Problem:
A contractor bought a transit concrete
mixer. There will be no maintenance cost
the first year as it was sold with oneyear free maintenance. In the second year,
the maintenance is estimated at P1,000. In
subsequent years, the maintenance cost
will increase P1,000 per year (that is the
3rd maintenance will be P2,000, 4th year
maintenance will be P3,000 and so forth).
What amount that must be set aside not at
6% interest to pay the maintenance costs
on the concrete mixer for the first 6
years of ownership?
a. 11,459.33
c. 19,559.00
b. 15,928.88
d. 7,927.09
Solution:
P (1+i)n
n
= 2
n
G i (1+i) i(1+i)n
(1+0.06)6
P
6
=
2
6
1000 (0.06) (1+0.06) 0.06(1+0.06)6
P=P11,459.35
Problem:
A machine is to be purchased for P155,000
it has an estimated life of 8 years and a
salvage value of P6000. A sinking fund is
to be established so money will be
available to purchase a replacement when
the first machine wears out at the end of
8 years. An amount of P13,030 is to be
deposited at the end of each year during
the lifetime of the first machine into
this sinking fund. What interest rate (%)
that this fund must earn to produce
sufficient fund to purchase the
replacement machine at the end of 8 years?
a. 13%
c. 15%
b. 10%
d. 7%
Solution:
F=155,000-6,000
F=149,000
A[(1+i)n -1]
F=
i
13,030[(1+i)8 -1]
149,000=
i
Solving for i:
i=10%
compounded annually, compute the
equivalent uniform annual cost of the
machine if it will cost P100,000 per year
to operate?
a. 165,098.16
c. 142,883.04
b. 210,950.82
d. 250,282.23
Solution:
EUAC=A1 +A2
A1 =100,000
A[(1+i)n -1]
P=
(1+i)n i
A2 [(1+0.1)10 -1]
400,000=
(1+0.1)10 (0.1)
A2 =P65,098.16
EUAC=A1 +A2
EUAC=100,000+65,098.16
EUAC=P165,098.16
Problem:
An electric replacement pump is being
considered for purchase. It is capable of
providing 200 hp. The pertinent data are
as follows:
Cost = P3200
Maintenance Cost per year = P50
Electric efficiency = 0.85
Life Expectancy = 14 years
The pump is used for 400 hours per year
and the cost of electricity is P0.04 per
kilowatt hour (1 horsepower = 0.746 kW).
Assuming the pump will have no salvage
value, what will be the monthly cost?
a. P238.20
c. P520.36
b. P304.55
d. 425.10
Solution:
400(0.04)(200)(0.746)
+50
0.85
Annual Cost=P2858.47
2858.47
Monthly Cost=
12
Monthly Cost=P238.20
Annual Cost=
Problem:
The first cost of a certain equipment is
P324,000 and a salvage value of P50,000 at
the end of its life for 4 years. Money is
worth 6% annually. If there is no salvage
value and the annual maintenance cost is
P18,000, find the capitalized cost of
perpetual service.
a. 562,000
c.
624,000
b. 358,000
d.
745,000
Solution:
OC
CC=FC+
i
CC=324,000+
18,000
0.06
CC=P624,000
Problem:
A printing machine SOLNA (single color)
costs P400,000 to purchase with a life of
10 years with no salvage value. If the
rate of interest is 10% per annum,
Problem:
A contractor can buy dump trucks for
P80,000 each (surplus) or rent them for
P1189 per truck per day. The truck has a
savage value of P100,000 at the end of its
useful life of 5 years. Annual cost of
maintenance is P20,000.00. If money is
worth 14% per annum. Determine the number
of days per year that a truck must be used
to warrant the purchase of the truck.
a. 180 days
c. 200
days
b. 250 days
d. 300
days
Solution:
(FC-SV)i
+OC
(1+i)n -1
Annual Cost=800,000(0.14)
(800,000-100,000)(0.14)
+
+20,000
(1+0.14)5 -1
Annual Cost=P237,898
Annual Cost if Rent=1189x
1189x=237,898
x=200 days
Annual Cost=(FC)i+
Problem:
A contractor has 50 men of the same
capacity at work on a job. They can
compete the job in 30 days, the working
day being 8 hours, but the contract expire
in 20 days. He decides to put 20
additional men. If all the men get P3.00
per day for a full or part day and if the
liquidated damages are P100 for every full
or part day he requires over his contract.
How many days could he finish the job?
a. 12
c. 10
b. 22
d. 17
Solution:
x=number of days to finish the job
50(30)=(50+20)x
x=21.4
x=22 days
Problem:
A new bridge connecting Dumaguete City and
Santander with a 100-year life expected to
have an initial cost of $20 million. This
bridge must be resurfaced every five
years, at a cost of 1 million. The annual
inspection and operating costs are
estimated to be $50,000. Determine the
present-worth cost of the bridge using the
capitalized equivalent approach (i.e. take
the life of the bridge as infinite). The
interest rate is 10% per year, compounded
annually.
a. $16,072,537
c. $30,873,767
b. $10,900,977
d. $22,137,970
Solution:
Present Worth (Capitalized Equivalent)=
Rebuilt Cost OC
FC+
+
(1+i)n -1
i
Present Worth
(Capitalized Equivalent)=20,000,000
1,000,000 50,000
+
+
(1-0.1)5 -1
0.1
Present Worth=$22,137,974.81
Problem:
The proposal of Mutual Savings Fund will
cost P500,000. The benefits at the end of
the first year are estimated to be
P100,000, increasing P10,000 per year in
subsequent years. Assuming a 12% interest
rate, with no salvage value and an 8-year
analysis period, compute the benefit cost
ratio.
Present worth gradient series factor (P/G,
12%, 8) = 14.471
Present worth factor (uniform series)
(P/A, 12%, 8) = 4.9676
Capital recovery factor (uniform series)
(A/P, 12%, 8) = 0.2013
Sinking fund factor (uniform series) (A/F,
12%, 8) = 0.0813
a. 1.05
c.
2.13
b. 1.28
d.
2.52
Solution:
Present=500,000
Present Worth Benefits=P1 +P2
P1
=4.9676
A
P1 =4.9676(100,000)
P1 =496,760
P2
=14.471
G
P2 =14.471(10,000)
P2 =144,710
PT =496,760+144,710
PT =641,740
641,470
Benefit Cost Ratio=
500,000
BCR=1.28
Problem:
ABC Corporation has decided to sell $1000
bonds, which will pay semiannual dividends
of $20 (2% per period) and will mature in
5 years. The bonds are sold at $830, but
after broker’s fees and the other expenses
the company ends up receiving $760. What
is the company’s cost of the capital
raised through the sale of these bonds?
a. 13.84%
c. 10.52%
b. 7.65%
d. 15.46%
Solution:
PT =P1 +P2
20[(1+i)10 -1]
100
760=
+
10
(1+i) i
(1+i)10
Solving for i:
i=0.05127 or 5.127% (semi-annual)
i=(1+i)n -1
i=(1+0.05127)2 -1
i=10.52%
Problem:
In a certain city in Philippines, all
seven digit telephone numbers begin with
350. How many telephone number maybe
assigned to that city if the last four
digits should not begin or end in zero?
a. 9300
c. 9800
b. 8100
d. 7600
Solution:
3, 5, 0, _, _, _, _,
3, 5, 0, (9 x 10 x 10 x 9)
No. of telephon numbers=9 x 10 x 10 x 9=8100
Problem:
The average score for Mathematics test is
77 and the standard deviation is 8.
Compute the probability that any one
student scored between 61 and 93.
a. 95.4%
c. 87.1%
b. 92.1%
d. 89.2%
Solution:
This is a special case. The difference
from the mean is 2 times the standard
deviation.
Therefore, Probability = 95.4%
Problem:
About 68.2% of the scores fall within the
normal curve, which ranges from 50 to 80
and is symmetric about the mean. What is
the standard deviation of the scores in
this distribution?
a. 22
c. 30
b. 10
d. 15
Solution:
Since the scores fall on a 68.2% range,
then the distance from the mean are equal
to one deviation from mean value.
80-50
Standard Deviation=
=15
2
Problem:
From the given data of a distribution of
breakdown per week of a certain machine,
determine the number of times this machine
is expected to breakdown per week over a
period of time.
Probability
Breakdown/Week
0.11
0.18
0.26
0.35
0.42
0
1
2
3
4
a. 5.02
c. 6.83
b. 4.75
d. 3.43
Solution:
P
0.11
0.18
0.26
0.35
0.42
B
0
1
2
3
4
PB
0
0.18
0.52
1.05
1.68
∑ 𝑃𝐵 = 3.43
No. of times this machine is expected to
breakdown/week = 3.43
Problem:
Samsung, a computer chip manufacturer, has
found that only 1 out of 2000 chips is
defective. A certain company ordered a
shipment of chips. How many chips will the
company order before the probability that
at least one chip is defective is 45%.
a. 1740
c. 2074
b. 1194
d. 3385
Solution:
Probability that a chip is defective:
1
P=
=0.005
2000
Probability that none is defective:
𝑃 = 1 − 0.0005
𝑃 = 0.9995
Probability that no chips are defective:
𝑃 = (0.9995)𝑛
Probability that at least one chip is
defective is 45%
1 − (0.9995)𝑛 = 0.45
(0.9995)𝑛 = 0.55
𝑛 = 1195
No. of chips that the company can order
before the probability that at least one
chip is defective reaches 45%.
No. of chips = 1195 – 1
No. of chips = 1194
Problem:
A box of mass 5.0 kg is pulled vertically
upward by a force of 68 N applied to a
rope attached to the box. Determine the
rate of acceleration of the box.
a. 5.59 m/s2
c. 3.79 m/s2
b. 4.92 m/s2
d. 6.46 m/s2
Solution:
About
to move (a=0)
∑ FH =0
∑ FV =0
REF+W=68
ma+mg=68
5(a)+5(9.81)=68
m
a=3.79
s
Problem:
A car travels over the crest of a hill at
10.0 m/s. The radius of curvature at the
crest is 12.0 m. Determine the minimum
speed required for the passenger to feel
momentarily “weightless”.
a. 27.36 m/s
c. 10.84 m/s
b. 6.19 m/s
d. 21.77 m/s
Solution:
∑ FV =0
mv2
r
mv2
mg=
r
W=
60V2
60(9.81)=
12
m
V=10.84
s
Problem:
A 28.0 kg block is connected to an empty
1.35 kg bucket by a cord running over a
frictionless pulley. The coefficient of
static friction between the table and the
block is 0.450 and the coefficient of
kinetic friction between the table and the
block is 0.320. Sand is gradually added to
the bucket until the system just begins to
move. Calculate the mass of sand added to
the bucket.
a. 23.09 kg
c. 6.99 kg
b. 11.25 kg
d. 14.24 kg
Solution:
T=fr +REF
T=μN+ma
T=0.45(28)(9.81)+28(0)
T=123.61 N
∑ FV =0
T+REF=W
123.61+(m+1.35)(0)=(m+1.35)(9.81)
m=11.25 kg(mass of sand)
Problem:
A student of mass 50 kg decides to test
Newton’s laws of motion by standing on a
bathroom scale placed on the floor of an
elevator. Assume that the scale reads in
Newton. Determine the scale reading when
the elevator is accelerating upward at
0.50 m/s 2 .
a. 515 N
c. 652 N
b. 708 N
d. 431 N
Solution:
∑ Fv =0
F=REF+W
F=ma+mg
F=50(0.5)+50(9.81)
F=515 N
Problem:
How large must the coefficient of static
friction be between the tires and the road
if a car is to round a level curve at of
radius 85 m at a speed of 95 km/h?
a. 0.91
c. 0.56
b. 0.84
d. 0.68
Solution:
2
tan θ = V ⁄gr
2
μ= V ⁄gr
2
(95⁄3.6) ⁄
μ=
9.8(85)
μ=0.84
Problem:
A car traveling 60 kph can brake to a step
in a distance of 20 m. If the car is going
twice as fast at 120 kph, what is the
stopping distance? Assume the maximum
braking force is approximately independent
on the speed.
a. 100 m
c. 80 m
b. 94 m
d. 76 m
Solution:
F•d= 1⁄2 m[(V1 )2 -(V2 )2 ]
F(20)= 1⁄2 m[(60)2 -(0)2 ]
F⁄m =90
if V is twice
F•d= 1⁄2 m[(V1 )2 -(V2 )2 ]
F(d)= 1⁄2 m[(120)2 -(0)2 ]
F⁄m = 7200⁄
d
90= 7200⁄d
d=80 m
Problem:
Calculate the power required of a 1400 kg
car when the car accelerates along a level
road from 90 to 110 km/h in 6.0 second to
pass another car. Assume the average
retarding force on the car is FR = 700 N,
throughout.
a. 90 hP
c. 82 hP
b. 77 hP
d. 100 hP
Solution:
V2 =V1 +at
110⁄
90
3.6 = ⁄3.6 +a(6)
a=0.93 m⁄s2
F=700+ma
F=700+1400(0.93)
F=2002 N
P=FV2
P=2002(110⁄3.6)
P=61172.22 watts
P= 61172.22⁄746
P=82 hP
Problem:
A 4.00 kg box is placed on the floor at
the edge of a merry-go-round of radius
3.00 m. The coefficient of static friction
between the box and the floor is 0.200.
The merry-go-round accelerates from the
rest and eventually the box slides off the
edge. Determine the speed at which this
occurs.
a. 2.08 m/s
c. 1.33 m/s
b. 0.95 m/s
d. 2.42 m/s
Solution:
∑ FH =0
mV2
=Fr
r
2
mV
=μN
r
4V2
=0.2(4)(9.81)
3
V=2.42𝐦/𝐬
Problem:
The tires of a car make 65 revolutions as
the car reduces its speed from 95 km/h to
45 km/h the tires have a diameter of 0.80
m, what was the angular acceleration of
the tires?
a. -4.1 rad/s 2
c. 5.2 rad/s 2
b. -3.5 rad/s 2
d. 3.9 rad/s 2
Solution:
w2 =w2o +2w(θ)
Vo = 95⁄3.6 =26.4 m/s
V= 45⁄3.6 =12.5 m/s
V=rw
12.5=0.4w
w=31.25 rad/s
26.4=0.4(wo )
wo =66 rad/s
w2 =w2o +2αθ
(31.25)2 =(66)2 +2α(65)(2π)
α=-4.1 rad/s2
Problem:
For airport, runways and taxiways, the
minimum vertical curve lengths are based
on:
a. Sight distance
b. Comfort standards involving vertical
acceleration
c. Appearance criteria
d. Stopping sight distance
Problem:
The harbor entrance should, if possible,
be located on what side of the harbor?
a. windward lee
c. lee side
b. outer end
d. inner end
Problem:
During the recent typhoon that hits the
Philippines, the wave due to the strong
winds produced a wave length at a point on
the surface of the water at depth of 4 m,
determine the wavelength for a period of
3.88 seconds for a shallow water wave.
a. 18 m
c. 20 m
b. 22 m
d. 24 m
Solution:
L= (9T ⁄2π) tanh ((2π(d)⁄L))
2
2
L= (9(3.88) ⁄2π) tanh ((2π(4)⁄L))
L=23.50 tanh ((25.133⁄L))
L=20 m
Problem:
A toll plaza at the entrance to the NLEX
consist of 3 booths each of which can
handle an average of one vehicle every 6
second. The peak traffic of a toll booth
is recorded as follows from 7:00 AM to
8:00 AM.
Time Period 10 minute
Comulative
volume
Value
7:00 – 7:10 200
200
7:10 – 7:20 400
600
7:20 – 7:30 500
1100
7:30 – 7:40 250
1350
7:40 – 7:50 200
1550
7:50 – 8:00 150
1700
Determine the service rate of one toll
booth every 10 minutes.
a. 150
c. 120
b. 180
d. 100
Solution:
Service rate = 10(60)/6
Service rate = 100 vehicle per 10 minutes
Problem:
The water level that guarantees about 98%
of tide is safe to ships seems to be
suitable from the expression of the
technical resolution of the International
Water Wave Congress. Such water level,
which is 0.15 m – 0.4 m below MLLW should
be used for design of ports.
a. Design Low Tide (DLT)
c. Residual Water Level (RWL)
b. Low Water Level (LWL)
d. High Water Level (HWL)
Problem:
The wave length of a shallow-water wave is
80 m long and its period of oscillation is
7.2 seconds. If it has a depth of water
equal to 30 m, compute the velocity of the
wave in m/s.
a. 20.45 m/s
c. 18.29 m/s
b. 11.04 m/s
d. 13.24 m/s
Solution:
gT
2πd
V= ( ) tanh (
)
2π
L
L
d< (shallow water)
2
9.81(7.2)
2π(30)
tanh (
)
2π
80
V=11.04𝐦/𝐬
V=
Problem:
What is the ratio of the area to the
perimeter of an ellipse 9𝑥 2 + 25𝑦 2 = 225?
a. 1.82
c. 1.28
b. 1.45
d. 1.52
Solution:
9x2 +25y2 =225
x2 y2
+ =1
25 9
a = 5
b = 3
A=πab
A=π(5)(3)=15π
Perimeter = 2π√
Perimeter=2π√
a2 +b2
2
25+9
2
Perimeter=8.246π
15π
Ratio=
=1.82
8.246π
Problem:
The vertices of a triangle are at A(1,2),
B(3,8) and C(8,-1). Locate the point of
intersection of its medians.
a. (3,4)
c. (2,-1)
b. (4,3)
d. (3,2)
Solution:
Note: Point of intersection of medians =
centroid of the triangle.
𝑥1 + 𝑥2 + 𝑥3
𝑥̅ =
3
1+3+8
𝑥̅ =
=4
3
𝑦1 + 𝑦2 + 𝑦3
3
2 + 8 + (−1)
𝑦̅ =
=3
3
Centroid is at (4,3)
𝑦̅ =
Problem:
An ellipse has an eccentricity of 1/3 and
the distance between the foci is equat to
4. Compute the length of the latus rectum.
a. 8.35
c. 13.22
b. 15.93
d. 10.67
Solution:
2c=4
C=2
c=ae
1
2=a ( )
3
a=6
a2 =b2 +c2
(6)2 =b2 +(2)2
b2 =32
2b2
L=
a
L=
2(32)
=10.67
6
Problem:
Mines in a harbour during wartime are so
placed that ship going to a certain port
within the harbor to avoid hitting them
must follow a path which is at all times
equidistant from a point 9km north of
another port and a line whose bearing is
due east. One point of this line is 3km.
south of the port. Considering the port as
the origin. Find the equation of the path
of the ship.
a. 𝑥 2 − 18𝑦 + 81 = 0
c. 𝒙𝟐 − 𝟐𝟒𝒚 + 𝟕𝟐 = 𝟎
b. 𝑥 2 + 18𝑦 − 81 = 0
d. 𝑥 2 + 24𝑦 − 72 = 0
Solution:
b. 2/3
d. 5/3
Solution:
4
V= πr3
3
dv 4
dr
= π(3r2 )
dt 3
dt
4
dr
27π= π[3(3)2 ]
3
dt
dr 3
=
dt 4
Problem:
A particle moves according to the
parametric equation 𝑥 = 𝑡 3 and 𝑦 = 2𝑡 2 , where
t is in seconds and x and y are in meters.
What is the distance covered by the
particle when t = sec. ?
a. 12.75m
c.
14.09m
b. 13.66m
d.
11.52m
Solution:
x=t3
dx
=3t2
dt
dx=3t2 dt
Equation of path of ship
d1 =d2
√(x-0)2 +(y-9)2 =y+3
x2 +y2 -18y+81=(y+3)2
x2 +y2 -18y+81=y2 +6y+9
x2 -24y+72=0
y=2t2
dy
=4t
dt
dy=4t dt
dS=√(dx)2 +(dy)2
dS=√(3t2 dt)2 +(4t dt)2
dS=√9t4 +16t2 dt
dS=√9t2 +16 t dt
1
Problem:
Sand is pouring from a spout at the rate of
25 cc/sec. It forms a cone whose height is
always 1/3 the radius of its base. At what
rate in cm/sec is the height increasing,
when the cone is 50 cm high?
a. 0.000354 cm/sec
c. 0.000213 cm/sec
b. 0.000488 cm/sec
d. 0.000571 cm/sec
Solution:
r = 3h
πr2 h
V=
3
π(3h)2 h
V=
3
V=πh3
dV
dh
=3π3h2
dt
dt
dh
25=9π(50)2
dt
dh
=0.000354 cm/sec
dt
Problem:
A spherical balloon is inflating at a rate
of 27𝜋 𝑖𝑛3 /𝑠𝑒𝑐. How fast is the radius of the
balloon increasing when the radius is 3
inches?
a. 3/4
c. 1/3
1
2
S=
∫ (9t2 +16) 18t dt
18
3 t
(9t2 +16)2
1
S= [ (2)
]
18
3
0
3 t
1
2
S=
[(9t +16)2 ]
0
27
When t = 2 sec.
3
3
1
S=
{[(9)(2)2 +16]2 -(0+16)2 }
27
3
3
1
S=
[(36+16)2 -(16)2 ]
27
S=11.5177m
Problem:
Oil spilled from a tanker spreads in a
circle whose circumference increases at a
rate of 40 ft/sec. How fast is the area of
the spill increasing when the
circumference of the circle is 100𝜋𝑓𝑡?
a. 1800
c. 2400
b. 2000
d. 2800
Solution:
C=πd
πd2
A=
4
dA π
dd
= (2d)
dt 4
dt
dC
dd
=π
dt
dt
dd
40=π ( )
dt
dd 40
=
dt π
100π=πd
d=100
dA π
40
= (2)(100) ( )
dt 4
π
dA
=2000ft/s
dt
Problem:
An inverted conical container has a
diameter of 42 inches and a depth of 15
inches. If water is flowing out the vertex
of the container at a rate of 35𝜋 𝑖𝑛3 /𝑠𝑒𝑐,
how fast is the depth of the water
dropping when the height is 5 inches?
a. -5/7
c. 5/7
b. -3/7
d. 3/7
A boat is being pulled toward a dock by a
rope attached to its bow through a pulley
on the dock 7 feet above the bow. If the
ropes hauled in at a rate of 4ft/sec, how
fast is the boat approaching the dock when
25 feet of rope is out?
a. 25/6
c.
21/8
b. 23/6
d.
23/8
Solution:
Solution:
r 21
=
h 15
21
r=
h
15
1 2
V= πr h
3
1 21 2
V= π ( h) h
3 15
49 3
V=
πh
75
dV 49 2 dh
=
πh
dt 25
dt
49
dh
-35π=
π(5)2
25
dt
dh
5
=dt
7
y2 =(7)2 +x2
dy
dx
2y
=0+2x
dt
dt
@ y=25
x=√(25)2 -(7)2 =24
dx
2(25)(4)=2(24)
dt
dx 25
=
dt 6
Problem: lumabas na ito (NOV. 2018)
A Toyota Landcruiser drives east from
point A at 30kph. Another car, BMW,
starting from B at the same time, drives
𝑆30°𝑊 toward A at 60kph. B is 30km away
from A. How fast in kph is the distance
between two cars changing after 30
minutes?
a. 55kph
c.
80kph
b. 70kph
d.
60kph
Problem:
The sides of an equilateral triangle are
increasing at the rate of 27 in/sec. How
fast is the triangle’s are increasing when
the sides of the triangle are each 18
inches long?
a. 412.14
c. 432.76
b. 420.89
d. 409.32
Solution:
1
(𝑥)2 sin 60°
2
𝑑𝐴 1
𝑑𝑥
= 2𝑥 sin 60°
𝑑𝑡 2
𝑑𝑡
𝑑𝐴 1
= (2)(18) sin 60° (27)
𝑑𝑡 2
𝒅𝑨
= 𝟒𝟐𝟎. 𝟖𝟖𝟖
𝒅𝒕
𝐴=
Problem:
Solution:
Using Cosine Law for ADC:
S2 =(30t)2 +(30-60t)2 -2(30t)(30-60t) cos 60°
After 30 min., t = 0.5hrs.
S2 =(15)2 +(0)2 -2(15)(0) cos 60°
S2 =(15)2
S=15 km
S2 =(30t)2 +(30-60t)2 -60t(30-60t)(0.5)
S2 =900t2 +900-3600t+3600t2 -900t+1800t2
S2 =6300t2 -4500t+900
dS
2S
=2(6300)t-4500
dt
dS
2(15)
=12600(0.5)-4500
dt
dS
=60kph
dt
Problem:
Find the moment of inertia with respect to
x – axis of the region in the first
quadrant, which bounded by the curve 𝑦 2 =
4𝑥, the line 𝑦 = 2 and the y – axis.
a. 1.6
c. 2.8
b 1.3
d. 2.1
Solution:
𝐼 = 𝐴𝑑 2
𝑦2
𝐼 = ∫ 𝑥𝑑𝑦(𝑦 2 )
𝑦1
2
𝑦2 2
(𝑦 )𝑑𝑦
0 4
𝑰 = 𝟏. 𝟔
𝐼=∫
Problem:
Find the area enclose by the curve 𝑥 2 +
8𝑦 + 16 = 0, the y – axis and the line
passing through (0, 0) and intersecting
the line x = 4.
a. 5
c. 4
b. 2
d. 3
Solution:
x = 4
10
y=
3
x2 +8y+16=0
x2 =-8y-16
x2 =-8(y+2)
y=kx
10
=k(4)
3
5
k=
6
5
y= x
6x
2
A= ∫ ydx
x1
4
5
-x2 -16
A= ∫ [( x) + (
)] dx
6
8
0
A=-4
A=4
Problem:
A circular hole 4 inches in diameter and 1
feet deep in a metal block is rebored to
increase the diameter to 4.12 inches.
Estimate the amount of metal removed.
a. 4.33π
c. 2.88π
b. 5.44π
d. 3.66π
Solution:
π(d21 -d22 )h
Vremoved =
4
π(4.122 -42 )(12)
Vremoved =
4
Vremoved =2.88π
Problem:
A projectile is shot vertically upward
from ground level with an initial velocity
of 49 m/s. What is the impact velocity
when it touches the ground?
a. -49 m/s
c. -52 m/s
b. -36 m/s
d. -24 m/s
Solution:
V2f =V2o ±2as
V2f =(49)2 -2(9.81)(0)
m
Vf =±49
s
∴Vf =-49 m/s