A close traverse has the following data:
LINE BEARING DISTANCE
AB S. 15°36’ W. 24.22
BC S. 69°11’ E. 15.92
CD N. 57°58’ E. ?
DA S. 80°43’ W. ?
Find the distance DA in meters.
Solution:
LINE BEARING DIST. LAT. DEP.
AB S. 15°36’ W. 24.22 -23.33 -6.51
BC S. 69°11’ E. 15.92 -5.66 +14.88
CA +28.99 -8.37
AD = 75.01 m. say 75m.
Using the following notes, what is the elevation of BM14?
STA. B.S F.S ELEV.
BM12 00.04.64 209.65
1 5.80 5.06
2 2.25 5.02
BM1300.0 6.02 5.85
3 8.96 4.34
4 8.06 3.22
5 9.45 3.71
6 12.32 2.02
BM14 00.01.98
STA. B.S H.I. F.S ELEV.
BM12 00.04.64 214.29 209.65
1 5.80 215.03 5.06
2 2.25 212.26 5.02
BM13 00.06.02 212.43 5.85
3 8.96 217.05 4.34
4 8.06 221.89 3.22
5 9.45 227.63 3.71
6 12.32 237.93 2.02
BM14 00.01.98 235.95
The area of a circle circumscribed about an equilateral triangle is 254.47 sq.m. What is the area of the triangle in sq.m.?
Solution:
A = 105.24 m2
In triangle ABC, AB =15m., BC = 18m., and CA =24m. How far is the point of intersection of the angular bisectors from vertex C in m.?
Solution:
OC = 14.30
What is the value of x in log 625 = 4?
Solution:
logx 625 = 4
625 = x4
x = 5
The correct distance between two points is 220.45 m. Using a 100 m. tape that is “x” m. too long, the length to be laid on the ground should be 220.406 m. What is the value of “x”?
Solution:
x = 0.02 m.
The volume of the frustum of a regular triangular pyramid is 135 cu.m. The lower base is an equilateral triangle with an edge of 9m. The upper base is 8m. above the lower base. What is the upper base edge in meters?
Solution:
x = 3 m.
A circular cylinder with a volume of 6.54 cu.m. is circumscribed about a right prism whose base is an equilateral triangle of side 1.25m. What is the altitude of the cylinder in m.?
Solution:
h = 4 m.
The ratio of the volume to the lateral area of a right circular cone in 2:1. If the altitude is 15 cm., What is the ratio of the silent height to the radius?
Solution:
The ratio is 5.2
The area bounded by the curve y2 = 12x and the line x=3 is revolved about the line x=3. What is the volume generated?
Solution:
Using 2nd Prop. of Pappus
V = 2π x A
V = 2π(1.2)(2/3)(3)(12)
V = 180.16 say 181
Check by integration:
V = 180.96 say 181
Find the moment of inertia, with respect to x-axis of the area bounded by the parabola y2 = 4x and the line x = 1.
Solution:
Ix=2.13
What is the integral of Sin5x Cos3xdx if the lower limit is zero and the upper limit is π/2?
Solution:
= 0.0417
A stone is thrown upward at an angle of 30° with the horizontal. It lands 60m. measured horizontally and 2m. below measured vertically from its point of release. Determine the initial velocity of the stone in m/s.
Solution:
V=25.35 m/s
From a speed of 20 mph a train accelerates at the rate of 0.395 m/s2. How many seconds will it take the train to travel one km. during the acceleration?
Solution:
t = 52.1 sec.
What is the differential equation of the family of lines passing through a fixed point(h, k)?
Solution:
(y – k)dx – (x – h)dy = 0
Find the present value in pesos of perpetuity of P15000 payable semi-annually if money is worth 8% compounded quarterly.
Solution:
P=371287 (semi-annually)
A machine costing P720,000 is estimated to have a book value of P40,545.73 when retired at the end of 10 years. Depreciation cost is computed using a constant percentage of the declining book value. What is the annual rate of depreciation in %.
Solution:
Book value = (F.C.)(1-d)k
40,545.73 = 720,000(1-d)10
(1-d)10=0.05631
1 – d = 0.75
d = 0.25
d = 25%
P200,000 was deposited at an interest rate of 24% compounded semi-annually. After how many years will the sum be P621,170?
Solution:
No. of years = 5 years
A bomber plane, flying horizontally 3.2 km. above the ground, is sighting on at a target on the ground directly ahead. The angle between the line of sight and the path of the plane is changing at the rate of 5/12 rad/min.. When the angle is 30°, what is the speed of the plane in mph?
Solution:
say 200 mph
A manufacturer can produce a commodity at a cost of P2 per unit. At a selling price of P5 each, consumers have been buying 4000 pieces a month. For each P1.00 increase in that price, 400 fewer pieces will be sold each month. At what price a piece will the total profit be maximized?
Solution:
Original selling price = P5 per unit
x = increase in selling price
5+x = new selling price
4000-400x = unit sold
Sales = (5+x)(4000 – 400x)
Production cost = 2 (4000 – 400x)
Profit = (5+x) (4000 – 400x) – 2(4000 – 400x)
P = (4000 – 400x)(5 + x – 2)
P = (4000 – 400x)(3 + x)
4000 – 400 = 400(3) + 400x
800x = 4000 – 1200
x = 3.5
New selling price = 5+3.5
New selling price = 8.50
What is the derivative with respect to x of 2 Cos2 (x2+2)?
Solution:
2 Cos2 (x2+2)
= 4 Cos (x2+2)[Sin (x2+2)(2x)]
2 Cos2 (x2+2)
= -8x Sin (x2+2) Cos (x2+2)
The area in cut of two irregular sections 35m. apart are 34 m2 respectively. The base width is 10m. and the side slope is 1:1. Find the corrected volume of cut in cu.m. using prismoidal correction formula.
Solution:
V = 1564.84 cu.m.
The common tangent CD of a reversed curve is 280.50m. and has an azimuth of 312°29’. BC is a tangent of the first curve whose azimuth is 252°45’. DE is a tangent of the second curve whose azimuth is 218°13’. The radius of the first curve is 180m. P.I.1 is at station 16 + 523.37. B is at P.C. and E is at P.T. What is the stationing of P.I.2?
Solution:
Stationing of P.I.2 = (16+420)+(187.66)+177.13
Stationing of P.I.2 = 16+784.79
say 16+784.80
What is the angle of intersection of the tangents of a simple curve if their bearings are N. 75°12’ E. and S. 78°36’E. respectively?
Solution:
I = 180 – (75°12’ + 78°36’)
I = 26.2°
How far from the x-axis is the focus F of the hyperbola x2 - 2y 2+ 4x + 4y + 4 = 0?
Solution:
Distance of focus from x-axis = 2.73
What is the center of the curve x2+y2-2x-4y-31=0?
Solution:
Center is at (1,2)
What is the slope of the line 3x + 2y + 1 = 0?
Solution:
In how many minutes after 7:00 o’clock will the hands be directly opposite each other for the first time?
Solution:
x = 5.455 min.
Find the value of y in the following equation:
3x – 2y + w = 11
x + 5y – 2w = – 9
2x + y – 3w = -6
Solution:
D = [(-45 + 8 + 1)]
D = -46
D = [(81-44 -6)-(-18+36-33)]
y = -1
How many four – digit numbers can be formed by the use of digits 1, 2, 3, 4, 6 and 7 if one digit is used only once in one number?
Solution:
What is the area in sq.cm. of the circle circumscribed about an equilateral triangle with a side 10 cm. long?
Solution:
Acircle = 104.7 cm2
A closed traverse has the following data.
Course Bearing Dimension
1-2 N 9.27°E. 58.7 m.
2-3 S 88.43°E. 27.3 m.
3-4 N 86.78° E. 35.2 m.
4-5 S 5.3° E. 35.0 m.
What is the bearing of line 5-1?
Solution:
LINES BEARING DIST. LAT DEP.
1-2 N 9.27°E. 58.7 m. +57.93 + 9.46
2-3 S 88.43°E. 27.3 m. – 0.75 +27.29
3-4 N 86.78°E. 35.2 m. + 1.98 +35.14
4-5 S 5.3° E. 35.0 m. –34.85 + 3.23
5-6 –24.31 –75.12
Note:
Latitude = Distance x Cos bearing
Departure = Distance x Sin bearing
Bearing = S 72.07° W.
Sin (270 + B) = ?
Solution:
Sin (270 + B) = Sin 270 Cos B
+ Sin B Cos 270
Sin (270 + B) = -Cos B
The base edge of the regular triangular pyramid 12 m. if the altitude is 9 m. What is the volume in cu.m?
Solution:
V = 187.06m3
Point A is in between point B and C. The distance Between B and C from point A are 1000 m. and 2000 m. respectively. Measured from point A, the angle of the elevation of B and C is 44.4 m. considering the effects of curvature and refraction, what is the value of θ?
Solution:
θ = 8°15’
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