CALCulator
TECHniques
Optimize your skills in using your
CASIO 991ES PLUS
Includes Calculator
Approach on Solving
Various Problems in
Algebra
Trigonometry
Analytic Geometry
Calculus
Differential Equation
Advance Engineering
Mathematics
Computer Fundamentals
Engineering Mechanics
Engineering Economy
Electrical Engineering
CHRISTIAN JAY V. ESPINOZA
2017
Christian Jay V.Espinoza
01/01/17
CALCulator TECHniques
by C.J. Espinoza
1. EE Board April 1992
Find the value of x in
a)
b)
c)
d)
16.47
12.87
18.27
20.17
Solution:
Note: The calculator should be in DEGREE MODE.
PRESS:
Then,
DISPLAY:
x=
16.47058824
Thus, the value of x is 16.47.
2. EE Board April 1997
Find the value of x and y from the equations:
x - 4y + 2 = 0
2x + y – 4 = 0
a)
b)
c)
d)
11/7, -5/7
14/9, 8/9
4/9, 8/9
3/2, 5/3
Solution:
PRESS:
Display:
[
]
Then,
PRESS:
1
CALCulator TECHniques
by C.J. Espinoza
Display:
x=
14/9
Then,
PRESS:
Display:
y=
8/9
Thus, x = 14/9 and y = 8/9.
3. ME Board October 1996
Solve the value of x:
x + y = -4
x+z-1=0
y+z+1=0
a)
b)
c)
d)
x = -1, y = -5, z = 3
x = 1, y = 2, z = -3
x = -1, y = -3, z = 2
x = -2, y = -3, z = -1
Solution:
PRESS:
Display:
Then,
PRESS:
Display:
x=
-1
y=
-3
Then,
PRESS:
Display:
Then,
PRESS:
2
CALCulator TECHniques
by C.J. Espinoza
Display:
z=
2
Thus, x = -1, y = -3 and z = 2.
4. EE Board April 1997
Multiply the following: (2x+5y) (5x -2y)
a)
b)
c)
d)
10x2 – 21xy + 10y2
-10x2 + 21xy -10y2
10x2 + 21xy – 10y2
-10x2 – 21xy – 10y2
Solution:
Step 1: Evaluate the given equation by substituting x and y with any values (less
than or greater than 0).
First, store values to the variables and y. Let’s say
, and y
.
PRESS:
Then,
Display:
(2x+5y) (5x -2y)
64
Step 2: Evaluate the given choices and compare the answer when the same
values of x and y are substituted.
Since values 2 and 3 are still stored in the variables x and y respectively, it is not
necessary to store again the values to the said variables. Thus,
PRESS:
Display:
10x2 – 21xy + 10y2
4
The remaining choices are evaluated as follows:
= -4
3
CALCulator TECHniques
b)-10x2 + 21xy -10y2
c) 10x2 + 21xy -10y2
d) -10x2 - 21xy -10y2
by C.J. Espinoza
= 64
= -256
Therefore, 10x2 + 21xy -10y2 is the correct answer.
5. ECE Board November 1993
Simplify the following equation
7a+2 – 8(7)a+1 + 5(7)a + 49(7)a-2
a)
b)
c)
d)
-5a
-3a
-7a
-4a
Solution:
Step 1: Evaluate the given equation by substituting a small value (less than or
greater than 0) to the constant a.
First, store any value to the constant a. Let’s say a
.
PRESS:
Then,
Display:
7a+2 – 8(7)a+1 + 5(7)a + 49(7)a-2
-49
Step 2: Evaluate the given choices and compare the answer when the same value
of a is substituted.
Since value of 2 is still stored in the constant a, it is not necessary to store again the
values to the said constant. Thus,
PRESS:
Display:
-5a
-25
4
CALCulator TECHniques
by C.J. Espinoza
The remaining choices are evaluated as follows:
b) -3a
c) -7a
d) -4a
= -9
= -49
= -16
Therefore, -7a is the correct answer.
6. EE Board April 1996
EE Board March 1998
The polynomial x3+4x2-3x+8 is divided by x-5, then the remainder is,
a)
b)
c)
d)
175
140
218
200
Solution:
PRESS:
Display:
x3+4x2-3x+8
218
Thus, the remainder is 218.
7. ME Board April 1996
Solve for x: 10x2 + 10x + 1 = 0
a)
b)
c)
d)
-0.113, -0.887
-0.331, -0.788
-0.113, -0.788
-0.311, -0.887
Solution:
PRESS:
Input the data:
PRESS:
Display:
a
b
PRESS:
5
CALCulator TECHniques
by C.J. Espinoza
Display:
x=
√
or – 0.1127
x=
√
or – 0.8873
PRESS:
Display:
Therefore, the values of x are -0.113 and
-0.887.
8. CE Board November 1997
Evaluate the log6845 = x.
a)
b)
c)
d)
3.76
5.84
4.48
2.98
Solution:
PRESS:
Display:
log6 (845)
3.761295388
Therefore, answer is 3.
9. EE Board October 1992
Given: log 6 + x log 4 = log 4 + log (32 + 4x). Find x.
a)
b)
c)
d)
2
3
4
6
Solution:
PRESS:
Display:
log6 + x log 4 = log 4 + log (32 + 4x)
3
6
CALCulator TECHniques
by C.J. Espinoza
Therefore, the value of x is 3.
10. ECE Board March 1996
A merchant has three items on sale, namely a radio for P 50, a clock for P 30 and a
flashlight for P 1. At the end of the day, he sold a total of 100 of the three items and has
taken exactly P 1,00 on the total sales. How many radios did he sell?
e)
f)
g)
h)
16
18
20
24
Solution:
Let x = number of radios sold out
y = number of clocks sold out
z = number of flashlights sold out
x + y + z = 100
z = 100 – x – y
50x + 30y + z = 1000
50x + 30y + (100 – x – y) = 1000
49x + 29y = 900
y = (900-49x)/29
Set your calculator to TABLE MODE.
PRESS:
Input f(x) = (900-49x)÷29
Start? 16
End? 24
Step? 2
Display:
0| X |
F(X)
1| 16 |
4
2| 18 | 0.6206
3| 20 | -2.758
4| 22 | -6.137
5| 24 | -9.517
|
|
|
|
|
|
f(x) = number of clock sold out
16 and 4 are the possible values.
Therefore, the number of radios did he sell is 16.
7
CALCulator TECHniques
by C.J. Espinoza
11. EE Board March 1998
Gravity causes a body to fall 16.1 ft in the first second, 48.3 in the 2nd second, 80.5 in the
3rd second. How far did the body fall during the 10th second?
i)
j)
k)
l)
248.7 ft
308.1 ft
241.5 ft
305.9 ft
Solution:
This is an example of arithmetic progression.
PRESS:
Display:
0|
1|
2|
X
1
2
| Y |
| 16.1 |
| 48.3 |
PRESS:
Display:
ŷ
305.9
Therefore, 305.9 ft is the correct answer.
12. CE Board May 1995
What is the sum of the progression , 9,
m)
n)
o)
p)
, 9 ….. up to
th
term?
1030
1035
1040
1045
Solution:
Set the calculator to STAT MODE 2 – for ARITHMETIC PROGRESSION
PRESS:
PRESS:
Display:
0|
1|
2|
X
1
2
|
|
|
Y
4
9
|
|
|
8
CALCulator TECHniques
by C.J. Espinoza
PRESS:
Display:
∑(Xŷ, ,
)
1030
Therefore, 1030 is the correct answer.
13. CE Board May 1995
The numbers 28, x+2, 112 form a G.P. What is the 10th term?
q)
r)
s)
t)
14336
13463
16433
16344
Solution:
PRESS:
Then, input only the first and the 3rd term.
PRESS:
Display:
0|
1|
2|
X
1
3
| Y |
| 28 |
| 112 |
PRESS:
Display:
ŷ
14336
Therefore, 14336 is the correct answer.
9
2017
CALCulator TECHniques
by C.J. Espinoza
1. ME Board October 1995
Simplify the expression
– (sec ) sin2
a) o
b) cos2
c) sin2
d) in
Solution:
Step 1: Evaluate the given equation by substituting a value to . Let x = .
L t’ ay x = 20.
PRESS:
Display:
in x
0.4080820618
Step 2: Evaluate the given choices and compare the answer when the same value of x is
substituted.
a)
o
PRESS:
Display:
o x
0.4080820618
The remaining choices are evaluated as follows:
b) cos2
c) sin2
d) sin
Therefore, o
= 0.1665309692
= 0.8334690308
= 0.9129452507
is the correct answer.
2. In a right triangle, the lengths of the sides are 2 and 3 meters respectively. Find the length of
the hypotenuse.
a) 2.61 meters
b) 3.61 meters
c) 4.61 meters
11
CALCulator TECHniques
by C.J. Espinoza
d) 5.61 meters
Solution:
?
2
3
PRESS:
Then,
PRESS:
Display:
Pol(2,3)
r=3.605551275,
Therefore, the length of the hypotenuse is 3.61 meters.
3. ECE Board April 1994
A pole cast a shadow 15 m long when the angle of elevation of the sun is 61°. If the pole is
leaned 15° from the vertical directly towards the sun, determine the length of the pole.
a)
b)
c)
d)
42.44 m
46.21 m
48.23 m
54.23 m
Solution:
This is an application of sine law.
Let x = length of the pole
61°
15°
x
15 m
PRESS:
12
CALCulator TECHniques
by C.J. Espinoza
Input this data:
[
o
in
o
in
]
PRESS:
Display:
x=
54.23
y=
59.89
Display:
Thus, the length of the pole is 54.23 m.
4. If the longitude in Tokyo is 139oE and that of Manila is 121oE, what is the time difference
between Tokyo and Manila.
a)
b)
c)
d)
1 hour and 5 minutes
1 hour and 8 minutes
1 hour and 10 minutes
1 hour and 12 minutes
Solution:
PRESS:
Input this data:
0| X
1| 24
2| 12
PRESS:
|
|
|
Y
360
180
|
|
|
Display:
(139-121) ̂ °
°
’ ”
Therefore, 1 hour and 12 minutes is the correct answer.
13
2017
CALCulator TECHniques
by C.J. Espinoza
1. EE Board April 1994
Find the distance between A (4, -3) and B(-2, 5).
a)
b)
c)
d)
11
9
10
8
Solution:
Set your calculator to COMPLEX MODE.
PRESS:
Then, type this equation: |A-B|
PRESS:
Next,
PRESS:
Display:
|A-B|
10
Therefore, the distance is 10.
2. ECE Board April 1999
Find the inclination of the line passing through (-5,3) and (10,7).
a)
b)
c)
d)
14.63
14.73
14.83
14.93
Solution:
PRESS:
PRESS:
Display:
0|
1|
2|
X
-5
3
|
|
|
Y |
10 |
7 |
To get the inclination of the line, get the arc tangent of slope of the line.
PRESS:
15
CALCulator TECHniques
by C.J. Espinoza
Display:
Tan-1(B
14.93141718
Therefore, the inclination of the line is 14.93.
3. EE Board April 1994
Given the three vertices of a triangle whose coordinates are A(1,1), B(3,-3), and C(5,-3). Find
the area of the triangle.
a)
b)
c)
d)
3
4
5
6
Solution:
2
A(1,1)
0
0
2
4
6
C(5,-3)
-2
B(3,-3)
-4
The area of a triangle with given vertices (x1, y1), (x2, y2) , (x3, y3) is
|
Atriangle = 0.5 det |
|
x1
x2
x3
y1
y2
y3
|
Atriangle = 0.5 det |
|
1
3
5
1
-3
-3
1
-3
-3
1
1
1
1
1
1
1
1
1
|
|
|
|
|
|
PRESS:
Display:
|
|
|
1
3
5
|
|
|
PRESS:
16
CALCulator TECHniques
by C.J. Espinoza
Display:
Abs( .5det(MatA
4
Therefore, 4 is the correct answer.
4. ECE Board April 1999
If the points (-2, 3), (x, y) and (-3, 5) lie on a straight line, then the equation of the line is _______.
a)
b)
c)
d)
x - 2y -1 = 0
2x + y -1 = 0
x + 2y -1 = 0
2x + y + 1 = 0
Solution:
PRESS:
Display:
0|
1|
2|
X
-2
-3
|
|
|
Y
3
5
|
|
|
PRESS:
Display:
A
-1
PRESS:
Display:
B
-2
From the SLOPE-INTERCEPT FORM:
y = mx + b
y = Bx + A
y = -2x - 1
We can rewrite it as,
2x + y + 1 = 0
Therefore, 2x + y + 1 = 0 is the correct answer.
5. ME Board April 1998
The equation of a line that intercepts the x-axis at x = 4 and the y – axis at y = -6 is
a) 3x + 2y = 12
b) 2x – 3y = 12
17
CALCulator TECHniques
by C.J. Espinoza
c) 3x – 2y = 12
d) 2x + 3y = 12
Solution:
PRESS:
Display:
0|
1|
2|
X
4
0
|
|
|
Y
0
-6
|
|
|
PRESS:
Display:
A
-6
PRESS:
Display:
B
5.5 or 3/2
From the SLOPE-INTERCEPT FORM:
y = mx + b
y = Bx + A
y = 3/2x - 6
We can rewrite it as,
3x – 2y = 12
Therefore, 3x – 2y = 12 is the correct answer.
6. A circle which equation is x2 + y2 - 6x + 10y + 18 = 0 has its center at :
a)
b)
c)
d)
(-3, 5)
(3, -5)
(-3, -5)
(3, 5)
Solution:
PRESS:
DISPLAY:
18
CALCulator TECHniques
by C.J. Espinoza
3
PRESS:
DISPLAY:
-5
Therefore, (3, -5) is the correct answer.
7. ME Board April 1998
What is the radius of a circle whit the ff. equation: x2 – 6x + y2 – 4y – 12 = 0
a)
b)
c)
d)
3.46
5
6
7
Solution:
First, get the coordinates of the center of the circle.
PRESS:
DISPLAY:
3
Store it to X.
PRESS:
PRESS:
DISPLAY:
2
Store it to Y.
PRESS:
Then, enter the given equation.
PRESS:
19
CALCulator TECHniques
by C.J. Espinoza
DISPLAY:
x2 – 6x + y2 – 4y – 12 = 0
-25
The radius is the square root of the absolute value of
PRESS:
DISPLAY:
√|
|
5
Therefore, the radius is 5.
8. A parabolic arc has a height of 20 ft and a width of 36 ft at the base. If the vertex of the
parabola is at the top of the arc, at what height above the base is it 18 ft wide?
a)
b)
c)
d)
5 ft
10 ft
15 ft
20 ft
Solution:
(0,0)
0
-20
0
(9,y)
-10
(-18,20) -20
20
(18,20)
-30
PRESS:
Input the data:
0| X |
1| 362 |
2| 0 |
Y |
0 |
20 |
X-column = square of the base or width
Y-column = height
Height at 18 ft wide = ?
PRESS:
20
CALCulator TECHniques
by C.J. Espinoza
Display:
182ŷ
15
Therefore, the height = 15 ft.
9. ME Board April 1997
What is the radius of the sphere center at the origin that passes the point 8, 1, 6?
a)
b)
c)
d)
9
10
√
10.5
Solution:
Set your calculator to VECTOR MODE. Then, input the coordinate of the point to VctA.
PRESS:
Display:
A
[
]
Last, get the absolute value of VctA.
PRESS:
Display:
Abs(VctA
10.04987562
10.04987562 = √
Therefore the radius of the sphere is √
.
21
2017
CALCulator TECHniques
by C.J. Espinoza
1. ECE Board Exam April1998
Evaluate:
a)
b)
c)
d)
(
)
1
0
2
Infinite
Solution:
Evaluate the given equation. Since there is no infinity button in the fx-991ESPLUS, substitute x
with a very large number, say x = 12x1012.
PRESS:
Display:
1
Therefore, 1 is the correct answer.
2. CE Board Exam November 1997
Evaluate:
a)
b)
c)
d)
(
)
1/5
2/5
3/5
4/5
Solution:
Evaluate the given equation. Substitute x with a value close to 1, say x = 0.999999999.
PRESS:
Display:
Therefore, 2/5 is the correct answer.
3. ME Board Exam April 1998
23
CALCulator TECHniques
by C.J. Espinoza
Given the function f(x) = x to the 3rd power – 6x + 2. Find the derivative at x = 2.
a)
b)
c)
d)
6
7
3x2 - 5
8
Solution:
PRESS:
Display:
|
6
Therefore, the answer is 6.
4. The distance of a body travels is a function of time, t and is defined by x(t) = 18t + 9t2. What
is the velocity at t = 3 sec?
a)
b)
c)
d)
18
36
54
72
Solution:
PRESS:
Display:
|
72
Therefore, 72 is the correct answer.
5. EE Board Exam October 1997
If y = 4 cos x + sin 2x, what is the slope of the curve when x = 2 radians?
a)
b)
c)
d)
-2.21
-4.94
-3.25
2.21
Solution:
To get the slope, simply get the first derivative of the given equation.
24
CALCulator TECHniques
by C.J. Espinoza
PRESS:
Display:
|
-4.94
Therefore, the answer is -4.94.
6. EE Board Exam October 1997
Differentiate (x2 + 2)1/2
a)
b)
c)
d)
(x2 + 2)3/2
Solution:
What we w
d
a ed “REVERSE ENGINEERING”
Step : D ffere t ate the g ve equat
at a y va ue f
D ’t forget to set your calculator to RADIAN MODE.
greater tha
Let’
ay = 3
PRESS:
Then,
Display:
0.9045340337
Step 2: Evaluate the given choices and compare the answer when the same value of x is
substituted.
PRESS:
25
CALCulator TECHniques
by C.J. Espinoza
Display:
1.658312395
The remaining choices are evaluated as follows:
b)
= 0.945340337
c)
= 1.809068067
d) (x2 + 2)3/2
=36.48287269
Therefore,
is the correct answer.
7. ME Board Exam October 1997
Find the second derivative of x3 – 5x2 + x = 0
a)
b)
c)
d)
6x – 10
3x +10
3x2 – 5x
10x – 5
Solution:
Solve for the first derivative with limit = 0.
Remember: When you are dealing with derivatives, always set your calculator to RADIAN
MODE.
PRESS:
Display:
|
1
Store this value to A.
PRESS:
Display:
Ans A
Solve for the first derivative and edit the limit to 1x10-6.
Display:
|
26
CALCulator TECHniques
by C.J. Espinoza
0.99999
Store this value to B.
PRESS:
Display:
Ans B
Now, we can compute the second derivative when x = 1x10-6.
PRESS:
Display:
-9.999997667
Next, evaluate the given choices with x =
and compare the answer.
a) 6x - 10
PRESS:
Display:
6x - 10
-9.999994
The remaining choices are evaluated as follows:
b) 3x+10
c) 3x2 – 5x
d) 10x-5
= 10.000003
= -4.999997x10-6
= -4.99999
-9.999994 is the closest value to -9.999997667. Therefore, the second derivative of x3 – 5x2 + x
is 6x - 10.
8. ECE Board Exam November 1991
A balloon is released from the ground 100 meters from an observer. The balloon rises directly
upward at the rate of 4 meters per second. How fast is the balloon receding from the observer
10 seconds later?
a) 1.36 m/sec
b) 1.49 m/sec
c) 1.55 m/sec
d) 1.68 m/sec
Solution:
27
CALCulator TECHniques
V1 = 4
by C.J. Espinoza
S
S1 = 4t
100
S2 = 1002 + (S1)2
S2 = 1002 + (4t)2
S2 = 1002 + 16t2
S=√
t
Set your calculator to RADIAN MODE.
PRESS:
Differentiate S = √
PRESS:
t w th t =
. Change t to x.
Display:
√
t
1.485562705
Therefore, 1.49 m/sec is the correct answer.
9. ME Board Exam April 1998
A box is to be constructed from a piece of zinc 20 sq. in by cutting equal squares from each
corner ad turning up the zinc to form the side. What is the volume of the largest box that can be
so constructed?
a)
b)
c)
d)
599.95 cu. in.
592.59 cu. in.
579.50 cu. in.
622.49 cu. in.
28
CALCulator TECHniques
by C.J. Espinoza
Solution:
x
20-2x
x
x
20-2x
x
20-2x
x
20-2x
First, get the equation of what is asked.
Maximum Volume of the box = ?
V = LxWxH
V = (20 – 2x)(20-2x)(x)
V = (20 – 2x)2(x)
Set your calculator to TABLE MODE.
PRESS:
Input f(x) = (20 – 2x)2(x)
PRESS:
Start? 1
End? 5
Step? 0.5
Incomplete Display:
0| X
4| 2.5
5| 3
6| 3.5
7| 4
|
|
|
|
|
F(X) |
562.5 |
588
|
591.5 |
576 |
Find the maximum value of f(x). It shows that the maximum volume of the box takes place
between 3 and 3.5.
Make another table. Start from 3 and end to 3.5. Step = 0.01
29
CALCulator TECHniques
by C.J. Espinoza
PRESS:
Incomplete Display:
0| X
3| 3.2
4| 3.3
5| 3.4
|
|
|
|
F(X)
591.87
592.54
592.41
|
|
|
|
It happens that the maximum volume of the box takes place at 3.3. Make another table to get the
exact value. Start from 3.3 and end to 3.4. Step = 0.01
Incomplete Display:
0| X |
3| 3.32 |
4| 3.33 |
5| 3.34 |
F(X)
592.58
592.59
592.59
|
|
|
|
It happens that the maximum volume of the box is 592.59 at 3.33 and 3.34.
Therefore, the answer is 592.59.
10. CE Board Exam May 1997
Find the minimum amount of tin sheet that can be made into a closed cylinder having a volume
of 108 cu. inches in square inches.
a)
b)
c)
d)
123.50
125.50
129.50
127.50
Solution:
First, get the equation of what is asked.
Minimum Surface Area = ?
Let:
x = radius of the base of the cylinder
y = height of the cylinder
V = x2y
108 = x2y
y=
The surface area of a cylinder is,
30
CALCulator TECHniques
by C.J. Espinoza
S. A. = 2 xy + 2 x2
S. A. = 2 x(
S. A. =
+ 2 x2
+ 2 x2
Set your calculator to TABLE MODE.
PRESS:
Input f(x) =
+ 2 x2
PRESS:
Start? 1
End? 5
Step? 0.5
Display:
0| X
1| 1
2| 1.5
3| 2
4| 2.5
5| 3
6| 3.5
7| 4
8| 4.5
9| 5
|
|
|
|
|
|
|
|
|
|
F(X)
222.2
158.13
133.13
125.66
128.54
138.68
154.53
175.23
200.27
|
|
|
|
|
|
|
|
|
|
Find the minimum value of f(x). It shows that the minimum amount of surface area takes place
between 2.5 and 3.
Make another table. Start from 2.5 and end to 3. Step = 0.02
PRESS:
Incomplete Display:
0| X |
3| 2.54 |
4| 2.56 |
5| 2.58 |
6| 2.6 |
F(X)
125.57
125.55
125.54
125.55
|
|
|
|
|
It happens that the minimum surface area is 125.54 at 2.58(radius).
31
CALCulator TECHniques
by C.J. Espinoza
125.50 is the closest value to 125.54.
Therefore, the answer is 125.50.
11. EE Board Exam October 1993
Water is flowing into a conical cistern at the rate of 8 m3/min. If the height of the inverted cone
is 12 m and the radius of its circular opening is 6 m, how fast is the water level rising when the
water is 4 m deep?
a)
b)
c)
d)
0.64 m/min
0.56 m/min
0.75 m/min
0.45 m/min
Solution:
dv/dt=8 m3/min
D = 2 x 6m
12 m
dx/dt=?
4m
PRESS:
PRESS:
Display:
0|
1|
2|
3|
X | Y |
0 | 0 |
12 |314.15 |
6
| 0 |
PRESS:
32
CALCulator TECHniques
by C.J. Espinoza
Display:
÷ ŷ
0.6366197724
Therefore, 0.64 m/min is the correct answer.
12. CE Board Exam November 1998
Water is running into a conical vessel 15cm deep and having a radius of 3.75 cm across the top.
If the rate at which the water rises is 2 cm/sec, how fast is the water flowing into the conical
vessel when the water is 4 cm deep?
a)
b)
c)
d)
2.37 m3/sec
4.57 m3/sec
5.73 m3/sec
6.28 m3/sec
Solution:
dv/dt = ?
D=2x3.75 cm
15 cm
2 cm/sec
4cm
PRESS:
PRESS:
Display:
0|
1|
2|
3|
X
0
15
7.5
| Y |
| 0 |
| 44.178|
|11.004|
33
CALCulator TECHniques
by C.J. Espinoza
PRESS:
Display:
ŷ
6.283185307
Therefore, the water flowing into the conical vessel is 6.28 m3/sec.
13. CE Board Exam May 1995
Water is running into a hemispherical bowl of radius 10 cm at a constant rate of 3 cm3/min.
When the water is x cm deep, the water level is rising at the rate of 0.0149 cm/min. What is the
value of x?
a)
b)
c)
d)
2
3
4
5
Solution:
3 cm3/min
R = 10cm
10 cm
x
R = 10 cm
20 cm
10 cm
PRESS:
34
CALCulator TECHniques
by C.J. Espinoza
PRESS:
Display:
0|
1|
2|
3|
X | Y |
0 | 0 |
10 |314.15 |
20 | 0 |
PRESS:
Display:
(3÷0.0149) ̂
4.007441183
Therefore, the value of x is 4.
14. CE Board Exam November 1996
Evaluate: ∫
a)
b)
c)
d)
0.011
0.022
0.033
0.044
Solution:
PRESS:
Display:
∫
3
Therefore, 0.022 is the correct answer.
15. Using power series, evaluate the ∫
d
a) 0.66722
35
CALCulator TECHniques
by C.J. Espinoza
b) 0.94608
c) 0.96622
d) 0.99611
Solution:
PRESS:
Display:
∫
d
0.9460830694
Therefore, 0.94608 is the correct answer.
16. EE Board Exam April 1996
Evaluate ∫ ∫ ∫
a)
b)
c)
d)
.
1/3
2/3
4/3
5/3
Solution:
In this problem, we have three integrands.
∫
⊳①
∫
⊳②
∫
⊳③
First, integrate∫
. Change z to x.
Remember: When you are dealing with integrals, always set your calculator to RADIAN MODE.
PRESS:
Display:
∫
d
2
Store this value to A.
PRESS:
Display:
Ans A
36
CALCulator TECHniques
by C.J. Espinoza
2
Then, integrate ∫
Change r to x.
PRESS:
Display:
∫
d
Store this value to B.
PRESS:
Display:
Ans B
Next, integrate ∫
PRESS:
Display:
∫
d
1
Store this value to C.
PRESS:
Display:
Ans C
1
Last, multiply the results.
PRESS:
Display:
ABC
Therefore, the answer is .
37
2017
CALCulator TECHniques
by C.J. Espinoza
1. EE Board April 1996
Solve xy’ (2y – 1) = y (1 – x)
a)
b)
c)
d)
ln (xy) = 2 (x – y) + C
ln (xy) = x + 2y + C
ln (xy) = 2y – x + C
ln (xy) = x – 2y + C
Solution:
What we will do here is “REVERSE ENGINEERING”.
Solution:
Step 1: Get the value of y’( ). Substitute x and y with any value (n ≠ o). Let’s say x = 2,
y = 3.
y’ =
(
)
(
)
PRESS:
PRESS:
Display:
(
)
(
)
or 0.3
Hence, y’ = -3/10
Step 2: Choose one from the choices and get its first derivative. See if it is = 0.3.
Use choice b.
Get the derivative of choice b with respect to x. Substitute y with 3.
Enter
(ln( x)
x
2( ))
Display:
(ln( x)
x
2( ))
-0.5
Store the value to X.
PRESS:
Display:
Ans X
-0.5
Get the derivative of choice b with respect to y. Change y to x and substitute x with 2.
39
CALCulator TECHniques
(ln(2x)
Enter
2
by C.J. Espinoza
2x)
Display:
(ln(2x)
2
2x)
-1.666666667
Store the value to Y.
PRESS:
Display:
Ans Y
-1.666666667
The first derivative of choice b = -X/Y
Input
PRESS:
Display:
0.3
It is equal to the value of the derivative( ) of the given equation.
Therefore, ln (xy) = x + 2y + C is the correct answer.
2. Find the half-life of a radioactive substance if 20 percent of it disappears in 40 years.
a)
b)
c)
d)
126.25 yrs.
125.25 yrs.
124.25 yrs.
123.25 yrs
Solution:
When
x=0
, y = 100%
x = 40 yrs , y = 100% – 20% = 80%
x=?
, y = 50%
PRESS:
Display:
0|
1|
2|
X |
0 |
40 |
Y
1
0.8
|
|
|
PRESS:
Display:
40
CALCulator TECHniques
by C.J. Espinoza
.5x̂
124.2513488
Therefore, 124.25 yrs. is the half-life of the radioactive substance.
3. A steel ball is heated to a temperature of 100ᵒC and then placed immediately in a place
which is maintained at a temperature of 40ᵒC. At the end of 2 minutes, the temperature of
the ball drops to 80ᵒC. When will the temperature of the ball be 60ᵒC?
a)
b)
c)
d)
3.4mins
4.4mins
5.4mins
6.4mins
Solution:
PRESS:
When
x=0
x=2
x =?
PRESS:
, y = 100 – 40 =60
, y = 80 – 40 = 40
, y = 60 – 40 = 20
Display:
0|
1|
2|
X
0
2
|
|
|
Y
60
40
|
|
|
PRESS:
Display:
20x̂
5.419022583
Therefore, the answer is 5.4 mins.
41
2017
Christian Jay V.Espinoza
01/01/17
CALCulator TECHniques
by C.J. Espinoza
1. EE Board April 1997
Simplify i29+ i21 + i
a)
b)
c)
d)
3i
1-i
1+i
2i
Solution:
Divide the exponents by 4 then get the remainder.
PRESS:
PRESS:
DISPLAY:
7
PRESS:
DISPLAY:
5
PRESS:
Display:
i+i+i
3i
Therefore, 3i is the correct answer.
2. EE Board October 1993
Write the polar form of the vector 3 + j4.
a)
b)
c)
d)
5 53.1
6 53.1
8 53.1
10 53.1
Solution:
PRESS:
Display:
3+4i r θ
5 53.13010235
43
CALCulator TECHniques
by C.J. Espinoza
Therefore, the polar form of the vector 3 + j4 is 5 53.1.
3. EE Board November 1998
Find the value of (1+i)5, where i is an imaginary number.
a)
b)
c)
d)
1-i
-4(1+i)
1+i
4(1+i)
Solution:
Rewrite the equation as follows: (1+i)2 (1+i)3
PRESS:
PRESS:
Display:
(1+i)2 (1+i)3
- 4-4i
We can rewrite the answer as:
- 4-4i = -4(1+i)
Therefore, -4(1+i) is the correct answer.
4. EE Board October 1997
Rationalize
a)
b)
c)
d)
Solution:
Set your calculator to COMPLEX MODE.
PRESS:
Input the given equation
PRESS:
Display:
44
CALCulator TECHniques
by C.J. Espinoza
1 + 2i
Therefore, 1 + 2i is the correct answer.
5. Express
a)
b)
c)
d)
in rectangular form.
1.193 + j1.163
1.352 – j0.315
1.684 – j1.462
1.167 + j0.732
Solution:
Set your calculator to RADIAN MODE.
PRESS:
PRESS:
Display:
e0.32 0 56
1.166778537+0.7315112612i
Therefore, 1.167 + j0.732 is the correct answer.
6. EE Board April 1999
Evaluate ln(2+3i)
a)
b)
c)
d)
1.34 + j0.32
2.54 + j0.866
2.23 + j0.21
1.28 + j0.98
Solution:
Set your calculator to RADIAN MODE.
PRESS:
PRESS:
Display:
Pol(2,3
r=3.605551275,
0 9827937232
Those values of r and θ are automatically saved to X and Y.
Then, set your calculator to COMPLEX MODE.
PRESS:
45
CALCulator TECHniques
by C.J. Espinoza
Next, input this equation: ln(X) + Yi
PRESS:
Display:
ln(X) + Yi
1.282474679 + 0.9827937232i
Therefore, 1.28 + j0.98 is the correct answer.
7. Find the principal value of the complex expression (1 + j2) raise to the power of (3 + j4).
a)
b)
c)
d)
2.236 – j6.541
0.129 + j0.034
-2.013 + j0.514
0.134 + j0.034
Solution:
First, evaluate ln (1 + j2).Then, multiply it to (3 + j4).
PRESS:
PRESS:
Display:
Pol (1, 2
r=2.236067977,
0
Those values of r and θ are automatically saved to X and Y.
Set your calculator to COMPLEX MODE.
PRESS:
After that, input this equation: [ln(X) + Yi](3+4i)
PRESS:
Display:
(ln(X) + Yi)(3+4i)
-2.014438003+6.540321978i
Last, input and evaluate e-2.0144 6 5 03
PRESS:
Display:
46
CALCulator TECHniques
by C.J. Espinoza
e0.32 0 56
0.1290152424+0.03392254661i
Therefore, the answer is 0.129 + j0.034.
8. EE Board October 1997
3
Transpose the matrix |
0
0|
Solution:
Input the given matrix.
PRESS:
To transpose a matrix,
PRESS:
Display:
Ans
|
3
0
|
0
Therefore, |
3
0
|is the correct answer.
0
9. The inverse of matrix |
0
|
Solution:
Input the given matrix to MatA.
PRESS:
To get the inverse of a matrix,
PRESS:
Display:
Ans
|
Therefore, |
0
0
|
| is the correct answer.
47
CALCulator TECHniques
by C.J. Espinoza
10. Find the laplace transform of
a)
b)
c)
d)
-16s/(s2 + 16)
8(3s2 – 16)/(s2 + 16)3
-81/(s2 + 16)
8s/(s2 + 16)2
Solution:
Step 1: Set your calculator to RADIAN MODE. Integrate the given f (t). For this
problem, use the equation Lf (t) =∫
. Change t with x.
PRESS:
Display:
∫
1.98975975x10-3
Step 2: Evaluate the given choices f(s) as s = 9.
Let s = x.
a) - 16s/(s2 + 16)
PRESS:
Display:
5360
The remaining choices are evaluated as follows:
b) 8(3s2 – 16)/(s2 + 16)3
c) -81/(s2 + 16)
d) 8s/(s2 + 16)2
= 1.98975975x10-3
= -0.8350515464
= 0.7422680412
Therefore, 8(3s2 – 16)/(s2 + 16)3 is the correct answer.
11. EE Board March 1998
Determine the inverse laplace of
I(s) =
a) 2e-25t sin100t
48
CALCulator TECHniques
by C.J. Espinoza
b) 2t e-25t sin100t
c) 2e-25t cos100t
d) 2t e-25t cos100t
Solution:
Step 1: Evaluate I(s) as s = 9.
Let s = x.
PRESS:
Display:
00
500
Step 2: Set your calculator to RADIAN MODE. Integrate the given choices f(t) and
compare the answer when the same value of x is substituted to t. For this problem,
use the equation Lf(t)=∫
.
a) 2e-25t sin100t
PRESS:
Display:
∫
00
0.0179275
The remaining choices are evaluated as follows:
b) 2t e-25t sin100t
c) 2e-25t cos100t
d) 2 t e-25t cos100t
= 0.0001092
= 0.0060953
= -1.4212213
Therefore, 2e-25t sin100t is the correct answer.
49
2017
CALCulator TECHniques
by C.J. Espinoza
1. A box is pushed along the floor by a force of 40 lb making an angle of 30 degrees with the
horizontal. Find the horizontal and vertical components of the force.
a)
b)
c)
d)
Fx = 32.6 lbs, Fy = 36.2 lbs
Fx = 34.6 lbs, Fy = 20 lbs
Fx = 36.6 lbs, Fy = 32.2 lbs
Fx = 20 lbs, Fy = 36.2 lbs
Solution:
Set your calculator to COMPLEX MODE, DEGREE MODE and set the complex result to rectangular
form.
PRESS:
Input and evaluate the given force.
PRESS:
Display:
40 30
34.64101615+20i
Therefore, Fx = 34.6 lbs, Fy = 20 lbs.
.
2. Find the resultant of the coplanar forces 80 lbs at -30 degrees and 60lbs at 60 degrees.
a)
b)
c)
d)
100 lbs at 6.87 degrees
100 lbs at -67.8 degrees
120 lbs at 111.3 degrees
120 lbs at 67.8 degrees
Solution:
Set your calculator to COMPLEX MODE, DEGREE MODE and set the complex result to polar form.
PRESS:
Input and add the given forces.
PRESS:
Display:
80 -30+
100 6.869897646
Therefore, the resultant of coplanar forces is 100 lbs at 6.87 degrees.
.
3. ME Board October 1996
Assume the three force vectors intersect at a single point.
F1 = i + 3j + 4k
F2 = 2i + 7j - k
51
CALCulator TECHniques
by C.J. Espinoza
F3 = -i + 4j + 2k
What is the magnitude of the resultant force vector, R?
a)
b)
c)
d)
13.23
14.73
15
16.16
Solution:
Input the 1st force vector to VctA.
PRESS:
Display:
A
[
]
Input the 2nd force vector to VctB.
PRESS:
Display:
B
[
]
Input the 3rd force vector to VctC.
PRESS:
Display:
C
[
]
Get the absolute value of vector sum of the three force vectors.
PRESS:
Display:
Abs(VctA+ VctB+ VctC
15
Therefore, the magnitude of the resultant force vector, R , is 15.
4. EE Board March 1998
Electrical loads are arranged on horizontal x, y axes as follows:
52
CALCulator TECHniques
Load
1
2
3
4
5
6
7
8
XYcoordinate coordinate
0
2
1
1
1
3
2
0
2
4
3
1
3
3
4
2
by C.J. Espinoza
Kilowatt
load
100
180
200
120
150
200
180
100
Locate the centroid of the load.
Solution:
PRESS:
To show the frequency,
PRESS:
Type the data:
X
0
1
1
2
2
3
3
4
1
2
3
4
5
6
7
8
Y
2
1
3
0
4
1
3
2
Frequency
100
180
200
120
150
200
180
100
To get the coordinates of the centroid, get the ̅ and ̅.
PRESS:
Display:
̅
2
PRESS:
Display:
̅
2.048780488
Therefore, x = 2, y = 2.049.
53
2017
CALCulator TECHniques
by C.J. Espinoza
1. ME Board April 1995
P 4,000 is borrowed for 75 days at 16% per annum simple interest. How much will be due at
the end of 75 days?
a)
b)
c)
d)
P 4,133.33
P 4,150.00
P 4,166.67
P 4,333.33
Solution:
Remember: Y = A + Bx is suitable to Simple Interest problems.
PRESS:
PRESS:
Display:
0| X |
1| 0 |
2| 360 |
Y
1
1.16
|
|
|
PRESS:
Display:
4x103×75ŷ
4133.333333
Therefore, the money will be P 4,133.33 at the end of 75 days.
.
2. ECE Board April 1998
The amount of P 12,800 in 4 years at 5% compounded quarterly is
a)
b)
c)
d)
P 14,785.34
P 15,614.59
P 15,847.33
P 16,311.26
Solution:
Set the calculator to STAT MODE 6.
PRESS:
Then, type the given data
0| X |
Y
|
1| 0 |
1
|
2| 1 | 1 + (5%/4) |
Future worth = Present worth x (years of interest period x mode of interest) ̂
55
CALCulator TECHniques
by C.J. Espinoza
PRESS:
Display:
12800x(4x4) ̂
15614.58621
Therefore, P 15,614.59 is the correct answer.
3. ME Board October 1996
Fifteen years ago, P1, 000.00 was deposited in a bank account, and today it is worth P 2, 370.00.
The bank pays interest semi-annually. What is the interest rate paid in this account?
a)
b)
c)
d)
3.8%
4.9%
5.8%
5.0%
Solution:
PRESS:
Then, type the given data
0|
X
|
Y
|
1|
0
| 1000 |
2| 15(2) | 2370 |
Interest rate (nominal rate) = (B – 1) x mode of interest
PRESS:
Display:
(B-1)x2
0.05836129643
Therefore, the interest rate paid is 5.8%.
4. ECE Board November 1998
The effective rate of 14% compounded semi-annually is
a)
b)
c)
d)
12.36%
14.49%
14.88%
14.49%
Solution:
PRESS:
Then, type the given data
56
CALCulator TECHniques
0|
1|
2|
X
0
1
|
|
|
by C.J. Espinoza
Y
|
1
|
1+14%/2 |
PRESS:
Display:
2ŷ-1
0.14.49
Therefore, the effective rate is 14.49%.
5. CE Board May 1995
How long (in years) will it take money to quadruple if it earns 7% compounded semi-annually?
a)
b)
c)
d)
40.30
33.15
26.30
20.15
Solution:
Set the calculator to STAT MODE 6.
PRESS:
From the formula of Future Worth:
F = P (1+i)n
Let money = P 1.00
X-column = year
Y-column = the worth of money per year
When x = 0 , y = 1
x = 1 , y = 1(1+ )2
PRESS:
F = 4P?
PRESS:
Display:
4̂
20.14879168
Therefore, the money will quadruple in 20.15 years.
57
CALCulator TECHniques
by C.J. Espinoza
6. EE Board April 1997
A machine has an initial cost of P 50,000 and a salvage value of P 10,000 after 10 years. Find the
book value after 5 years using straight-line depreciation.
a)
b)
c)
d)
P 12,500
P 30,000
P 16,400
P 22,300
Solution:
Remember: Y = A + Bx is suitable to Straight-Line method.
PRESS:
PRESS:
Display:
0|
1|
2|
X |
Y
|
0 | 50000 |
10 | 10000 |
PRESS:
Display:
5ŷ
30000
Therefore, the book value after 5 years is P 30,000.
7. ME Board April 1998
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 3rd year? Use SOYD method.
a)
b)
c)
d)
P 10,000.00
P 857.00
P 937.00
P 747.00
Solution:
Remember: The parabolic function Y = A + Bx + Cx2 is suitable to SOYD method of depreciation.
PRESS:
Then, type the given data
0| X |
Y
1| 0 | 10000
2| 20 |
0
3| 21 |
0
|
|
|
|
58
CALCulator TECHniques
by C.J. Espinoza
PRESS:
Next,
PRESS:
Display:
2ŷ-3ŷ
857.1428571
Therefore, the depreciation in the 3rd year is P 857.00.
8. CE Board May 1999
A machine costing P45, 000 is estimated to have a book value of P 4,350 when retired at the end
of 6 years. Depreciation cost is computed using a constant percentage of the declining book
value. What is the annual rate of depreciation in %?
a)
b)
c)
d)
33.25%
32.25%
35.25%
34.25%
Solution:
Remember: The function Y = ABx is suitable to Declining Balance method.
PRESS:
Then, type the given data
0| X |
Y
|
1| 0 | 45000 |
2| 6 | 4350 |
PRESS:
Next,
PRESS:
Display:
1-B
0.3225465525
Therefore, the annual rate of depreciation in % is 32.25%.
59
CALCulator TECHniques
by C.J. Espinoza
10. P 1500 was deposited in a bank at an interest rate of 10% compounded annually. Tabularize
the yearly future worth of the money from the 1st year to 10th year.
Solution:
Set your calculator to TABLE MODE.
PRESS:
Then,
From the formula of future worth with compounded interest: F = P (1+i)n
where: P = 1500, i = 0.10 and n = 1 to 10
Input the equation: (x) = 1500(1+0.10)x
PRESS:
Display:
0| X
1| 1
2| 2
3| 3
4| 4
5| 5
6| 6
7| 7
8| 8
9| 9
10| 10
|
|
|
|
|
|
|
|
|
|
|
F(X)
1650
1815
1996.5
2196.1
2415.1
2657.3
2923
3215.3
3536.9
3890.6
|
|
|
|
|
|
|
|
|
|
|
60
2017
CALCulator TECHniques
by C.J. Espinoza
1. Convert 160910 to hexadecimal
a)
b)
c)
d)
5A516
A9516
C4116
64916
Solution:
Set your calculator to BASE-N MODE.
PRESS:
PRESS:
Display:
1609
PRESS:
Display:
12
Therefore, 64916 is the correct answer.
2. Convert 1210 to base 2
a)
b)
c)
d)
1100
1001
1101
1010
Solution:
Set your calculator to BASE-N MODE.
PRESS:
PRESS:
Display:
12
PRESS:
Display:
62
CALCulator TECHniques
by C.J. Espinoza
12
Therefore, 1100 is the correct answer.
3. Convert F116 to Octal
a)
b)
c)
d)
3318
3618
7418
6628
Solution:
Set your calculator to BASE-N MODE.
PRESS:
PRESS:
Display:
F1
PRESS:
Display:
F1
Therefore, 3618 is the correct answer.
4. Add: 47816 + 79216
a)
b)
c)
d)
C1116
B0016
C0A16
BFA16
Solution:
Set your calculator to BASE-N MODE.
PRESS:
PRESS:
Display:
63
CALCulator TECHniques
by C.J. Espinoza
478+792
Therefore, C0A16 is the correct answer.
5. Find the sum of 6F54016, 10101111002, 7251523008 and 8910. The result must be in base
10.
a)
b)
c)
d)
123546897
123654987
123489576
123456789
Solution:
Set your calculator to BASE-N MODE.
PRESS:
Then, convert all values to Decimal.
PRESS:
Display:
6F540
PRESS:
Display:
6F540
PRESS:
Display:
1010111100
PRESS:
Display:
725152300
Sum 123000000, 456000, 700 and 89
123000000 + 456000 + 700 + 89 = 123456789
64
CALCulator TECHniques
by C.J. Espinoza
Therefore, 123456789 is the correct answer.
65
2017
CALCulator TECHniques
by C.J. Espinoza
1. The resistance of a wire is 130 ohms at 100 degrees C and 100 ohms at 30 degrees C.
Determine the temperature when the resistance is 109 ohms.
a)
b)
c)
d)
41
51
61
71
Solution:
Set the calculator to STAT MODE 2
PRESS:
X column = resistance
Y column = temperature
Enter this data:
0|
X
|
1| 130 |
2| 100 |
Y
100
30
|
|
|
Temperature at 109 ohms?
PRESS:
Display:
109ŷ
51
Therefore, the temperature when the resistance is 109 ohms is 51.
67