Intermediate Algebra

Intermediate Algebra



Problems and Solutions



with your fx-991ES or fx-


115ES Calculator












Dr Allen Brown








Cambridge 
Paperbacks

Cambridge Paperbacks

www.CambridgePaperbacks.com

First published by Cambridge Paperbacks 2018



© Allen Brown 2018

All rights reserved. No part of this publication may be reproduced or
transmitted in any form or by any means, electronic or mechanical, including
photocopy, recording, or any information storage and retrieval system without
permission in writing from the author.



Disclaimer

Although the author and publisher have made every effort to ensure that the
information in this book was correct during preparation and printing, the
author and publisher hereby disclaim any liability to any party for any errors
or omissions.

Read this First


As you progress through your education you will be

pursuing a maths course and the basic skill you need
to acquire is algebra. This ebook will help you to build

you maths skill set at the same time teach you to make
effective use of your scientific calculator.


In every example the keystrokes for your calculator

are given, all you have to do is repeat the same
keystrokes and you will get the correct answers. The

more times you do this, the more confident you will

become at using your calculator as a real resource in
helping you to develop your algebraic skills.


There are several examples on deriving equations for
calculating the volume of geometric shapes and

composites. These are valuable lessons as they

introduce you to the world of mathematical
modelling. Today there is a great need for these skills

and this is an excellent place to start.



Dr Allen Brown
Cambridgeshire

Contents

Example 1: ............................................................................ 2

Example 2: .......................................................................... 11

Example 3: .......................................................................... 14

Example 4: .......................................................................... 17

Example 5: .......................................................................... 20

Example 6: .......................................................................... 23

Example 7: .......................................................................... 27

Example 8: .......................................................................... 29
Example 9: .......................................................................... 32

Example 10: ........................................................................ 34

Example 11: ........................................................................ 37

Example 12: ........................................................................ 39

Example 13: ........................................................................ 43

Example 14: ........................................................................ 45












1

Example 1:

Calculating the volume of four solid objects. A cuboid,

a cylinder, a sphere and a cone.























First looking at your fx-991ES PLUS calculator you will

see letters that are used as symbols.














2

The range from A to F, also M, X and Y. Values can be

assigned to these letters which means they can be
used in equations.

The Cuboid















The cuboid has six surfaces. Three parallel pairs. The

lengths of the sides are A, B and C. Volume of cuboid

is
=

The keystrokes for your fx-991ES calculator are,
Q(A)Q(B)Q(C)


Calculate the volume of the cuboid.


















3

r


A? 2.82=
B? 5.83=

C? 3.77=








3
Volume of cuboid is 61.98 cubic metres of m . What
is the volume of the following cuboid?
















r


A? 1.44=
B? 3.37=

C? 0.87=







3
The volume is 4.22 m .

4

The cylinder
















Height of cylinder is A, the diameter of cylinder is B.
Volume of cylinder is


2
= ( )
2

The keystrokes for your fx-991ES to calculate the

volume are,
CqL(Q(B)a2$)dQ(A)


Here is a couple of examples.















r

B? 1.58=

5

A? 5.2=








3
The volume of the cylinder is 10.19 m . Here is
another example,




















r

B? 0.47=

A? 2.28=









3
The volume of this cylinder is 0.395 m .



6

The Sphere














A sphere has a distinctive shape – round. Volume of

sphere,
4
3
=
3
It is difficult to measure the radius of a sphere, much

easier to measure its diameter using digital callipers.





















The volume is,
4 3 4 1 3
3
= ( ) = × =
3 2 3 8 6

7

The keystrokes for this expression,
CqLQ(D)q(x ) a6
3


Now use this to calculate the volume of the sphere as

measured by the digital callipers.




















r

D? 22.01=








3
The volume of this sphere is 5,582.8 mm . In terms of
3
cubic cm, divide by 1,000 to give 5.5828 cm .





8

The Cone





























The cone has height A and base diameter D. The

volume is given by,

1 2
2
= ( ) =
3 2 12

The keystrokes for this expression are,

CqLa12$Q(D)dQ(A)


We are now going to use this to measure the volume

of a cone.



9

r

D? 1.75=

A? 3.62=









3
Volume of the cone is 2.9 m .

Whenever you are required to calculate the volume of

a cuboid, or cylinder, or sphere or cone, you now

know how to use your fx calculator to help you.


10

Example 2:

A storage tank comprises a cylinder of height A and

diameter B and a hollow

hemisphere attached to the
top of the cylinder. Derive an

expression of the volume of

the structure.





The diagram shows the storage tank. The volume of

the cylinder is given by,

2
( )
2


The volume of the hemisphere is given by,

1 4 3 2 3
× ( ) = ( )
2 3 2 3 2


2 1 1
3
= 3 =
3 8 12
The total volume of the composite is given by,





11

1 1
2
3
= +
4 12
Or
2 1
= ( + )
4 3


The keystrokes for you fx-991ES calculator are,
aqLQ(B)dR4$(Q(A)+1

a3$J(B))



A cylinder height of a storage tank (as shown on page
11) is 4.68 metres and its base diameter is 1.83

metres, what is its volume?

r
Diameter B? 1.83=

Height A? 4.68=







3
The volume of the tank is 13.91 m .

The cylinder height of another Storage tank is 5.77 m

and its base diameter is 1.49m, what is its volume?


12

r


Diameter B? 1.49=
Height A? 5.77=











3
The volume of the second tank is 10.926 m .


You will find your fx-991ES calculator very useful for
finding the volume of composite structures.




















13

Example 3:

A composite structure is made up from two cones.

Derive an expression for the volume of the structure.





















Here is an image of the composite structure. The base

diameter for both cones is A . The length of one cone
is B and the length of the other cone is C. The volume

of each cone is,

1 2
( )
3 2


1 2
( )
3 2
Total volume is given by

1 2 1 2
= ( ) + ( )
3 2 3 2

14

Simplify this expression

1 2
= ( ) ( + )
3 2


2
= ( + )
12


The keystrokes for your fx-991ES calculator for

performing this calculation are,


CaqLQ(A)dR12$(Q(B)
+Q(C))

















Calculate the volume of the composite.
r

Cone diameter A? 30.8=

First cone length B? 53.7=

Second cone length C? 84.22=

15

3
The volume of composite is 34,252.9 cm . Find the
volume of a second composite,













r

Cone diameter A? 22.6=

First cone length B? 41.7=

Second cone length C? 72.92=









3
The volume of the second composite is 15,326 cm .







16

Example 4:

A composite structure is made up from a cone and a

frustum. Derive an expression for the volume of the

structure.


















The base of the frustum has diameter A and the top

of the frustum has a diameter D. The volume of the
frustum is,


2
2
( + + )
12
The volume of the cone is,
1 2 2
( ) =
3 2 12

Total volume of composite is,
2
2
2
+ ( + + )
12 12



17

Simplify,

2
2
2
= [ + ( + + )]
12

The keystrokes for your fx-991ES calculator are,

CqLa12$(Q(B)Q(A)d+
Q(C)(J(A)d+J(A)Q(D)+J(D)d

))



Calculate the volume of this composite.




















r

cone length B? 23.6=

cone diameter A? 12.5=

Frustum length C? 50.46=

Frustum diameter D? 4.83=



18

3
The volume of the composite is 4,135.27 cm . You
have probably seen a frustum before,






















The Rolo has a frustum shape. The manufacturers
need to know the volume of each chocolate.







19

Example 5:

A solid cylinder has a section removed; derive an

expression for the volume of the remaining cylinder.

















The image shown here is of a cylinder with a section
removed. Volume of the whole cylinder,

2
( )
2
Volume of the section cylinder removed is,

1 2
( )
2 2
Volume of remaining cylinder,

2 1 2
= ( ) − ( )
2 2 2

Which can be expressed as,

2 1
= ( ) ( − )
2 2



20

The keystrokes for your fx-991ES calculator are,

CqL(Q(C)a2$)d(Q(A)
pQ(B)a2$)



Find the volume of cut away cylinder.


















r

cylinder diameter C? 3.62=

cylinder length A? 15.86=
length of cut away section B? 5.53=











3
The volume of the structure is 134.77 cm .


21

Calculate the volume of another cylinder with a cut

away section.











r

cylinder diameter C? 1.94=
cylinder length A? 21.15=

length of cut away section B? 13.71=









3
The volume of this structure is 42.26 cm .











22

Example 6:

Plot the following function,

2
3
( ) = − 4 + 5
in the range -2.0 ≤ x ≤ 4.5


To perform this task you can use your fx-991ES

calculator. First set the calculator in its TABLE mode,
(w7). When you see f(x)= in the display, enter

the following keystrokes,
Q(X)q(x )p4J(X)d+5=
3


Start? z2.0=

End? 4.5=

Step? 0.5=



















23

Here are the first 12 data values. Minimum F(X) is -19

Maximum F(X) is 15.125. Select graph axis with
F(X)min = -20 and F(X)max = 16. Plot data points from

your fx-991ES onto graph paper.
































From the graph estimate the approximate root values
for the function. x = -1, x = 1.4 and x = 3.6. you can

determine how close these values are by using your

fx-991ES calculator. Enter the following keystrokes,
w1
Q(X)q(x )p4J(X)d+5
3


24

r X? z1=








A very good estimate of a root value.

r X? 1.4=n







A reasonable estimate of the root value.



r X? 3.6=n







Not a really good estimate of the root value.


For the exact roots, use the following keystrokes,
w5









25

Option 4 allows you to solve cubic equations.
4 1=z4=0=5=



=







=








=








Compare these results with the values obtained from

the graph – they are close.











26

Example 7:

In the following equation, make B the subject of the

equation.



= 2 √




Divide both sides by 2π,


= √
2


Square both sides,
2
( ) =
2

Multiply both side by g,

2
= ( )
2
The keystrokes for your fx-991ES calculator for
performing this calculation are (w1),

Q(B)Q(=)q(CONST)35(Q(A)a
2qL$)d

The gravitational constant g is stored in your fx-991ES

calculator – use CONST to access it.


27

We can use this to calculate the

length (in metres) of a pendulum to
give a required oscillation period.

Calculate the value of B when

A is
1: 4.3 seconds

2: 2.2 seconds
3: 0.55 seconds



1:
r A? 4.3=






2:

r A? 2.2=







3:
r A? 0.55=









28

Example 8:

In the following equation, make x the subject of the

equation.

3 −
= ln ( )
3 +


3 −
ln( ) = ln ( )
3 +
Use the log rule,

ln( ) = ln( )

3 −

ln( ) = ln ( )
3 +
Remove the logs

3 −

=
3 +
Multiply both sides by (3 + x)
(3 − )

(3 + ) = (3 + )
(3 + )


(3 + ) = 3 −
Expand


3 + = 3 −


29

Subtract 3e y




3 − 3 + = 3 − 3 −



= 3 − 3 −
Add x to both sides,


+ = 3 − 3 − +



+ = 3 − 3
Collect common terms


(1 + ) = 3(1 − )

y
Divide both sides by (1 + e )
1 1


(1 + ) = 3(1 − )


(1 + ) (1 + )


3(1 − )
=
1 +
To confirm you have the right solution, consider the
original equation,

3 −
= ln ( )
3 +

The range of x is 0 ≤ < 3. If x is outside this range
will cause an error (why?). Calculate the value of y

when = 1.6.

30

The keystrokes for your fx-991ES calculator for

performing this calculation are,

3 − 1.6
= ln ( )
3 + 1.6

Cha3p1.6R3+1.6

$)=








The value -1.18958 is stored as M, now consider,

3(1 − )
=
1 +
Substituting M for y should give the value 1.6.

3(1 − )
=
1 +

The keystrokes for this calculation,
a3(1pHM$)R1+H

M=n









Thus confirming the correct result.

31

Example 9:

The expression for compound interest bank loan is

given by,


= (1 + ) :
100 12
where A is the final sum to be paid back to the bank,

B the amount borrowed, C the % interest rate and D
the duration of the loan. A/12D is the amount to be

paid back every month.


The keystrokes for your fx-991ES Calculator are
CQ(A)Q(=)Q(B)(1+Q(C)a1

00$)^Q(D)$Q(:)J(A)a1

2J(D)



Paul Twister is taking out a bank loan to buy a car.
He needs £6,000 and he intends to pay it back over a

period of 3 years. The interest rate is 3.9%. How

much will he pay back to the bank and what are his
monthly payments?
r


The amount to borrow B? 6000=


32

The interest rate C? 3.9=

Duration of the loan D? 3=










He will be paying back to the bank £6,729.74p for the
loan.

=










His monthly payments will be £186.94p. In reality Paul

will be paying interest on money he has already paid
back to the bank. Towards the end of the load, he will

be paying interest on money he paid some 35 months

ago. Meanwhile the bank has loaned that money to
someone else and is now receiving interest payments

from at least two lenders.






33

Example 10:

Make x the subject of the equation,



= √ 1− 3 1
1+ 3


Square both sides,

1 − 3
2
=
1 + 3

3
Multiply both sides by (1 + x ),

3
(1 − )
2
3
3
(1 + ) = (1 + )
3
(1 + )

2
3
3
(1 + ) = (1 − )
Expand,
2 3
2
3
+ = 1 −

3
Add x to both sides
3
3
3
2 3
2
+ + = 1 − +

2
Subtract y from both sides,
2
3
2 3
2
2
+ + − = 1 −
34

Collect terms
3
3
2
(1 + ) = 1 −
3
Divide by (1 + y ),
3
(1 + ) 1 − 2
3 =
1 + 3 1 + 3

1 − 2
3
=
1 + 3
Leaving,

3 1− 2
= √ 2
1+ 3

To test whether this is correct, consider the original
equation,


1 − 3
= √
1 + 3

3
Calculate y for a test value of say x = 0.256. Since x
occurs twice, we can make this an M with the

following keystrokes,
C0.256q(x )=
3







35

This value now becomes M, re-writing the original

equation,


1 −
= √
1 +


The keystrokes for calculating y are,
sa1pMR1+M=









This can now be fed back into Eq:2 as M,


3 1 − 3
= √
1 + 3

The keystrokes for this are,
Sa1pMdR1+Md=n








Since the test value has bee recovered, this shows

that Eq:2 is in fact correct. A straight forward method

to determine if your answer is correct.

36

Example 11:

A composite comprises a cuboid and a cone as shown

in the diagram below. Derive an equation for its

volume.



The volume of the cuboid is

given by,




The volume of the cone is,


1 2
( )
3 2

The total volume is therefore,


1 2
= + ( )
3 2

2

= +
12


or

2
= ( + )
12



37

The keystrokes for your fx calculator are,


CQ(C)(Q(A)Q(B)+aqLQ(D)
dR12$)


Calculate the volume of the composite when:


A = 23.81 cm, B =25.94 cm, C = 19.65 cm, D = 7.38 cm

r

C? 19.65=

A? 23.81=
B? 25.94=

D? 7.38=








3
The volume of the composite is 12,416.6 cm . If the
diameter of the cone base is increased to 10.43 cm,
what is the new volume? r===


D? 10.43=








3
New volume is 12,696 cm .
38

Example 12:

A cuboid structure has a cylindrical section removed

from it as shown in the diagram below.
















The density of the material is 4.8 g/cc. Derive an

expression for the mass of the structure.


The volume of the cuboid is and the hole is depth

E, the volume of the hole is given by,


2
( )
2

The volume of the structure is therefore,


2
− ( )
2

2
= −
4

39

Density = mass/volume, the mass is given by,

2
= 4.8 ( − )
4


the keystrokes for your fx calculator for this equation

are,

C4.8(Q(A)Q(B)Q(C)pqL

Q(E)Q(D)da4$)


Calculate the mass if: A = 12.5 cm, B = 10.3 cm, C =

11.8 cm, D = 4.2cm and E = 4.68 cm.

A? 12.5=


B? 10.3=


C? 11.8=


E? 4.68=

D? 4.2









The mass of the structure is 6,981.17 grams or 6.981

kg.

40

The structure is filled with a fluid of density 1.4 g/cc to

a height F. What is the new mass of the structure.





















The volume of the fluid is given by,


2
( )
2

The mass of the fluid is therefore,

2
2 1.4
1.4 ( ) =
2 4


The total mass is,

2
1.4
= 6981.17 +
4

The keystrokes for this equation are,




41

C6981.17+1.4qL

Q(D)dQ(F)a4



Calculate the total mass when F = 2.82 cm.
r

Diameter D? 4.2=

Fluid height F? 2.82=







The mass of the structure has increased to 7,035.87

grams by adding the fluid.




This example shows modelling of a structure; deriving

an equation for first volume and then mass. By using
symbols you are able to see the change in the volume

or mass easily as the dimensions are adjusted.












42

Example 13:

A composite is constructed from attaching a cylinder

to a frustum as shown in the diagram below. Derive

an expression for its volume.
























The volume of a frustum is given by,



2
2
( + + )
12
The volume of the cylinder is given by,

2
( )
2

Volume of structure






43

2
2
2
( + + ) + ( )
12 2
This can be rearranged to give,


2
= ( + ) + ( + )
4 3 12

The keystrokes for your fx calculator this equation are,

CqLQ(C)da4$(Q(A)a3

$+Q(B))+qLJ(A)J(D)a1

2$(J(D)+J(C))


Calculate the volume of the composite when,

A = 12.6 cm, B = 7.3 cm, C = 2.46 cm, D = 15.39 cm.

r


C? 2.46=
A? 12.6=

B? 7.3=

D? 15.39=







3
The volume of the composite is 960.8 cm .


44

Example 14:

A composite is constructed

from stacking four cylinders,

each of the height E as shown
in the diagram on the left (not

drawn to scale). The diameter

of the base cylinder is A and
the successive cylinders are B,

C and D. Derive an expression

for the total surface area of the
composite.







When you look down from above you

will see a pattern as shown on the left.
You do not need to calculate the area of

each annulus as the total annulus area
2

is simply ( ) .
2

Area of composite,


2
= 2 ( ) + 2 ( ) + 2 ( ) + 2 ( ) + 2 ( )
2 2 2 2 2
45

2
= ( + + + ) +
2

The keystrokes for this equation are,


CqLQ(E)(Q(A)+Q(B)+Q(C)
+Q(D))+qLJ(A)da2


Calculate the area of the composite if :


Diameter of red (A) = 13.56 cm
Diameter of green (B) = 11.48 cm

Diameter of blue (C)= 8.41 cm

Diameter of yellow (D) =3.56 cm
Height (E) = 5.73 cm
r


E? 5.73=
A? 13.56=

B? 11.48=
C? 8.41=

D? 3.56=








2
Surface area of composite is 955 cm .


46