Prepared by: Engr. Jan Nathan A. Anastacio
1. Solve for a in the equation: a = (64^x)(4^y)
A. 4^(x+3y)
B. 4^(3xy)
C. 256^(xy)
D. 4^(3x+y)
E. None of the above
a = 64x 4y = 43
a = 43x+y (Ans.)
x
4
y
= 43x 4y
2. Simplify 3^x – 3^(x-1) – 3^(x-2)
A. 3^(x-2)
B. 3^(3x-3)
3x − 3x−1 − 3x−2
3x − 3x 3−1 − 3x 3−2
1
1
5
9
C. 5×3^(x-2)
3x 1 − 3 − 9 = 3x
= 3x
D. 13×3^x
3x 5 3−2 = 5 3x 3−2
E. None of the above
5 3x−2 (Ans. )
5
32
3. During the seventh stage of the 2010 Paris–
Nice bicycle race, Thomas Voeckler posted the
fastest average speed, but Alberto Contador won
the race. The seventh stage was 119 kilometers
long. Voeckler’s average speed was 0.0034
meters per second faster than Contador’s.
Traveling at these average speeds, Contador took
3 seconds longer than Voeckler to complete the
race stage. Find Thomas Voeckler’s average
speed during the seventh stage of the 2010
Paris–Nice cycle race in kph.
A. 11.62
B. 16.12
C. 58.03
D. 41.81
E. None of the above
Vv = Vc + 0.0034
119000 = Vv t v = Vc t v + 3
(Vc + 0.0034) t v = Vc (t v + 3)
Vc t v + 0.0034t v = Vc t v + 3Vc
Vc =
0.0034tv
3
119000 =
0.0034tv
3
tv + 3
t v = 10245.45s
119000 = Vv 10245.25
Vv = 11.6149m/s
Vv = 41.81kph (Ans.)
4. Brenan had a job interview in a nearby city 72
miles away. On the first leg of the trip he drove an
average of 30 mph through a long construction
zone, but was able to drive 60 mph after passing
through this zone. If driving time for the trip was
1-1/2 hr, how long was he driving on the
construction zone?
A. 24 mins
B. 12 mins
C. 36 mins
D. 42 mins
E. None of the above
d = vt
72 = 30 t + 60 1.5 − t
t = 0.6hrs
t = 36mins (Ans. )
5. Gabby saves 20% of his income. If his expenditure is
increased by 35%, how many percent must his income
be increased so that he may save 10% of it?
A. 10%
B. 18%
C. 20%
D. 25%
E. None of the above
I=E+S
S = 0.2I
E = 0.8I
xI = 0.8I 1.35 + 0.1 xI
x = 1.2
%∆ =
1.2−1
1
100% = 20%
%∆ = 20% (Ans. )
6. A year ago, Trixie got a crush with Kenneth, where
the product of their ages was 156. Fortunately, they are
now both sweethearts; and they plan to marry 12
years after, where the product of their ages will be 650.
Knowing that Jacob is older than Trixie, Find the
present age of Trixie.
A. 14
B. 13
C. 12
D. 15
E. None of the above
x − 1 y − 1 = 156
x + 12 y + 12 = 650
156
y = x−1 + 1
x + 12
156
+
x−1
x = 13
156
y = 13−1 + 1
y = 14
x = 13 (Ans. )
1 + 12 = 650
7. Solve for w from the following equations:
Mode 5 2
3x – 2y + w = 11
3
-2
1
11
x + 5y - 2w = -9
1
5
-2
-9
2x + y - 3w = -6
2
1
-3
-6
A. 1
B. 2
C. 3
D. 4
E. None of the above
x=2
y = −1
w = 3 (Ans. )
8. When (x+3)(x-4) + 4 is divided by x – k, the
remainder is k. Find the value of k.
A. 4 or 2
B. 2 or -4
C. 4 or -2
D. -4 or -2
E. None of the above
Remainder Theorem:
If f(x) is divided by x − r , the
remainder is f r .
Try 4: 4 + 3 4 − 4 + 4 = 4 ∴ OK
Try 2: 2 + 3 2 − 4 + 4 = −6
Try − 2: −2 + 3 −2 − 4 + 4 = −2 ∴ OK
Ans: 4 or -2
9. The term involving x^9 in the expansion of
(x^2 + 2/x)^12 is:
A. 25434x9
B. 52344x^9
C. 25344x^9
Binomial Theorem:
r th term = nCr a
a = x2
n− r−1
b
r−1
b = 2x −1
Find the term involving x 9
D. 23544x^9
x9 = x2
E. None of the above
x9 = x
2(12− r−1 )
x9 = x2
12−r+1 −r+1
12−(r−1)
x −1
x
r−1
−r+1
9 = 2 12 − r + 1 − r + 1
r=6
r th term =
12C5
x2
12−5
2x −1
r th term = 25344x 9 (Ans. )
5
10. Find the sum of the coefficients in the
expansion of (x + 2y –z)^8.
A. 256
B. 1024
C. 1
D. 6
E. None of the above
To get the sum of the coefficients in the
expansion, substitute 1 to each of the
variables.
Sum = 1 − 2 1 − 1
Sum = 256 (Ans. )
8
11. Two thousand kilogram of steel containing 8% of
nickel is to be made by mixing steel containing 14%
nickel with another steel containing 6% nickel. How
much of the steel containing 6% nickel is needed?
A. 1500 kg
B. 800 kg
C. 750 kg
D. 500kg
E. None of the above
x + y = 2000
0.14x + 0.06y = 0.08 2000
Mode 5 1
1
1
2000
0.14
0.06
160
x = 500kg
y = 1500kg (Ans. )
12. A and B can do a piece of work in 42 days, B and C in
31 days, and A and C in 20 days. Working together, how
many days can all of them finish the work?
A. 18.9
B. 19.4
C. 17.8
D. 20.9
E. None of the above
1
1
+
A
B
42 = 1
1
1
+
B
C
31 = 1
1
1
+
A
C
20 = 1
Mode 5 2
42
42
0
1
0
31
31
1
20
0
20
1
1
A
541
= 26040
1
1
1
+
+
A
B
C
1
B
79
= 26040
x =1
x = 18.86days (Ans. )
1
C
761
= 26040
13. How many minutes after 10:00 o’clock will the
hands of the clock be opposite of the other for the first
time?
A. 21.41
B. 22.31
C. 21.81
𝑥
5 10 + 12 = 𝑥 + 6(5)
x
12
x
D. 22.61
E. None of the above
5(10)
13. How many minutes after 10:00 o’clock will the
hands of the clock be opposite of the other for the first
time?
A. 21.41
B. 22.31
C. 21.81
D. 22.61
E. None of the above
x
5 10 + 12 = x + 6(5)
x
5 10 − 6 5 = x − 12 =
5 10 − 6 =
11x
12
12
x = 11 5(10 − 6)
60
x = 11 10 − 6
x = 21.818mins (Ans. )
11x
12
14. A man fires a target 420 m away hears the bullet
strikes to 2 second after he pulled the trigger. An
observer 525 m away from the target and 455 m from
the man heard the bullet strike the target one second
after he heard the report of the rifle. Find the velocity of
the bullet.
A. 525 m/s
B. 360 m/s
C. 350 m/s
D. 336 m/s
E. None of the above
Let t1 be the time the man heard the bullet strike,
t 2 be the time the observer heard the report of
the rifle and t 3 be the time the observer heard the
bullet strike.
t1 =
420
420
+
Vb
Vs
t2 =
455
Vs
420
525
+
Vb
Vs
t3 =
420
525
+
Vb
Vs
Vs =
420
420
+
525−455
Vb
420
=2
=2
Vb −1
Vb = 525m/s (Ans.)
t3 = t2 + 1
=
455
+
Vs
525−455
420
1− V
b
1
15. A boat takes 2/3 as much time to travel downstream
from C to D, as to return, If the rate of the river’s current
is 8 kph, what is the speed of the boat in still water?
A. 38
B. 39
C. 40
D. 41
E. None of the above
2
td = 3 tu
d
Vb +Vw
1
Vb +8
2
=3
2
=3
d
Vb −Vw
1
Vb −8
Vb = 40kph (Ans. )
16. The time required for an elevator to lift a weight
varies directly with the weight and the distance through
which it is to be lifted and inversely as the power of the
motors. If it takes 20 seconds for a 5-hp motor to lift 50
lbs. through 40 feet, what weight can an 80-hp motor lift
through a distance of 40 feet within 30 seconds?
A. 1000 lbs.
B. 1150 lbs.
C. 1175 lbs.
D. 1200 lbs.
E. None of the above
t∝
wd
P
20 =
→ t=
kwd
P
k 50 40
5
k = 0.05
30 =
0.05 w 40
80
w = 1200lbs (Ans. )
17. The price of 8 calculators ranges from P200 to
P1000.If their average price is P950,what is the lowest
possible price of any one of the calculators?
A. 500
B. 550
C. 600
D. 650
E. None of the above
To get the minimum price possible for one
calculator that will result to the given average, we
have to assume the maximum cost for all the other
calculators.
8 950 = 7 1000 + 1 x
x = P600 (Ans. )
18. Find the 2020th digit in the decimal equivalent
of 1785/9999 starting from the decimal point.
A. 1
B. 7
C. 8
D. 5
E. None of the above
1785
9999
= 0.178517851785 …
2020
4
= 505 ∴ 2020 is divisible by 4
2020th digit = 5 (Ans. )
19. Find the 100th term of the sequence 1.01,
1.00, 0.99….
A. 0.05
The sequence is arithmetic.
an = a1 + d n − 1
B. 0.04
a100 = 1.01 + 1 − 1.01 100 − 1
C. 0.03
a100 = 0.02 (Ans. )
D. 0.02
E. None of the above
20. A rubber ball is dropped from a height of 15m. On
each rebound, it rises 2/3 of the height from which it
last fell. Find the distance traveled by the ball before it
becomes to rest.
A. 75m
B. 96m
C. 100m
D. 85m
E. None of the above
The sequence is geometric because the next height
multiplied by 2/3 after every rebound.
Sn =
a1 1−rn
1−r
a
1
S∞ = 1−r
The number of bounce approaches infinity..
2∞
S=
15 1−3
2
1−3
S = 45
d = 2 45 − 15
d = 75m (Ans. )
21. Given the equation y = 7sin(5x-10)+3.
Determine the amplitude.
A. 7
B. 3
C. 10
D. 2
E. None of the above
y = Asin Bx + C + D
A = amplitude
P=
2π
B
= period
1
B
F = P = 2π = frequency
C
PS = − B = phase/horizontal shift
VS = D = vertical shift
A = 7 (Ans. )
22. Two sides of a triangle measure 36m and 49m. One
possible dimension of the third side is:
A. 84m
B. 12m
C. 85m
D. 13m
E. None of the above
lmax < 36 + 49
lmax < 85m
lmin > 49 − 36
lmin > 13m
l = 84m (Ans. )
23. By how much would the perimeter of a square be
increased if its area is tripled?
A. 73.21%
Assuming s = 1
A = si2 = 1
B. 141.42%
3A = sf2
C. 173.21%
sf = 3
D. 41.42%
E. None of the above
%∆ =
final −initial
initial
100% =
%∆ = 73.21% (Ans. )
4 3−4 1
4 1
100%
24. The perimeter of a rectangle is 252cm. If the sides
are in the ratio 2:5, find its area.
A. 3109.72cm2
B. 3240.00cm2
C. 2769.04cm2
D. 3567.11cm2
E. None of the above
2L + 2W = 2(5x) + 2(2x) = 252
x = 18
L = 5x = 5 18
L = 90
W = 2x = 2 18
W = 36
A = LW = 90 36
A = 3240cm2 (Ans. )
25. A 23m high mast is placed on the top of the building.
An observer sees that the top of the mast is at an elevation
51°05’ and its bottom 35°10’. How high is the building?
A. 34.01m
B. 29.66m
C. 27.18m
D. 30.34m
E. None of the above
H + h = dtan 51°05′
H = dtan 35°10′
dtan 35°10′ + 23 = dtan 51°05′
d = 43.07m
H = 43.07tan 35°10′
H = 30.34m (Ans. )
26. One edge of a regular hexahedron is 24cm long.
Determine the ratio of the volume to the surface area.
A. 4
B. 6
C. 3
D. 2
E. None of the above
s3
6s2
243
6 24 2
V
SA
=
V
SA
= 4 (Ans. )
=
27. A donut has a cross sectional area of 9mm2 and a
mean radius of 32mm. Determine the volume of the
donut.
A. 675πmm3
Pappus Theorem
1st Proposition:
SA = PRθ
2nd Proposition:
V = ARθ
B. 567πmm3
V = 9(32)(2π)
C. 657πmm3
V = 576πmm3 (Ans. )
D. 576πmm3
E. None of the above
28. Determine the area in the first quadrant closed by the
parabola (x-5)^2 = 6(6-y).
A. 73.5sq. units
B. 47.1sq. units
C. 77.0sq. units
D. 75.0sq. units
E. None of the above
y=6−
x−5 2
6
At y = 0:
0=6−
x−5 2
6
x = 11
11
A = ‫׬‬0 ydx
A=
11
‫׬‬0
6−
x−5 2
6
dx
A = 47.06units 2 (Ans. )
29. Determine the perimeter in the first quadrant
closed by the parabola (x-5)^2 = 6(6-y).
A. 15.66 units
B. 16.21 units
C. 18.77 units
x−5 2
6
y=6−
y′ = −
x−5
3
At y = 0:
D. 19.04 units
0=6−
E. None of the above
x = 11
A=
11
‫׬‬0
x−5 2
6
1+
(x−5) 2
− 3
dx
A = 15.66units 2 (Ans. )
30. A certain horizontal curve has a central angle of
51.23. If the STA PC is 5+244 and STA PI is 5+562,
determine STA PT.
A. 5+873
B. 5+837
C. 5+845
STA PI = STA PC + T
5562 = 5244 + Rtan
51.23°
2
R = 663.272m
STA PT = STA PC + Lc
D. 5+854
STA PT = 5244 + 663.272 51.23°
E. None of the above
STA PT = 5 + 837 (Ans. )
π
180
31. An ascending grade of 3% intersects another grade
at STA 12+598 whose elevation is 533m. The change in
grade is -0.15% per 20m station. The two grades are to
be connected by a parabolic curve 800m long.
Determine the elevation of the midpoint on the curve.
A. 523m
B. 524m
C. 525m
D. 526m
E. None of the above
−
0.15%
20
=
g2 −g1
800
=
g2 −3%
800
g 2 = −3%
s1
g1
=
s1
3%
L−s1
g2
=
800−s1
−3%
s1 = 400
1
El. MP = El. PI − 0.03 400 + 2 400 0.03
1
El. MP = 533 − 0.03 400 + 2 400 0.03
El. MP = 527m (Ans. )
32. The tangents of a spiral curve intersect at an angle
of 30.1° at STA 12+598. The degree of the central curve
is 4.2° and the length of the spiral curve is 51.05m.
Determine the distance from the intersection of the
tangents at TS and ST and the midpoint of the long
chord joining SC and CS.
A. 12m
B. 13m
C. 14m
D. 15m
E. None of the above
32. The tangents of a spiral curve intersect at an angle
of 30.1° at STA 12+598. The degree of the central curve
is 4.2° and the length of the spiral curve is 51.05m.
Determine the distance from the intersection of the
tangents at TS and ST and the midpoint of the long
chord joining SC and CS.
A. 12m
B. 13m
C. 14m
D. 15m
Es + m
Es =
R+p
I
2
cos
−R
Ic
2
m = R − Rcos
p=
L2s
24R
20 = R(D)
R=
E. None of the above
p=
20
π
4.2 180
π
180
= 272.837m
51.052
24(272.837)
= 0.398m
32. The tangents of a spiral curve intersect at an angle
of 30.1° at STA 12+598. The degree of the central curve
is 4.2° and the length of the spiral curve is 51.05m.
Determine the distance from the intersection of the
tangents at TS and ST and the midpoint of the long
chord joining SC and CS.
A. 12m
B. 13m
C. 14m
D. 15m
E. None of the above
Es + m
Es =
R+p
cos
I
2
−R
R = 272.837m
p = 0.398m
Es =
272.837+0.398
cos
30.1
2
Es = 10.10m
m = R − Rcos
Ic
2
− 272.837
32. The tangents of a spiral curve intersect at an angle
of 30.1° at STA 12+598. The degree of the central curve
is 4.2° and the length of the spiral curve is 51.05m.
Determine the distance from the intersection of the
tangents at TS and ST and the midpoint of the long
chord joining SC and CS.
A. 12m
B. 13m
C. 14m
D. 15m
E. None of the above
32. The tangents of a spiral curve intersect at an angle
of 30.1° at STA 12+598. The degree of the central curve
is 4.2° and the length of the spiral curve is 51.05m.
Determine the distance from the intersection of the
tangents at TS and ST and the midpoint of the long
chord joining SC and CS.
A. 12m
B. 13m
C. 14m
I = Ic + 2θs
L2
51.052
θs = 2RLs = 2(272.837)(51.05)
s
180
π
θs = 5.36°
30.1° = Ic + 2 5.36
Ic = 19.3795°
m = 272.837 1 − cos
D. 15m
E + m = 13.995m
E. None of the above
E + m = 14m (Ans. )
19.3795
2
33. Given a block 300N lying on a horizontal floor, and
coefficient of friction to be 0.3, determine the frictional
force if a 50N force is applied parallel to the floor.
A. 90N
B. 50N
C. 100N
D. 300N
E. None of the above
fmax = 0.3 300N = 90N
P = 50N < 90N ∴ The block will not move.
σ Fx = 0
P−f=0
50 − f = 0
f = 50N (Ans. )
34. Given a block 300N lying on a horizontal floor, and
coefficient of friction to be 0.3, determine the frictional
force if a 100N force is applied parallel to the floor.
A. 90N
B. 50N
C. 100N
D. 300N
E. None of the above
fmax = 0.3 300N = 90N
P = 100N > 90N ∴ The block will move.
σ Fx = ma
f = 90N (Ans. )
35. A stone is dropped down a well and 5 sec later, the
sound of the splash is heard. If the velocity of sound is
342m/s, what is the depth of the well?
A. 107.67m
B. 122.63m
C. 1710.00m
D. 1602.33m
E. None of the above
d = Vo +
gt2
2
d = Vs 5 − t
gt2
2
= Vs 5 − t
9.81t2
2
= 342 5 − t
t = 4.6852s
d = 342 5 − 4.6852
d = 107.67m (Ans. )
36. A man deposited P5000 at the end of every year
earning 7.5% compounded continuously. Determine the
worth of his money after 15 years.
A. P3127540.18
B. P3172540.18
C. P3217540.18
D. P3271540.18
E. None of the above
F = Pert
rt
F = σ14
0 Pe
0.075t
F = σ14
0 5000e
F = P133545.48 (Ans. )
37. A self-employed engineer wants to get a lump sum of
P5000000 when he retires at the end of 25 years. How
much should he deposit every end of 3 months in a fund
that gives an interest of 10% compounded quarterly to
satisfy his desire?
A. P11559.39
B. P12416.28
C. P13269.14
D. P10815.45
E. None of the above
A
F=
i nt
1+
−1
n
i
0.1 25 4
5000000 =
A 1+ 4
0.1/4
A = P11559.39 (Ans. )
−1
38. Find the equivalent of 6% compounded semi-annually F = P 1 + i nt
n
in a rate compounded quarterly.
i nt
i
A. 5,56%
P 1+n
=P 1+n
B. 5.96%
0.06 2
i 4
C. 6.21%
1+
= 1+
2
4
D. 6.05%
i = 5.96% (Ans. )
E. None of the above
nt