Republic of thePhilippines

BATANGAS STATE UNIVERSITY

Pablo Borbon Main II, Alangilan Batangas City
College of Engineering, Architecture & Fine Arts
www.batstate-u.edu.ph Tel. No. (043) 425-0139 loc. 118

Mechanical/Petroleum Engineering Department

ME 529 – MECHANICAL ENGINEERING PRACTICE WITH COMPREHENSIVE

EXAMINATION
First Semester, AY 2020-2021

MATHEMATICS, ENGINEERING SCIENCES, ENGINEERING ECONOMY AND ME LAWS

& ETHICS

INSTRUCTION: Provide accurate and logical solutions to the following problems.

1. How many liters of a 60% solution of nitric acid should be added to 40 L of a 30% solution to obtain a
50 % solution of the acid?
A. 50 B. 80 C.65 D. 95
Given:
60% solution of nitric acid
40L of a 30% solution
50% solution of the acid
Required:
Liters of a 60% solution needed to add to the mixture.
Solution:

60% X liters 30% 40Liters 50% of acid

60%(X) + 40 L (30%) = 50%(X+40)
(0.6)(X)+ 40L (0.3) =(0.5)(X+40)
X= 80 L

∴ The liters required to be added to the mixture is 80 L. The answer is letter B.

Discussion:
First, it is best to illustrate the problem so we can comprehend better what solution we will be
making in order to solve the problem. Then, we can equate an equation based on the known values. After
producing the equation which is 0.6X + 40(0.3) = 0.5(x+40), we can now find the number of liters required
to be added to the mixture which is at 80 liters.

2. A tank can be filled by the pipe in 9 hrs. and by another pipe by 12 hrs. starting empty, how long it
will take to fill the tank if water is being taken out by a third pipe at rate per hour equal to the one-sixth
the capacity of the tank?
A. 18 hrs. B. 15 hrs. C. 21 hrs. D. 36 hrs.
Given:
Pipe 1 = 9hrs. filling time.
Pipe 2 = 12hrs. filling time.
Pipe 3 = one-sixth of the capacity of tank filling time.

Required:
Pipe Total, Total time to fill the tank
Solution:

Pipe 1
Pipe 2 Pipe 3

1 1 1 1
𝑃𝑖𝑝𝑒𝑇𝑜𝑡𝑎𝑙 = + ℎ𝑟𝑠. − ℎ𝑟𝑠. =
9 ℎ𝑟𝑠. 12 6 𝑡
12 + 9 − 18
𝑃𝑖𝑝𝑒𝑇𝑜𝑡𝑎𝑙 =
108

3 1
𝑃𝑖𝑝𝑒𝑇𝑜𝑡𝑎𝑙 = =
108 𝑡

𝑷𝒊𝒑𝒆𝑻𝒐𝒕𝒂𝒍 = 𝟑𝟔 𝒉𝒓𝒔.

∴ The time required to fill the pipe is 36 hrs. The answer is letter D
Discussion:
First, it is best to illustrate the problem so we can comprehend better what solution we will be
making in order to solve the problem. Then, we can make an equation based on the known values. In order
to solve the total time to fill the pipe, we will add the time it took pipe 1 and pipe 2 then will be subtracted
to the pipe 3. The explanation the time of pipe 3 will be subtracted is because the third pipe is emptying the
water. Then, the solution will be 1/total time of pipes is equal to 1/time of pipe 1 plus 1/time of pipe 2 minus
1/time of pipe 3. Thus, the time required to fill the pipes is equal to 36 hours.

3. Find the remainder if we divide 4y3+18y2+8y-4 by (2y+3).

A.10 B. 15 C. 11 D. 13
Given:
4y3+18y-4
2y+3

Required:
Remainder
Solution:
4y3 + 18y2 + 8y – 4
2y + 3
y = -3/2
3 3 3
4(− 2)3 + 18(− 2)2 + 8(− 2)– 4

R= 11

∴ The remainder is 11. The answer is letter C.

Discussion:
To solve the problem, we will use the remainder theorem. First, we will simplify the dividend to
another equation. Then, after arriving at a new equation, we can now substitute this to the old equation and
after calculating the values, we can now arrive at the remainder. Thus, for this problem, the remainder is
11.

4. The sum of two numbers is 35 and their product is 15. Find the sum of their reciprocal.
A. 2/7 B. 2/3 C. 7/3 d. 5/2
Given:
Two numbers, 35 and 15.
Required:
Sum of the reciprocal of the two numbers
Solution:
X+Y =35
XY = 15
1 1 𝑋+𝑌
+ =
𝑋 𝑌 𝑋𝑌
35
=
15
𝟕
=
𝟑
∴The sum of the reciprocal of two numbers is 7/3. The answer is letter C.

Discussion:
First, we will make 2 equations based on the given statements. The first equation will be that sum
of two numbers is equal to 35. The second equation will be that the product of two numbers is equal to 15.
Finding the sum of the reciprocal numbers, we will substitute the two equations made beforehand and the
resulting sum of the reciprocal of two numbers is equal to 7/3.

5. The seating section in a coliseum has 30 seats in the first row, 32 seats in the second row, 34 seats in
the third row, and so on until the tenth row is reach after which there are ten rows each containing 50
seats. Find the total number of seats in this section.
A. 900 B. 890 C. 810 D. 390
Given:
First row = 30 seats
Second row = 32 seats
Third row = 34 seats
Ten rows with 50 seats each
Required:
Total number of seats in the section
Solution:
From first row up to tenth row:
𝑛
𝑆𝑛 = (2𝑎1 + (𝑛 − 1)𝑑)
2
10
𝑆10 = ((2)(30) + (10 − 1)(2)
2
 𝑆10 = 390 𝑠𝑒𝑎𝑡𝑠
Another 10 seating section row 11 to 20:
10 rows x 50 seats = 500 seats
Total seats = 390 +500
Total seats = 890 seats

∴The total number of seats in this section is 890. The answer is letter B.

Discussion:
In order to solve the problem, we will use the arithmetic sequence. For the seats 1-10, the resulting
total number of seats is equal to 390 seats. Solving for another ten seating rows with 50 seats each result to
a total of 500 seats. Finding the overall total number of seats, it is equal to 890 seats.
6. A chemist of distillery experimented on two alcohol solution of different strength 30% and 60%
alcohol, respectively. How many cubic meters of each strength must be used in order to produce mixture
of 50 cubic meters that contains 40% alcohol?
A. 25,25 B. 30,20 C. 23 2/3, 26 1/3 D. 33 1/3, 16 2/3
Given:
50 m3 of 40 L alcohol
Two solution with 30% and 60% alcohol
Required:
Cubic meters of each solution to produce the mixture.
Solution:

30%X m3 60% 50-x m3 40% 50m3

30 % (X) + 60% (50-X) = 40% (50)

0.3(X) + 0.6(50-X)= .40 (50)
0.3 (X)= 10
X= 33.33m3 - for 30% alcohol
For 60% alcohol
50-X= 50-33.33
= 16.67 m3
 ∴ The required m3 for 30% alcohol is 33.33m3 and 16.67m3 for 60% alcohol solution. The
answer is letter D.

Discussion:
First, it is best to illustrate the problem so we can comprehend better what solution we will be
making in order to solve the problem. Then, we can equate an equation based on the known values. After
producing the equation which is 0.3X + 0.6(50-x) = 40(5), we can now find the volume measured in cubic
meter for each solution. For the 30% solution, 33.33m3 will be needed while for the 60% alcohol solution,
16.67m3 will be needed.

7.What is the angle between two vectors A and B? A= 2i +10k +4k, B= 20i– 4j +4k
A. 85.98 deg B. 61.60 deg C. 74.49deg D. 42.25 deg

Given:
Vector A= 2i + 10 j + 4k
Vector B= 20i - 4j + 4k
Required:
Angle between the vectors
Solution:
Finding the magnitude of vector, A
A= √(22 ) + (102 ) + (42 )
A= 2√30
Finding the magnitude of vector B
B= √(202 ) + (−42 ) + (42 )
B = 20
Using the formula,
i+j+k
α = cos-1
/A/x/B/

2i + 10 j + 4k
20i - 4j + 4k
(2)(20)+(10)(-4)+(4)(4)
α = cos-1
/2√30/x/20/
𝛂 = 𝟖𝟓. 𝟗𝟕𝟎𝟑 𝐝𝐞𝐠

∴ The angle between the vectors A and B is 85.97 deg. The answer is letter A.
Discussion:
First, we will find the magnitude for both vectors A and B. Then, by using the formula, we can now
substitute all the known values and we can now find the angle between the vectors A and B which is
calculated as 85.97 degrees.

8. Box A has 4 white balls, 4 blue balls and 6 orange balls. Box B has 2 white balls ,6 blue balls and 6
orange balls. If one ball is drawn from each box, what is the probability that the one of the two ball will
be white?
A. 0.985 B. 0.437 C. 0.743 D. 0.347
Given:
Box A: 4 white balls, 4 blue balls, 6 orange balls,
Box B: 2white balls, 6 blue balls, 6 orange balls
Required:
Probability of that one of the two ball will be white.
Solution:
Probability 1: White Ball on Box A, Box B not white
4 12
𝑃𝑟 = 𝑥
14 14
Probability 2: Box A not a White ball, Box B will white
2 10
𝑃𝑟 = 14 𝑥 14
12 5
𝑃𝑟 = +
49 49
7
𝑃𝑟 = = 𝟎. 𝟑𝟒𝟕
49
∴ The probability of getting white ball is 0.3469. The answer is letter D.

Discussion:
To solve the probability that of the two balls that will be drawn will be white, there will be two
possible scenarios. First is that the ball drawn in Box A is white and the ball drawn in Box B is not white.
The second scenario will be that the ball drawn in Box A is not white and the ball drawn in Box B is white.
Adding the probabilities of the two possible scenarios leads to the result of the overall probability which is
0.3469.

9. Find the sum of even two integers between 1 to 31.

A. 210 B. 240 C. 220 D. 242
Given:
Even integers between 1 to 31
1st term =1
Last Term = 31
Required:
Sum of even integers.
Solution:
No. of terms( 1st term + Last Term)
Sum of even integers =
2
15( 1+31)
Sum of even integers =
2
Sum of even integers = 240

∴ The sum of even integers is 240. The answer is letter B.

Discussion:
To calculate for the sum of even integers, the formula to be used is the total number of terms times
the sum of the first and the last term divided by 2. Then, the resulting sum of the even integers in the
problem is calculated as 240.

10. Find the sum of odd integers from 1 to 41.

A. 361 B. 240 C. 220 D. 441
Given:
Odd integers between 1 to 41.
1st term =1
Last Term = 41

Required:
Sum of the odd integers.
Solution:
No. of terms( 1st term + Last Term)
Sum of odd integers =
2
21( 1+41)
Sum of odd integers =
2
Sum of odd integers = 441
 ∴ The sum of odd integers is 441. The answer is letter D.

Discussion:
To find the sum of odd integers from 1 to 41, we will use the formula same as the previous problem.
The sum of the odd integers is equal to the total number of terms multiplied by the sum of the first term and
the last term divided by 2. Thus, the sum of odd integers from 1 to 41 is equal to 441.

11. The sides of a triangle are 8cm, 10 cm, and 14cm. Find the radius of inscribed and circumscribed
circle.
A. 3.45 ,7.14 B. 2.45, 8.14 C. 2.45, 7.14 D. 3.45, 8.14
Given:
Sides of the triangle 8cm, 10 cm and 14cm
Required:
Radius of inscribed and circumscribed circle
Solution:
10 8

14

For circumscribe triangle,

a = 8 b = 10 c=14
𝑎+𝑏+𝑐
S=
2
8+10+14
S= = 16
2

(s-a)(s-b)(s-c)
r= √
s

(16-8)(16-10)(16-14)
r= √
16

r= 2.4494 cm
For inscribe triangle,
a = 8, b = 10 c= 14
A= √s(s-a)(s-b)(s-c)
 a+b+c
S=
2
S = 16
A= √16(16-8)(16-10)(16-14)
A= 16√6
(8)(10)(14)
At =
4r
r= 7.144

∴ The radius are 2.45 and 7.14. The answer is letter C.

Discussion:
First, we will calculate for the radius of the inscribed circle. Inscribed circle means that the circle
is inside the triangle. To solve the problem, we will use the Heron’s formula. Calculating for the area, we
can now also solve the radius of the inscribed circle through calculating the circumscribed triangle. Also,
by using again the Heron’s formula, we can now also solve the radius of the circumscribed circle through
calculating the inscribed triangle.

12. Given of diameter x and altitude of h. What percent is the volume of the largest cylinder which can be
inscribe in the cone to the volume of the cone?
A. 44% B. 46% C. 56% D. 65%
Given:
Diameter = x
Altitude = h
Required:
Percent of volume of the largest cylinder inscribe in the cone to the volume of the cone
Solution:
Let: R = radius of the cone
r = radius of the inscribed cylinder
H = height of the cone
h = height of the inscribed cylinder
Equation 1: V = πr2h
By ratio and proportion:
R r
H
= H−h

Hr = RH – Rh
 HR−Hr
Equation 2: h=
R

Substitute h in equation 1:
V = πr2h
HR−Hr πr2 HR πr2 Hr πr3 H
V = πr2( R
) =( R
) –( R
) = πr2H – R

Get the derivative of the volume to solve for r:

dV πr3 H
dr
= ∫ πr 2 H – R
dV 3πr2 H
dr
= 2πrH – R
=0
3πr2 H
2πrH = R
2R
r=
3

Substitute r to equation 2:
HR−Hr
h= R
HR H 2R
h= R
– R( 3 )
2H H
h=H– 3
= 3

Substitute r and h to equation 1:

V = πr2h
2R 2 H 4πR2 H
V = π( 3 ) ( 3 ) = 27

Let Vc = volume of cone

1 π
Vc = 3
πR2 H = 3 R2 H

Getting the ratio:

4πR2 H
V 27 4πR2 H 3 4
(V ) = π 2 = 27
× πR2 H = 9 = 0.4444444444 = 44%
c R H
3

Therefore, the percent of volume of the largest cylinder inscribed in a cone to the volume of the
cone is about 44%, letter A.
Discussion:
We can solve the problem by forming the necessary equations needed to get the percentage of the
volume of the largest cylinder inscribed in a cone to the volume of the cone. The first equation will be based
on the volume of the cone and the second equation is the derived equation which is simplified in terms of
h. Then, by getting the derivative form, we can now calculate for the radius. Substituting the radius yields
the height. Substituting both the radius and height to the original equations, we can now calculate for the
volume of the largest cylinder inscribed in a cone and the volume of the cone. Thus, the total percent is
equal to 44%.

13. What is the area of triangle with sides of 5, 7 and 10?

A. 16.248 B. 25.248 C. 18.248 D. 30.248
Given:
a=5
b=7
c = 10
Required:
Area of triangle
Solution:
A = √S (S − a)(S − b) (S − c)
a+b+c
Where S = 2
5+7+10
S= 2
= 11

A = √S (S − a)(S − b) (S − c)

A = √11 (11 − 5)(11 − 7) (11 − 10)

A = 2√66 =16.24807681
Therefore, the area of the triangle is 16.248 sq. units, letter A.

Discussion:
By using the Heron’s formula, we can now calculate for the area of the triangle. The area of the
triangle calculated is equal to 16.248 square units.

14. What is the interior angle (in radians) of a dodecagon?

A. 2.361 B. 2.9 C. 2.618 D. 2.75
Given:
Dodecagon (12sides)
Required:
Interior angle in radians
Solution:
(𝑛−2)𝜋
β= 𝑛
, rad
 (12−2)𝜋 5𝜋
β= =
12 6

β = 2.617993878
Therefore, the interior angle of a dodecagon in radians is 2.618, letter C.

Discussion:
The dodecagon is polygon with a total number of sides equal to 12. Then, by using the formula for
getting the interior angle, we can now determine the answer in terms of radians which is equal to 2.618
radians.

15. A circle with a radius of 6 has half of its area removed by cutting a border of uniform width. Find the
width of the border.
A. 2.2 B. 1.76 C. 1.35 D. 3.75
Given:
Radius = 6
Required:
Width of the border
Solution:
A = πr2
Solving for A when half of the area is removed:
1 1
A = 2 𝜋R2 = 2 𝜋(6)2 = 18π

18π = πr2
r = 3√2 = 4.242640687
Solving for the width, x:
x=R–r
x = 6 – 3√2
x = 1.757359313 ≈ 1.76
Therefore, the width of the border is 1.76, letter B.

Discussion:
First, we should analyze what is stated in the problem. A circle with a radius of 6 has half of its
area removed by cutting a border of uniform width. Then, we can form equations and from then, we can
compute for the inner and outer radii. After computing, we can calculate the width of the border through
the differences of the two radii. Thus, the width of the border is measured as 1.76 units. to be added to the
mixture which is at 80 liters.

16. Compute the surface area of the cone having a slant height of 5 cm and a diameter of 6 cm.
A. 47.12 cm2 B. 38.86 cm2 C. 25.64 cm2 D. 30.24 cm2
Given:
Slant height = 5cm
Diameter = 6cm
Required:
Surface area of cone
Solution:
Solving for h using the given slant height:
l 2 = r 2 + h2
h = √𝑙 2 − r 2 = √(5cm)2 − (3cm)2
h = 4cm
Solving for surface area:
SA = πr√r 2 + h2
SA = π(3cm)√(3cm)2 + (4cm)2
SA = 15π = 47.1238898cm2 ≈ 47.12 cm2
Therefore, the surface area of the cone is 47.12 cm2, letter A.

Discussion:
First, we should analyze what is stated in the problem. Then, we can now solve for the height with
the known values of slant height and the radius from the diameter. Then, we can now solve for the surface
area of the cone which is calculated through pi times the radius times the square root of the sum of the
squared of radius and the squared of the height. Thus, the surface area of the cone is 47.12 square
centimeters.

17. 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 m3. Find its altitude in m.
A. 4.5 B. 5.5 C. 4 D. 5
Given:
Cylinder is circumscribed about a right prism
Side of square = 1m
 Volume of cylinder = 6.283m3
Required:
Altitude in meter
Solution:
For a given circumscribed circle, triangle OAB is right angled triangle. By Pythagorean theorem,
OA2 = OB2 +AB2
r2 = (0.5m)2 + (0.5m)2
r2 = 0.5m2
volume of cylinder = area of base × height
V = πr2h
Volume 6.283m3
h= πr2
= π(0.5m2 ) = 3.99988203m ≈ 4m

Therefore, the altitude of the cylinder is approximately 4m, letter C.

Discussion:
First, we should analyze what is stated in the problem. Then, we can solve for the radius using the
Pythagorean theorem. Then, we can now calculate for the altitude or height by dividing the volume by pi
times the squared of the radius. Thus, the altitude of the cylinder is 4 meters.

18. A pipe lining material of silicon carbide used in the conveyance of pulverized coal to fuel a boiler, has
a thickness of 2 cm and inside diameter of 10 cm. Find the volume of the material with a pipe length of 6
meters.
A. 45 239 cm3 B. 42 539 cm3 C. 49 532 cm3 D. 43 932 cm3
Given:
Thickness = 2cm
Inside diameter = 10cm
Required:
Volume of material with pipe length of 6m
Solution:
volume of cylinder = πr2h
volume of pipe = πh (𝑟𝑜𝑢𝑡 2 − 𝑟𝑖𝑛 2 )
where: rin = 5cm
rout = 5cm + 2cm = 7cm
Solving for the volume of the pipe:
𝑉𝑝𝑖𝑝𝑒 = πh (𝑟𝑜𝑢𝑡 2 − 𝑟𝑖𝑛 2 )
 100cm
𝑉𝑝𝑖𝑝𝑒 = π (6m × ) [(7cm)2 − (5cm)2 ] = 14400π
1m

𝑉𝑝𝑖𝑝𝑒 = 45238.93421cm3 ≈ 45239 cm3

Therefore, the volume of the pipe is 45239 cm3, letter A.

Discussion:
First, we should analyze what is stated in the problem. Then, we should solve for the outer radius
through the known values. Then, we can now solve for the volume of the pipe through multiplying the pi
to the height and then multiplying it to the difference of the squared outer radius and the squared inner
radius. Thus, the volume of the pipe is equal to 45329 cubic centimeters.

19. A group of children playing with marbles places 50 pieces of the marbles inside a cylindrical container
with water filled to a height of 20 cm. If the diameter of each marble is 1.5 cm and that of the cylindrical
container 6 cm. What would be the new height of water inside the cylindrical container after the marbles
were placed inside?
A. 23.125 cm B. 22.125 cm C. 24.125 cm D. 25.125 cm
Given:
50 pieces of marbles
Height of water without marbles = 20cm
Diameter of marble = 1.5cm
Diameter of container = 6cm
Required:
new height of water inside the container after the marbles were placed
Solution:
4
Volume of sphere = 3 𝜋r 3

Volume of cylinder = πr2h

Total volume = Vwater + Vmarbles
Solving for Vtotal:
4
Vtotal = πr2h + ( 𝜋r 3 ) (50)
3
4
Vtotal = [𝜋(3cm)2 (20cm)] + [50 × 3 𝜋(0.75cm)3 ]

Vtotal = 653.843971cm3
Solving for height using Vtotal:
Vtotal = πr2h
 Vtotal 653.843971cm3
h= πr2
= π(3cm)2
= 23.125cm

Therefore, the height of the water after the marbles were placed is 23.125cm, letter A.

Discussion:
First, we should analyze what is stated in the problem. Then, we should solve for the total volume
which is the sum of the volume of the sphere and the volume of the cylinder. Then, we can now solve for
the height by dividing the total volume with pi multiplied by the squared of the radius. The height of the
water is 23.125 centimeters.

20. The corners of a cubical block touch the closed spherical shell that encloses it. The volume of the box
is 2744 cm3. What volume in cm3 inside shell is not occupied by the block?
A. 4 713.56 B. 3 360.12 C. 4 133.25 D. 5 346.42
Given:
volume of the box = 2744cm3
Required:
volume in cm3 inside the shell that is not occupied by the block
Solution:
Let: V = volume inside the sphere but outside the box
VS = volume of the sphere
VB = volume of the box
Getting the side of the box using VB = x3:
3
x = √2744cm3
x = 14cm
Getting the radius using the diagonal of the cube:
d = √x 2 + x 2 + x 2
d = √3(14cm)2
d = 14√3 cm
d 14√3 cm
r=2= 2
= 7√3 cm = 12.12435565cm

Getting the volume outside the box using V = VS – VB:

V = VS – VB
4
V = 3 𝜋r 3 − VB
4
V = 3 𝜋(12.12 cm)3 − 2744cm3
 V = 4713.555065cm3 ≈ 4713.56 cm3
Therefore, the volume not occupied by the block is 4713.56 cm3, letter A.

Discussion:
First, we should analyze what is stated in the problem. Then, we should calculate for the side of the
box using VB = x3. After that, we can now get the radius using the diagonal of the cube. Then, we can now
get the volume outside the box the difference of the volume of the sphere and the volume of the box. Thus,
the volume outside the box is 4713.56 cubic centimeters.

21. Determine B such that 3x + 2y – 7 = 0 is perpendicular to 2x – By + 2 = 0.

A. 5 B. 3 C. 4 D. 2

Given:
3x + 2y – 7 = 0
2x – By + 2 = 0
Required:
Value of B
Solution:
3x + 2y – 7 = 0
3x 7
y=− 2
+2
3
By inspection, m1 = − 2

2x – By + 2 = 0
2x 2
y= B
+B
2
By inspection, m2 = B
1
Since perpendicular, m2 = − m
1

2 1
=− 3
B −
2

2 2
=
B 3

2B = (2)(3)
B=3
Therefore, the value of B to make the two equations perpendicular is 3, letter B.
Discussion:
First, we should get the slope of the first equation of the line. Then by inspection, we can also get
the slope for the second equation. Since perpendicular, the 2nd slope is equal to the negative reciprocal of
the 1st slope. Then, calculating both equations, we can now get the value of B which is equal to 3.

22. Find the slope of the equation x2 = y when x = 1.

A. 2 B. 6 C. 4 D. 1
Given:
x2 = y
x=1
Required:
Slope of the equation
Solution:
To solve for the slope, differentiate the equation in terms of y with respect to x:
y = x2
dy
= 2x
dx

When x = 1,
dy
= 2(1) = 2
dx

Therefore, the slope of the equation is 2, letter A.

Discussion:
First, we should analyze what is stated in the problem. To solve for the slope, differentiate the
equation in terms of y with respect to x. Thus, substituting x which is equal to 1, the slope of the equation
is equal to 2.

23. Find the equation of the directrix of the parabola y2 =16x.

A. x = -4 B. x = 4 C. x = -8 D. x = 8
Given: y2 =16x
Required: equation of the directrix of the parabola
Solution:
General equation,
y2 =4ax
By inspection,
 y2 =16x
4a = 16
a=4
Equation of the directrix,
x = -a
x = -4
Therefore, the equation of the directrix of the parabola y2 =16x is x = -4, letter A.

Discussion:
First, we should know the general equation of a parabola. Then, by inspection, we can now solve
for a. After that, we can now get the equation of the directrix which is x = -a. Thus, the equation of the
directrix of the given parabola is equal to x = -4.

24. What is the equation of the normal to the curve y = 3x2 – 2x + 7 at the point (1,8)?
A. x + 4y – 33 = 0 B. x + 2y + 30 = 0 C. x - 4y – 33 = 0 D. x - 2y + 30 = 0
Given:
y = 3x2 – 2x + 7
point (1,8)
Required: equation of the normal to the curve
Solution:
y = 3x2 – 2x + 7
Get the derivative,
y’ = 6x – 2
When x =1,
y’ = 4
Since y’ is equal to slope,
m=4
Solving for n,
n = -1/m
n = -1/4
Using the general equation,
(y – y1) = n (x – x1)
(y – 8) = -1/4 (x – 1)
Simplifying,
 4y – 32 = -x + 1
4y + x – 33 = 0
Therefore, the equation of the normal to the curve is 4y + x – 33 = 0, letter A.

Discussion:
First, we should analyze what is stated in the problem. We should get first the derivative of the
given curve. Then, we can now get the slope of the given curve. After that, the equation normal to the curve
is given as 4y + x -33 = 0.

25. A triangle is defined by the points (0,0), (3,1), (-3,1). Determine what type of triangle?
A. Vertical B. Equilateral C. Angular D. Isosceles

Given:
A (0,0)
B (3,1)
C (-3,1)
Required: type of triangle
Solution:
Solving the distance from A to B,
2
D = √(x2 - x1 )2 + (y2 - y1 )

D = √(3 - 0)2 + (1 - 0)2

D = √10
Solving the distance from B to C,
2
D = √(x2 - x1 )2 + (y2 - y1 )
√10 √10
D = √(-3 - 3)2 + (1 - 1)2
D=6
Solving the distance from C to A,
2
D = √(x2 - x1 )2 + (y2 - y1 ) 6

D = √(0 + 3)2 + (0 - 1)2

D = √10
Since two sides, A to B and C to A, are equal in length, the triangle is said to be isosceles because
isosceles triangle is defined as a triangle having two equal sides.

Discussion:
First, we should analyze what is stated in the problem. In order to solve the problem, we should
solve for the distances starting from A to B, then B to C and lastly C to A. Since only two sides are equal,
the triangle is defined to be as an isosceles triangle.

26. The point (5,2) is the midpoint of the segment of a line connecting A (-3,-5) and B(x,y). Find point B.
A. 13, 9 B. -2, 7 C. 5, -9 D. -1, -3
Given:
M (5,2)
A (-3,-5)
B (x,y)
Required: values point B
Solution:
General form,
x1 + x2 y1 + y2
( , ) = M (x.y)
2 2
Solving for x component,
-3 + x2
=5
2
-3 + x2 = 10
x2 = 13
Solving for y component,
-5 + y2
=2
2
-5 + y2 = 4
y2 = 9
Therefore, the values of B are (13,9), letter A.

Discussion:
First, we should analyze what is stated in the problem. In order to solve the coordinates of B, we
will use the midpoint formula. By manipulating the equation with the known values, the coordinates of B
are calculated as (13,9).
27. Determine the coordinates of the point which is three fifths of the way from the point (2, -5) to the (-3,
5).
A. (-1, 1) B. (-2, -1) C. (-1, -2) D. (1, 1)
Given:
A (2,-5)
B (-3,5)
r1 = 3/5
Required: coordinates of the point which is three fifths of the way from the point A to B
Solution:
Using the formula for x,
x1 r2 + x2 r1
x=
r2 + r1
2 3
(2) (5) + (-3) (5)
x=
2 3
(5) + (5)

x = -1 P(x,y)
Using the formula for y,
y1 r2 + y2 r1
y=
r2 + r1
2 3
(-5) (5) + (5) (5)
y=
2 3
(5) + (5)

y=1
Therefore, the coordinates of the point which is three fifths of the way from the point A to B are
(-1,1), letter A.

Discussion:
First, we should analyze what is stated in the problem. We should solve for the value of x through
the known values. After getting the value of x, we can now also get the value of y with the known values.
Thus, the coordinates of the point which is three fifths of the way from the point A to B are (-1,1).

28. What conic section is represented by x2 + 4xy + 4y2 + 2x = 10?

A. Circle B. Parabola C. Ellipse D. Hyperbola
Given: x2 + 4xy + 4y2 + 2x = 10
Required: kind of conic section
Solution:
Using the formula,
B2 – 4AC = 0
where B = 4, A = 1, and C = 4,
(4)2 – 4(1)(4) = 0
16 – 16 = 0
0=0
Since, it equals to zero, the conic section represented by x2 + 4xy + 4y2 + 2x = 10 is a parabola, letter
B.

Discussion:
First, we should analyze what is stated in the problem. We are asked to find the kind of conic
section. To find this, we will use the formula for determinants. Since it is equal to zero, the conic section
represented by x2 + 4xy + 4y2 + 2x = 10 is a parabola.

29. What conic section is represented by 4x2 – y2 + 8x + 4y = 15?

A. Parabola B. Ellipse C. Hyperbola D. Circle
Given: 4x – y + 8x + 4y = 15
2 2

Required: kind of conic section

Solution:
Using the formula,
B2 – 4AC > 0
where B = 0, A = 4, and C = -1,
(0)2 – 4(4)(-1) > 0
16 > 0
Since, it is greater than zero, the conic section represented by 4x2 – y2 + 8x + 4y = 15 is a hyperbola,
letter C.

Discussion:
First, we should analyze what is stated in the problem. We are asked to find the kind of conic
section. To find this, we will use the formula for determinants. Since it is greater than zero, the conic section
represented by 4x2 – y2 + 8x + 4y = 15 is a hyperbola.
30. What conic section is represented by x2 + y2 – 4x +2y – 20 = 0.
A. Circle B. Ellipse C. Parabola D. Hyperbola
Given: x2 + y2 – 4x +2y – 20 = 0
Required: kind of conic section
Solution:
Using the formula,
B2 – 4AC < 0
where B = 0, A = 1, and C = 1,
(0)2 – 4(1)(1) < 0
-4 < 0
Since, it is less than zero and considering that A and C are equal, the conic section represented by
x2 + y2 – 4x +2y – 20 = 0 is a circle, letter A.

Discussion:
First, we should analyze what is stated in the problem. We are asked to find the kind of conic
section. To find this, we will use the formula for determinants. Since it is less than zero, and considering
that the coefficient of A and C are equal, then the conic section represented by x2 + y2 – 4x +2y – 20 = 0 is
a circle.

31. If the distance x from the point of departure at time t is defined by the equation x = -16t2 + 5000t +
5000, what is the initial velocity?
A. 2000 B. 5000 C. 0 D. 3000
2
Given: x = -16t + 5000t + 5000
Required: initial velocity
Solution:
Getting the first derivative of distance leads to velocity,
x’ = -32t + 5000
Therefore, the initial velocity is 5000, letter B. If you consider the first term and substitute a given
time, that will be the final velocity.

Discussion:
First, we should analyze what is stated in the problem. We are asked to find the initial velocity. To
find this, we will get the first derivative of the distance to solve the velocity. Thus, the velocity is equal to
5000.
32. A man on wharf 3.6m above sea level is pulling a rope tied to a raft at 0.60 m/s. How fast is the raft
approaching the wharf when there are 6m of rope out?
A. -0.95 m/s B. -0.75 m/s C. -0.85 m/s D. -0.65 m/s
Given:
d = 3.6 m
D’(t) = 0.60 m/s
l=6m
Required: velocity of the raft approaching the wharf when there are 6m of rope out

Solution:
Let the position of the wharf be,
x1 = 0
y1 = 3.6
For the boat,
x2 = distance from the wharf to the boat
y2 = 0, since the boat is on the sea level
Solving for x2,
2
D = √(x2 - x1 )2 + (y2 - y1 )
6m

6 = √(x2 - 0 )2 + (0 - 3.6)2 3.6 m

36 = (x2 - 0 )2 2
+ (0 - 3.6)
36 = (x2 )2 + 12.96
36 = (x2 )2 + 12.96
(x2 )2 = 23.04
x2 = 4.8
Distance formula in terms of time,

D(t)= √(x2 )(t)2 + 12.96

D(t)2 = (x2 )(t)2 + 12.96
Take the derivative,
2D(t)*D’(t) = 2x2 (t)*x’(t)
Substitute the values,
(2)(6)(0.6) = 2(4.8)*x’(t)
x’(t) = 0.75 m/s
Therefore, the velocity of the raft approaching the wharf when there are 6m of rope out is 0.75 m/s,
none in the choices.

Discussion:
First, we should analyze what is stated in the problem. We are asked to find the velocity of the raft.
To find this, we will use the distance formula to solve for x 2. Then, we will also use distance formula in
terms of time. After which we will take the derivative and substitute the values in order to arrive at the
velocity. Thus, the velocity is equal to 0.75 m/s.

33. Water is pouring into a pool. After n hours, there are 2n + √n gallons in the pool. At what rate is the
water pouring in the pool when n = 4 hours?
A. 3/4 gph B. 1 gph C. 9/4 gph D. 7/4 gph
Given: v = 2n + √n
Required: rate of the water pouring in the pool when n = 4 hours
Solution:
Take the derivative,
1 1
v' = 2 + (n)- 2
2
Substituting n = 4,
1 1
v' = 2 + (4)- 2
2
9
v' = gph
4
9
Therefore, the rate of the water pouring in the pool when n = 4 hours is 4 gph, letter C.

Discussion:
First, we should analyze what is stated in the problem. We are asked to find the rate of the water
pouring in the pool when the time is 4 hours. To find this, we will find the derivative using power rule.
Thus, solving for the rate, the final answer is calculated as 9/4 gallons per hour.

34. The derivative of ln cos x is

A. sec x B. –sec x C. –tan x D. tan x
Given: ln cos x
Required: derivative of ln cos x
Solution:
d dx
(ln cos x) =
dx cos x
d 1
(ln cos x) = dx
dx cos x
Since 1/cos x = sec x:
d
(ln cos x) = sec x
dx
Therefore, the derivative of ln cos x calculated is equal to sec x. The final answer is letter A.
Discussion:
To get the derivative of a function involving ln, we need to know first what its derivative is. The
derivative form of any function involving ln is du/dx all over u. Then, after solving for the derivative of the
function, we can see that the derivative form 1/cos x can be simplified into sec x. Thus, the final answer of
sec x is computed.

35. Integrate x cos(2x2+7) dx

1 1
A. 4 sin(𝟐𝐱 𝟐 +7) + C B. sin(2x2+7) + C C. 4 cos(2x 2 +7) + C D. sin(x2+7) + C

Given: x cos(2x2+7) dx
Required: integral of x cos(2x2+7) dx
Solution:
ʃx cos(2x2+7) dx
Let u = 2x2+7
du = 4xdx
du/4 = xdx
ʃx cos(2x2+7) dx = ʃcos u (du/4)
ʃx cos(2x2+7) dx = 1/4ʃcos u du
1
ʃx cos(2x2+7) dx = 4
sin u+C
1
ʃx cos(2x2+7) dx = 4
sin(2x 2 +7) +C
1
Therefore, the integral of x cos(2x2+7) dx calculated is equal to 4 sin(𝟐𝐱 𝟐 +7) + C. The final answer is

letter A.

Discussion:
To solve the problem, let us denote and find u and du. The term u is equal to 2x2+7 and the term
du is equal to 4xdx. It can be then simplified into du/4 = x dx. Now, we have the simplified form of cos u
du wherein the constant 1/4 is being expresses outside the integral sign. Solving for the derivative form of
cos u du which is sin u du, we can now arrive at the simplified form or answer. Thus, the final answer is
1
4
sin(2x 2 +7) + C.

36. A box is to be constructed from a piece of zinc 20 sq. in by cutting equal squares from each corner and
turning up the zinc to form the side. What is the volume of the largest box that can be so constructed?
A. 599.95 cu. in B. 579.50 cu. in C. 592.59 cu. in D. 622.49 cu. in
Given: A = 20 sq. in
l = (20-2x) in
w = (20-2x) in
h = x in
Required: volume of the largest box that can be constructed
Schematic Diagram:

20-2x

x
x 20-2x
20

Solution:
V = lwh
V = (20-2x)(20-2x)(x)
V = (20-2x)2(x)
V = (400-80x+4x2)(x)
V = 4x3-80x2+400x
Getting the derivative of V:
d
= 12x 2 − 160x + 400
dx
12x2 – 160x + 400 = 0
Simplifying:
 Dividing both sides by 4:
3x2 – 40x + 100 = 0
(3x – 10) (x – 10) = 0
x = 10/3, 10
Using x = 10
V = (20-2x)2(x)
V = [20-2(10)]2(10)
V = 0 (not applicable or acceptable)
Using x = 10/3
V = (20-2x)2(x)
V = [20-2(10/3)]2(10/3)
V = 592.5925926 cu. in
Therefore, the volume of the largest box that can be constructed is equal 592.5925926 cu. in. The final
answer is letter C.

Discussion:
First, we should make a schematic diagram to see clearly and to picture out how we will solve the
problem. After knowing the known terms, we can now find the volume of the box with a simplified
equation. Then, we will get the derivative form of that equation and we will come up with two roots. Using
the first root which is 10, the volume calculated is 0 which is not applicable or acceptable. Then, by using
the second term which is 10/3, we have arrive at the volume of the largest box that can be constructed which
is equal to 592.5925926 cu. in.

37. What is the derivative of the function with respect to x of (x + 1)3 – x3?
A. 3x + 3 B. 6x – 3 C. 3x – 3 D. 6x + 3
Given: (x + 1)3 – x3
Required: derivative of the function with respect to x of (x + 1)3 – x3?
Solution:
d
[(x + 1)3 – x3]= d(x3+3x2+3x+1-x3)
dx
d
[(x + 1)3 – x3]= d(3x2+3x+1)
dx
d
[(x + 1)3 – x3]= 6x+3
dx
Therefore, derivative of the function with respect to x of (x + 1)3 – x3 calculated is equal to 6x+3. The
answer is letter D.

Discussion:
To solve for the problem, first, we should expand the given terms to its simplest form. After arriving
at the simplified form, we can now simply differentiate the given function. Thus, the derivative of the
function with respect to x of (x + 1)3 – x3 is being computed as 6x+3.

38. Find the area bounded by the parabola x2 = 16(y – 1) and its latus rectum.
A. 56.27 B. 42.67 C. 46.27 D. 52.67
Given: Parabola, x2 = 16(y-1)
Required: area bounded by the parabola x2 = 16(y – 1) and its latus rectum.
Schematic Diagram:
x2 = 16(y – 1)

LR

Solution:
x2 = 16(y – 1)
LR = 4a
LR =16
16 = 4a
Solving for a:
a=4
Solving for the bounds:
y = 1 and y = a+1
y = a+1 = 4+1 = 5
Bounds: y = 1 and y = 5
x2 = 16(y – 1)
x = 4√(y-1)
x = 4(y-1)1/2
Solving for the area bounded:
5
A = 2∫1 4(y-1)1/2dy
5
A = 8∫1 (y-1)1/2dy
A = 8 [2/3] (y-1)3/2 (bounded from 1 to 5)
A = [16/3](5-1)3/2 – [16/3](1-1)3/2
A = 42.66666667 sq. units
Therefore, area bounded by the parabola, x2 = 16(y – 1) is equal to 42.66666667 sq. units and its
latus rectum is 16. The answer is letter B.

Discussion:
First, we should graph the given parabola, x2 = 16(y – 1) and determine the parts of the graph such
as the latus rectum and the points being bounded. Then we can solve for the latus rectum which is equal to
16. Then, we can determine a which is equal to 4. After determining a, we can now find the bounds that
will be used for integrating which are 1 up to 5. Then, we can now solve the area bounded by the parabola,
x2 = 16(y – 1) which is then equal to 42.66666667 sq. units.

39. Evaluate the integral of (3t − 1)3dt

1 1 1 1
A. 12(3t − 1)4+C B. 3(3t − 1)4+C C. 4(3t − 1)4+C D. 12(3t − 1)3+C

Given: (3t − 1)3 dt

Required: integral of (3t − 1)3 dt
Solution:
ʃ(3t − 1)3 dt
Let u = 3t − 1
du = 3dt
du/3 = dt
ʃ(3t − 1)3 dt = ʃu3(du/3)
ʃ(3t − 1)3 dt = 1/3ʃu3du
1 1 4
ʃ(3t − 1)3 dt = ( )u +C
3 4
1
ʃ(3t − 1)3 dt = 12
(3t − 1)3+C
1
Therefore, the integral of (3t − 1)3dt calculated is equal to 12(3t − 1)3+C. The answer is letter D.
Discussion:
To solve the problem, let us denote and find u and du. The term u is equal to 3t-1 and the term du
is equal to 3dt. It can be then simplified into du/3 = dt. Now, we have the simplified form of un du wherein
the constant 1/3 is being expresses outside the integral sign. Solving for the derivative form of u n du, we
1
can now arrive at the simplified form or answer. Thus, the final answer is 12(3t − 1)3+C.

40. Find the area bounded by the parabola, x2 = 4y and y=4.

A. 21.33 B. 31.32 C. 33.21 D. 13.23
Given: Parabola, x2 = 4y and line, y=4
Required: area bounded by the parabola, x2 = 4y and y=4.
Schematic Diagram:
x2 = 4y

y=4

y=0

Solution:
x2 = 4y
x = 2√y
x = 2y1/2
Solving for the bounds:
y = 0 and y = 4
Solving for the area bounded:
4
A = 2∫0 2y1/2dy
4
A = 4∫0 y1/2dy
A = 4 [2/3] (y)3/2 (bounded from 0 to 4)
A = [8/3](4)3/2 – [8/3](0)3/2
A = 21.33333333 sq. units
Therefore, area bounded by area bounded by the parabola, x2 = 4y and y=4 is equal to 21.33333333
sq. units. The answer is letter A.
Discussion:
First, we should graph the given the parabola, x2 = 4y and the line y=4. Then, we can now find the
bounds that will be used for integrating which are 0 up to 4. Then, we can now solve the area bounded by
the parabola, x2 = 4y and the line y=4 which is then equal to 21.33333333 sq. units

41. How many years will P100, 000 earned a compound interest of P50,000 if the interest rate is 9%
compounded quarterly?
A. 4.55 B. 5.68 C. 3.55 D. 2.35
Given: P = P100,000
F = P+I = P100,000 + P50,000 = P150,000
i = 9% = 0.09 compounded quarterly
Required: number of years will P100,000 earned a compound interest of P50,000 at an interest rate of 9%
compounded quarterly
Solution:
i mn
F = P[(1 + m) ]

0.09 4n
P150,000 = P100,000[(1 + 4
)

Solving for n:
n = 4.555663053 years
Therefore, the number of years will P100,000 earned a compound interest of P50,000 at an interest
rate of 9% compounded quarterly is equal to 4.555663053 years. The answer is letter A.

Discussion:
First, we should determine the given values. Then, we should analyzed what formula to be used in
the problem. To find the future amount, we will add the principal with the interest which sums up to
P150,000. For the principal amount, it is given by P100,000. The interest rate is 9% to be compounded
quarterly. Solving for the number of years will P100,000 earned a compound interest of P50,000 at an
interest rate of 9% compounded quarterly, the final answer is equal to 4.555663053 years.

42. By the condition of a will, the sum of P20,000 his left to a girl to be held in trust fund by her guardian
until it amounts to P50,000 when the girl received the money, if the fund is to be invested at 8%
compounded quarterly?
A. 7.98 yrs B. 11.57 yrs C. 10.34 yrs D. 10.45 yrs
Given: P = P20,000
 F = P50,000
i = 8% = 0.08 compounded quarterly
Required: number of years when will the girl received the money
Solution:
i mn
F = P[(1 + ) ]
m

0.08 4n
P50,000 = P20,000[(1 + 4
) ]

Solving for n:
n = 11.56779247 years
Therefore, the number of years the girl will receive the money is calculated as 11.56779247 years.
The answer is letter B

Discussion:
First, we should determine the given values. The future amount is given by P50,000, the principal
amount given by P20,000 and an interest rate of 8% compounded quarterly. Then, we should analyzed what
formula to be used in the problem. To find the future amount, we will add the principal with the interest
which sums up to P150,000. For the principal amount, it is given by P100,000. The interest rate is 9% to
be compounded quarterly. Solving for the number of years the girl will receive the money, it is calculated
as 11.56779247 years.

43. A P10,000 loan is to be paid-off in 10 equal payments. The annual interest rate is 15%. How much
interest will be paid in the first two (2) years?
A. P3,005.10 B. P2,940.20 C. P2,970.20 D. P2,926.10
Given: P = P10,000
i = 15% = 0.15 annually
10 equal payments
n=2
Required: interest that will be paid in two years
Solution:
i(1+i)n
A = P[
(1+i)n −1
]
0.15(1+0.15)10
A = P10,000[ (1+0.15)10 −1 ]

A = P1,992.52
 Payment Periodic Principal
Interest Balance
Period Payment Repaid
0 --- --- --- P10,000

1 P1,992.52 P1,500.00 P492.52 P9,507.48

2 P1,992.52 P1,426.122 P566.398 P8,941.082

Interest for 2 years = P1,500 + P1,426.122

Interest for 2 years = P2,926.12
Therefore, the interest that will be paid in two years amounts to P2,926.12. The answer is letter D.

Discussion:
First, we should determine the given values. The principal amount is P10,000 having an annual
interest of 15% that will be paid in 10 equal payments. Then, we should analyzed what formula to be used
in the problem. We can now solve for the annuity/periodic payment which is P1,992.52. Then to find the
total interest for two years, a table was made to further understand and to simply calculate the total interest
for the two years which amounts to P2,926.12.

44. Find the nominal rate which if converted quarterly could be used instead of 12% compounded semi-
annually.
A. 10.58% B. 11.28% C. 9.38% D. 11.82%
Given: compounded quarterly
12% compounded semi-annually
Required: nominal rate if compounded quarterly
Solution:
NRquarterly = NRsemi-annually
i n i n
(1 + n) = (1 + n)

i 4 0.12 2
(1 + 4) = (1 + )
2

Solving for i:
i = 0.1182520564 x 100%
i = 11.82520564%
Therefore, the nominal rate if compounded quarterly is equal to 11.82520564%. The answer is letter
D.
Discussion:
The problem requires what will be the nominal rate is if it is converted quarterly instead of 12%
compounded semi-annually. By using the formula of nominal rate of interest, we can equate nominal rate
of interest for compounded quarterly, to the nominal rate of interest compounded semi-annually by 12%.
By solving the nominal rate of interest for compounded quarterly, the final answer is 11.82520564%.

45. A book store purchased a best-selling book at P200 per copy. At what price should this book be sold so
that, giving 20% discount, the profit is 30%.
A. P450 B. P357 C. P500 D. P400
Given:
P= P200
Discount= 20%
Profit= 30%
Required:
Final Price
Solution:
y(0.8) = y(0.3) + P200 → y(0.8) − y(0.3) = P200
y(0.5) = P200
y = P400
∴ 𝐓𝐡𝐞 𝐛𝐨𝐨𝐤 𝐬𝐡𝐨𝐮𝐥𝐝 𝐛𝐞 𝐬𝐨𝐥𝐝 𝐚𝐭 𝐏𝟒𝟎𝟎. The answer is letter D.

Discussion:
To solve the problem, we will make an equation involving the known values and the required value
which is the final price. Having a discount of 20% and profit of 30% yields the equation and the final price
calculated is equal to P400. Thus, the book should be sold at a price of P400.
46. What annuity is required over 12 years to equate with a future amount of P20,000? Assume i = 6%
annually.
A. P4,185.54 B. P2,185.54 C. P3,185.54 D. P1,185.54
Given:
n=12 years
i= 6%
F= P20,000
Required:
Annuity
Solution:
(1 + 𝑟)𝑛 − 1 0.06
F = P( ) → P = P20,000 ( )
𝑟 (1 + 0.06)12 − 1
𝐏 = 𝐏𝟏𝟏𝟖𝟓. 𝟓𝟒𝟎𝟔
∴ 𝐓𝐡𝐞 𝐚𝐧𝐧𝐮𝐢𝐭𝐲 𝐫𝐞𝐪𝐮𝐢𝐫𝐞𝐝 𝐢𝐬 𝐏𝟏𝟏𝟖𝟓. 𝟓𝟒. The answer is letter D.

Discussion:
First, we should determine the given values. The future amount which is given at P20,000, interest
rate at 6% and the total number of years which is 12. Solving for the annuity, the value is calculated as
P1,1185.14

47. The corporation purchased a machine for 1million. Freight installation charges amount to 3% of the
purchased price. If the machine shall be depreciated over a period of 8 years with a salvage value of 12%,
determine the depreciation charge during the fifth year using the sum of the years digit method.
A. P101, 107.11 B. P107, 110.11 C. P170, 110.11 D. P100, 711.11
Given: Cash-flow Diagram:
Useful life= 8 years
F= P1 000 000+P30 000
F = P1 030 000
SV at 12% = P 123 600
Total Depreciation= P 906 400
Required:
Depreciation charge during the fifth year
Solution:
SVD Method
8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 36
4
At the fifth year, ;
36
4
Fifth depreciation charge = (P906 400) = P100 711.11
36
𝐏 = 𝐏𝟏𝟎𝟎 𝟕𝟏𝟏. 𝟏𝟏
∴ 𝐓𝐡𝐞 𝐝𝐞𝐩𝐫𝐞𝐜𝐢𝐚𝐭𝐢𝐨𝐧 𝐜𝐡𝐚𝐫𝐠𝐞 𝐝𝐮𝐫𝐢𝐧𝐠 𝐭𝐡𝐞 𝐟𝐢𝐟𝐭𝐡 𝐲𝐞𝐚𝐫 𝐢𝐬 𝐏 𝟏𝟎𝟎 𝟕𝟏𝟏. 𝟏𝟏. The answer is letter D.
Discussion:
First, we should determine the given values. The first cost includes the initial cost plus the freight
charges. The salvage value is determined through the 12% of the first cost. Solving for the depreciation
charge during the fifth year, the value is equal to P100,711.11.

48. What is the annual rate of simple interest if $265 is earned on four months on an investment of
$15 000?
A. 5.3% B. 4.5% C. 3.5% D. 7.5%
Given:
n=4 months
R= $265
P=$15 000
Required:
Annual rate of simple interest
Solution:
$265
𝑖4 = = 0.1767
$15000
4 ∗ 3 = 𝑖12 (𝑎𝑛𝑛𝑢𝑎𝑙 𝑖𝑛𝑡𝑒𝑟𝑒𝑠𝑡)
𝑖12 = 0.053 = 5.3%
𝒊𝟏𝟐 = 𝟓. 𝟑%
∴ 𝐓𝐡𝐞 𝐚𝐧𝐧𝐮𝐚𝐥 𝐫𝐚𝐭𝐞 𝐨𝐟 𝐢𝐧𝐭𝐞𝐫𝐞𝐬𝐭 𝐢𝐬 𝟓. 𝟑%. The answer is letter A.

Discussion:
First, we should determine the given values. The principal amount is $15,000 having an interest of
$265 for a period of four months. Solving for the annual rate of interest, the calculated value is equal to
5.3%.

49. Mr. Suave borrowed P50,000 from SSS, in the form of calamity loan, with interest at 8%, compounded
quarterly, payable in equal quarterly installments for 10 years. Find the quarterly payments.
A. P4,827.80 B. P2,827.80 C. P3,827.80 D. P1,827.80
Given:
P= P 50 000
r=8%
n= 4
Required:
Quarterly payments
Solution:
r 𝑛𝑡
A = P (1 + )
n
0.08 4(10)
A = P 50 000 (1 + )
4
P110401.98
A= ;(quarterly amounts in 10 years)
4(10)
𝐀 = 𝐏 𝟐 𝟕𝟔𝟎. 𝟎𝟓
∴ 𝐓𝐡𝐞 𝐪𝐮𝐚𝐫𝐭𝐞𝐫𝐥𝐲 𝐢𝐧𝐬𝐭𝐚𝐥𝐥𝐦𝐞𝐧𝐭 𝐚𝐦𝐨𝐮𝐧𝐭 𝐩𝐚𝐲𝐚𝐛𝐥𝐞 𝐢𝐧 𝟏𝟎 𝐲𝐞𝐚𝐫𝐬 𝐢𝐬 𝐏𝟐𝟕𝟔𝟎. 𝟎𝟓.

Discussion:
First, we should determine the given values. The principal amount is P50,000 having an an interest
of 8% that will be in quarterly payments. Then, we should analyzed what formula to be used in the problem.
We can now solve for the quarterly installment amount payable in 10 years which is equal to P2760.05.

50. Alexander Michael owes P25,000 due in one year and P75,000 due in four years. He agrees to pay
P50,000 today and the balance in 2 years. How much must he pay at the end of two years if money is worth
5% compounded semi-annually?
A. another P50,000 B. only P39,026.30 C. only P25,000 D. only P30,000.00
Given: Cash-flow Diagram:
F1= P25,000 due in one year
F2= P75,000 due in four years
P= P50,000
n= 2 years
m= 2 (semi-annually)
i= 5%
Required:
Final payment at the end of two years
Solution:
0.05 2
i = (1 + ) −1
2
i = 0.050625
F = P(1 + i)2
F P25 000 P75 000
P50 000 + = +
(1 + 0.050625) (1 + 0.050625) (1 + 0.050625)
𝐅 = 𝐏𝟑𝟗 𝟎𝟐𝟏. 𝟐𝟖
∴ 𝐇𝐞 𝐦𝐮𝐬𝐭 𝐩𝐚𝐲 𝐏𝟑𝟗 𝟎𝟐𝟏. 𝟐𝟖 𝐚𝐭 𝐭𝐡𝐞 𝐞𝐧𝐝 𝐨𝐟 𝐭𝐰𝐨 𝐲𝐞𝐚𝐫𝐬. 𝐓𝐡𝐞 𝐚𝐧𝐬𝐰𝐞𝐫 𝐢𝐬 𝐁.

Discussion:
First, we should determine the given values. The future amount in 1 year which is equal to P25,000
and the future amount in 4 years which is equal to P75,000. The principal amount is P50,000 having an
interest rate of 5% compounded semi-annually. Then, we should analyzed what formula to be used in the
problem. We can now solve for the final payment at the end of two years, thus Mr. Alexander Michael must
pay P39,021.28 at the end of two years.

51. Section 22 of the Republic Act No. 8495 "The Philippine Mechanical Engineering Act of 1988 " is
about?
A. oath
B. Issuance of Certificates of Registration and Professional License
C. Re-examination
D. Seal of a Professional Mechanical Engineer
Discussion:
The issuance of Certificates of Registration and Professional License is written in Section 22 of the
Republic Act No. 8495.

52. The new Mechanical Engineering Law was signed last __________.
A. February 12, 1998 C. January 18, 1998
B. February 6, 1998 D. February 12, 1998
Discussion:
As stated in the Mechanical Engineering Laws, the new Mechanical Engineering Law was signed
last February 12, 1998.

53. 100 kw or over but not more than 300 kw, the personnel required in one CPM or one RME or one PME.
Provided that every mechanical work in this category operating on more than __________ shift every 24-
hours
A. Three B. two C. four D. one
Discussion:
Only one shift every 24 hours is allowed in a plant with 100KW or over but not more than
300KW, where one CPM or RME or one PME is required.
54. The new Mechanical Engineering Law has how many articles?
A. Two B. three C. five D. six
Discussion:
The new Mechanical Engineering Law has five articles namely (Article I Title, Statement of
Policy, And Definition Of Terms, Article II Board Of Mechanical Engineering, Article III Examination,
Registration And License, Article IV Practice Of The Profession, Article V Penal And Concluding
Provisions).

55. What are the contents in Article II?

A. Board of Mechanical Engineering C. Term of Office
B. Statement Policy D. Examination, Registration and License
Discussion:
Article II of the Mechanical Engineering Law includes the contents regarding the Board of
Mechanical Engineering.

56. What are the contents in Article III?

A. Board of Mechanical Engineering C. Examination, Registration and
License
B. Practice of the Profession D. Term of Office
Discussion:
Article III of the Republic Act No. 8495 is about Examination, Registration and License. It is
composed of 21 sections which are from section 12 up to section 32.

57. One of the qualification of applicants for Professional Mechanical Engineer must be: He has specific
record of a total of __________ or more than 1 year of active mechanical engineering practice.
A. Two B. Three C. four D. five
Discussion:
A total of four years or more of active mechanical engineering practice is required in order to be
qualified as an applicant for Professional Mechanical Engineer as indicated in Article III, Section 14 of
R.A. 8495.
58. One of the qualification of applicants for Professional Mechanical Engineer must be: He is competent
to practice, attested to by at least __________ Professional Mechanical Engineers.
A. One B. Two C. three D. many
Discussion:
According to Article III, Section 14, to be qualified on the application for Professional Mechanical
Engineer, he must be competent to practice, as attested by at least two PME’s.

59. Section 34 in the Mechanical Engineering Law is about:

A. Preparation of Plans C. Field of Action
B. Personnel required in Mechanical Plant D. Renewal of License
Discussion:
Section 34, which is indicated on Article IV of R.A. 8495, is about the Personnel Required in
Mechanical Plant.

60. Section 24 in the Mechanical Engineering Law is about:

A. Seal of Professional Mechanical Engineer C. Integration and Accreditation of ME
B. Revocation and Suspension of Certificate D. Grounds for Suspension
Discussion:
Section 24, indicated on Article III of the Mechanical Engineering Law, is about the Seal of
Professional Mechanical Engineer. It says that a PME shall obtain a seal of such design prescribed by the
board which is upon registration.

61. A rubber ball is thrown from a building at an angle of 60 deg with the horizontal at an initial velocity
of 38 m/s. After hitting level ground at the base of the building, it has a covered a total distance of 200
m. How tall is the building in meters?
A. 183 m B. 197 m C. 383.68 m D. 483.68 m
Given:
θ = 60 degree
Vo = 38 m/s
x = 200 m

Required:
Height of the building, y
Solution:
 y = x tan θ – [gx2 / (2Vo2 cos2 θ)]
y = 200 tan 60 – {9.81(200)2 / [2(38)2 cos2 60]}
y = 197.0801 m
Therefore, the height of the building is B, 197 m.

Discussion:
First, we should determine the given values. To find the height of the building, we will use the
formula stated above. Substituting all the known values to the formula, the height of the building is
calculated as 197 meters.

62. A highway curve has a super elevation of 10 deg. What is the radius in feet of the curve such that there
will be no lateral pressure between the tires and the roadway at a speed of 60 mph?
A. 1363.92 ft B. 2363.92 ft C. 3363.92 ft D. 4363.93 ft
Given:
θ = 10 degree
V = 60mph
Required:
Radius of the curve
Solution:
tan θ = V2 / gr; r = V2 / g (tan θ)
r = [(60mi/hr) (1609m/mi) (3.281ft/m) (1hr/3600s)]2 / (32.2ft/s2) (tan 10)
r = 1363.5757 ft
Therefore, the radius of the curve is A, 1363.92 ft.

Discussion:
First, we should determine the given values. The bank angle is 10 degrees and has a velocity of 60
miles per hour. Converting 60 miles per hour to feet per second for consistency of units. To solve for
the radius of the curve, r is equal to the squared of velocity all over gravity time tangent theta. Thus,
the radius of the curve is 1,363.92 feet.

63. A ball is thrown upward with an initial velocity of 20 m/s. How high does it go?
A. 10.38 m B. 20.38 m C. 30.38 m D. 40.38 m
Given:
V = 20m/s
 Required:
Maximum height
Solution:
V = Vo – at; V = 0 at maximum height
0 = 20m/s – (9.81m/s2) t
t = 2.04 sec
at t = 2.04s
y = 20m/s (2.04s) – ½ (9.81m/s2) (2.04s)2
y = 20.3874 m
Therefore, the maximum height it reached is B, 20.38 m.

Discussion:
First, we should determine the given values. The velocity is equal to 20 meter per second. To solve
for the maximum height, it is equal to the velocity times the time minus the resulting value of half of
the gravity multiplied by the time being squared. Thus, the maximum height reached is equal to 20.38
meters.

64. A highway curve has a super elevation of 7 deg. What is the radius of the curve such that there will be
no lateral pressure between the tires and roadway at a speed of 40 mph?
A. 265.71 m B. 345.34 m C. 438.34 m D. 330.78 m

Given:
θ = 7 degree
V = 40mph
Required:
Radius of the curve
Solution:
tan θ = V2 / gr; r = V2 / g (tan θ)
r = [(40mi/hr) (1609m/mi) (1hr/3600s)]2 / (9.81m/s2) (tan 7)
r = 265.3471 m
Therefore, the radius of the curve is A, 265.71 m.
 Discussion:
First, we should determine the given values. The bank angle is equal to 7 degrees and the velocity
is equal to 40 miles per hour. Then, we should convert the velocity in terms of feet per second for
consistency of units. To solve for the radius of the curve, r is equal to the squared of velocity all over
gravity time tangent theta. Thus, radius of the curve is calculated as 265.71 meters.

65. A marksman fires a rifle horizontally at a target. How much does the bullet dropped in flight if the
target is 150m away and the bullet has a muzzle velocity of 500 m/s?
A. 0.34 m B. 0.64 m C. 0.44 m D. 0.54 m
Given:
Distance of the bullet = 150 m
Velocity = 500 m/s
Required:
Sdropped = the bullet dropped in flight
Solution:
S = Vt; t = S/V
t = 150m / 500m/s
t = 0.3 sec
S = V(t)+(1/2)(g)(t)2
S = (500 m/s)(0.3s)+(1/2)(9.81 m/s2)(0.3s)2
S = 150.44 m
Sdropped = 150.44 m -150 m
Sdropped = 0.44 m
Therefore, the bullet dropped at 0.44 m.

Discussion:
First, we should determine the given values. The distance of the bullet is equal to 150 meters while
the velocity is equal to 500 meter per second. Through this, we can solve for the time which is at 0.3
seconds. Then, we can now solved for the maximum distance which is calculated as 150.44 meters. To
find the distance the bullet dropped, we will subtract the maximum height of 150.44 meters by 150
meters which results to a total of 0.44 meters.

66. A rubber ball was dropped from a height of 36 m, and each time it strikes the ground it rebounds to a
height of 2/3 of the distance from which it fell. Find the total distance travelled by the ball before if
comes to rest.
 A. 72 m B. 150 m C. 120 m D. 180 m
Given:
Height = 36 m
Rebounds at 2/3 of the height
Required:
d = total distance travelled by the ball before it comes to rest
Solution:
The total distance travelled by the ball is:
d = 36down + 36(2/3) up + 36(2/3) down + 36(2/3)2up + 36(2/3)2down +36(2/3)3up + …...
= 36 + 2[36(2/3)] + 2[36(2/3)2] + 2[36(2/3)3] + …...
= 36 + [72(2/3)] / (1 - 2/3) Excluding the 1st term, finding the sum of the series
= 36 + 144
d = 180 m
Therefore, the total distance travelled by the ball is D, 180m.

Discussion:
First, we should determine the given values. The height which is equal to 36 meters and rebounds
at 2/3 of the height. Then, substituting the known values to the formula, we can now determine the total
distance travelled by the ball which yields 180 meters.

67. If a train passed as many telegraph poles in one minute as it goes miles per hour, how far apart are the
poles?
A. 18 feet B. 88 feet C. 38 feet D. 96 feet
Given: t = 1 min
v = 1 mi/hr
Required: distance (s)
Solution:
t = 1 min (1 hr/60 mins) = 1/60 hr
v = 1 mi/hr (5280 ft/1 mi) = 5280 ft/hr
s = vt = (5280 ft/hr) (1/60 hr)
s = 88 ft
Thus, the distance between the poles is 88 ft. Letter B.
Discussion:
First, we should determine the given values. The time is 1 minute and the velocity is 1 miles per
hour. We should convert the velocity in terms of feet per second for consistency of units. Thus, solving for
the distance of between the poles yields 88 feet.

68. A truck travels at 110 kph for 2 hours at 70 kph for the next 2 hours, and finally at 90 kph for 1 hour.
What is the car’s average velocity for the entire journey?
A. 40 kph B. 60 kph C. 80 kph D. 90 kph
Given: v1 = 110 kph t1 = 2 hr
v2 = 70 kph t2 = 2 hr
v3 = 90 kph t3 = 1 hr
Required: Average Velocity (vave)
Solution:
s1 = v1 t1 = (110 km/hr) (2 hr) = 220 km
s2 = v2 t2 = (70 km/hr) (2 hr) = 140 km
s3 = v3 t3 = (90 km/hr) (1 hr) = 90 km
Solving for vave:
velocity = distance/time
vave = stotal/ttotal = (220 km + 140 km + 90 km) / (2 hr + 2 hr + 1 hr)
vave = 90 km/hr
Thus, the average velocity is 90 kph. Letter D.

Discussion:
First, we should determine the given values. Then, we will simply use the formula of S=vt in order
to obtain the three distances. Then, to solve for the average velocity, it is the total distance divided by the
total time. Thus, the average velocity is 90 kilometers per hour.

69. A car travels from 20 to 50 kph in 2 sec? At the same acceleration, how long will it take the car to go
from 20 to 45 kph?
A. 1.66 sec B. 1.88 sec C. 2.22 sec D. 3.55 sec
Given: @ Point 1 @ Point 2
vO1 = 20 kph vO2 = 20 kph
vF1 = 50 kph vF2 = 45 kph
t1 = 2 sec
 a1 = a 2
Required: time @ point 2 (t2)
Solution:
Solving for a1;
vO1 = 20 km/hr (1 hr/60 min)(1 min/60s)(1000 m/1 km) = 5.56 m/s
vF1 = 50 km/hr (1 hr/60 min)(1 min/60s)(1000 m/1 km) = 13.89 m/s
vF1 = vO1 + a1 t1
a1 = (vF1 - vO1) / t1
a1 = (13.89 m/s – 5.56 m/s) / 2 s
a1 = 4.165 m/s2
Solving for t2:
vO2 = 20 km/hr (1 hr/60 min)(1 min/60s)(1000 m/1 km) = 5.56 m/s
vF2 = 45 km/hr (1 hr/60 min)(1 min/60s)(1000 m/1 km) = 12.5 m/s
vF2 = vO2 + a2 t2
t2 = (vF2 - vO2) / a2
t2 = (12.5 m/s – 5.56 m/s) / 4.165 m/s2
t2 = 1.67 s
Thus, the car takes 1.67 s to go from 20 kph to 45 kph. Letter A

Discussion:
First, we should determine the given values. Then, we can now use the formulas necessary to solve
the problem. To solve for the acceleration, the formula is given by final velocity minus initial velocity all
over the time. After obtaining, we can now get the value of the time. Thus, the car takes 1.67 seconds to go
from 20 kilometer per hour to 45 kilometer per hour.

70. The brakes of a certain truck can produce an acceleration of 5 m/s2. How far does the truck travel during
the time the brakes applied from a velocity of 40 m/s?
A. 220 m B. 120 m C. 90 m D. 50 m
Given: a = 5 m/s2
vO = 40 m/s
vF = 0
Required: distance (s)
Solution:
Solving for the distance (s):
vF2 = vO2 - 2as
s = (vF2 - vO2) / 2a = ((0)2 – (40)2) / -2 (5)
s = 160 m
Thus, the distance the truck travels during the time the brakes are applied is 160 m.

Discussion:
First, we should determine the given values. Then, by using the necessary formulas to solve the
problem, the distance can be calculated by the difference of the squared final velocity and the squared initial
velocity all over 2 times the acceleration. Thus, the distance the truck travels during the time the brakes are
applied is 160 m.

71. A PLDT tower a monument stand on a level plane. The angles of depression of the top and bottom of
the monument viewed from the top of the PLDT tower are 13° and 35°, respectively. The height of the
tower is 50 meters. Find the height of the monument.
A. 16.5 m B. 71.5 m C. 33.5 m D. 95.5 m
Given: Hbuilding= 50 m ɵBottom monument = 35°
ɵTop monument = 13°
Required: Height of the monument (h)
Solution:
Using Similar Triangles;

For triangle 1;
tan ɵ = (50 – h) / x
x = (50 – h) / tan ɵ
x = (50 – h) / tan 13° Equation 1

For triangle 2;
tan ɵ = 50 / x
x = 50 / tan ɵ
x = 50 / tan 35° Equation 2

Substitute Equation 2 to Equation 1

50 / tan 35° = (50 – h) / tan 13°
(50 tan13°) / tan 35° = 50 – h
h = 50 - (50 tan13°) / tan 35°
h = 33.5 m
Thus, the height of the monument is 33.5 m. Letter C

Discussion:
First, we should determine the given values. Then, by using the necessary formulas to solve the
problem, through similar triangles, we can determine the height of the monument which is at 33.5 meters.

72. A man standing on a 48.5 m building high, has an eyesight height of 1.5 m from the top of the building
and nearest wall, which are 50 deg and 80 deg respectively. Fin the height of another building in meters.
The man is standing at the edge of the building and horizontal plane.
A. 59.5 m B. 49.5 m C. 69.5 m D. 39.5 m
Given: Hbuilding = 48.5 m Hhuman = 1.5 m
ɵTop wall = 50° ɵBottom horizontal plane= 80°
Required: height of another building (h)
Solution:
Using Similar Triangles;
For triangle 1
tan ɵ = (50 – h) / x
x = (50 – h) / tan
x = (50 – h) / tan 50° Equation 1

For triangle 2
tan ɵ = 50 / x
x = 50 / tan ɵ
x = 50 / tan 80° Equation 2

Subtitute Equation 2 to Equation 1

50 / tan 80° = (50 – h) / tan 50°
50 tan 50° / tan 80° = 50 – h
h = 50 – (50 tan 50° / tan 80°)
h = 39.5 m
Thus, the height of another building is 39.5 m. Letter D.

Discussion:
First, we should determine the given values. Then, by using the necessary formulas to solve the
problem, through similar triangles, the height of the another building is measured at 39.5 meters.

73. From an oblique triangle in which side a = 6 cm, side b = 9 cm and angle C = 32 degree. Solve for angle
A.
A. 19.1 deg B. 39.1 deg C. 29.1 deg D. 109 deg
Given: a = 6 cm angle C = 32 deg
b = 9 cm
Required: angle A
Solution:
Using Cosine Law to solve for side c;
c2 = a2 + b2 – 2abcos(C)
c = √(6)2 + (9)2 − 2(6)(9)cos (32)
c = 5.0409 cm

Using Cosine Law to solve for angle A;

a2 = b2 + c2 – 2bccos(A)
cos A= (b2 + c2 – a2) / 2bc
cos A= ((9)2 + (5.0409)2 – (6)2) / 2 (9) (5.0409)
A = 39.1 deg.
Thus, angle A is 39.1 degree. Letter B

Discussion:
First, we should determine the given values. Then, by using the necessary formulas to solve the
problem, through using the Cosine Law, the angle A is determined and has a value of 39.1 degrees.
74. The Philippine Long Distance Telephone Company has 183 m straight vertical tower. There are two
cables in the same vertical plane anchored at two points on a level with the base of the tower. The angles
made by the cables with the horizontal are 44 deg and 58 deg respectively. Find the lengths of the cables.
A. 236 m and 251 m B. 263 m and 215 m C. 245 m and 263 m D. 289 m and 315 m
Given: h = 183 cm ɵ = 58 deg.
ɑ = 44 deg.
Required: Length of cables (x and y)
Solution:
sin ɑ = 183cm / x
x = 183cm/ sin 44
x = 263.44 cm

sin ɵ = 183cm / y
y = 183cm/sin 58
y = 215.79 cm
Thus, the length of the cables are 263 cm and 215 cm. Letter B

Discussion:
First, we should determine the given values. Then, we can have a schematic diagram in order to
fully understand the problem. Then, by using the necessary formulas to solve the problem, the values of x
and y can be determined through the use of the triangles as shown above. Thus, the length of the cables are
263 centimeters and 215 centimeters.

75. The value of Tan (A + B), where tan A = 1/3 and tan B = 1⁄4 is:
A. 7/12 B. 7/11 C. 1/11 D. 7/13
Given: tan A = 1/3
tan B = 1/4
Required: tan (A+B)
Solution:
tan (A+B) = (tan A + tan B) / 1 – tan A tan B
tan (A+B) = (1/3 + 1/4) / (1 – (1/3)(1/4))
tan (A + B) = 7/11
Thus, the value of tan (A+ B) is 7/11. Letter B
Discussion:
First, we should determine the given values. Then, by solving the trigonometric identities for
tan(A+B) = (tan A + tan B) / 1 – tan A tan B), the value of the given trigonometric function is equal to 7/11.

76. If loga 10 = 0.250, log10 a equals:

A. 4 B. 2 C. 0.50 D. 10,000
Given: loga 10 = 0.250
Required: log10 a
Solution:
For loga 10 = 0.250
log 10 / log a = 0.250/1
Using proportion:
log a/log 10 = 1/ 0.250
log a/log 10 = 4
log10 a = 4
Thus, log10 a is equal to 4. Letter A

Discussion:
First, we should determine the given values. Then, to solve for a, we will simplify the logarithmic
functions. Thus, through simplifying, we have arrived at the final answer which is equal to 4.

77. If sin A=3/4 and A is in quadrant II while cos B=7/25 and B is in quadrant 1, find sin (A+B).
A. -3/5 B. 3/5 C. 4/5 D. 3⁄4
Given: sin A = ¾
sin B = 7/25
Required: sin (A+B)
Solution:
Solving for cos A;
sin A = 3 /4
Using Pythagorean theorem;
c 2 = a 2 + b2
a2 = c2 - b2
a 2 = 4 2 – 32
a = √42 − 32
a = √7
√7
cos A = 4

Solving for sin B;

cos B = 7 / 25
Using Pythagorean theorem;
c 2 = a 2 + b2
b2 = c 2 - a 2
b2 = 252 – 72
b = √252 − 72
b = 24
sin B = 24/25

sin (A+B) = sinA cosB + cosA sinB

3 7 √7 24
sin (A+B) = (4) (25) + ( 4 ) (25)

sin (A+B) = 0.8

sin (A+B) = 4/5
Thus, sin (A+B) is 4/5. Letter C

Discussion:
First, we should determine the given values. Then, by using the pythagorean theorem, we can
determine the values of A and B. For the trigometric identities of sin(A+B), it is equal to sinA cosB + cosA
sinB. Thus, the final value is equal to 4/5.

78. The hypotenuse of a right triangle is 34 cm. Find the length of two legs, if one leg is 14cm longer than
the other.
A. 15 and 29 B. 18 and 32 C. 16 and 30 D. 17 and 31
Given: h = 34 cm = c
a = 14 + b
b
Required: length of two legs.
Schematic Diagram:
34
b

14 + b

Solution:
𝑎2 +𝑏2 = 𝑐 2
(14 + 𝑏)2 +𝑏2 = 342
𝑏 = 16 𝑐𝑚
𝑎 = 30 𝑐𝑚
Final Answer: 16 cm and 30 cm

Discussion:
First, we should determine the given values. Then, we can now make equations to find the unknown
value. We can find the two legs of the triangle through the use of the Pythagorean theorem. Thus, the two
legs of the triangle are equal to 16 centimeters and 30 centimeters.

79. The identity sec2 A – sec2 A sin2 A is equal to:

A. sin A B. sec A C. zero D. one
Given: sec 2 A – sec 2 A sin2 A
Required: identity
Solution:
sec 2 A – sec 2 A sin2 A
sec 2 A (1 − sin2 𝐴)
sec 2 A (cos2 𝐴)
1
cos2 𝐴
(cos 2 𝐴)

sec 2 A – sec 2 A sin2 A = 𝟏

Discussion:
First, we should determine the given values. Then, by solving the trigonometric identities given
and simplifying all the terms, the simplied trigonometric form is cos2A/cos2A which is then equal to the
value of 1.
80. Simplify the identity: (1+tan 2A)/ (1+cot2 A).
A. 1 B. tan2 A C. sin A D. tan A
Given: (1+tan2A)/ (1+cot2A).
Required: identity.
Solution:
1+tan2 𝐴 1+tan2 𝐴 1+tan2 𝐴
= 1 = tan2 𝐴 + 1
1+cot2 𝐴 1+ 2
tan 𝐴 tan2 𝐴

(1+tan2 𝐴) (tan2 𝐴)
= = 𝐭𝐚𝐧𝟐 𝑨
tan2 𝐴+1

Discussion:
First, we should determine the given values. Then, by solving the trigonometric identities for
(1+tan A)/ (1+cot2A), the value of the given trigonometric function is equal to tan2A.
2

81. A person draws 3 balls in succession from a box containing 5 red balls, 6 yellow balls and 7 green
balls. Find the probability of drawing the balls in the order red, yellow and green.
A. 0.3894 B. 0.0894 C. 0.03489 D. 0.04289
Given: 5 Red Balls
6 Yellow Balls
7 Green Balls
Required: probability of drawing the balls in the order red, yellow and green - P(RYG)
Schematic Diagram:
5R
6Y
7G
Solution:
P(RYG) = P(R) * P(Y|R) * P(G|Y|R)
5 6 7
P(RYG) = 18 (17) (16)
35
P(RYG) = 816

P(RYG) = 0.04289215686

Discussion:
To find probability of drawing the balls in the order red, yellow and green, we will multiply the
probability of red being drawn to the probability of yellow to be drawn next and lastly to the probability
of green to be drawn last.
82. If five coins are tossed simultaneously find the probability that they will just have two head.
A. 5/32 B. 5/16 C. 7/32 D. 1/16
Given: 5 coins.
Required: probability that they will just have two head – P(2H)

Solution:
Let p = probability of getting head
q = probability of getting tail
1
p=q=2

n=5
r=2
P(2H) = (𝑛𝐶𝑟) 𝑝𝑟 𝑞𝑛−𝑟
1 2 1 5−2
P(2H) = (5𝐶2) (2) (2)
𝟓
P(2H) =
𝟏𝟔

Discussion:
To find probability of getting 2 heads, we can use the formula presented above. Thus, the
probability of getting two heads is equal to 5/16.

83. If five coin are tossed simultaneously, find the probability of getting at least two tails.
A. ¼ B. 3/16 C. 4/5 D. 6/31
Given: 5 coins.
Required: probability of getting at least two tails – P(2T)
Solution:
P(2T) = 5C2 (1/2)2 (1/2)3 = 5/16
P(3T) = 5C3 (1/2)3 (1/2)2 = 5/16
P(4T) = 5C4 (1/2)4 (1/2)1 = 5/32
P(5T) = 5C5 (1/2)5 (1/2)0 = 1/32
To find the total probability of getting at least two tails:
PTotal = P(2T) + P(3T) + P(4T) + P(5T)
PTotal = 5/16 + 5/16 + 5/32 + 1/32
PTotal = 13/16
Discussion:
To find probability of getting at least two tails, we can use the formula presented above. We will
solve for the probability of getting two tails, three tails, four tails and lastly five tails. Thus, the probability
of getting at least two tails is equal to 13/16.

84. Two cards are drawn at random from an ordinary deck of 52 cards. Find the probability P that one is a
spade and one is a heart.
A. 3/51 B. 1/26 C. 13/102 D. 2/52
Given: 52 ordinary deck of cards.
Required: probability that one is a spade and one is a heart – P(S&H)
Solution:
P (Spade and Heart) = P(S) * P(H|S) + P(H) * P(S|H)
13 13 13 13
P (Spade and Heart) = ( )+ ( )
52 51 52 51
13 13
P (Spade and Heart) = 2 (52) (51)
𝟏𝟑
P (Spade and Heart) = 𝟏𝟎𝟐

Discussion:
To find probability of drawing a card that one is a spade and the other is a heart can be calculated
as presented above. Since there are 13 each of the spade and hearts and the total number of cards in a deck
is 52, it can be calculated through picking first either the spade or heart. The calculation will be 13/52
multiplied by 13/51. 13/52 since the possibility of drawing either a spade or heart is 13/52. It will then be
multiplied to 13/51 since one card is already drawn and the possibility of getting either a spade or heart is
at 13/51. Since there will be two possible scenarios, the resulting value of 13/52 multiplied by 13/51 will
be multiplied to 2 to get the total probability which is 13/102.

85. The arithmetic mean of 6 numbers is 17. If two numbers are added to the progression, the new set of
numbers will have an arithmetic mean of 19. What are the two numbers if their difference is 4?
A. 21, 29 B. 23, 27 C. 24, 26 D. 22, 28
Given: A.M = 17 A.M = 19
n=6 n=8
d=4
Required: two numbers if their difference is 4.
Solution:
Sn = A.M (n)
𝑆6 = 17(6) 𝑆8 = 19(8)
𝑆6 = 102 𝑆8 = 152
Let the two numbers be n and n + 4.
The sum of two numbers:
n + n + 4 = 𝑆8 − 𝑆6
2n + 4 = 152 – 102
2𝑛+4 50
=
2 2

n + 2 = 25
n = 23
n + 4 = 27
Final Answer: The two numbers are 23 and 27.
Discussion:
To find the two numbers having a difference of 4, we can use the formula presented above, Thus,
by calculating all the necessary equations, the resulting two numbers are 24 and 27.

86. From a bag containing 6 red balls, 8 white balls and 10 blue balls, one ball is drawn at random.
Determine the probability that is red or white
A. 7/12 B. ¾ C. ¼ D. 1/3
Given: 6 Red Balls
8 White Balls
10 Blue Balls
Required: probability that is red or white – P (R or W)
Solution:
P (R or W) = P(R) + P(W)
6 8
P (R or W) = +
24 24
𝟕
P (R or W) = 𝟏𝟐

Discussion:
To determine the probability of getting either or white ball from a bag containing 6 red balls, 8
white balls and 10 blue balls, first we will sum up the total amount of balls in the bag. Then, to find the
probability, we will add the probability of getting either red or white. Thus, the resulting probability is equal
to 7/12.

87. From a bag containing 4 black balls and 5 white balls, two balls are drawn one at a time. Find the
probability that one ball is white and one ball is black. Assume that the first ball is returned before the
second ball is drawn.
A. 40/81 B. 2/9 C. 20/81 D. 1/9
Given: 4 Black Balls
5 White Balls
Required: probability that one ball is white and one ball is black – P (B, W)
Schematic Diagram:

4B
5W

Solution:
* Since the first ball is returned before the second ball is drawn,
the total number of balls in the bag each draw is constant *
P (B, W) = P (5) * P(W|B) + P(W) * P(B|W)
P (B, W) = 2 P(B) P(W|B)
4 5
P (B, W) = 2 ( 𝑋 )
9 9
𝟒𝟎
P (B, W) =
𝟖𝟏

Discussion:
To determine the probability of getting either or white ball from a bag containing 6 red balls, 8
white balls and 10 blue balls, first we will sum up the total amount of balls in the bag. Then, to find the
probability, we will add the probability of getting either red or white. Thus, the resulting probability is equal
to 7/12.

88. A fair coin is tossed 5 times. What is the probability of getting at least one tail.
A. 0.64 B. 0.97 C. 0.58 D. 0.86
Given:
5 Times- coin was tossed
Required:
Probability of getting at least one tail
Solution:
Since the sum of all the probabilities is equal to 1,
P (at least one tail) = 1 – P (no tail)
P (no tail) = P (all heads)
Let H= heads
P (no tail) = P (all heads) = P (H,H,H,H,H)
For every toss, the probability of getting a tail or a head is ½. Since, there are 5 tosses,
P (H,H,H,H,H) = (½)(½)(½)(½)(½) = (½)5 = 1/ 32
P (at least one tail) = 1 – P (H,H,H,H,H)
P (at least one tail) = 1 - 1/ 32
P (at least one tail) = 31/ 32 = 0.96875 = 0.97
P (at least one tail) = 0.97
Thus, the probability of getting at least one tail is 0.97, Final Answer: B.

Discussion:
The problem requires the probability of getting at least one tail from tossing 5 times. “At least one
tail” means that there can be one, two or three or four or five tails. Having five heads is the only option
which is not included. Since the sum of all probabilities is equal to 1, the probability of getting at least one
tail is equal to 1 minus the probability of all heads. For every toss, the probability of getting a head is ½.
Since, there are 5 tosses, five times the probability of getting a head is equal to 1/ 32. This answer will be
subtracted from 1 because the problem only requires the probability of getting at least one tail. Therefore,
1- 31/32 is equal to 31/32 which is equivalent to 0.97.

89. From a bag containing 4 black balls and 5 white balls, two balls are drawn one at a time. Find the
probability that both balls are black. Assume that the 1st ball is returned before the second ball is drawn.
A. 4/9 B. 16/81 C. 1/6 D. 5/9
Given:
4 black balls
5 white balls
Required:
Probability that both balls, which are drawn, are black
Solution:
Let PB = Probability of having both black balls
Let PB1 = probability of having black ball in the first draw
Let PB2 = probability of having black ball in the second draw
There are a total of 9 balls
For PB1,
PB1 = 4/ 9
For PB2, the first ball was returned before the second ball is drawn
PB2 = 4/ 9
Therefore, PB = PB1 x PB2
PB = (4/ 9)( 4/ 9)
PB = 16/81
Thus, the probability that both balls are black is 16/81, Final Answer: B.

Discussion:
The bag contains a total of 9 balls in which 4 balls are black and 5 balls are white. The problem
requires the probability of having both black balls. There are 2 draws and before the second draw, the first
ball is returned. Therefore, in the first the probability of having a black ball is 4/ 9. In the draw, the
probability of having black ball again is 4/9 since the first ball is returned. 4/9 multiply by 4/ 9, the answer
is 16/81.

90. Find the probability that a couple having 3 children will have at least one girl.
A. 7/8 B. ½ C. ¾ D. 5/8
Given:
3 children
Required:
Probability that a couple having 3 children will have at least one girl
Solution:
Since the sum of all the probabilities is equal to 1,
P (at least one girl) = 1 – P (no girl)
P (no girl) = P (all boys)
Let B= boys
P (no girl) = P (all boys) = P (B,B,B)
The probability of having a boy is ½. Since, there are 3 children,
P (B,B,B) = (½)(½)(½)= (½)3 = 1/ 8
P (at least one girl) = 1 – P (B,B,B)
P (at least one girl) = 1 – 1/8
P (at least one girl) = 7/8
Thus, the probability that a couple having 3 children will have at least one girl is 7/8, Final Answer:
A

Discussion:
The problem states that a couple has 3 children and requires the probability of having at least one
girl. “At least one girl” means that there can be one, two or three. Since the sum of all the probabilities is
equal to 1, the probability of having at least one girl is equal to 1 minus the probability of having no girl at
all. No girl is equal to all boys. The probability of having a boy is ½. Since, there are 3 children, the
probability is 1/8. 1- 1/8 is equal to 7/8. Therefore, the probability that a couple having 3 children will have
at least one girl is 7/8.

91. What is the equivalent resistance, in ohms, of the circuit where there are 1 of 20 ohms and 1 of 10 ohms
connected in series to the circuit?
A. 55.5 ohms B.33.3 ohms C. 50.5 ohms D. 60.0 ohms
Given:
1 of 20 ohms
1 of 10 ohms
Series Circuit
Required:
Equivalent Resistance in Ohms
Schematic Diagram:

Solution:
Since the resistors are connected in series,
Req = R1 + R2
Substituting the values,
Req = 20 Ω + 10 Ω
Req = 30 Ω
Thus, the equivalent resistance is 30 Ω, Final Answer: none of the choices.

Discussion:
The resistors are connected in series in the problem. Therefore, equivalent resistance is equal to the
sum of all the values of the resistors. The equivalent resistance in the problem is 30 Ω.

92. A solid copper wire conductor at 50°C has the following characteristics. Resistivity of 1.77x10-8 ohm-
meter, wire diameter of 0.4 inch and wire total length of 10000 m. What is the resistance of the wire
conductor?
A. 2.18 ohm B. 5.5 ohm C. 3.25 ohm D. 6.55 ohm
Given:
Temperature of the wire = 50°C
Resistivity = 1.77 x 10-8 ohm-meter
Wire Diameter = 0.4 inch
Wire Total Length = 10, 000 m
Required:
Resistance of the Wire Conductor
Schematic Diagram:

Solution:
Given the resistivity, wire diameter and wire length, the formula to use is,
R = ρ (L/ A)
Where ρ – resistivity in ohm -meter
L – length in meter
A = area in meter2
R =Resistance in ohms
Substituting the given in the formula,
R = ρ (L/ A)
A = π/4 (d2)
2.54 cm 1m 2
π (0.4 inch x x )
1 inch 100 cm
A=
4
2
π (0.01016 m)
A=
4

A = 8.1073 m2
10,000 m
R = 1.77 x 10-8 Ω – m ( )
8.1073 m2

R = 2.183212298 Ω = 2.18 Ω
R = 2.18 Ω
Thus, the resistance of the wire conductor is 2.18 Ω, Final Answer: A.

Discussion:
The temperature given has nothing to do to get the resistance of the circuit since the resistivity for
that specific temperature of the wire is already given. To get the resistance, the resistivity must be multiplied
with length which is divided by the area. The area which is being pertained is the area of a circle. The units
must all be in SI units to be canceled. Substituting all the given values and the area, the answer is 2.18 Ω
for the resistance of the wire conductor.

93. An electric motor has a trade man. Label indicating 2 hp, 240 V, 15 amp. Calculate the motor power
factor. Assume motor efficiency to be 85%.
A. 0.987 B. 0.657 C. 0.867 D. 0.487
Given:
Power = 2 HP
Voltage = 240 V
Current = 15 A
Motor Efficiency = 85 %
Required:
Motor Power Factor
Solution
To get the motor power factor,
Power Factor = Real Power/ Apparent Power
0.746 KW
Real Power = 2 HP x 1 HP
= 1.492 KW
Since the problem provide motor efficiency,
Real Power = 1.492 KW/ 0.85
Real Power = 1.755294118 KW
To solve for the apparent power,
Apparent Power = Voltage x Current
Apparent Power = 240 V (15 A)
1 KVA
Apparent Power = 3600 VA x 1000 VA

Apparent Power = 3.6 KVA

Power Factor = 1. 755294118 KW/ 3.6 KVA
Power Factor = 0.4875816993 = 0.487
Power Factor = 0.487
Thus, the motor power factor is 0.487, Final Answer: D.

Discussion:
The problem is requiring the motor power factor. The apparent power can be solved from the given
values since the formula is voltage times the current. The real power is given however there is a motor
efficiency. Real power will be divided by the efficiency and the answer will be used to solve the power
factor.

94. If three resistors of 5, 10, 15 ohms respectively are connected in parallel. What is the equivalent
resistance of the combination?
A. 30 ohms B. 2.73 ohms C. 2.07 ohms D. none of these
Given:
R1 = 5 Ω
R2 = 10 Ω
R3 = 15 Ω
Resistors are connected in parallel
Required:
Equivalent Resistance of the Combination
Schematic Diagram:

Solution:
Given 3 resistors are connected in parallel, the formula to use is,
Let Req = equivalent resistance
1 1 1 1
= + +
Req R1 R2 R3
Substituting the values,
1 1 1 1
= + +
Req 5Ω 10 Ω 15 Ω
1 11
=
Req 30Ω
Req = 30Ω / 11
Req = 2.727272727 Ω = 2.73 Ω
Req = 2.73 Ω
Thus, the equivalent resistance of the combination is 2.73 Ω, Final Answer: B.

Discussion:
The circuit is composed of three resistors connected in parallel and the problem only requires the
equivalent resistance. To get the equivalent resistance of the combination, the sum of the reciprocal of each
value of resistors will be raised to negative 1. Substituting the values of the resistors, the answer will be
2.727272727 Ω or simply 2.73 Ω.
95. Three 100-ohms resistors are connected in series-parallel. What is the equivalent resistance?
A. 300 ohms B. 150 ohms C. 67 ohms D. none of these
Given:
Three 100-ohms
Resistors are connected in series-parallel
Required:
Equivalent Resistance
Schematic Diagram:

Solution:
Two resistors are connected in parallel and to get the equivalent resistance of these two resistors,
R2 R3
Req2-3 =
R2 + R3
(100 Ω)(100 Ω)
Req2-3 = 100 Ω + 100 Ω

10000Ω2
Req2-3 = 200 Ω

Req2-3 = 50 Ω

Since, R1 and Req2-3 are series connected,

ReqTotal = R1 + Req2-3
ReqTotal = 100 Ω + 50 Ω
ReqTotal = 150 Ω
Thus, the equivalent resistance is 150 Ω, Final Answer: B.
Discussion:
A series- parallel connection is a combination of series and parallel connections. Since the problem
is all 100 ohms, it is not confusing where to put the values. To eliminate the parallel connection, the formula
R2 R3
to be used is Req2-3 = . The Req2-3 is the equivalent resistance of R2 and R3. R1 and Req2-3 is series
R2 + R3

connected already, therefore, they will be added according to the formula for series circuit. And the final
answer is 150 Ω.
96. In a circuit, three resistors of 10, 15, and 20 ohms are connected in series. Find the potential at the
source if the current flowing is 4 amp.
A. 90 V B. 135 V C. 180 V D. 45 V
Given:
R1 = 10 Ω
R2 =15 Ω
R3 =20 Ω
Resistors are connected in series
Current = I = 4 A
Required:
Potential at the Source
Schematic Diagram:

Solution:
To get the total resistance,
RTotal = R1 + R2 + R3
RTotal = 10 Ω + 15 Ω + 20 Ω
RTotal = 45 Ω
To get the potential source or the voltage,
V = IRtotal
Where V = Potential Source
 I = Current in Amperes
Rtotal = Total Resistance in Ohms
V = IRtotal
V = (4 A)(45 Ω)
C J/C
V = 180 ( )( )
S C/S

J/C = Volts
V = 180 V
Thus, the potential at the source if the current flowing is 4 A is 180 V, Final Answer: C.

Discussion:
There are three resistors connected in series, and to get the total resistance, the values of the
resistance must be added. The value of the total resistance is needed to get the potential source in the circuit
since the formula for this is current multiplied by total resistance. Therefore, after substituting the values in
the equation, the answer is 180 V.

97. What total current is drawn by a circuit composed of a 10 ohm resistor in series with two 10 ohm parallel
combination? The supply voltage is 120 V.
A. 4 A B. 10 A C. 6 A D. 8 A
Given:
R1 = 10 ohms
Two 10 ohms parallel combination
Voltage = 120 V
Required:
Total current is drawn by a circuit
Schematic Diagram:

Solution:
First, solve for the equivalent resistance of R2 and R3.
To solve for equivalent resistance connected in parallel,
R2 R3
Req2-3 = R
2 + R3

(10 Ω)(10 Ω)
Req2-3 = 10 Ω + 10 Ω

100Ω2
Req2-3 = 20 Ω

Req2-3 = 5 Ω

Since, R1 and Req2-3 are series connected,

ReqTotal = R1 + Req2-3
ReqTotal = 10 Ω + 5 Ω
ReqTotal = 15 Ω
To get the current,
V
I=
ReqTotal
Where V = Voltage in Volts
I = Current in Amperes
ReqTotal = Total Resistance in Ohms
I = 120 V/ 15 Ω
I=8A
Thus, total current is drawn by the circuit is 8 A, Final Answer: D

Discussion:
The two resistors connected in parallel must be solved first to get their equivalent resistance. Their
equivalent resistance must be added to the remaining resistor since they are connected in series. The total
resistance will be the sum of the remaining resistor and the equivalent resistance of resistor 2 and 3. To get
the current, the voltage which is given in problem will be divided by the total resistance. The answer for
the total current drawn by the circuit is 8 A.
98. What is the power required to transfer 97,000 coulombs of charge through a potential rise of 50 volts
in one hour?
A. 0.5 KW B. 0.9 KW C. 1.3 KW D. 2.85 KW
Given:
Charge = 97, 000 coulombs of charge
Potential Rise = 50 volts
Time = 1 hour
Required:
Power
Solution:
Let Q = the charge in coulombs
1 hour = 3600 sec.
To get the current, the formula of charge will be used.
Q=Ixt
where I = current in Amperes
t = time in seconds

Substituting the values,

97,000 coulombs = I (3600 sec)
I = 97, 000/ 3600 (C/S), Coulombs / s is equal to Amperes
I = 97, 000/ 3600 A
I = 26.94 A
Given the value of voltage, power can be obtained.
Power = Current x Voltage
Power = 26.94 A x 50 V
1 KW
Power = 1347.222222 Watts x 1000 Watts

Power = 1.3472 KW = 1.3 KW

Power = 1.3 KW
Thus, power required is 1.3 KW, Final Answer: C

Discussion:
Given the charge in coulombs and the time which is 1hour, the value of the current can be obtained.
Current is equal to the charge divided by the time in seconds. The current obtained is needed to compute
for the power, since the formula for the power is current multiplied by the voltage. Therefore, the power
required is 1.3 kW.

99. Two coils connected in a parallel to a 6 volts battery. One resistance is 17 and the other is 5. Find the
current in each.
A. 1.2 A & 0.353 A B. 1.1 A & 0.453A C. 1.0 A & 0.553 A D. 0.9 A & 0.653 A
Given:
Voltage = 6 Volts
R1 = 17 ohms
R2 = 5 ohms
Required:
Current in each resistor
Schematic Diagram:

Solution:
To solve for the current,
I=V/R
Since the resistors are connected in parallel, the value of the voltage that passes through resistors is equal.
Let I1= the current in the 17 Ω- resistor
I2= the current in the 5 Ω- resistor
I1 = V / R1
I1 = 6 V/ 17 Ω
I1 = 0.3529411765 A = 0.353 A
I1 = 0.353 A
I2 = V / R2
I2 = 6 V/ 5 Ω
I2 = 1.2 Ω
Thus, the value of the current for R1 is 0.353 A and R2 is 1.2 Ω, Final Answer: A.
Discussion:
Resistors connected in parallel have equal voltages. Therefore, in this problem, 6 V is passing
through R1 and R2 separately. To solve for the value of the current in each resistance, voltage will be divided
by the value of the resistance. Thus, the value of the current for R1 is 0.353 A and R2 is 1.2 Ω.

100. How many kilowatts of electricity must be supplied to run a motor that is rated at 10 hp of electrical
energy is transformed into mechanical energy with an efficiency of 80%?
A. 9.33 kw B. 10 kw C. 5.97 kw D. 12 kw
Given:
Power rated = 10 HP
Efficiency = 80%
Required:
Kilowatts of Electricity to be supplied
Solution:
Using the formula of efficiency,
Efficiency = Pout/ Pin
The power output is equal to,
Pout = Efficiency x Pin
Convert the power in HP into KW.
0.746 KW
Pin = 10 HP x
1 HP

Pin = 7.46 KW
Pout = 0.80 (7.46 KW)
Pout = 5.968 KW = 5.97 KW
Pout = 5.97 KW
Thus, the power output is 5.97 KW, Final Answer: C.