Review Innovations
Civil Engineering Nov 2023
Solution 1.
x = speed of the boat in still water
y = speed of the water (direction is to the left)
y(10/60)
𝐿 = >𝑥 ! +
Œ
(x - y)(10/60)
56
54
the cap fell here

1!1,3!'!
4!
𝑦=
(x + y)t = (x – y)(10/60) + y(10/60) + yt
xt = (10/60)x
t = 10/60 hr
$!$,"!&!
("1",!'$ 89: '1."<°
(!
=0
1!1,3!'
<'3.!'
= 651.25 yd
Solution 5.
Solution 3.
n = number of houses in my road
m = my number
!("#$)(&#')
𝜃 = 54.86°
yd
C
y
yd
855
97
5
E
L
𝑛=
The least number of questions that the next student has
to answer correctly is…
(35M + 35) - 32M = 3M + 35
answer
Solution 7.
N = 10C2 = 45
−1 + √1 + 8𝑚!
2
Solution 8.
Hence, n = 288 when m = 204
q
D
x
10!
= 210
6! × 4!
Solution 6.
64% of 50 by M students = 32M
70% of 50 by (M + 1) students = 35M + 35
1 + 2 + ⋯ + (𝑚 − 1) = (𝑚 + 1) + (𝑚 + 2) + ⋯ + 𝑛
𝑚−1
𝑛−𝑚
[1 + (𝑚 − 1)] =
[(𝑚 + 1) + 𝑛]
2
2
(𝑚 − 1)𝑚 = (𝑛 − 𝑚)(𝑚 + 𝑛 + 1)
!
𝑚 − 𝑚 = 𝑛𝑚 + 𝑛! + 𝑛 − 𝑚! − 𝑚𝑛 − 𝑚
2𝑚! = 𝑛! + 𝑛
𝑛! + 𝑛 − 2𝑚! = 0
"#$! %&#'! ("''!
=L
= 1.5L
= 2L
= 2.5L
=…
Number of days = 198
𝑁=
Solution 2.
855! = 870! + 975! − 2(870)(975) cos 𝜃
dx
= - kxo
dt
B
x = xo e - kt
870 yd
𝐴+,- = 2𝐴+./
"
(870)(975) sin 𝜃 = 27"! 𝑥𝑦 sin 𝜃:
!
2𝑥𝑦 = 848,250
𝑥𝑦 = 424,125
1!1,3!'
𝑦= 4
Manila
=L
= L + L/2
= 1.5L + (1.5L)/3
= 2L + (2L)/4
=…
L, 1.5L, 2L, 2.5L … ß AP
an = a1 + (n – 1)d
100L = L + (n – 1)(0.5L)
n = 199
𝐿>?@ = B651.25! + 651.25! − 848,250 cos 54.86°
𝐿>?@ = 600 yd
yt + y(10/60) = 1
y(10/60) + y(10/60) = 1
y = 3 mph
A
!74 ! %
#$!$,"!&! (!()
((! )!
1!1,3!'!
Π= Location of bat when he discovered his cap was missing
 = Location of cap when 10 minutes after it fell
Ž = Location where he recovered his cap
cos 𝜃 =
=
!4 %
− 848,250 cos 54.86°
𝑥 − 4* = 0
𝑥 = 651.25 yd
(x + y)t
Ž
Solution 4.
Initial height
First day
Second day
Third day
…
𝐿! = 𝑥 ! + 𝑦 ! − 2𝑥𝑦 cos 𝜃
1 mi
yt
MSTE 1
When t = 25 yrs., x = (100% - 1.1%)xo = 0.989xo
0.989 xo = xo e -25 k
e - k = 0.9891/25
Cebu
Baguio
Davao
Review Innovations
Civil Engineering Nov 2023
Thus,
x = 0.989
t /25
xo
By De Moivre's Theorem
A × B = 6(2) – 2(5) – 1(2) = 0 ß orthogonal
P × Q = 2(-6) – 4(12) + 1(-3) = -63 ß not orthogonal
(r cis q)n = r n cis nq
0.5xo = 0.989t /25 xo
0.525 = 0.989t
25ln 0.5
t=
= 1566.65 yrs.
ln 0.989
(2 cis 20°)3 = 2 3 cis 60°
= 8(cos60° + i sin 60°)
answer
éi
j k i
jù
ê
P × Q = ê 2 -4 1 2 -4 úú
ê -6 12 -3 -6 12 ú
ë
û
P × Q = (12i – 6j + 24k) – (24k + 12i – 6j)
P × Q = 0 ß parallel
=4+4 3i
By Calculator (MODE à 2:CMPLX)
Note: rcis q = r Ð q
(2 cis 20°)3 = 23 Ð (3 ´ 20) = 4 + 4 3 i
Solution 9.
Mode …. 3 …. 5
Y (salt)
100
100e–0.05
Solution 15.
cos q =
Solution 12.
P = P4 + P5 + P6 + P7
P = 0.554 + (0.553)(0.451)(4C3) ´ 0.55
+ (0.553)(0.452)(5C3) ´ 0.55
+ (0.553)(0.453)(6C3) ´ 0.55
P = 0.6083
salt = 30yˆ = 22.31 lbs
Solution 10.
(𝑛𝐶𝑚)(𝑎A(B )(𝑏B )
Solution 14.
§ If A × B = 0, then A and B are orthogonal
§ If A × B = 0, then A and B are parallel
Solution 11.
Note: cis q = cos q + i sin q
When x = 0.5xo
X (time)
0
1
MSTE 1
B
2
= (12𝐶𝑚)(𝑥 ! )3!(B O P
𝑥
= (12𝐶𝑚)(2B )𝑥 !1(CB
cos q =
( 2 + 2 + 1 )( 6 + 3 + 2 )
2
2
2
2
2
2
=
4
3(7)
Solution 16.
V=|A×B´C|
#
𝑃 = Y [0.55C (0.454 ( 1 )(𝑥 − 1)𝐶3] × 0.55
é 2 -3 4 2 -3 ù
V = ê 1 2 -1 1 2 ú
ê
ú
ê 3 -1 2 3 -1 ú
ë
û
V = | (8 + 9 - 4) - (24 + 2 - 6) | = 7
GH1
Solution 13.
Evaporation rate, q = 0.002 m3/hr per square meter
For a surface 6-m in diameter
Q = 0.002 [ 1 p(6 2 ) ]
The 9th term is constant
9DE term = (12C8)(2" )𝑥 $ = 126,720
2(6) + 2( -3) + ( -1)(2)
q = 79.02°
Another Solution
For constant term,
24 – 3m = 0
m=8
r=m+1=9
A ×B
AB
4
Q = 0.018p m3/hr
For sum of coefficients, set x = 1
𝑆F = (1! + 2/1)3! − 126,720 = 404,721
Manila
V = Qt
0.03 = 0.018pt
t = 0.53 hr = 31.8 min
Cebu
**** ~o0o~ ****
Nothing Follows
answer
Baguio
Davao