CEERS REVIEW CENTER
EE REVIEW APRIL 2017
NOTES: CALCULATOR TECHNIQUES
Doña Esperanza Village, Tisa, Cebu City
Tel. No. 261-2244
email add: ceers.philippines@yahoo.com
1.
Given
4
3
1
6



. Find x.
x  1 x  1 2x x2  1
–1/12
1/12
A.
B.
B.
C. –1/13 *
D. 1/13
Solution:
Input in the calculator:
4
3
1
6



x  1 x  1 2x x 2  1
CALC
4.
 x  4  x  9 
B.
x
C.
 x  4  x  9 
D.
 x  2  x  2  x  3  x  3 
2
Express the logarithm of
A.
B.
C.
D.
a3
c5b2
in terms of log
3/2 log a – 5 log c – 2 log b *
3/2 log a – 2/5 log c – 5/2 log b
3 log a – 5 log c – 2 log b
3 log a – 5/2 log c – log b

B.
5.

2  619
15
1
, find x.
2
7.
555
16
C. x = - 2
A.
B.
1
3
C. 2
5
D.
1
*
2
Solution:
Substitute value of x from choices to equation 1
and solve for y. Then substitute both x and y to the
second equation. The 2nd equation should be
satisfied.
A.
B.
C.
D.
5x
18x4  10x  18
5x
*
18x 4  10x  18
x
18x 4  5x  18
5x
9x 4  10x  18
0.5X4  1  5X  0.5  SHIFT  SOLVE
X  2.187
0.5
0.5Y  9Y  
,Y  SHIFT  SOLVE
2
Y  0.0263
Substitute x to choices:
5x
 0.0263
18x 4  10x  18
Answer : B
8.
11
4
2
D.
13
C.
Given x 4 z  2z  5x  z and yz  9y  
y = f(x).
2x  1 x  8
, then

x 5
7x
x
16
9
4

11

Find the sum of the roots of x 2  4x  5  0.
A.
B.
6
3
C. 4 *
D. 0
Solution:
MODE  EQN  ax2  bx  c  0
Solution:
assign C as z in the equation.
Input in calculator:
CX
X4 Y  3XY 2  3  5X  16C
Y
CALC
X  2, Y  1, C  choices
Result should be zero.
z  16 / 9
Given y  2x  1 and xy  x4 y3  2x 
 9  X  2  X  2 
Answer : B
A.
Given x = 2 and y = -1 from the
zx
equation x 4 y  3xy2  3  5x  16z , find z.
y
0.6    SHIFT  STO  B  '"
Press   Result:
If
in calculator. Input the 2nd equation:
1
XY+X4 Y3  2X   Press 
2
Re sult is zero.
Thus, D. 1/ 2 is the answer.
A.
Press   Result: 0.7667
3
logA  5logC  2logB
2
Press   Result: 0.7667
0.7666
X4  5x 2  36
3.
x and y.
Ex: z  0.5
0.5    SHIFT  STO  A   
 A3 
log  5 2 
C B 


0.5    SHIFT  STO  X )
Press   Result:
Substitute : 1/ 2   
Y2
X  1/ 2 and Y=2 are already stored
0.7    SHIFT  STO  C hyp
Input in the calculator:
2
Solution:
Assign any value for z and solve for
6.
Input in the calculator:
Solution:
Assign a value of x. Ex: x  0.5 and
store the value. Substitute to given
and choices.
X
Y  2X  1, Y
SHIFT  SOLVE  X?
Solution:
Same strategy with #2. But this time assign
values to a, b and c. Ex. a = 0.5, b = 0.6 and c
= 0.7.
 9  x  2 x  2 *
555
16
AC  Input the choices to calcu
D. x = -2, 4/3 *
a, log b and log c.
Factor x4  5x2  36 .
A.
7  2449
30
Solution:
Same strategy with #1.
x ?  Substitute choices
The result should be zero.
Answer: -1/13
2.
x
a  1, b  4, c  5
roots : 5 and -1
sum  5   1  4
9.
2log4 x  logx 9  11 , find x.
A.
B.
z
, find
2
0.823 *
0.134
C. 0.567
D. 0.333
Solution:
Same strategy with #1.
Answer: 0.823
10. Given the system of equations:
x  2y  3z  4
2x  y  4z  3
3x  4y  z  2
Find z.
A.
B.
2*
4
C. 3
D. 5
Solution:
[1/17]
CEERS REVIEW CENTER
EE REVIEW APRIL 2017
NOTES: CALCULATOR TECHNIQUES
Doña Esperanza Village, Tisa, Cebu City
Tel. No. 261-2244
email add: ceers.philippines@yahoo.com
MODE  EQN  an x  bn y  cn z  dn
1 2
3
4
2
1
4
3
3 4
1
2
Result: x  4, y  3, z  2
Answer : z  2
A.
B.
a2
 (a  2)
a
11. Reduce to simplest form:  2
.
3a  12
a
a2
A.
B.
C. a – 2
a2
D.
*
a4
a+2
a2
a2
13. If 3 x  9 y and 27 y  81z , find x/z.
3/8
8/5
C. 5/8
D. 8/3 *
Solution:
3x  9 y  xlog3  ylog9
x  2y
27 y  81z  ylog27  zlog81
3
z y
4
x
2

8/3
z 3/4
1
1
14. Given x   5 , find x2  2 .
x
x
Solution:
Same strategy with #2.
12. Find the value of k so that the
equation 2x 2  3kx  9  0 will have only one
root.
A.
2 2
C. 2 3
B.
2 2*
D. 3 3
Solution:
Substitute the choices for k. Then solve for the
roots. It should have 1 root only.
try k  2 2
MODE  EQN  ax  bx  c  0
A.
B.
25
65
C. 23 *
D. 42
Solution:
1
x   5  SHIFT  SOLVE
x
x  4.791
1
Input : x 2  2  Result: 23
x
15. The first term of an arithmetic progression is 6
and the 10th term is 3 times the first term.
What is the 5th term?
2


a  2, b  3 2 2 , c  9
3 2
2
try k  2 2
root :
B.


a  2, b  3 2 2 , c  9
3 2
2
Answer: Either A or B
root : 
A.
11
3
11
4
Solution:
MODE  STAT  A  BX
x y
1 6
10 3  6 
AC  5yˆ  11.33 or 34/3
34
*
3
33
D.
2
C.
16. The sum of the five arithmetic means between
34 and 42 is
A.
B.
150
266
Reciprocal of harmonic progression is
arithmetic progression.
1 1 1
Harmonic progression: 1, , ,
4 x 10
Arithmetic progression: 1,4,x,10
MODE  STAT  A  BX
x y
1 1
2 4
AC  3yˆ  7
C. 190 *
D. 232
Solution:
MODE  STAT  A  BX
x y
1 34
7 42
AC 
ˆ
  190
  Xy,2,6
Answer : 7
17. There are 6 geometric means between 4 and
8748. Find the sum of all the terms.
A.
B.
4368
13116
C. 13120 *
D. 4372
Solution:
MODE  STAT  A  B ^ X
x y
1 4
8 8748
AC 
19. How many terms of the arithmetic sequence 9, -6, -3… must be taken so that the sum is
66?
A.
B.
11 *
12
C. 13
D. 10
Solution:
MODE  STAT  A  BX
x y
1 9
2 6
ˆ
  13,120
  Xy,1,8
AC 
1 1 1
18. What is the value of x if 1, , , ... form a
4 x 10
harmonic progression?
ˆ

  Xy,1,choice
Substitute the choices: 11
ˆ
  66
  Xy,1,11
Answer : 11
A.
B.
8
6
Solution:
C. 7 *
D. 2
20. Given the 5th and 9th term of the geometric
progression, 11 and 25 respectively, 56.82 is
nth term. n =
A. 14
B. 13 *
C. 17
D. 16
Solution:
MODE  STAT  A  B ^ X
x
5
9
y
11
25
AC  56.82xˆ  13
[2/17]
CEERS REVIEW CENTER
EE REVIEW APRIL 2017
NOTES: CALCULATOR TECHNIQUES
Doña Esperanza Village, Tisa, Cebu City
Tel. No. 261-2244
email add: ceers.philippines@yahoo.com
MODE  STAT  A  BX
x y
1 4
4 1
AC  8yˆ  3
21. What is 150o in gradians?
A.
500
*
3
B. 100
400
3
200
D.
3
C.
22. What is 66 gradians to radians?
B. 
33
*
100
11
D.

2
C.
A. 4:24:33 *
B. 4:19:34
C. 4:20:33
D. 4:29:32
Solution:
MODE  STAT  A  BX
x y
4  120
5 210
AC  15xˆ  5.409   ' ''
Solution:
SHIFT  SET  UP
 Angle unit  Rad
66  SHIFT  DRG  g
Re sult :
25. In how many minutes after 12 noon will the
hands of the clock forms an angle of 30o?
33

100
23. What time after 5 o’clock will the hands of the
clock form an angle of 5o for the first time?
A. 5:27:22
B. 5:26:22 *
C. 5:12:22
D. 5:29:22
A.
B.
5.454 min *
6.001 min
C. 5.364 min
D. 5.909 min
Solution:
t  1 5.4545   5.4545 min
26. Given a line passing through points (1, 4), (4,
1) and (k, 8). What is k?
A.
B.
-4
5
A.
61 *
C.
60
B.
33
D.
32
Solution:
MODE  CMPLX
 2  5i   0  11i 
C. -3 *
D. 0
61
28. The distance between (5, -7, z) and (3, -1, 6)
is 8.062. Find the value of z.
A.
B.
12
3
C. 11 *
D. 8
61
29. Find the equation of the parabola passing
through the points (1, 8), (6, 53) and (-2, 29).
A.
y  3x 2  2x  10
B.
y  2x 2  5x  11 *
C.
y  2x 2  4x  1
D.
y  2x 2  5x  11
Solution:
30. Given the equation of the
circle 2x 2  2y 2  3x  8y  3  0 , find the
coordinates of its center.
A.
B.
(1, -2)
(0.75, -2) *
C. (0.75, -2.5)
D. (1, 2.5)
Solution:
h
B
3
3

  0.75
2A
2  2 4
k
B
8

 2
2A
2  2
Center :  0.75,  2 
Solution:
MODE  CMPLX
 2  5i   0  11i 
Result: 4o24'32.73"
Interpreted as 4 : 24 : 33

y  Cx 2  Bx  A
y  2x 2  5x  11
24. What time after 4 PM will the hands of the
clock forms and angle of 15o for the second
time?
500
3
3

10
27. Find the distance between (5, -5) and (0, -11).
Result: 5o 26'21.82"
Interpreted as 5:26:22

A.
6 180
MODE  STAT  _  Cx 2
x
y
1
8
6 53
2 29
AC  Identify the A, B and C
AC  5xˆ  5.4394   ' ''
Solution:
SHIFT  SET  UP
 Angle unit  Gra
150  SHIFT  DRG  
Re sult :
Solution:
MODE  STAT  A  BX
x y
5  150
31. Find the equation of the line that passes
through the points (1, -9) and (0, 5).
A.
B.
14x + y = 5 *
5x + 14y = 1
C. 14x – y = 5
D. 5x – 14y = 1
Solution:
MODE  STAT  A  BX
x
y
1
9
0
5
AC  Identify the A, B
y  Bx  A
y  14x  5
14x  y  5
Solution:
[3/17]
CEERS REVIEW CENTER
EE REVIEW APRIL 2017
NOTES: CALCULATOR TECHNIQUES
Doña Esperanza Village, Tisa, Cebu City
Tel. No. 261-2244
email add: ceers.philippines@yahoo.com
32. Locate the vertex of the
parabola 5y  16y  15x  23  0.
A.
B.
2
7 5
 5 , 8 


 17 8 
 25 ,  5  *


A.
B.
 8 17 
C.   , 
 5 25 
 5 7
D.   , 
 8 5
Solution:
k
Solution:
MODE  CMPLX
17
25
 17 8 
h,k    25 , 5 


7
8
13
y
12
o
35o
A.
B.
34.23
34.09o *
C. 35.11
D. 32.10o
8  12choices  Result : 7
A.
B.
y + z = -1 *
y + 2z = -1
C. 2x + y = 2
D. x + y = 1
Solution:
Create 2 vectors:
VctA:  4  1 i   2  2  j   3  3  k
VctB :  4  1 i   7  2  j   8  3  k
Cross Product of VctA and VctB:
0i  15j  15k
Plane equation using (1,2,-3):
0  x  1  15  y  2   15  z  3   0
15y  30  15z  45  0
y  z  1
84 m * *
88 m
8  1234.09  7
36. A pole casts a shadow 15m long when the
angle of elevation of the sun is 61o. If the pole
is leaned 15o from the vertical directly towards
the sun, determine the length of the pole.
59.9m
54.2m *
C. 55.6m
D. 58.7m
Solution:
L
15

sin61o sin 180  61  90  15 
C. 86 m
D. 90 m
37. Find the angle in the figure:

7.5 *
8.3
Solution:
Rec 13,35   x  __, y  7.456
A.
B.
43 *
129
C. 86
D. 22
Solution:
MODE  MATRIX  MatA  3x3
9 1 1


 6 0 1  AC
5  5 1
1
det MatA   43
2
40. Given (2, 6, -8), find the spherical coordinates.
A.
B.
C.
D.
(10.9, 78.9o, 114.5o)
(11.3, 61.5o, 123.9o)
(10.2, 71.6o, 141.7o) *
(12.0, 66.0o, 111.9o)
10
A.
B.
C. 11.9
D. 10.6
39. What is the area of the triangle given the
vertices (9, 1), (-6, 0) and (5, -5).
Answer :34.09o
A.
B.
A.
B.
Answer : 7.5
o
L  54.2 m
34. A surveyor wishes to calculate the distance
between two signs. The distance from the
surveyor to the first sign is 70 m and the
distance to the second sign is 100 m. The
angle between the two lines of sight measures
56º. Find the distance between the two signs.
A.
B.
Try 34.09
41. Find the value of the determinant of A.
 6 0 3 11
 9 7 5 3 

A
 11 4 7 8 


 2 1 1 8 
o
Solution:
MODE  CMPLX
33. Find the equation of the plane that contains
the points (1, 2, -3), (4, 2, -3) and (4, 7, -8).
r, ,    10.198, 71.6o, 141.7o 
38. Find the unknown side in the figure:
2
x  0.68 
if reference is north: =90+51.67=141.7 o
Answer : 21.8
35. Given the triangle below, find the unknown
angle.

r  6.3246 was stored in X by calcu
Pol  X,  8   r  10.198,   51.67o
o
Answer :84.09
Substitute k  1.6 to y.
Pol  2,6   r  6.3246,   71.565
C. 21.8o *
D. 22.5o
Solution:
Pol 10,4   r  __,   21.8o
70  10056
B
16
8

  1.6 
2A
2 5
5
5  1.6   16  1.6   15x  23  0
22.9o
23.1o
133
154 *
C. 109
D. 122
Solution:
Modify the matrix:
6 6 0 6 3 6 11 6 
 9
7
5
3 
A
 11
4
7
8 


2

1
1
8 

Store in calculator the three matrices:
MODE  MATRIX  MatA  3x3
7 5
4 7

 1 1
3
8 
8 
AC  SHIFT  MATRIX  4 
Dim  MatB  3x1 
 9 
11 
 
2 
AC  SHIFT  MATRIX  4 
Dim  MatC  1x3 
0
3/6
11/ 6
det MatA  MatB xMatC  *  6 
Re sult : 154
Solution:
4
[4/17]
CEERS REVIEW CENTER
EE REVIEW APRIL 2017
NOTES: CALCULATOR TECHNIQUES
Doña Esperanza Village, Tisa, Cebu City
Tel. No. 261-2244
email add: ceers.philippines@yahoo.com
42. A solid has a circular base with radius 8. Find
the volume of the solid if every plane
perpendicular to a fixed diameter is a square.
8192
*
3
8111
3
A.
B.
8192
C.
5
8913
D.
4
MODE  STAT  _  Cx 2
x
y
0
0
16 ^ 2 
y  Cx 2  Bx  A
y  4x 2  64x

16
0
8192
4x 2  64x dx 
3
43. Determine the volume of the water inside a
spherical container, if the sphere has radius 5m
and the water is 8m deep.
A.
B.
Solution:
10 0
AC  Identify the A, B and C
y  Cx 2  Bx  A
445
4
440
3
V

8
0
 3.14x 2  31.416x dx  469.42
444
C.
3
448
D.
*
3
 1 3 3 
10 1 8 


A   9 3 0  and B   2 2 7 
 2 3 8 
9 3 8
14491
14391 *
1 3  3 
9 3 0 


 2 3 8 
AC  SHIFT  MATRIX  4 
Dim  MatB  3x3 
A.
B.
0.53 *
1.49
C. 0.54
D. 0.12
45. Find the determinant of the inverse
 1 3 3 
of A   9 3 0  .
 2 3 8 
C. -3.44 x 10-3 *
D. 3.89 x 10-3
C. -9
D. 12
 5

6
1

2
AC  SHIFT  MATRIX  4 
Dim  MatB  2x2 
3

9
0

 1
AC  SHIFT  MATRIX  4 
Dim  MatC  2x2 
2

3

4 

 9
det MatA  MatCxMatB 1
1 3  3 
9 3 0 


 2 3 8 

Re sult :  11
AC  SHIFT  MATRIX  4 

10
-11 *
Solution:
Store in calculator the 3 matrices:
MODE  MATRIX  MatA  2x2
Solution:
Store in calculator the 2 matrices:
MODE  MATRIX  MatA  3x3
8
7 
8 
det MatA xMatB 1
Re sult : 14391
-2.44 x 10-3
2.56 x 10-3

 1 3 3 
10 1 8 
46. Given A   9 3 0  and B   2 2 7  ,
 2 3 8 
9 3 8
A
find
.
B
10 1
2
2

 9
3
det MatA  3MatB 
A.
B.

A.
B.
Dim  MatB  3x3 
8

7
8 
2 4 
C
.
3 9 
 1 3 3 
9 3 0


 2 3 8 
C. 14991
D. 14099
Solution:
Store in calculator the 2 matrices:
MODE  MATRIX  MatA  3x3
10 1

2
2
 9
3
Input the matrix in the calculator:
MODE  MATRIX  MatA  3x3
det MatA 1
C
,
B
 1 5
3 0 
given A  
 , B  9 1 and
2 6 


Solution:
Answer : 3.44x10 3
Answer : D
A.
B.
47. Find A 
2
44. Find A  3B
16 0
AC  Identify the A, B and C
V
 5
5
y  3.14x 2  31.416x
Solution:
8
MODE  STAT  _  Cx 2
x
y
0
0
Re sult : 0.53
48. Find slope of the
curve 2x 2  5y 2  6x  8y  11  0
A.
B.

7/5
7/6 *
C. 5/6
D. 5/7
No Answer. Erroneous question.
 1 6 11
49. Given A   3 8 3  and its inverse
23 9 0 
x
70 
 27
1 
is B 
69 253 30 . Find x and y.
1340 
y 
 157 129
[5/17]
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A.
B.
-99, -10 *
-107, 101
C. -89, 100
D. -66, -25
A.
B.
Solution:
Store in calculator the matrix:
MODE  MATRIX  MatA  3x3
6
1
3 8

23 9
MatA
C. 32ei53.13
40 *
o
i 43.13o
D. 24e
24
Solution:
MODE  CMPLX
 3  4i 853.13 
 11
 3 
0 
Result : 40
52. Expressed  2  3i in polar form.
4
5 1 11
50. Given matrix A = 1 9 2 , find the
3 0 2
A.
B.
13e3.9312 i
B.
0.8831 i
13e
1 3
9 0  *
2 2 
1 3
9 0 
2 2
5
C.  1
11
5
D.  1
10
1 3 
9 0 
2 2 
1 3 
9 0 
2 2 
Solution:
Store in calculator the matrix:
MODE  MATRIX  MatA  3x3
5 1 11
1 9 2
3 0 2
AC  SHIFT   VECTOR  5 
Dim  VctB  3
24  8 6
AC  calculate :
AC  SHIFT   VECTOR  5 
abs  VctA  VctB  VctC 
54. What is the cross product A  B of the vectors?
A  i  4j  6k , B  2i  3j  5k
169  134.76o
Since angles in choices are positive:
  134.76  360  225.24
change angle unit to radian :
 
225.24 
  3.9311
 180o 
o
A.
2i  7j  5k
C. 2i  7j  5k *
B.
i  7j  5k
D.  i  j  k
Input the 2 vectors in calculator:
MODE  VECTOR  VctA  3
1 4
6
53. Assume the three force vectors intersect at a
single point: F1  i  3j  4k , F2  2i  7j  k ,
AC  SHIFT   VECTOR  5 
Dim  VctB  3
2 3 5
AC  calculate :
VctA x VctB
Result : 2 7  5
Answer : 2i  7j  5k
F3  i  4j  2k . What is the magnitude of the
resultant force vector, R?
14.73
15 *
C. 13.23
D. 16.16
Solution:
A.
B.
2
3*
C. 1
D. 4
Solution:
If vectors are perpendicular, their
dot products is zero.
A  B  0  2  4   b  2   1 2 
Solution:
3.9311i
Answer : 84.32o
56. Compute the value of “b” if A and B are
perpendicular. A  2i  bj  k , B  4i  2j  2k
Answer : 15
 2  3i3  2  3i  r
A.
B.


VctA  VctB
cos1 
 abs  VctA  x abs  VctA  


D. 169e0.8831 i
Solution:
MODE  CMPLX
1693.9311  169e
AC  SHIFT   VECTOR  5 
C. 169e3.9312 i *
o
transpose of A.
5
1

11
5
1

11
A.
Input the 2 vectors in calculator:
MODE  VECTOR  VctA  3
4 12 6
Dim  VctB  3
2 7 1
Dim  VctC  3
1 4 2
AC  calculate :
1
0.201  99 / 1340 0.0522 
Re sult :  0.051
0.1888  0.022 
 0.117 0.0962  1/ 134 
Therefore, x  99, y  10
Input the 3 vectors in calculator:
MODE  VECTOR  VctA  3
1 3
4
b3
57. Find the area of the triangle whose vertices
are P(2, -1, 3), Q(1, 2, 4) and R(3, 1, 1).
A.
B.
4.745 *
10.3
C. 8.38
D. 7.74
Solution:
Input the 2 vectors in calculator:
MODE  VECTOR  VctA  3
1  2  2  1
4  3
AC  SHIFT   VECTOR  5 
Dim  VctB  3
Trn MatA 
55. What is the angle between two vectors A and
B? A  4i  12j  6k , B  24i  8j  6k
5 1  3 
Re sult : 1 9 0 
11 2 2 
A.
B.
51. What is the product of the complex numbers 3
84.32° *
85.28°
C. 75.28°
D. 70.32°
 3  2 1  1
1  3 
AC  calculate :
abs  0.5 * VctA x VctB 
Answer : 4.743
Solution:
o
– 4i and 8 ei53.13 ?
[6/17]
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58. Find the volume of the parallelepiped whose
edges are represented by A  2i  3j  4k ,
B  i  2j  k and C  3i  j  2k
C.
D.
5
6
C. 7 *
D. 8
Solution:

B.


Input the 3 vectors in calculator:
MODE  VECTOR  VctA  3
2 3
4
AC  SHIFT   VECTOR  5 

Dim  VctB  3
1 2 1
AC  SHIFT   VECTOR  5 
Dim  VctC  3
3 1 2
AC  calculate :
Answer : 7
tan x
.
x 0 x
C. infinity
D. 1 *
Solution:
MODE  COMP
Solution:
MODE  COMP
A.
2
B.
–1
dy
60. Find
if y  5 2x 1 .
dx
A.
B.
5 ln25 *
5 ln2x  1
2x 1
2x 1
 ln5
D.  5
ln10
2x 1
C. 5
2x 1
1
2
D. – 2 *
C.
6*
6
A.
e3t
12t  1
C.
9e3t
*
12t  1
B.
3e3t
12t  2
D.
9e3t
12t
Solution:
Assign t  0.5
dy
d 3t
e
t  0.5
dt
dt
y' 

 5.762
dx
d 2 t
2t  

dt dt 
3 t 0.5
Substitute t  0.5 to choices
64.
x 11

x
lim
x 0
– 1/2
1/2 *
C. – ¼
D. 1/4
9e3t
9e  

 5.762
12t  1 12  0.5   1
3 0.5
C.
67. Find the partial derivative of
z  3x 2 tan y  y 3 cos x with respect to y.
65. What is the first derivative of y 2 x  yx 2  5x if x
= 2?
A.
B.
– 0.2
– 0.3 *
C. – 0.5
D. – 0.4

Solution:
Y 2 X  YX2  5X, Y
SHIFT  SOLVE
X? 2  
Y1  ___  SHIFT  STO  A
SHIFT  SOLVE
X? 2.00001  
Y2  ___  SHIFT  STO  B
63. What is the slope of the curve y = 1 + x2 at the
point where y = 10?
A.
x 3
X 11
 CALC  X  0
X
Re sult : 0.5
Solution:
x2y  8  0
8
y 2
x
d  8 
y' 
x  2  2
dx  x 2 
 CALC  X  0.01
X
Result : 1

t
and y  e3t .
3
Solution:
SHIFT  SET  UP  Angle unit  Rad
tan  X 
66. Find y’ if x  2t 2 
 
A.
B.
62. Find the slope of the curve defined by the
equation x2y – 8 = 0 at the point (2, 2).
59. Evaluate lim
0
5
C. 25, 35
D. 10, 50
P  x1y 2
1
x
 60   20
1 2
2
y
 60   40
1 2
 VctA x VctB   VctC

61. Divide the number 60 into two parts so that
the product of one part and square of the
other is maximum. Find the two numbers.
15, 45
20, 40 *
D. 8
10  1  x 2
x3
d
y' 
1  x2
dx
A. 52x 1 ln25  80.47
A.
B.
7
Solution:
y  1  x2
Answer : A
V   AxB   C
A.
B.
Solution:
MODE  COMP
d 2X 1
5
x  0.5  80.471
dx
Substitute x =0.5 to choices
y' 
y
BA

 0.2958
x 0.00001
A.
3x 2 sec 2 y  3y 2 cos x *
B.
3x sec y  3y 2 cos x
C.
3x 2 sec y  3y 2 sin x
D.
3x 2 tan y  y 3 cos x
Solution:
SHIFT  SET  UP  Angle unit  Rad
x will be considered constant.
Assing x  1,
z  3 tan y  y3 cos 1
z d

3 tan y  y 3 cos 1
 4.3
y  0.5
y dy
Substitute x=1, y=0.5 to choices


A. 3x 2 sec 2 y  3y 2 cos x 
1
2
2
3 1
 3  0.5  cos 1  4.3
2
cos  0.5 
C. 9
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68. Find the third derivative of
y
1
3x 2  8x 2
 cos x.
d
 sin  3x   ln 7x  x 0.5  __  A
dx
d
y '2 
 sin  3x   ln 7x  x 0.50001  __  B
dx
BA
y" 
 12.98
0.00001
Substitute x  0.5 to choices
y '1 
A.
y '''  3x

5
2
5

3x 2
B.
y ''' 
C.
y '''  3x
D.
y '''  3x
 sin x *
 ex

3
2
 sin x

5
2
 cos x
 9 sin  3  0.5    0.5 
2
2
2
 7

 0   1.936


 2

2
 5

 0   2.667


 3

Therefore, C has the nearest satisfying
point.
5

d    4
3

 12.98
2
Therefore, C is the answer.
1

d  2
 3x  8x 2  cos x 
y '1 
 __  A


dx

x 0.5
1

d  2
 3x  8x 2  cos x 
 __  B

dx 

 x  0.50001
1

d  2
 3x  8x 2  cos x 
y '3 
 __  C

dx 

 x  0.50002
BA
y "1 
 ___  D
0.00001
CB
y "2 
 ___  E
0.00001
E D
y "' 
 17.4902
0.00001
Substitute x  0.5 to choices
A. y '''  3x

5
2
 sin x  3  0.5 

5
2
y '''  17.449
Therefore, A is the answer.
 sin  0.5 
70. Find all points on the graph of
y  x 3  3x where the slope is equal to zero.
A.
B.
C.
D.
A.
B.
C.
D.
Solution:
y'  m  0
Substitute x-coordinate of choices:




d 3
x  3x
0
x 1
dx
d 3
y' 
x  3x
0
x 1
dx
Substitute x to the equation to
get the correct y:
y' 
69. Find the second derivative
of y  sin 3x   ln 7x .
B.
9sinx  x2
9 sin  3x   x
C.
9sin  3x   x 2 *
D.
27cos  3x   2x
3
72. Build a rectangular pen with three parallel
partitions using 500 feet of fencing. What
dimensions will maximize the total area of the
pen?
(-2, 2) and (1, 2)
(1, -2) and (-1, -2)
(-1, 2) and (1, -2) *
(-2, -2) and (1, 2)
y  1  3 1  2  point: 1,  2 
3
50 ft by 125 ft *
25 ft by 187.5 ft
60 ft by 100 ft
55 ft by 112.5 ft
5x  2y  500
A  xy
 500  5x 
A  x

2


Mode  Table
 500  5x 
f x  x 

2


(Based on choices: x is multiple of 5)
Start ? 5 
End? 60 
Step? 5 
f x
x
45
6187.5
50
6250  max.
55
6187.5
x  50
500  5x 500  5  50 

 125
2
2
Answer : 50 ft x 125 ft
y
Solution:
73. A trough 10m long as it ends isosceles
trapezoids, altitude 2 m, lower base 2 m.,
upper base 3 m. If water is let in at a rate of 3
m3/min, how fast is the water level rising when
the water is 1 m deep?
x
y
A.
B.
0.18 m/min
0.12 m/min *
C. 0.21 m/min
D. 0.28 m/min
Solution:
y   1  3  1  2  point:  1, 2 
3
Answer : C
A.
7

d=   4 
2

C.  9 sin  3x   x 2
Solution:
y '2 
Solution:
Sustitute the choices if it satisfy the
equation: C and D satisfy
Solve for distances:
0.5
71. Find the point (x, y) on the graph
of y  x nearest the point (4, 0).
A.
7 7
 2, 2 


7 7 
C.  ,
*
 2 2 


B.
5 5
 3,3 


5 5 
D.  ,
 3 3 


x
1
[8/17]
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dh dV / dt 3 m3 / min


dt
A
A
0.5 2

x
1
x  0.25
A   0.25  0.25  2 10  25
dh 3

 0.12 m/min
dt 25
certain instant is:  2t  5  m/s. Find how far it
2
A.
B.
62.67 m *
63.33 m
C. 4.28 ft3/min
D. 5.79 ft3/min
Solution:
5 ft
75. At a certain instant, car A is 60 miles north of
car B. A is travelling south at a rate of 20 mi/h
while B is travelling east at 30 mi/h. How fast
is the distance between them changing 1 hour
later?
A.
B.
xy 2dxdy
A.

B.
1
8
3 mi/h
2 mi/h *
C. 5 mi/h
D. 6 mi/h
Solution:
1

2
0 3

s  vdt 

4
2t2  5 dt  62.67 m
 y  
16 ft
2
A. sec 4x  cos4 4x  C
1
B.
sec 4 4x  C *
16
1
C. sec 4 4x  C
4
79.
 
D. tan 4x sec 4 4x  C
Solution:
77. Evaluate
 cos
sin 4x
5
30t
ds d 
 
dt dt 
  30t 
 60  20t 
2
2
5
dx
4x
ln 2
1
0
1
A.
B.
d  1

sec 4 4x 
dx  16


2
2
  30t    2
t 1

d 
1

 72.85
dx  16cos  4x 4 

 x 0.5
sin4x
cos5 4x

e y  1tan xdxdy
0.123
0.034
ln 2
1
0
1
C. 0 *
D. 1.267
ey  1tan xdxdy
 ln 2 ey  1 dy   1 tan xdx   0
 0
  1

Differentiate the choices and substitute
the value of x used in the given.
s
2
D. 
2
3
B.
 60  20t 
4
3
xdx  dy

1 5
5
2
0  2 y dy   6
1
0
0
Solution:
20 ft
s
C.
xy 2dxdy
 
60  20t
5
*
6
Solution:
1 ft
dV / dt

12 min
16
dV 16

 4.188 ft 3 /min
dt
12
dV
leaking  8  4.188  3.811 ft 3 / min
dt
r
2
0 3
C. 65.00 m
D. 69.89 m
2
3.18 ft3/min
3.81 ft3/min *
1

78.
moves in the interval from t = 0 to t = 4
seconds.
A  r 2    4   16
74. Water is poured at the rate of 8 ft3/min into a
conical shaped tank, 20 ft deep and 10 ft
diameter at the top. If the tank has leak in the
bottom and the water level is rising at the rate
of 1 in/min, when the water is 16 ft deep, how
fast is the water leaking?
A.
B.
76. The velocity of a body “t” seconds after a
dh dV / dt

dt
A
1 ft
dV / dt

12 min
A
5 20

r 16
r4
sin  4  0.5  
cos  4  0.5  
5
80. Find the area of the region in the first
quadrant, bounded by the curves y = x3 and y
= 4x.
A.
B.
A=2
A=4*
C. A = 8
D. A = 6
Solution:
 72.85
[9/17]
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NOTES: CALCULATOR TECHNIQUES
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83. Solve the equation xdy – ydx = 2x3dx.
y  x 3 , y  4x
x  4x
3
C.
D.
x  4x  0
x  0,x  2
Trial and error:
Using vertical strip:
3
X? 0.5 
x1
  4x  x dx
3
0
81. Determine the area under the curve
y  x3  3x 2 and the x-axis between x = 1
and x = 3.
C. 50
D. 52
A   y dx
3
  x 3  3x 2 dx
1
A  46
82. Find the area of the region enclosed by the
given set of curves y  x 2 and y  x  2.
C.6.50
D.4.5 *
dC d  1  x3 

 5


dx dx  x 
X  0.5
dC
dy
5
  dx  
 2.5
dC
dx
2
dy
Therefore, C is the answer.
84. Solve for the equation xdy + ydx = 2x ydx.
Solution:
y  x2, y  x  2
B.
ln x  y  x2  c
x2  x  2
C.
ln xy  x2  c *
x2  x  2  0
x  1,x  2
Trial and error:
Using vertical strip:
D.
ln x  y  2x2  c
x1
2
  x  2  x 2 dx
1
A  4.5
[10/17]
e
y

0.5  0.5  2 
e1
 0.2759
1 3
x  x2  c
3
1 3
x  x2  C
3
dC d  1 1 3


e  x  x2 
 0.75
dx dx 
3
 X 0.5


dy x(x  2)
.

dx
ey
A.
y  ln x3  x2  c
B.
y  ln
C.
1
y  ln x3  x 2  c
2
D.
1
y  ln x3  x 2  c *
3
Solution:
x  x  2
ey 
85. Solve the equation
2
ln xy  x3  c
A   y 2  y1 dx
Try : D. y  ln


3
A.
x2
X? 0.5 
dC d

ln x 1  x 2
1
X  0.5
dx dx
dC d

ln 0.5y  0.52
1
Y 1
dy dy
dC
dy
1
  dx    1
dC
dx
1
dy
Therefore, C is the answer.
dC d  y  0.53 

2


dy dy  0.5 
Y 1
Solution:
A. 8.55
B. 7.32
yx
x
A
c  ln xy  x 2
Try : C. y  x 3  cx
c
XA  Y  2X Y, A
SHIFT  SOLVE
Try : C. ln xy  x 2  c
Y? 1 
dy
A
 2.5
dx
A4
Assign : x  0.5,y  1, dx  1, dy  A
2
Y? 1 
dy
A
 1
dx
XA  Y  2X3 , A
SHIFT  SOLVE
A   y 2  y1 dx
A. 46 *
B. 42
C. y = x3 + cx *
D. y = x3 + c
Assign : x  0.5,y  1, dx  1, dy  A
x2
2
y = x2 + cx
y = x2 + c
Assign : x  0.5,y  1, dx  1, dy  A
1 3
x  x2  c
3
dC d  y 1
3
2

e   0.5    0.5  
e
dy dy 
3
 Y 1
dy
0.75
1

   0.2759
dx
e
1
Therefore, D is the answer.
86. Solve the differential equation
(1  2xy)dx  (4y3  x2 )dy  0
a.
b.
c.
d.
x + x2y + y4 = c
x – x2y + y4 = c *
x – xy + y4 = c
x – x2y2 + y4 = c
Solution:
Assign : x  0.5,y  0.5, dx  1, dy  A


1  2XY  4Y 3  X2 A  0,A
SHIFT  SOLVE
X? 0.5 
Y? 1 
dy
A
 2
dx
Try : B. x  x 2 y  y 4  C
dC d
4

x  x 2  0.5    0.5 
 0.5
dx dx
X  0.5
dC d
2

0.5   0.5  y  y 4
 0.25
dy dy
Y 1
dy
0.5

 2
dx
0.25
Therefore, B is the answer.




CEERS REVIEW CENTER
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NOTES: CALCULATOR TECHNIQUES
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87. The general solution of y” + 2y’ + y = 0 is:
x
y  c1e
B.
 c2xe
x
y  c1ex  c2xex *
D.
y  c1ex  c2xex
Y? 1 
dy
A
 1.25
dx
m  2m  1  0
m  1  repeated 2
y   C1  C2 x  e1x
 C2 xe
x


88. Solve the equation x 2  y 2 dx  2xydy  0
A.
x2  y2  cx
C. x  y  c
B.
x2  y2  cx *
D. x2  y2  c
dC d  0.52  y 2 

 4


dy dy  0.5 
Y 1
dC
dy
5
  dx  
 1.25
dC
dx
4
dy
Therefore, B is the answer.
x
3
x 4 x2

xC*
4
2
B.
x 4 x2

C
4
2

90. Evaluate
A.
B.

0
A.
B.
1
2
lnx   C *
2
D. 2 (ln x)2 + C
C.
x2  4
dx .
x2
4
4
8
0
x2  4
dx 
x2
C. 2
D. 16 *

4
0
x
x

xC
4
2
D.
x 4 x3

xC
4
3
1.21
3.29
10
2
C. 1.61 *
D. 2.39
1
dy  1.61
y
3
1

4
2
4
2xydydx
3
 112
dy
 10  0.05y ; x =0, y=20,
dx
find the y if x = 5.
95. Given the D.E.
A.
B.
39.2
28.3
C. 59.8 *
D. 89.5
5
dy
 dx
10  0.05y 0
Try : 59.8
59.8
5
dy
20 10  0.05y  0 dx
55

choices
20
the y if x = 5.
A.
B.
17.2 *
71.2

C. 20.8
D. 30.2
5
ydy   e x dx
0
Try : 17.2
6
4
0
0.0231
0.0184 *
choices
1
93. Evaluate sin  cos d .
[11/17]
3


2
A.
B.
40dydx  
Solution:
 x dy  
y1
4
2
96. Given the D.E. e x dx  ydy  0; y  0   1, find
Solution:
y2
3
1
x  2 dx  16
92. Find the area under the curve y = 1/x between
the limits y = 2 and y = 10.
A.
B.
4
2
Solution:
Solution:
Simplify first to avoid math error.

C. 100
D. 120
 40   1dy    1dx   2   y dy    x dx 
 2
  1

 2
  1

Solution:
Same strategy with number 89.
91. Evaluate
110
112 *
4
ln2 x
C
x
2 ln x + C
4
4
2
Solution:
1
2
C.
Solution:
3
1
A.
B.
3
 x  1 dx
A.
4
   40  2xy  dydx
94.
   40  2xy  dydx   
lnx
dx .
x
Solution:
89. Evaluate
Solution:
Use calculator and in Radian mode for angle.
Answer: 0.0184
3
x2  y2
c
x
2

dC d x  12 

5


dx dx  x 
X  0.5
2

d  x4 x2

 x



dx  4
2


d  x4 x2


 x
 0.625


dx  4
2
 x 0.5
x3  x  1   0.5    0.5   1  0.625
Try : B. x 2  y 2  cx
Solution:
y " 2y ' y  0
y  C1e
A.
X? 0.5 
C.
x
Differentiate the choices and substitute
the value of x used in the given.
X2  Y 2  2XYA  0, A
SHIFT  SOLVE
y  c1ex  c2xex
A.
Assign : x  0.5,y  1, dx  1, dy  A
C. 0.0312
D. 0.0481

17.2
1
5
ydy   e x dx
0
147.42  147.41
CEERS REVIEW CENTER
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NOTES: CALCULATOR TECHNIQUES
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email add: ceers.philippines@yahoo.com
97. Given the D.E. y ' 
3yx 2
; y(1) = 1 find the
x  2y 4
3
ynew
y if x = 3.
A.
B.
5.2
1.4
3yx 2
ynew  y  3
 x 
x  2y 4
Assign : x  0.01, x1  1,y1  1
3yx 2
 0.01 : x  x  0.01
x  2y 4
X?1   ....
If x = 3, the last value of y is the answer.
Answer: 2.5
98. Given the D.E.  y  xy  dx   x  x y  dy  0 ;
2
2
y(1) = 1 find the y if x = 2.
Solution:
0
3
xy 2  y
ynew  y 
 0.01 : x  x  0.01
x  x2y2
CALC
 Reg  xˆ  
Answer : 8.6 s
Y ?1 
101. A certain radioactive element dissipates at a
rate proportional to its amount instantaneously
present. If one-half of any given amount of the
element dissipates in 1,600 years, what
fraction will dissipate during the second
century?
A. 8.3% *
C. 5.6%
B. 3.8%
D. 9.2%
3
Y ?1 
0.02
1.4
C. 1.11 *
D. 1.077
Solution:
MODE  STAT  e ^ X
x
y
xy 2  y
 x 
x  x2y2
Assign : x  0.01, x1  1,y1  1
X?1   ....
If x = 2, the last value of y is the answer.
Answer: 1.11
CALC
A.
B.
ynew  y
x
 y  y '  x 
ynew  y 
C. 2.5 *
D. 1.077
Solution:
y y
y '  new
x
ynew  y  y '  x 
ynew  y 
y' 
99. If “a fool and his money are soon parted,” the
rate at which it leaves is probably proportional
to the amount remaining. If a certain fool
starting with $20,000 starts gambling his
money 8away and after 2 hr has lost $2000,
how long will it take for him to lose 90% of the
original amount?
A. 44 hours *
C. 35 hours
B. 46 hours
D. 48 hours
2
Solution:
MODE  STAT  e ^ X
x
y
0 20,000
2
18,000
AC   0.1 20,000    SHIFT  (STAT) 1
AC  1  SHIFT  (STAT) 1
Solution:
MODE  STAT  e ^ X
x
y
AC  200  SHIFT  (STAT) 1
 Reg  yˆ    91.7% remain
Dissipate  100  91.7  8.3%
102. A certain metal has a temperature of 80°C
and is placed in a room with a temperature of
20°C. After 20 minutes, the temperature of the
metal is now 50°C. How long will it take the
metal to reach a temperature of 40°C?
A.
B.
28.5 min
34.5 min
C. 31.7 min *
D. 36.8 min
Solution:
MODE  STAT  e ^ X
x
y
(time) (T  Ts)
0
20
80  20
50  20
AC  (40  20)  SHIFT  (STAT) 1
 Reg  xˆ    31.7 min
[12/17]
Solution:
9
9
o
F  oC  32   37   32  98.6
5
5
MODE  STAT  e ^ X
x
y
(time) (T  Ts)
choices 98.6  82
12
90  82
AC   87  82   SHIFT  (STAT) 1
0 100
1600 50
 Re g  xˆ  
Answer : 43.7 hrs  44hours
100. An RC circuit is observed during discharge to
have an initial capacitor potential of 100 V and
after 3.0 s to have a potential of 20 V. How
long will it take for the capacitor to discharge
to 1.0 V?
A. 7.3 s
C. 6.5 s
B. 8.6 s *
D. 9.1 s
100
20
103. Police arrive at the scene of a murder at 12
am. They immediately take and record the
body's temperature, which is 90oF; and
thoroughly inspect the area. By the time they
finish the inspection, it is 1:30 am. They again
take the temperature of the body, which has
dropped to 87oF; and have it sent to the
morgue. The temperature at the crime scene
has remained steady at 82oF. Determine the
time the murder occurred if temperature of
body if alive is 37 oC.
A. 9:35 PM
C. 9:40 PM *
B. 9:22 PM
D. 9:34 PM
 Re g  xˆ  
Re sult should be: 13.5
40
Try: 9:40 as 9
above.
60
Answer: 9:40
104. Consider a large tank holding 1000 L of pure
water into which a brine solution of salt begins
to flow at a constant rate of 6 L/min. The
solution inside the tank is kept well stirred,
and is flowing out of the tank at a rate of 6
L/min. If the concentration of salt in the brine
solution entering the tank is 0.1 Kg/L,
determine when the salt reaches 50 kg.
A.
B.
110.50 min
112.67 min
Solution:
C. 115.52 min *
D. 98.70 min
CEERS REVIEW CENTER
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dQ
 R A  RR
dt
dQ
Q
  0.1 6 
6
dt
1000
50

0
t

dQ
 dt
0.6  0.006Q
Assign x  0.5, y  1
Assign x  0.5, y  1
x2  y2
Given : c 
x
2
dC d  x  12 

 3


dx dx  x 
X  0.5
Given : c  x 2  y 2
dC d 2 2

x 1
1
X  0.5
dx dx
dC d

0.52  y 2
 2
Y 1
dy dy
dy
dC / dx
1 1



dx
dC / dy
2 2
Try : C. xy  c
c  xy
dC d

 x 1  X0.5  1
dx dx
dC d

 0.5y  Y 1  0.5
dy dy
dy
1

 2
dx
0.5
dC d  0.5  y


dy dy  0.5
2
0
115.52 min  t
X  19.53 kg
105. A tank contains 200 liters of fresh water. Brine
containing 2 kg/liter of salt enters the tank at
the rate of 4 liters per min. and the mixture
kept uniform by stirring, runs out at 3 liters per
min. Find the amount of salt in the tank after
30 min.
A. 197 kg *
C. 100 kg
B. 179 kg
D. 57.3 kg
Solution:
Formula :

VR
 200  t 
Q   0  2  200   

 200 
at t = 30min

3
4 3
 200  30 
Q   0  2  200   

 200 
Q  197


4


 Y 1
dy
3

3/4
dx
4
Try : B. x 2  y 2  cy
c
x2  y2
y
dC d  x  1


dx dx  1
2
2

1


 X 0.5
dC d  0.5  y


dy dy 
y
2
 V  t   VA  VR
Q   Qo  Cin Vo  
 Cin V  t 

 Vo 
V  t   volume at any time
2


2

 0.75


 Y 1
dy
1
4


dx
0.75
3
107. Find the orthogonal trajectories of the family
 2  200  t 
3
4 3
 2  200  30 
106. Find the orthogonal trajectories of the family
of curves x 2  y 2  cx
of curves x 2  y 2  c 2
A.
x 2  y 2  cx
C. xy  c *
B.
x2  y2  c
D. x 2  cy 2  y
Solution:
Solve for the slope or y’ of given and choices. Their
value should be negative reciprocal.
Answer: C

C.
cos
D.
 
 
5
7t  sin 7t
7
1
7t 
sin 7t
7
D.
3
2
s
4
s2

4
s

3
s

2
C. t 2 sint
D. t sint *
t cos t
t cost
A.
B.
Solution:
Assign s =5,
F  s 
2s
s
2

1
2

2 5
5  1
2
2

5
338
Evalute the choices using the
 
 
3cos  7t   sin  7t 

Laplace formula:  f  t  e  st dt
0
Try D.
 t
8
Solution:
0
2

sin t e5t dt 
5
338
Assign s =5,
3s  5 3  5   5 5


s2  7  5 2  7 8
x 2  y 2  cx
C. x 2  cy 2  x
F  s 
B.
x 2  y 2  cy *
D. x 2  cy 2  y
Evalute the choices using the

Laplace formula:  f  t  e  stdt
111. Find the Laplace transform
 2 t  0,2 
of f  t   
0 otherwise
A.
2 2 2s
 e *
s s
B.
2s  2e 2s
0
Try A.

0  3cos
8
 7t  
5
7
sin
 
5

7t e 5t dt 
8

109. What is the Laplace transform of f(t) = 3t + 4?
[13/17]
C.
4
*

s2 s

A.
Solution:
Solve for the slope or y’ of given and choices. Their
value should be negative reciprocal.
Answer: B
s
3
3
s
110. What is the inverse Laplace transform of
2s
?
f  s 
2
2
s 1
 
 

Solution:
Assign s =5, and infinity as 8.
8
23
5t
0  3t  4 e dt  25
Substitute to choices: Try B.
3 4 3 4 23
B. 2   2  
s
s 5
5 25
108. Find the inverse Laplace transform
3s  5
of F  s   2
s 7
5
A. 3cos 7t 
sin 7t *
7
cos
2
B.

B.
4
A.
1 2s
e
s
2
D. 2  e2s
s
C.
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Solution:
Assign s =5, the integration is up
to 2 only due to the given function
definition.

2
0
Assign s =5, and infinity as 8.
2
2 5t
43
0 cos  3t  e dt  305
Substitute to choices: Try B.
1
s
B.

2s 2 s2  36
5t
2e dt  0.3999

Substitute to choices: Try A.
2 2
2 2 2 5
A.  e2s   e    0.3999
s s
5 5
112. Find the inverse Laplace transform
2
of F  s 
sk
A. 2e2kt
C. e kt
B.
2e
 kt
*
D. e

Laplace formula:  f  t  e dt
 st
0
Try B. Substitute s and k.
8
1
1t
5t
0  2e e dt  3


114. Determine the Fourier coefficient at n=0
0x 3
2
of f  x   
if the period is 6.

2

3x0

A.
B.
1
kt
2
Solution:
Assign s =5, k  1
2
2
1
F  s 


s  k 5 1 3
Evalute the choices using the

5
1
43



2  5  2  5 2  36
305
a0 = 1
a0 = 0.5
C. a0 = 0.25
D. a0 = 0 *
of f  t   cos 3t  
A.
B.
C.
D.
Solution:
[14/17]
1
s

s s2  36
1
s

*
2s 2  s2  36 
2
s

s  3 2 s2  36

2s 
2s
3  s2  36 

1 L
 nx 
an   f  x  cos 
 dx
L L
 L 
1 L
 nx 
f  x  sin 
 dx


L
L
 L 
where : L is half-period
int egration should be one cycle
n is the order
bn 
a10 
1 
 10x   0
cos  x  cos 
 dx
 0
  
b10 
1 
 10x   0.0643
cos  x  sin 
 dx
 0
  
116. Determine the Fourier coefficient at n = 2 of
function f  x   x 10  x  , 0  x  10, if the
period is 10.
A. a3 = 0, b3 = - 1.533
B. a3 = -1.533, b2 = 0
C. a3 = 0, b2 = - 2.533
D. a2 = - 2.533, b2 = 0 *
Solution:
Formula :
an 
1 L
 nx 
f  x  cos 
 dx
L L
 L 
1 L
 nx 
bn   f  x  sin 
 dx
L L
 L 
where : L is half-period
int egration should be one cycle
n is the order
1 0
1 3
a0   2dx   2dx  0
3 3
3 0
113. Find the Laplace transform
2
Formula :
115. Determine the Fourier coefficient at n = 10
0x 
cos x
of f  x   
if the period
0

 x  2

is 2 .
A. a10 = 0, b10 = 0.0643 *
B. a10 = 0.0643, b2 = 0
C. a10 = 0, b10 = 0.1064
D. a210 = 0.1064, b2 = 0 *
Solution:
Formula :
1 L
 nx 
an   f  x  cos 
 dx
L L
 L 
1 L
 nx 
f  x  sin 
 dx
L L
 L 
where : L is half-period
int egration should be one cycle
n is the order
bn 
1 10
2x 
a2   x 10  x  cos 
 dx  2.533
5 0
 5 
an 
1 L
 nx 
f  x  cos 
 dx
L L
 L 
1 L
 nx 
f  x  sin 
 dx
L L
 L 
where : L is half-period
int egration should be one cycle
n is the order
bn 
a3 
1  2
 3x   0.11
x cos 
 dx
 / 2 0
  / 2
b2 
 / 2 0
1

 3x   1.047
x 2 sin 
 dx
  / 2
118. Find the 1’s complement of 11001012.
A.
B.
00110102 *
00110112
C. 00010102
D. 00111002
Solution:
MODE  BASE  N  BIN log
 SHIFT  BASE  3  Not 
1100101  )  
Don't include the leading ones.
Answer : 0011010 2
119. Find 548 + 110112.
C.
D.
81
71 *
C. 23
D. 55
Solution:
1 10
2x 
b2   x 10  x  sin 
 dx  0
5 0
 5 
MODE  BASE  N  DEC  x3
117. Determine the Fourier coefficient at n =3 of
 press the pointer down  o  54
function f  x   x , 0  x   .
2
Solution:
Solution:
Formula :
 SHIFT  BASE  3
   SHIFT  BASE  3
 press the pointer down  b  11011
A.
B.
C.
D.
a3 = 0.11, b3 = - 1.047 *
a3 = - 0.11, b3 = 1.047
a3 = 0.21, b3 = - 2.047
a3 = - 0.21, b3 = 2.047
 
Answer : 71
CEERS REVIEW CENTER
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120. Find the product of 6616 and 1123.
B.
123. Hexadecimal equivalent of 1668.
A.
B.
3375668
1AF7216
C. 114544
D. 3375628 *
Solution:
MODE  BASE  N   OCT  ln
 SHIFT  BASE  3
A.
B.
76 *
33
Solution:
MODE  BASE  N   OCT  ln
 press the pointer down  h  66
 66    HEX  x
 x  SHIFT  BASE  3
Answer : 76
 press the pointer down  d  1123
 
Answer : 3375628
A.
B.
0111002 *
1D16
C. 1111002
D. 1E16
5018
14116
A.
B.
– 1.5
2
C. 1.5 *
D. 2.5
Solution:
x  x 93  78
z

 1.5

10
124. Find 658 + 34 – 10012 + F216.
A.
B.
121. Find the bitwise complement of 2316.
C. 67
D. 90
126. On a statistics examination the mean was 78
and the standard was 10. Determine the
standard score of the student whose grade
was 93.
C. 320 *
D. 1110000002
Solution:
MODE  BASE  N  DEC  x
2
 SHIFT  BASE  3
0.8767 *
0.0067
C. 0.983
D. 1.5
Solution:
MODE  STAT  AC  SHIFT
 press the pointer down  o  65
  STAT  1  7 : Distr  1: P(  2.40
   SHIFT  BASE  3
 )     STAT  1  7 : Distr  1: P(
 SHIFT  BASE  3  Not
 press the pointer down  d  34
 1.20  )  
   SHIFT  BASE  3
Display : P  2.4   P  1.2
 press the pointer down  h  23
 press the pointer down  b  1001
Result : 0.87673
 
Answer : 011100 2
 press the pointer down  h  F2
 SHIFT  BASE  3
122. Find the logic AND of 11012 and 12.
A.
B.
12 *
11012
C. 14
D. 10012
Solution:
MODE  BASE  N  DEC  x 2
 SHIFT  BASE  3
 press the pointer down  b  1101
   SHIFT  BASE  3
 
Answer : 320
125. Find the 2’s complement of 11011112.
A.
B.
0100012 *
1110012
C. 1100012
D. 0010012
Solution:
MODE  BASE  N  BIN log
 SHIFT  BASE  3  and
 SHIFT  BASE  3  Neg
 12  
Answer : 12
 1101111  )  
Answer : 00100012
128. Find the area under the normal curve to the
left of z = - 1.78.
A.
B.
  STAT  1  7 : Distr  3 : R(  1.333
C. 0.083
D. 0.391
Re sult : 0.091266  Pr obability
No. of students  0.091266  300   27
130. The mean grade on a final examination was
72, and the standard deviation was 9. The top
10% of the students are to receive A’s. What
is the minimum grade a student must get in
order to receive an A?
A.
B.
90
84 *
C. 87
D. 75
Solution:
Use reverse. Convert the choices to z-values
84  72
Try: B. 84  z 
 1.333
9
Area to the right <0.10
MODE  STAT  AC  SHIFT
  STAT  1  7 : Distr  3 : R(  1.333
 )  
Solution:
MODE  STAT  AC  SHIFT
Display : R 1.333 
  STAT  1  7 : Distr  1: P(  1.78
Re sult : 0.091266  Pr obability
 )  
131. A man purchased on monthly installment a P
100,000 worth of land. The interest rate is 12
% compounded monthly and payable in 20
years. What is the monthly amortization?
Display : P  1.78 
Result : 0.0375
129. If the heights of 300 students are normally
distributed with mean 68.0 inches and
standard deviation 3.0 inches, how many
students heights greater than 72 inches.
A.
[15/17]
0.0375 *
0.0567
Solution:
72  68
z
 1.333
3
Area to the right is needed:
MODE  STAT  AC  SHIFT
Display : R 1.333 
MODE  BASE  N  BIN log
Solution:
D. 40
 )  
127. Find the area under the normal curve between
z = -1.20 and z =2.40.
A.
B.
25
20
C. 27 *
A.
B.
P1,101.08 *
P293.01
C. P2,201.1
D. P201
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email add: ceers.philippines@yahoo.com
Solution:
0.12
i
 0.01 per month
12
100,000 
20x12
 A 1  0.01
x
X 1
A
100,000
  1  0.01
20x12
X
X 1

 1101.09
A.
B.
132. What is the accumulated amount of five-year
annuity paying P 6,000 at the end of each
year, with interest at 15 % compounded
annually?
A.
B.
P39,302.23
P40500.02
C. P30,301.24
D. P40,454.29*
Solution:
i  0.15 per year
Present worth:
  6,000 1  0.15 
5
X
X 1
  20,112.94
20,112.94 1  0.15   40,454.29
5
133. A certain amount was borrowed and was to
be repaid in installments at the end of every
year thereafter with each payment P500 more
than the preceding for 8 yrs. The first payment
is P1500. If interest is 10% compounded
annually, determine the amount loaned?
P17,501
P16,017 *
C. P14,302
D. P19,204
Solution:
i  0.10 per year
Present worth:
  1500   x  1 500  1  0.10 
8
X 1
 16,016
X

P11,756 *
P11,576
C. P14,312
D. P19,123
Solution:
i  0.06 per year
Present worth:
  1500 1  .12
7
X 1
X 1
 1  0.06 
X
 11,756
135. A machine which costs P 10,000 was sold as
a scrap after being used for 10 years. If the
scrap value was P 500, determine the book
value at the end of the 5th year. Use straight
line method.
A.
B.
Future worth:
A.
B.
MODE  STAT  e ^ X
x
y
0
50,000
5
15,000
AC  3yˆ  24,279.67
134. Annual maintenance costs for a machine are
P1500 this year and are estimated to increase
12% each year every year. What is the
present worth of the maintenance costs for 7
years if interest rate is 6%.
P5,250*
P3,250
136. What is the book value of equipment
purchased 3 years ago at P 50,000 with a
salvage value of P 15,000 at the end of 5
years using declining balance method.
P30,023
P40,000
Solution:
A.
B.
C. P24,279.67 *
D. P32,302.57
P3,3021
P1,208,000
x
y
0
6,000,000 1.03 
8
0.05 6,000,000 1.03  
9
0.05 6,000,000 1.03  
AC  5yˆ  1,287,500
138. A VOM has a selling price of P 400. If its
selling price is expected to decline at a rate of
10% per annum due to obsolescence, what
will be its selling price after 5 years?
(Declining balance Method)
A.
B.
P236.2 *
P320.1
C. P302.12
D. P868.97
Solution:
MODE  STAT  e ^ X
x
y
0
400
1
400 1  0.10 
AC  5yˆ  236.2
[16/17]
C. P1,280,000 *
D. P1,203l,000
Solution:
MODE  STAT  _  cX2
C. P301.2
D. P302.3
Solution:
MODE  STAT  A  BX
x
y
0
10,000
10
500
AC  5yˆ  5,250
A.
B.
137. A telephone company purchased a microwave
radio equipment for P 6M. Freight and
installation charges amounted to 3% of the
purchase price. If the equipment is
depreciated at a period of 8 years with a
salvage value of 5%, determine the book
value at the end of the 5th year. Use SYD
method.
1
139. A machine costs P 8,000 and an estimated
life of 10 years with a salvage value of P 500.
What is its book value after 8 years using
straight line method?
A.
B.
P3500
P2000*
C. P2500
D. P1500
Solution:
MODE  STAT  A  BX
x
y
0
8,000
10 500
AC  8yˆ  2000
140. A company purchases an asset for P
10,000.00 and plans to keep it for 20 years. If
the salvage value is zero at the end of 20th
year, what is the depreciation in the third
year? Use SYD method.
A.
B.
P857.14 *
P643.14
C. P694.35
D. P640.32
Solution:
MODE  STAT  _  cX2
x
y
0
10,000
20
0
21
0
AC  2yˆ  3yˆ  857.14
141. An asset is purchased for P 500,000.00. The
economic life of the asset is 25years. What is
the total depreciation in the first three years
using double declining balance method?
A.
B.
P135,173
P120,653
Solution:
C. P110,656 *
D. P148,653
CEERS REVIEW CENTER
Doña Esperanza Village, Tisa, Cebu City
Tel. No. 261-2244
email add: ceers.philippines@yahoo.com
MODE  STAT  e ^ X
x
y
0
500,000
1
2 

500,000  1 
25 

AC  500,000  3yˆ  110,656
1
142. An asset is purchased for P 9,000.00. Its
estimated life is 10 years after which it will be
sold for P 1,000.00. Find the book value
during the 5th year if double declining balance
method is used.
A.
B.
P7545.45
P2521.45
C. P2949 *
D. P3500
Solution:
MODE  STAT  e ^ X
x
y
0
9000
1
2 

9000  1 

10


AC  5yˆ  2949
1
[17/17]
EE REVIEW APRIL 2017
NOTES: CALCULATOR TECHNIQUES