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Question and Solutions Preboard June 2, 2022 Final Exam

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Question and Answers Preboard June 2, 2022 Final Exam

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Voorbeeld van de inhoud

CE 4218L
Final Examination Part 1
June 2, 2022

1. Find the equation of the line whose y-intercept is -2 and is parallel to the line 2y = 6x + 3.

a.) y = -3x + 2 b.) y = 6x -2 c.) y = 3x +5 d.) y = 3x-2 e) None

2. The diameter of a circle has endpoints at (– 2, 3) and (6, 3). What is an equation of the
circle?
a.) (x – 2)2 + (y – 3)2 = 16 b.) (x + 2)2 + (y + 3)2 = 16
c.) (x – 2)2 + (y – 3)2 = 4 d.) (x + 2)2 + (y + 3)2 = 4 e) None

3. An ellipse has an equation of x2+3y2+2x-6y=0. Find the length of its diameter which has
a slope of 1.

. a.) 2 2 b.) 3 2 c.) 2 d.) 4 e) None

x2
4. Determine the equation of the line tangent to the graph y = + 1 , at point (-2,2).
4

a) y+x=0 b) y + 2x=0 c) y - 4x =0 d) y – x=1 e) None

5. Find the equation of the line tangent to the curve x2+y2-34=0 through point (3,5).

a) 3x+5y-34=0 b) 5x+3y-34=0 c) 3x-5y+34=0 d) 3x+4y+34=0 e) None

6. Determine the distance between the lines 3 y − 4 x − 5 = 0 and 3 y − 4 x + 25 = 0 .
a) 4 b) 6 c) 3 d) 5 e) None
7. Locate the focus of the parabola y − 4 x + 4 y -12 = 0 .
2


a) (-2,3) b) (-4,-2) c) (-3,-2) d) (-2,-3) e) none
8. Find tan  , where  is the acute angle between the lines 3x − 4 y + 15 = 0 and x + 2 y − 6 = 0 .
a) 1/11 b) 2 c) 1/2 d) 4 e) none
Determine the shortest distance from a point (-2,5) to the circle x + y - 6 x + 4 y = 12
2 2
9.
a) 5.0 b) 6.1 c) 2.5 d) 3.6 e) None


10. An semi-ellipse, major axis 8 and minor axis 6 is revolved about its major axis. Find the volume of the solid
of revolution in terms of π.

a) 64π b) 28π c) 60π d) 48π e) None

11. The diameters of two spheres are in the ratio 2:3 while the sum of their volumes is 1260 cu.m.
Find the surface area of the larger sphere in sq.m..

, CE 4218L
Final Examination Part 1
June 2, 2022

a) 972.6 b) 474.5 c) 380.7 d) 760.5 e) None

12. A circle shown in Figure A has chord length AB equal to 1.2m. Find the Fig A
perimeter of the circle in meters.

4 3 3 5 0.3m
a) b) c) d) e) None
3 2 4 4 A B
13. Which of the following is equal to sin 𝑦 + sin 𝑦 cot 2 𝑦?
a) sin 𝑦 b) csc 2 𝑦 c) sec 𝑦 d) csc 𝑦 e( None



14. If logMN=8x and logM/N = 2y, then logM is equal to:
a) 8x+2y b) 8x-2y c) 4x+y d) 4x-y e) None
15. Which line is a vertical asymptote of the equation y = ln( x − 1) ?
a) x = −2 b) x = −1 c) x = 0 d) x = 1
16. The positive value of sin x in the equation 6 cos 2 x + sin x − 5 = 0 is:
a) 1 b) 1/2 c) 1/3 d) 1/5 e) None

17. A cantilever beam (that is one end is fixed and the other end free), carries a uniform load of
4kN/m throughout its entire length of 3 m. The beam has a rectangular shape 100 mm wide
and 200 mm high. Find the maximum bending stress developed at a section 2 m from the free
end of the beam.
a) 8 MPa b) 10 MPa c) 12 MPa d) 14 MPa e) None
18. A 10-m long rod made up of steel (E=200 GPa) is hanged vertically by fixing its upper end to a
strong horizontal beam. The rod is supporting two vertically downward concentrated loads of
20 kN each, one at mid length and the other at the lower end. Disregarding the weight of the
rod, calculate the vertical movement of the lower end of the rod due to axial elongation
knowing the cross sectional area of the rod is 110 sq mm.
a) 9.09 mm b) 11.62 mm c) 13.64 mm d) 15.43 mm e) None
19. A simply supported beam having a span of 6 m is carrying a concentrated load of 90 kN at a
point 2 m to the righ from the left support of the beam. Compute the bending moment at the
midspan of the beam.
a) 120 kN.m b) 90 kN.m c) 60 kN.m d) 30 kN.m e) None
20. A hollow steel tube with an inside diameter of 100 mm must carry a tensile load of 400 kN.
Determine the outside diameter of the tube if the stress is limited to 120 MN/m2.
a) 131.24mm b) 119.35mm c) 113.65mm d) 126.82 mm e) None
21. A simply supported beam having a span of 6 m is carrying a uniform load of 8 kN/m throughout
its entire span. Compute the bending moment at a point 2 m from the left support of the
beam.
a) 32 kN.m b) 44 kN.m c) 18 kN.m d) 60 kN.m e) None

, CE 4218L
Final Examination Part 1
June 2, 2022

22. Determine the value of 𝜃 ( in degrees), in the figure shown,
so that the moment of the 10-kN force about point O is
maximum.
a) 56.31 b) 33.69 c) 22.57 d) 67.43 e) None
23. Find area of the smaller segment formed by cutting a circle
along a 10-cm long chord knowing that the radius of the circle is 10 cm.
a) 8.78 sq. cm b)7.82 sq. cm c) 9.06 sq cm d) 10.07 sq cm e) Nota

24. Find the numerical coefficient of 8th term in the expansion of (3x
3
− y 2 )12 .
a) 792 b) -243 c) -192456 d) 192456 e) None

25. A monument consists of a statue mounted on top of a concrete pedestal. A boy, standing 9m away from the
monument, notices that the angle subtended by the statue is equal to the angle of elevation of the top of the
concrete pedestal. If the statue is 1.5 high, what is the vertical distance between the eye level of the boy and
the top of the concrete pedestal?
a) 1.274m b) 1.427 m c) 2.147m d) 4.127m

26. Find the area of the region in the first quadrant which is bounded by y = 3 x − 8 and y = 4 − x
a) 20/3 b) 7 c) 22/3 d) 7 e) Nota

27. An arch light hangs at a height of 10 m above the center of a street 20 m wide. A man 2 m tall walks along
the sidewalk at a rate of 2 m/s. How fast is his shadow lengthening when he is 15 meters up the street?
a) 0.523m/s b) 0.416 m/s c) 0.325 m/s d) 0.112m/s e) Nota

28. A point moves on the curve y3 = 8x in such a way that the rate of change of the ordinate is always 5
units/sec. How fast is the abscissa changing when the ordinate is 2?
a) 7.5 b) 6.8 c) 5.7 d) 4.3 e) Nota
29. A homogeneous rod of constant cross section has length of 6 m and weight of 2500 N. The rod
is resting on smooth surfaces at A and B. A cable CD is holding the rod in the position shown.
Determine the tension in the cable.
a) 612 b) 426 c) 518 d) 382 e) None

B


3m

C D
1m
A
4m
30. Find the area of the region in the 2nd quadrant which is bounded by the parabola 𝑦 = 𝑥 2 + 1 and
the line 𝑦 = −𝑥 + 3.
a) 16/3 b) 5/3 c) 7/3 d) 10/3 e ) none

, G


① b= 1 -




2y = Gx + 3

y = 3x + 3/2
3x 2
y
=
-





O

(6 3)
,
(2 , 3)


n
=




R
2 (2 b) + (3 3)2
=
= = - -




1 + 3
=
4
k =
=
3
2
(x 2)2+ (X 3)
- - =
16





-




xi + 3y2 + 2x by 0- =




(x2 + 2x 1) + (3yz Gy + 9) 0
+ -
=




(X + 1) + 3(y 1) 1 + 3
- =




[(x+ 1) + 3(y 1) 4) - =




#
a = 2
,
b =F


* =
I
X

X
y =

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