Mathematics “CE Board Exam May 1996″
What is the equation of the line that passes thru (4, 0) and is parallel to the line x- y – 2 = 0?
Solution:
x – y + 4 = 0
What is the equation of the line that passes thru (4, 0) and is parallel to the line x- y – 2 = 0?
Solution:
x – y + 4 = 0
Two sides of a triangle are 50 m. and 60 m. long. The angle included between these sides is 30 degrees. What is the interior angle (in degrees) opposite the longest side?
Solution:
Using Cosine Law:
a2 = (50)2 + (60)2 -2(50)(60) Cos 30°
a = 30.06 m.
Cosine Law again:
(60)2 = (50)2 + (30.06)2 -2 (50) (30.06) Cos θ
θ = 93.75°
The following notes were taken during a differential leveling. What is the difference in elevation between BM1 and BM2?
STA B.S. F.S. ELEV.
BM1**** 7.11 751.05
1 8.83 1.24
2 11.72 1.11
BM2 ***10.21
Solution:
STA B.S. F.S. ELEV.
BM1 **** 7.11 751.05
1 8.83 1.24 756.92
2 11.72 1.11 764.64
BM2 ***10.21 766.15
Diff. in elev. = 766.15 – 751.05
Diff. in elev. = 15.1
How far from the y-axis is the center of the curve 2x2+2y2+10x-6y-55 = 0?
Solution:
The center of the curve is at -2.5 from the y-axis.
A mixture compound from equal parts of two liquids, one white and the other black was placed in a hemi-spherical bowl. The total depth of the two liquids is 6”. After standing for a short time the mixture separated the white liquid setting below the black. If the thickness of the segment of the black liquid is 2”, Find the radius of the bowl in inches.
Solution:
r = 7.33
Find the elements of the product of the two matrices, matrix BC.
Solution:
Find the value in degree of arc cos (tan 24°).
Solution:
tan 24° = 0.44522
Let θ = arc Cos tan 24°
θ = arc Cos(0.44522)
Cos θ = 0.44522
θ = 63.56°
The distance from D to E, as measured, is 165.2 m. if the 50 m. tape used is 0.01 m. too short, what is the correct distance in m.?
Solution:
Correct distance = 165.167
A 3-degree curve has an external distance of 8.53m.. What is the central angle? Use chord basis.
Solution:
I = 24°
Find the value of A in the equation
Solution:
A = 2
From a speed of 75 kph, a car decelerates at the rate of 500 m/min2. along a straight path. How far in meters will it travel in 45 sec?
Solution:
S = 796.875 m.
What is the equation of the tangent to the curve 9x2 + 25y2 -225 = 0 at (0,3)?
Solution:
9x2 + 25y2 – 225 = 0
9(xx1) + 25yy1 – 255 = 0
9x(0) + 25y(3) – 225 =0
3y = 9
y = 3
y – 3 = 0 (equation of the tangent)
The radius of the moon is 1080 mi. the gravitational acceleration at the moon’s surface is 0.165 time the gravitational acceleration at the earth’s surface. What is the velocity of escape from the moon in miles/sec?
Solution:
V = 1.47 miles/sec.
P200,000 was deposited on Jan. 1, 1988 at an interest rate of 24% compounded semi-annually. How much would the sum be on Jan. 1, 1993?
Solution:
A = P621,170.00
Find the slope of the ellipse x2 + 4x2 - 10x – 16y + 5 = 0 at the point where y =2 + 80.5 and x = 7.
Solution:
y’ = – 0.1768 (slope of ellipse)
The speed of the traffic flowing past a certain downtown exit between the hours of 1:00 P.M. and 6:00 P.M. is approximately
V = t3 - 10.5t2 + 30t + 20 miles per hour where t = number of hours past noon. What is the fastest speed of the traffic between 1:00 P.M. and 6:00 P.M. in mph?
Solution:
Therefore the fastest speed is 46 mph
Two railroad tracks are perpendicular to each other. AT 12:00 P.M. there is a train at each track approaching the crossing at 50 kph., one being 100km., the other 150 km. away from the crossing. How fast in kph is the distance between the two trains changing at 4:00 P.M.?
Solution:
The longitudinal ground profile and the grade line shows that the length of cut is 540m. while the length of fill is 690m. The width of the road is 10m. for both cut and fill. The profile areas between the ground line and the grade line are 4800 m2 for cut and 2592 m2 for fill. Find the difference between the volume of cut and the volume of fill in cu.m., if the side slopes are 1.5:1 for cut and 2:1 for fill.
Solution:
Diff. in vol. = 66568.11 cu. m.
A circular cylinder is circumscribed about a right prism having a square base one meter on an edge. The volume of the cylinder is 6.283 cu.m. Find its altitude in m.
Solution:
(2r)2 = (1) 2 + (1) 2
r = 0.707
V = π r2 h
6.283 = π(0.707)2 h
h = 4m.
What is the area bounded by the curves y2=4x and x2 =4y?
Solution:
A = 5.333 sq. units
A machine costing P45,000 is estimated to have a salvage value of P4,350 when retired at the end of 6 years. Depreciation cost is computed using a constant percentage of the declining book value. What is the annual rate of depreciation in %?
Solution:
B.V. = F.C. (1-k)n
4350 = 45000 (1-k)6
(1-k)6 = 0.0967
1-k = 0.677
k = 0.3225
k = 32.25%
A bag contains 3 white and 5 black balls. If two balls are drawn in succession with replacement, what is the probability that both balls are black?
Solution:
A conical vessel has a height of 24cm. and a base diameter of 12cm. it holds water to a depth of 18cm. above its vertex. Find the volume of its content in cc.
Solution:
V = 381.70 cu.cm.
Find the area of a piece of land with an irregular boundary as follows:
STA OFFSET DISTANCE (m.)
0+000 5.59
0+010 3.38
0+020 2.30
0+030 3.96
0+040 4.80
The stations are on straight line boundary. Find the area of the land in m2 by Trapezoidal Rule.
Solution:
A = 148.35 m2 say 148.4 m2
A compound curve has the following data:
I1 = 28° D1 = 3°
I2 = 31° D2 = 4°
STA P.I. 30 + 120.5
Find the stationing of P.C.C.
Use P.I. 30 + 120.5
Solution:
P.C. = 30 + 106.97