 0  3  3x  

y=400-2x Replacing by 400-2x in the area equation, A=xy, we obtain the area function A=x(400-2x) =400x-2x 2 dA =400-4x dt 0=400-4x 4x=400

18. A baseball diamond has a shape of a square with sides 90 meters long. A player 30 m. from the third base and 60 m. from the 2nd base is running at a speed of 28 m / sec. At what rate is the player’s distance from the home plate changing? a. -8.85 m/s b. 8.85 m/s c. -4.40 m/s d. 17.9 m/s S2  302   90 

x=100 ft y=400-2x=400-2(100)=200 ft

15. A trapezoid has an area of 360 m 2 and an altitude of 20 m. Its two bases have ratio of 4:5. What are the lengths of the bases? a. 12,15 b. 7,11 c. 8,10 d. 16,20

2

S  x 2  8100 dx 2x dS dt  dt 2 x 2  8100 when : x  30 dx  28 m / sec dt 30  28  dS   8.85 m / s 2 dt  30   8100

b 19. How many sides has an equilateral polygon if each of its interior angles is 165 o? a. 24 b. 12 c. 18 d. 10

20

(n  2)180  n( ) (n  2)180  n(165)

a From the formula for area : 1 A  a  b h 2 1 360   a  b  20  2 36  a  b  eq.1 From the condition : a b  4 5 a 4  b 5 4 a  b  eq.2 5 Substitute eq.2 to eq.1: 4b 36  b 5 180  4b  5b 9b  180 b  20  longer base With b  20; 4 a   20   16  shorter base 5 Thus, the bases are 16 & 20.

16. What is the equation of the line having a slope of 2 and passing through the point (-1, 1). a. 2x – y + 3 = 0 b. 3x +y – 3 = 0 c. 2x + y – 3 = 0 d. 2x + y – 3 = 0

180n  360  165n 15n  360 n  24 sides 20. Pedro can paint a fence 50% faster than Juan and 20% faster than Pilar, and together they can paint a given fence in 4 hours. How long will it take Pedro to paint the same fence if he had to work alone? a. 6 b. 8 c. 10 d. 12 Let : A  number of hours,Pedro can paint the house B  number of hours,Juan can paint the house C  number of hours, Pilar can paint the house 1 1 1 1     eq.1 A B C 4 1  1 1  1  1.5   ;  0.666    eq.2 A B B A 1  1 1 1  1.2   ;  0.833    eq.3 A C C A Substitute eq.2 and eq.3 in eq.1: 1  1  1 1  0.666    0.833    A A A 4 A  10 hours

21. A rectangle is inscribed in a right triangle whose sides are 5, 12, and 13 inches. Two adjacent sides of the rectangle lie along the legs of the triangle. What is the maximum area of the rectangle? a. 15 sq. in b. 20 sq. in

c. 25 sq. in

 2cos15o  j2sin15o 3cos75o  j3 sin75o  cos 60o  jsin60o  4cos30o  j4 sin30o 

d. 30 sq. in

2cos15o  j2sin15o  215o 3cos75o  j3 sin75o  375o cos 60o  jsin60o  1  60o

A=xy since ABC is similar to DE,

4cos30o  j4 sin30o  430o

y 12  x  . 5 12

 215o 375o    3 1  j 3  1  60o  430o  4

5 12-x  . 12 The area as a function of x, becomes

It follows that y=

5  A=x  (12  x)   12  5 2 = (12 x  x ) 12 differentiating, we obtain dA 5 0 (12  2 x) dx 12 x6 Then,

25. A truck travels from point M northward for 30 min. then eastward for one hour, then shifted N 30W. if the constant speed is 40kph, how far directly from M, in km. will be it after 2 hours? a. 37.3 b. 47.9 c. 45.2 d. 41.6

5  5  A=x  (12  x)   (6)  (12  6)   15 sq.inches  12   12 

22. A parabola has an equation of x2 = 20y. Locate the coordinates of the focus of the parabola. a. (5,0) b. (0,5) c. (4,5) d. (5,4) 4a  20 a5 F  0,a 

MA=40(30 /60)=20Km AB=40(1) = 40 km After 2 hours, tBC =30 min BC=40 (30 / 60) = 20 Km

Focus is at  0,5 

23. From the top of tower A, the angle of elevation of the top of the tower B is 46 . From the foot of a tower B the angle of elevation of the top of tower A is 28. Both towers are on a level ground. If the height of tower B is 120 m., How far is A from the building? a. 42.3 m b. 40.7 m. c. 38.6 m. d. 44.1 m

In triangle MDC: MD = AB – BC cos 30 MD = 40 – 20 sin 30 = 30 km CD = MA + BC cos 30 CD = 20 + 20 cos 30 = 37.32 km

CM  MD2  CD2 

302  37.3122  47.88km

26. What is the minimum possible perimeter for a rectangle whose area is 100 in2? a. 20 in b. 30 in c. 40 in d. 60 in A  xy P  2 x  2y  100  P  2   2y  y  200 P  2y y dP 200 0 2 dy y2

BC 120  sin 44 sin74 BC  86.718m In right triangle BAC: h = BC sin B h = 86.718 sin 28 h = 40.71 m

2cos15o  j2sin15o 3cos75o  j3 sin75o  cos60o  jsin60o  4cos30o  j4 sin30o 

24. Simplify:

1 1 j 3 2 3 c.  1  j 3 4 a.







200

3 1 j 3 4 1 d.  1  j 3 2 b.











2 y2 100=10x P=2(10)+2(10)

y=10 x=10

P=40

27. A man sold a book by mistake at 120% of the marked price instead of discounting the market price by 20%. If he sold the book for P14.40, what was the price for which he has sold the book? a. P10.20 b. P7.80 c. P9.60 d. P9.60 x  marked price of the book 0.80x  selling price of the book 1.20x  14.40 x  P12 selling price  0.80(12)  P9.6

28. Two lines have an equation of 2x – y + 2 = 0 and 2x + y – 4 = 0. Find the smallest angle between the two lines. a. 36.87° b. 42.80° c. 53.13° d. 126.87° tan  

m2  m1 1  m2m1

2x  y  4  0 y  2x  4 m1  2 2x  y  2  0 y  2x  2 m2  2 tan  

2   2 

1   2  2 

4 tan   1 4 4 tan    3   53.13

29. The square of a number increased by 16 is the same as 10 times the number. Find the numbers. a. 2,8 b. 3,6 c. 3,8 d. 5/8,0 2 x  16  10x x2  10x  16  0 x2

A1  A 2  21

 eq. 3 1 1 bh  bh  21 2 1 2 2 Substitute (1) and ( 2 ) to equation 3: 1 1 b  b  3   b b  3   21 2 2  b2  3b b2  3b    21 2  2  2  b2  3b  b2  3b  42 6b  42 b7 Thus: h1  7  3  10units

h2  7  3  4units

32. A train travels 2.5 miles up on a straight track with a grade of 110’. What is the vertical rise of the train in that distance? a. 0.716 miles b. 0.051 miles c. 0.279 miles d. 0.045 miles

sin1 10' 

y 2.5

y  2.5 sin1 10'

x8

y  0.051 miles

30. A particle moves along a path whose parametric equations are x = t3 and y = 2t2. What is the acceleration of the particle when t = 5 seconds? a. 30.26 m/s2 b. 18.56 m/s2 c. 20.62 m/s2 d. 23.37 m/s2 Solving for the acceleration along the x direction : dx  3t 2 dt d2 x  6t  acceleration along x dt 2 when,t  5; d2 x  6  5   30 dt 2 Solving for the acceleration along the y direction : dy  4t dt d2 y 4 dt 2 Thus, solving for the resul tan t acceleration :

33. Find the moment of inertia of the area bounded by the curve x2 = 4y, the line y = 1 and the y-axis on the first quadrant with respect to x-axis. a. 6/5 b. 7/2 c. 4/7 d. 8/7

a  302  42  30.26m / s2

31. Two triangles have equal bases. The altitude of one triangle is 3 units more than its base and the altitude of the other triangle is 3 units less than its base. Find the altitudes, if the areas of the triangle differ by 21 square units. a. 4 and 10 b. 4 and 26 c. 6 and 12 d. 7 and 23

1 Ix   x dy y2 0

x2  4y  x  2y1 2 1 Ix  2y1 2 y2 dy



0 1 Ix  2  y5 2 dy 0 1  2y7 2   Ix  2   7   0 4 Ix  7 34. The polynomial x3 + 4x2 -3x + 8 is divided by x-5, then the remainder is a. 175 b. 140 c. 218 d. 200 f x  x3  4x 2  3x  8

 

divisor  x  5 Note :Using remainder theorem, remainder  f  5  3 2 Remainder   5   4  5   3  5   8  218

35. What is the largest possible volume a right circular cylinder can have if it is inscribed in a sphere of radius 5? a. 302 cu. units b. 261 cu. units c. 394 cu. units d. 211 cu. units Re call : For max imum volume of cylinder inscribed in a sphere of radius R : r

6 R 3

and

h

2R 3

Thus, 6  5   4.08 units 3 2(5) and h=  5.77 units 3 r

h1  b  3  eq. 1 h2  b  3  eq. 2

Vcylinder  r 2h   (5.77)  302 cubic units

36. Find the real values of x and y in the equation 2x  j4y  3  j3x  y  j10 a. x=-1, y=2 b. x=1, y=-2 c. x=-2, y=1 d. x=2, y=-1 Rewrite the equation as (2x+y)+j(-3x+4y)=3-j10 separate the real from the imaginary parts thus, 2x+y=3 -3x+4y=-10 x=2 and y=-1 37. The axis of the hyperbola that passes through the foci, vertices, and center is called a. minor axis b. conjugate axis c. major axis d. transverse axis 38. Sand is being dumped from a dump truck at the rate of 10 ft3/min and forms a pile in the shape of a cone whose height is always half its radius. How fast is its height rising when the pile is 5 ft high? a. 1/(5π) ft/min b. 1/(10π) ft/min c. 1/(15π) ft/min d. 1/(20π) ft/min

abc 2 48 S  24 2 65  r(24) S

r  2.71

41. A plane, P, flies horizontally at an altitude of 2 miles with a speed of 480 miles/hour. At a certain moment it passes directly over a radar station, R. How fast is the distance between the plane and the radar station increasing 1 minute later? a. 287 mi/hr b. 318i/hr c. 466 mi/hr d. 524 mi/hr

dV dh  10 Find: when h=5. dt dt The volume of the cone is related to its height and radius by the

Given :

 2 1 r h. Since h= r , it follows that r=2h, so 3 2  4 3 V  (2h )2 h  h  3 Differentiating with respect to t, equation V=

dV dh  4h 2 dt dt Substituting the given information,

dx  480 dt dz Find : 1/ 60 hour later dt

Given :

By the theorem of Pythagoras, z=x 2  22 dz dx  2x 0 dt dt dz dx z x dt dt Since the plane travels 480 mi/hr, it will have flown 8 miles 2z

dh dt dh 10  100  dt dh 1  ft / min dt 10 10  4

1 1   hour  480x  8  . Since x=8, the value of z is easily 60 60   determined by the Pythagorean theorem.

in

z2  x 2  22 = 64+4

39. Find the distance between the points (2,5) and the line x – 2y + 3 = 0. a.

5

b.

8

c.

6

d.

9

Ax1  By1  C d A 2  B2 d d

d d

=68 z=

68 =2 17 dz 2 17 =8(480) dt dz 1920 = mi/hr or 466 m i/hr dt 17

x  2y  3 A2  B2   x  2y  3 

12   22  x1  2y1  3 5   2  2 5   3 5

d

5

d

5 5

5 5 5

 5

d  5  positive if the pt. is above the line 

40. The area of a triangle is 65 sq. cm. and its perimeter is 48 cm. compute the radius of the inscribed circle. a. 2.71 cm. b. 2.16 cm. c. 1.42 cm d. 2.49 cm

42. A church window is in the shape of a rectangle surmounted by a semicircle. If the perimeter of the window is 20 ft, what is its maximum area? a. 28 ft2 b. 49 ft2 c. 36 ft2 d. 52 ft2

r 2  2rx 2 P  20  2 x  2r  r 20=2x+(2+r 20  (2+r x 2 A

A

r 2  20  (2+r   2r   2 2  

A

r 2  20r  (2  r 2 2

dA  0  r  20  2(2+r dr r  2.8 20  2.8(2   x  2.8  A

  2.8  2

A  28ft

2

 2  2.8  2.8 

3 A  2x  dS 1 2  dx  dS  1     dy  2 x  4y 2x dx  4dy dx 2  dy x

2

dS  1 

43. How many diagonals can be drawn for a 12 sided polygon? a. 54 b. 48 c. 36 d. 62 For a Dodecagon: x  12 sides No. of diagonals x 12  (x  3)  (12  3) 2 2 No. of diagonals  54 diagonals 44. Find the equation of the axis of symmetry of the function y = 2x2 – 7x + 5. a. 7x + 4 = 0 b. 4x + 7 = 0 c. 4x – 7 = 0 d. 4x – 2 = 0 2 y  ax  bx  c

The axis of symmetry is the line pas sing through the x  coordinate of the vertex. Solving for the x  coordinate of the vertex : b x  formula 2a 7 x 2  2 x

4 dy  x2

x2  4 dy x2

3 x2  4 A  2  x dy x2 1 3 A  4  4y  4 dy 1 3 A  4  y  1 dy 1 A  4  2 

 y  13 2  3

3

 1

8  3 2  4   (2)3 / 2  3   A  43.32 A

48. Find the area of the hexagon ABCDEF formed by joining the points A(1,4), B(0,-3), C(2,3), D(-1,2), E(-2,-1) and F(3,0). a. 24 b. 20 c. 22 d. 15

7  equation of the axis of symmetry 4

Or, 4x  7  0 45. The two adjacent sides of a triangle are 5 and 8 meters respectively. If the included angle is changing at the rate of 2 rad/sec. at what rate is the area of the triangle changing if the included angle is 60 degrees? a. 23 sq.m/sec b. 15 sq.m/sec c. 20 sq.m/sec d. 25 sq.m/sec 5h ; h  8 sin  2  5  8 sin  A 2 dA d  20cos   20cos60  2   20 sq. m / sec dt dt A

46. If x varies directly as y and inversely as z; and x=14, when y=7 and z=2, find x when y=16 and z=8. a. 14 b. 4 c. 16 d. 8 y xk z 7 14  k 2 k4 x

4 16  8

8

47. Find the surface area of the portion of the curve x2 =4y from y =1 to y =3 when it is revolved about the y-axis. a. 19.84 b. 37.86 c. 16.75 d. 43.32

3 2 A

A

0 3

1 4 1 1 2 2 2 1 0 3 3 0 1  9  8  2  1  6  0      2  9  0  4  4  3  0  

1  26  14  2 A  20 square units A

49. Log of negative one to the base ten is written as log10 -1. Its rectangular form is: a. 0 – j1.36 b. zero c. 0 + j1.36 d. infinity

 

log  1  log  1 0j  log ej  jloge  0  1.364j

50. The area of a hexagon inscribed in a circle is 158 cm2. Find the difference in area between the hexagon and a circle. a. 33.05 cm2 b. 44.02 cm2 c. 28.41 cm2 d. 51.42 cm2

By formula: A

Acn 360o sin 2 n

A c (6) 360o sin 2 6 A c  191.05 cm2 (area of circle) Diff. in area  191.05  158 Diff. in area  33.05 cm2

158 

Using the figure: Area of the hexagon: r 2 sin60o (6) A 2 2 158  r sin60o (3)

r 2  60.81 A c  r 2 A c  (60.81) A c  191.05 cm2 (area of circle) DIff. in area  191.05  158  33.05 cm2

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