Subtest II Study Guide + Review
Find
below the different styles of Qs one is very likely
to encounter on the CSET Subtest II.
In
general, time constraints might preclude attempt / mastery of every Q
emphasized! In which case, make sure a couple of Qs from each "style"
are attempted.
–> Try
the 1st 5 SUBTEST II GeometryPractice
Questions on my site. Here are the Solutions [the other Qs
are in the Q Bank]:
#5. Sketch rectangle ABCD inscribed in a
circle. Let diagonals AC and BD intersect at the centre of the
circle, P. Drop a perpendicular PX onto AB.
NOTE: An ~
Angle.
Given: An (AOB) = 120° so that An(BPX) =
60°
Also, AB = 12 cm so that XB = 6
Using
properties of 30° –60° – 90 ° Δs
on ΔBPX:
PX = 2√3 cm so that BC = 2PX =
4√3 cm.
Also, radius of the circle, PB = 4√3
cm.
Finally, Area of the rectangle ABCD = AB·BC =
48√3 cm2.
Area
of the circle = π (4√3)2 =
48π cm2.
Required
Area = 48(π – √3) cm2.
#4.
Since Circumference of the circle = 36π, radius, r =
18cm.
Join the centre of the circle to 2 of the vertices of
the equilateral triangle. The central angle so formed is
120 °.
Drop the perpendicular onto the side
joining those 2 vertices to form a 30 °–60 ° –
90° Δ with the hypotenuse ≈ radius = 18cm so
that the side of the triangle opposite the 60° angle =
9√3 cm and the side of
the equilateral triangle, s = 18√3.
Area of the
equilateral triangle, A = (√3s2)/4
= 420.88 cm2.
#3.
Proceed as in #4 above and use properties of 30° –60° –
90 ° Δs.
Solutions check: side of
the equilateral triangle, s = 48√3 cm.
Using (again!)
properties of 30° –60° – 90 ° Δs,
r adius of the circle, r = 24cm.
Area of the circle, A =
576π cm2.
#2.
a) Draw 1 iscosceles triangle equivalent to one of the 12 congruent
triangles formed by joining the centre of the regular polygon /
circle to successive vertices of the polygon.
The base of the
triangle, b ≈
side
of the polygon, s =
10cm.
Central angle = 360/12 = 30°
Drop a
perpendicular from the apex of the triangle onto its base to form a
15°–75°–90° Δ and splitting the
base in two.
Using trigonometry, altitude of the
triangle, a =
18.66cm and the hypotenuse, h =
19.32cm.
Area of the isosceles triangle, At =
½ ab =
93.3 cm2.
Therefore,
Area of the 12–sided polygon, Ap =
12·At =
1119.6 cm2.
Also,
radius of the circumscribed circle, r ≈ hypotenuse, h =
19.32cm.
Therefore, Area of the circle, Acc =
1172.44 cm2.
Required
Area inside the
circumscribed circle but outside the
polygon, A = Acc –
Ap
= 52.84
cm2.
b)
If a circle were inscribed inside the polygon, radius of this circle,
r = altitude of the polygon, a =
18.66 cm so that Area of the inscribed circle, Aic =
1093.88 cm 2.
Required
Area inside the
polygon but outside the
inscribed circle, Ap –
Aic =
25.71 cm2.
#1.
Sketch and label the figure as instructed. Join O and C.
Since
both AO and CO are radii of the circle, let AO = CO ≈ x cm
and DC ≈ y cm.
Then,
AD = x +
1 cm.
Now, since ΔODC is a right triangle,
using Pythagoras' Theorem:
OD2 +
DC2 =
OC2.
=>
12 +
y2 =
x2
=> y2 =
x2 –
1 –––––––––––––––––––––––––
(*)
And since ΔADC is a right triangle, using
Pythagoras' Theorem:
AD2 +
DC2 =
AC2.
=>
(x + 1)2 +
y2 =
15/2 = 7.5 ––––––––
(**)
Substituting (*) in (**):
(x + 1)2 +
x2 –
1 = 7.5
Simplifying: 2x2 +
2x – 7.5 = 0
or 4x2 +
4x – 15 = 0
Solving the quadratic equation: x =
(–4 ± 16)/8.
The only feasible solution is x =
12/8 = 1.5.
Therefore, radius of the circle, r = 1.5 cm.
–>
PROOFS: Of
the 4 FR Qs, you may expect at
least 1
Geometric Proof. Likewise, at
least a
couple of the MCQs shall be Proof–based wherein one is required
to supply a missing step.
a) The middle section of the Q
Bank consists of Fill In The Blanks Qs to simulate the MCQ–style
Proof–related Qs. While many of these are basic and assess
fundamental skills, the most important
amongst these are:
#4, #5, #75, #76: relating to the
Triangle Sum Theorem
#16, #19, #20, #22–#25, #27–#33,
#46, #52, #53: relating to Parallelograms
#36, #58,
#59, #64, #66, #67: relating to the Inscribed Angle Theorem of
circles
#56, #62: about Pythagoras' Theorem, whose proof relies on
the Geometric Mean Theorems [SET 29 #5 & 6; SET 30 #4]
b)
The FR Proof Q customarily deals
with Triangle Congruence [SSS, SAS, ASA, AAS, RHL].
At times,
one is simply expected to prove 2 triangles in
the 3–dimensional
space are
congruent. This is a bloody scam! Th same rules
that relate to 2 dimensions [ie. in a plane, as one is accustomed to
to!] apply in 3–dimensions.
c) The other perennial
favourite relates to Parallelogram Theorems and their Converse. Qs in
SET 10 and 12 [I believe!] deal with this. These
are vital
Theorems! Attempt
these at
least a
couple of times!
d)
More "complex" [oh, there all quite simple!] Triangle
Congruence and Parallelogram–related Proofs are of the
type:
SET 12: #4 [very
important!]
SET
15: #7–#10
SET 16: #8–#12
SET 22: #1 [very
important!], #2,
#5, #6–#12
SET 29: #1, #9
SET 32 [very
important!]
SET
33 [ very
important!]
SET
34: #1–#3
SET 36 [ very
important!]
SET
37: #1–#4
e) You ought to
be familiar with properties and Proofs relating to Special
Quadrilaterals – Rhombus, Kite, Rectangle, Square.
f)
Knowledge of Proof by Contradiction [also referred to as: Indirect
Proof] is important. SET 28 is about this "concept"! One
Q is assured on this topic.
g)
Proofs concerning Similar Triangles haven't been emphasized so
far!
However, rules for Triangle Similarity [AA, SAS, SSS], the Geometric
Mean Theorems and the Pythagoras' Theorem are vital
to know! SET
29: #2–#6 and SET 30: #4 deal with these.
h)
Finally, I have hand–written 2 very
important proofs
relating to a Triangle in a Semicircle; and the Sum of Exterior
Angles of a Convex Polygon: these are back–to–back
either right
after the
SET Solutions OR right
before the
Statistics NOTES / Q bank at
the end.
[Sorry, the position has changed with editions!]
i) If time
permits, I strongly encourage studying the long (but easy!) proof of
the Inscribed Angle Theorem for Circles. Any regular Geometry
textbook should have it – or consult the Fill In The Blank
portion of the Q Bank: #58.
–> Elementary
Qs on Permutations & Combinations and simple Probability are
tested in the MCQ section OR as 1 FR Q. SETs 19–21 address this
important area! More complex Qs on Probability Rules can be found in
the Statistics Q Bank at the rear: #11–#20, for instance.
For
additional practice, the chapter on Permutations, Combinations and
Probability in a Precalculus textbook may be consulted.
–> Qs
on Coordinate Geometry are fair game! This deals with equations of
lines, the intersection of lines, use of the distance and mid–point
formulae, and under the guise of Locus, the extensive topic of Conic
Sections: Parabolas, Ellipses and Hyperbolas.
a) Proofs of
simple Geometric Propositions using coordinate geometry using
concepts of distance [length of segments], slopes of lines / segments
are not rare.
The Mid–Segment
Theorem is
a favourite. SET 13 #1 is about that! Choosing the coordinates of the
points and the orientation of the figure is key: for triangles,
choose the points to be A(0, 0), B(6a, 0) and C(12b, 12c). The
significance of the numbers 6 and 12 is that being multiples of 2 and
3, any division by those numbers shall yield an integer. Use the
notion of slopes [parallel lines have the same slope] and the
distance formula to quickly prove hypotheses!
b) Properties of
Parallelograms and those dealing with Triangle Congruence may also be
proven using Coordinate Geometry. Mid–point and Distance
Formulae are the primary tools.
c) Other Qs have related to
finding the point of intersection of the perpendicular bisectors,
medians and altitudes of triangles. SET 38: #5–#13 and SET 39
#1–#5 are important
Qs!
d)
SET 39: #8 and #9 deal with the Mid–Segment Theorem.
e)
SETs 24 and 26 address the topic of Conic Sections. Several Qs
[especially in the MCQ portion] are assured on this.
f)
In the FR category, especially
important are
Qs dealing with derivation of the Equation of Parabolas, Ellipses and
Hyperbolas in Standard Form
* y2 =
4ax;
* x2 /
a2 +
y2 /
b2 =
1;
* x2 /
a2 –
y2 /
b2 =
1
from
their definitions using
Locus [see SET 41 #5–#7]. A Precalculus textbook should have
the derivations.
–> Re Trigonometry,
solving a Right Triangle using Trigonometric Ratios is a basic
skill!
a) More complex Qs relate to finding the Areas
and Perimeters of Regular Convex n–sided
Polygons. Qs #1–#6 in SET 15 and SET 16 deal with this vital
area !
As
is the case with these Qs, regular polygons may be inscribed /
circumscribed in / around a circle.
b) Another favourite
"theme" is to use trigonometry to determine the coordinates
of vertices of a regular convex polygon inscribed in a circle. One
MCQ of this type is guaranteed!
For
instance: In
the diagram, a regular octagon is inscribed a circle of radius r.
what are the coordinates of point P?
Well,
it depends on how the
octagon is oriented and where P
is.
For instance, if the octagon is aligned as
in this figure
where r ~ white segments, then
since
for each congruent isosceles triangle – formed by the 2 white
segments and 1 blue segment – the "central" angle is
45°, well, starting on the positive x–axis and moving
anti–clockwise, the coordinates of vertices of the octagon
are:
on the positive x–axis: (r, 0)
[technically,
(r cos 0°, r sin 0 °) ~ (r, 0) but cos 0° =
1 and sin 0° = 0...],
then: (r cos 45°,
r sin 45 °),
on the positive y–axis (on top):
(0, r)
[technically, (r cos 90°, r sin 90 °)
but cos 90° = 0 and sin 90° =
1...],
then: (r cos 135°, r sin 135 °),
then,
on the negative x–axis: (–r, 0)
[technically,
(r cos 180°, r sin 180 °) since
since cos 180° = –1 and sin 180° =
0...],
then: (r cos 225°, r sin 225 °),
on
the negative y–axis (bottom): (0, –r)
[technically,
(r cos 270°, r sin 270 °) but cos 270° =
0 and sin 270° = –1...],
and finally:
(r cos 315°, r sin 315 °).
The
"pattern" is simply to add 45° successively to
(r cos 0 °, r sin 0°)...Hope it makes sense
why!
On the other hand, were the octagon oriented as
the stop sign
–
simply
draw a circumscribed circle so that the octagon is inscribed!
– then going about in
the anti–clockwise direction from the positive x–axis,
the general coordinates of the vertices are:
(r cos 22.5°,
r sin 22.5 °),
(r cos 67.5°, r sin
67.5 °) and so on...
The "pattern" is
simply to add 45 ° successively to (r cos 22.5°,
r sin 22.5°)...Hope it makes sense why!
Similar
"rules" apply for other inscribed polygons too...you may
expect hexagons, pentagons...
c) Test your competence by
attempting SET 35 #1 and #2. You may, additionally, rework both Qs
with a circumscribed
polygon.
–> At
least 1 Q on Spherical [non–Euclidean] Geometry is
guaranteed! SET 34 #4 is about this vital
topic!
In spherical [non–Euclidean]
Geometry, which of the following are True?
a) If a
point lies on a line, the point opposite it on the sphere may not lie
on the line.
b) Two lines on a sphere always intersect.
c)
There are no parallel lines on a sphere.
d) Two lines on the
sphere can intersect at right angles.
e) The sum of the angles of
a triangle is not less than 180º.
f) If an equiangular
triangle is drawn, the smallest angle can be between 60º and
180º.
g) The sum of the angles of a triangle is between 180º
and 540º.
h) It is possible to draw a 60º–60º–60º
triangle.
i) It is not possible to draw a 120º–120º–120º
triangle.
j) It is possible to draw a 90º–90º–90º
triangle.
Simply remember [~ "memorize"!] the following precepts and one ought to be able to resolve those Qs!
It
is not possible to draw non–intersecting
'lines' on a sphere. (A line is any 'equator' on a
sphere).
There are no parallel lines on a
sphere.
2 lines can intersect at Right Angles
(any line of longitude and the equator on a globe, for example!).
Any
triangle drawn on a sphere shall have the sum of its angles lie in
the range 180° to 540 ° (ie. 180+ to 540–).
Yes, the sum of the angles vary!!
Any equiangular
triangle (using d.) can have its angular measures between 60
and 180°.
From above, then, it not possible
to have a 60–60–60 triangle; it is possible
to have a 90–90–90 triangle; it is possible
to have a 120–120–120 triangle on a sphere!
–> Lengths
of Arcs of Circles and Areas of Sectors of Circles merit 1 Q in
the MCQ section. Keep in mind:
For Arc Length, l:
θ / 360° = l /
2πr
For Area of Sector, AS: θ
/ 360° = AS /
πr2
a)
A useful relation is derived when the 2 relationships above are
"merged":
l /
2πr = AS /
π r2
so
that l
=
2*AS /
r [upon simplification!].
SET 9 #7 employs this
"principle"!
b)
Qs on Arc Lengths usually relate to finding the distance between 2
cities on the same Longitude of the earth: SET 14 #
8–11.
Finding distances between cities on the same
Latitude requires elementary trigonometry.
To understand the
difference between the two [?!], click here!
b)
Another style of Q is about finding the area swept by
a rigid length, like say, a car wiper, as depicted here!
The
required area may be found as the difference in
the Areas of the Outer [~ Big] and Inner [~ Small "Cut–Off"]
Sector!
For
instance, assume a
windshield wiper traverses an angle of 115 degrees.
[Examine the 4th graph: 2nd row, 2nd column!]
If
you observe, the actual wiper with
the rubber component is
at the end of a short arm.
Given:
length of the actual wiper = 20cm, say.
Length
of the short arm = 10cm, say.
Find
the area of the potion of the windshield swept by
the wiper.
Solution. Total
length of the wiper [arm + rubber tip] = 30cm
Area
of the large sector swept by the entire wiper
=
115/360·π· 302 ––––––––––––––––––
(1)
Area
of the small sector [arm
alone]
= 115/360·π · 102 –––––––––––––––––––(2)
Area
swept by the wiper [rubber end alone] = (1) – (2 ) =
802.85 cm2
c)
Sector problems may insinuate themselves in 3–dimensional
solids; for instance, a sector rolled up forms a cone! [See below,
for this style of Qs!]
–> A very
important concept
on the test is Transformations.
a) One FR Q of the type
in SET 25 #1–#7 is de
riguer!
b)
The MCQs are more straightforward of the type in SET 25 #8, #9; SET
38 #1–#4 and SET 41: #2, #8 and #9.
c) In general, Qs
are of the form: what would happen to the Area / Perimeter
of any Polygon
[Square / Triangle / Rectangle, etc.] if it underwent a
transformation (x, y) –> (ax, b y)?
The trick is to "invent" the coordinates of the geometric
figure [choose easy–to–work–with numbers!], find
the old Area
/ Perimeter, and then use the transformation rule provided to find
the new coordinates
of the image, and finally, determine and compare the
Area / Perimeter of the transformed figure!
d)
Consult a standard Geometry textbook for how Reflections, Rotations,
Translations and Dilations "work"! In particular,
you ought to
be able to transform [rotation / dilation] an object about a point
P( c, d)
that is not the
Origin O (0, 0).
A cunning artifice is to first translate the
axes so that the point P( c,
d)
is the new Origin
using
X
= x – c
Y = y – d
Then
find the coordinates of the object in the new coordinate
system (X, Y) with (c, d) as the origin. Rotate / dilate the
object w.r.t the new origin, and then, finally, transform the
image back into the old axes using:
x =
X + c
y = Y + d
–> Qs
on Pythagoras' Theorem and Special Right Triangles [30° –60° –
90 ° and 45° –45° –
90 °] are usually embedded inside a more complex problem.
The 5 Qs on
my site are
representative of this [see links above!].
SET 22 #3 and #4
assess the provenance of the side ratios of Special Right
Triangles.
–> If
you encounter a figure with 2 polygons
[usually, irregular triangles, irregular quadrilaterals
/ pentagons, etc] "merged" together in some way [ie.
sharing sides / angles], then the Q likely deals
with Polygon Similarity!
So that corresponding sides must be proportional and
corresponding angles congruent.
Use this to solve for missing angles / sides.
A good
example is SET 42 #8.
–> It's
a very good idea to memorize the formulae of Areas of simple
Geometric figures. At
least a
couple of MCQ is devoted to this topic.
Qs from SETS 1 through
5 – and the handwritten Q
mentioned previously – address this. So does SET 30 #5 and SET
31 #1.
–> One
MCQ relates to Geometric Probability. SET 27 affords adequate
practice.
–> Finding
Volumes and / or Surface Area of Three–dimensional Solids
merit at
least 1
MCQ. Naturally, it is wise to commit the relevant formulae to
memory!
a) These Qs take the form of an applied problem such
as water flowing through a pipe, say; cutting a tree trunk at a 30°
/ 45 ° / 60° angle so that the Volume of a certain
section may be 1/3 or 1/2 or 2/3 of the Volume of the entire
"cylindrical section"; sand flowing through a pipe to form
a conical "pile"; lengths of 'oblique' diagonals of boxes;
and so on.
SET 6 #3–#5,
SET 7 #1–#4,
#6–#8
SET 8
SET 9 #3–#6, #8
SET 10 #10–#12
SET
13 #9, #10
SET 40 [ very
important!]
#8, #9, #11
SET 42 #4, #7
b) Another style of Qs has to do
with finding Ratios of
Volumes / Surface Area when the dimensions [radii / heights] of 2
solids bear a certain relationship.
SET 8 #4
SET 9
#1, #2, #8, #10
SET 10 #3
SET 11 #6–#11
SET 37 #5,
#6
SET 42 #2 [very
important!]
capture
the essence of these problems.
c) A Final type
relates to Ratios of Surface Areas and Volumes
of Similar Polyhedrons.
Keep in mind that if 2 similar solids bear a scale factor of k,
then the Ratio of their Surface Areas is k2,
and Ratio of their Volumes is k3 .
The
following Qs deal with this motif:
SET
39 #10
SET 40 #10
SET 41 #1
SET 42 #6
–> Relationships
between sides and Interior and Exterior angles of a Convex n–sided
[Regular or Irregular]
Polygon are routinely assessed.
SET 3 #9, #10
SET 4 #1
SET
40 #1–#6
are not atypical Qs.
–> One
MCQ is devoted to Constructions. One is expected to follow a set of
instructions to divine what figure would be yielded as a result.
SET
6 #1, #2, #6 and SET 7 #5 seek to capture the flavour of this
type of problem.
–> With
regard to Statistics,
a) Study / Master the Notes that
precede the Qs. Become familiar with the calculator commands
associated with the TI–series of calculators. For the TIs –
in general – click here for
a list of various commands and operations, and instructions to access
them. For the TI 83 / TI 83 + / TI 84 / TI 84+, in particular, click
here.
b) attempt
the following styles of
Qs no
matter what!
#1,
#2: dealing with Numerical measures of Univariate Distributions;
speculating how transformations affect
these measures
#3–#7, #10: relating to the Normal
Distribution
#9, #28: about Correlation and the Least Squares
Regression Line
#11–#20, #25, #26: which allude to
Probability Rules, Independent Events, and Conditional
Probability
#23, #24: dealing with Testing Hypothesis
[Goodness of Fit] using Chi–Square distributions
c) For
Notes for the topic of Sampling
and Surveys, consult
an elementary Statistics textbook . Otherwise, all
the
links in the site
http://www.socialresearchmethods.net/kb/sampling.htm
together, give an adequate coverage of the CSET
curriculum for the topic:
http://www.socialresearchmethods.net/kb/sampterm.htm
http://www.socialresearchmethods.net/kb/sampstat.htm
http://www.socialresearchmethods.net/kb/sampprob.htm
and
http://www.socialresearchmethods.net/kb/sampnon.htm
Likewise,
http://www.tardis.ed.ac.uk/~kate/qmcweb/scont.htm
and
all
the
links under
http://www.statpac.com/surveys/index.htm#toc
UPDATE
1. Equation of the sphere
(x
– a)2 + (y
– b)2 + (z
– c)2 =
r 2 ............
(*)
where the Centre of the sphere is P(a, b, c) and
r ~ radius.
[It's essentially an "extension"
of the equation of the Circle:
(x – h) 2 + (y
– k)2 =
r 2]
Example. Find
the side of the largest cube that can be inscribed in
the sphere: x 2 +
y2 +
z2 –
4x + 6z – 3 = 0
Solution.
Completing the square to resemble (*) above:
(x – 2)2 +
y2 +
(z + 3)2 =
16
=> r = 4
Sketch
a cube inscribed in a sphere, and label r = 4. [Do your best,
will you?! Yes, drawing this can be
hard but at the very least just imagine the bloody thing!]
Let O be the centre of the sphere, and p be the side of the cube. Join O with any corner of the cube. Let this segment ≈ radius of the sphere, be OF, where F is a corner of the cube on the sphere. Let X be a point [not shown] at the centre of a surface of the cube such that OX is perpendicular to the surface of the cube, and OX = p/2.
We can form right triangle OXF, right–angled at X where OF is the hypotenuse [being the segment from the centre of the sphere / cube to one of the corners of the cube].
To determine XF, we form another right–triangle FXY, Y being a mid–point on the edge of the cube [FG] so that FY = p/2 and using Pythagoras' Theorem, we find hypotenuse XF to be p/√2.
Also, OX = p/√2.
Therefore
using Pythagoras' Theorem for ΔFXO:
[FX]2 +
[XO]2 = 42
[p/√ 2]2 +
[p /2 ]2 = 42
=>
p2/2 + p2/4 =
16
=> 3p2/4 = 16
=>
p2 = 64/3
=> Edge, p
= √(64/3) ≈ 8 /√ 3 ≈ 4.61
Another – faster! – approach is to realize that the diameter of the sphere FD [~ 8] is the the longest diagonal of the cube and the hypotenuse of right ΔFDC where FC is the diagonal of any surface of the cube. If p is the edge of the cube, then, FO can be determined easily using Pythagoras' Theorem to be p√2. DC, of course, is the edge of the cube, p!
Finally, applying Pythagoras' Theorem to right ΔFDC:
[FC]2 + [DC]2 = 82
=>[p√2]2 + [p]2 = 82
=> 3p2 = 64 => p = 4.61 [as before!]
Here's a "back–wards" problem suitably illustrated!
UPDATE 2. Speculating
upon a figure formed by constructions
Example: Let
CA be the radius of a circle where C ~ Centre & A ~ pt. on the
circle. Using CA as radius mark (with a compass) another point on
the circle, say B; then from B, using the same radius, mark
another point C, and so on...
Join AB, BC, CD etc.
What
is the shape of the figure ABCD...so formed?
Solution.
Join CB to form triangle ABC. Since any triangle with 2 sides as the
radius is isosceles [base angles theorem] and here since CA = CB,
triangle ABC is equilateral. This implies that figure ABCD... is a
hexagon since only 6 segments can be marked on the circle...
UPDATE
3. Shape of Distributions and Relative
Positions of Mean, Median and Mode [in
Statistics]
In the Notes on Statistics in the Q Bank,
study DESCRIBING
DISTRIBUTIONS concept
#9.
Otherwise, click here and
view the 3 slides #39, #40 and #41 – by clicking the > key
on the screen! – for a visual representation of the
concept.
This website
also illustrates the concept!
It's
a smashing idea, in general, to master the Statistics Notes in the Q
Bank.
UPDATE
4. Calculator Proficiency
In
particular, one must be able to use a graphing calculator
to
a) find the Mean, Median, Quartiles, Standard
Deviation, etc. for a data–set [univariate: with variable
values and corresponding
frequencies] as in Qs #1 and #2 of the Statistics Q Bank, using
the STAT
–> CALC –> 1–Var Stats command
[with the TI 83+].
b) find the constants of the Least
Squares Regression Line: y = a + bx as
in Q #9 and #28 of the Statistics Q bank, using the STAT –>
CALC –> LinReg command
[with the TI 83+].
c) perform calculations relating to the
Normal Distribution as in Qs #3, #5, #7 and #10 in the Statistics Q
bank, using the 2nd
VARS –> Normalcdf and 2nd
VARS –> InvNorm commands
[with the TI 83+].
It is very useful also
to
know how to "read" a Normal Distribution table [this would
be in the front / back of a Statistics text book].
d)
calculate Chi–Square probabilities as in STATISTICS Q Bank #23
and #24 in the Statistics Q bank, using the 2nd VARS –>
χ2cdfcommand [with the TI 83+].
It
is very useful also to know how to "read"
a Chi–Square Distribution table [this would be in the
front / back of a Statistics text book] .
UPDATE
5. Transformations
In
the Subtest II Final Review document [sent
earlier!], I mentioned Qs in the form: what would happen to the
Area / Perimeter of any Polygon [Square / Triangle /
Rectangle, etc.] if it underwent a transformation (x, y) –>
( ax, by)?
The trick is to
"invent" the coordinates of the geometric figure [choose
easy–to–work–with numbers!], find the old Area
/ Perimeter, and then use the transformation rule provided to find
the new coordinates of the image, and finally,
determine and compare the Area / Perimeter of
the transformed figure!
We could
extend this concept to polyhedrons [3–dimensional solids] too
as in SET 38 #1.
Example. An icosahedron undergoes
a transformation such that (x, y, z) –> (2x, ¼y, 3z).
Find the ratio of Surface Areas and Volumes of the transformed object
to the original.
Solution. An icosahedron is a
20–sided figure. So what?!
We shall simply consider a
6–sided ordinary prism [~ box!] since the concept can extend to
any 3–dimensional figure!
Let the dimensions of
the box be 4units X 8 units X 6 units.
Old Surface Area: S1 =
208 unit2
Old Volume, V1 = 192
unit3
Upon transformation: (x, y, z) –>
(2x, ¼y, 3z), the new dimensions are 8units
X 2 units X 18 units.
New Surface Area: S2 =
392 unit2
New Volume, V2 = 288
unit3
Required ratios:
S2/S1 =
392/208 = 49/26
V2/V1 = 288/192 =
3/2
These ratios apply to the icosahedron,
too!
Example. A regular polygon is dilated with a
scale factor, k. Find the ratio of their
a)
areas
b) slopes of corresponding sides
c) areas
of inscribed circles
d) interior
angles
Solution. For this style of problem, simply
consider a square with coordinates: (0, 0), (0, 12), (12, 0) and (12,
12) with k = 3, say. Execute the transformation to
find the new vertices.
Then, perform the necessary
calculations!
UPDATE 6. Equations of
Lines
Be familiar with finding equations of lines,
a)
given a point and slope: y – y1 =
m(x – x 1)
b) given 2 points:
first, find the slope using m = y2 –
y1 / x2 –
x1
Be familiar with finding equations of
a)
Parallel lines
b) Perpendicular Lines
UPDATE
7. Translations of Simple Geometric figures
Example. To
shift the ellipse 4x 2 + 9y2 =
16x – 54y – 61 into a congruent ellipse
centred at (0, 0), what translation need be
performed?
Solution. Getting the terms to 1
side:
4x2 + 9y2 – 16x +
54y = – 61
and Completing the Square:
(x
– 2)2 / 9 + ( y + 3)2 / 4 =
1
which is an ellipse centred at (2, –3).
To
shift this to (0, 0), the required translation is: 2 units left [or
–2 units in the x–direction] and 3 units up [+3 units in
the y–direction.]
UPDATE 8. Permutations
A
recent style is the following Q.
Example: In how
many way can the letters p, q, r, s, t, u, v and w be
arranged without repetition if p must be the
1st or last letter?
Solution. Consider p being
the 1st letter. The number of arrangements
is:
1 X 7 X 6 X 5 X 4 X 3 X 2 X 1 =
5040.
Since p could be the last letter, too,
the required # is: 10,080.
UPDATE 9: Conditional
Probability using Tree Diagrams.
Throughout
the Subtest II Q Bank Solutions for Statistics,
problems involving conditional probability have been resolved using a
Probability Table, which is an extremely nifty approach. However, of
late, the test has required that these Qs be
addressed using a Conditional Probability Tree Diagram.
For
this, the following formulae need be known and mastered:
The
conditional probability that an event A occurs, given that an
event B has, is written as: P(A | B) = P(A and B)/P(B).
Similarly,
P(B | A) = P(A and B)/P(A)
Cross–multiplying in each
expression: P(A and B) = P(A) X P(A | B) = P(B) X P(A |
B).
Example. Suppose there are 4 red and 6 blue
marbles in a bag. One marble is drawn at random, and then
another, without replacement. Find the probability
that
a) both marbles are red.
b) only one of the marbles is
red.
c) the 2nd is red, given that the 1st is
blue.
Solution. Depicting the situation via a Tree
Diagram:
Here are the notations to accompany the probabilities
above:
P(1R) = 4/10 [~ P(1st is R)]
P(1B) = 6/10
P(2R
| 1R) = 3/9 [~ P(2nd is R, given that 1st is
R)]
P(2B | 1R) = 6/9
P(2R | 1B) = 4/9
P(2B | 1B) = 5/9
P(1R
and 2R) = P(1R) P(2R | 1R) = 4/10 X 3/9
P(1R and 2B) = P(1R) P(2B
| 1R) = 4/10 X 6/9
P(1B and 2R) = P(1B) P(2R | 1B) = 6/10 X
4/9
P(1B and 2B) = P(1B) P(2B | 1B) = 6/10 X 5/9
a)
P(both marbles are red) = P(1R and 2R) = 4/10 X 3/9
b) P(one
of the marbles is red) = P(1R and 2B) or P(1B and 2R) =
(4/10 X 6/9) + 6/10 X 4/9
c) P(the 2nd is red, given
that the 1st is blue) = P(2R | 1B) = 4/9
The
general principle behind the Tree
Diagram is this:
Of
course, there is no reason why more primary branches cannot exist
i.e. instead of A and A' alone, we can have outcomes: P1, P2, P3, P4,
etc.; likewise, more secondary branches, so that corresponding to
each primary outcome, we have several conditional secondary
outcomes: S1, S2, S3, S4, etc. yielding,
for primary
outcome, P1, the secondary branches: P(S1 | P1), P(S2 | P1), P(S3 |
P1), etc.
for primary outcome, P2, the secondary branches:
P(S1 | P2), P(S2 | P2), P(S3 | P2), etc.
and so
on.
Example. Suppose that only 0.1% of the
population suffers from tuberculosis (TB). A test is available such
that a patient with TB would (correctly) test positive 99% of the
time, whereas a patient not suffering from TB would
(incorrectly) test positive 10% of the time. Create a Tree Diagram
and Table to depict the situation. Calculate the probability
that:
a) given that a patient tests positive for TB, he
actually suffers from it.
b) given that a patient tests
negative for TB, he actually does not suffer from it.
Solution. Choosing 100,000 patients [so that we don't have to grapple with messy fractions / decimals!], construct a Tree Diagram as follows:
The notation for the values
in the Tree Diagram are as:
Given: P(TB) =
1/1000 (1)
and
P(TB') = 999/1000
(2)
so that
From the TOP branch:
Of
100,000 people, Number of individuals with TB, N(TB) = 100 and N(TB')
= 99,900 (3)
Given: P(+ | TB) =
99/100 (4)
and
P(– | TB) =
1/100 (5)
Using (3) and (4), of the 100 [out
of the 100,000] with TB,
Number of individuals who shall
test positive, N(TB and +) = 99/100 X 100 = 99
(6)
In probability notation, using (1) and
(4) this would be:
P(TB and +) = P(TB) P(+ | TB) = 1/1000 X
99/100 = 0.00099 (7)
Likewise, using (3) and
(5), of the 100 [out of the 100,000] with TB,
Number of
individuals who shall not test positive, N(TB and –)
= 1/100 X 100 = 1 (8)
In probability notation,
using (1) and (5) this would be:
P(TB and –) = P(TB)
P(– | TB) = 1/1000 X 1/100 = 0.00001
(9)
Similarly, the BOTTOM branch can be
determined.
Given: P(+ | TB') =
1/10 (10)
and
P(– | TB') =
9/10 (11)
Using (3) and (10), of the 99,900 [out
of the 100,000] without TB,
Number of individuals who shall
test positive, N(TB' and +) = 1/10 X 99,900 = 9,990
(12)
In probability notation, using (2) and
(10) this would be:
P(TB' and +) = P(TB') P(+ | TB') =
999/1000 X 1/10 = 0.0999 (13)
Likewise, using (3)
and (11), of the 99,900 [out of the 100,000] without TB,
Number
of individuals who shall not test positive, N(TB'
and –) = 9/10 X 99,900 = 89,910
(14)
In probability notation, using (2) and
(11) this would be:
P(TB' and –) = P(TB') P(–
| TB') = 999/1000 X 9/10 = 0.8991 (15)
In
table format, with numbers:
|
Positive |
Negative |
Total |
Has TB |
99 |
1 |
100 |
Hasn't TB |
9990 |
89910 |
99900 |
Total |
10089 |
89911 |
100000 |
In table format, with
probabilities:
|
Positive |
Negative |
Total |
Has TB |
0.00099 |
0.00001 |
0.001 |
Hasn't TB |
0.0999 |
0.8991 |
0.999 |
Total |
0.10089 |
0.89911 |
1 |
The 2 Qs can now be easily answered!
a) P(TB| +) = P(TB and +) /
P(+) = 99/10089 ~ 0.00099/0.10089 = 0.009812.
b) P( TB' | –
)= P(TB' and –) / P(–) = 8991/ 89911 ~ 0.8991/ 0.89911 =
0.9999.
UPDATE 10: Conditional
Probability with Sampling without Replacement.
A committee
of 7 is to be chosen consisting of Males and Females. If there are 20
Males and 10 Females, find
a) the probability that the Female is
chosen 2nd if the Male is selected 1st.
Solution.
P(Male is 1st, Female is 2nd) = 20C1*10C1 / 30C2
or more simply,
P(Male is 1st, Female is 2nd) = 20/30*10/29 [simplify]
b)
2 Males or 2 Females are chosen as the 1st two selections.
Solution. P(1st 2 are M) OR P(1st 2 are F) = P(M, M) +
P(F, F)
= 20C2 / 30C2 + 10C2 / 30C2
or more simply,
P(1st
2 are M) OR P(1st 2 are F)
= 20/30*19/29 + 10/30*9/29
[simplify]
c) the probability that exactly 3 Females are
chosen.
Solution. The relevant outcomes are FFFMM,
MFFFM, ...and so on. There are 5C3 ways in which 3 F can be
"arranged" in 5 "position".
Required: P
=
5C3 * 10C3 [~ choosing 3 F from 10] * 20C2 [~ choosing 2 M from 20] /
30C5 [~ choosing 5 students from 30]
UPDATE 11: Prove that
if a Point P(a, b) is
reflected over the line y = x, then the reflected point is P'(a,
b).
Solution. One
approach is: let the reflected point be R(m,
n). We shall prove that
m = b and
n = a.
Since y = x is
the line of symmetry, it is the perpendicular bisector of segment PR.
If S is the point of intersection of PR [sketch!]
and line y = x on
the line, then S may be denoted as S(c, c) [since it
lies on the y = x line!]. Since S is the mid–point of PR
, using the mid–point formula: c = ½ (a + m) and
c = ½ (b + n). Setting them equal and
simplifying: a + m = b + n (*).
Also, since PR and y = x are perpendicular, slope of PR = (n – b)/(m – a) = –1 [since slope of y = x is 1!]. Simplifying this: n – b = a – m (**)
Solve (*) and (**) simultaneously to get, m = b and n = a!
UPDATE
12: Given 4 coordinates of a quadrilateral – say P(a, b), Q(c,
d), R(e, f) and S(g, h), prove that the diagonals bisect each other.
Solution.
The
straightforward approach is to find the point of intersection, X say,
of the diagonals PR and QS, and use distance formula to show PX
= RX and QX = SX.
UPDATE
13: Given a 3
event Venn Diagram. Find a conditional probability.
Example.
Suppose 55% of adults drink coffee, 25% drink tea and 45% drink cola.
Also, 15% drink coffee and tea, 5% all three of them, 25% drink
coffee and cola and 5% drink only tea. Find the probability that an
adult drinks
a) exactly 1 beverage.
b) none of the
beverages.
c) coffee, if it is known that he drinks
tea.
Solution.
Let A ~
coffee, B ~ tea and C ~ cola
Given: P(A) = 0.55, P(B) = 0.25,
P(C) = 0.45, P(A and B) = 0.15, P(A and B and C) = 0.05, P(A and C) =
0.25, , P(B and A' and C') = 0.5 [Fill in the Venn
diagram below suitably!]
We
find EACH individual region – there are 7 of them – like
a simple puzzle:
We know P(A and B and C) = 0.05 and P(A and
B) = 0.15 => P(A and B and C') = 0.10
We know P(A and B and
C) = 0.05 and P(A and C) = 0.25 => P(A and C and B') = 0.20
We
know P(A and B and C') = 0.10 [found], P(and B and C) = 0.05 [given],
P(B) = 0.25 [given] and P(A' and B and C') = 0.05 [given] => P(B
and C and A') = 0.05 [subtracting the others from P(B)]
We
know P(A and C and B') = 0.20 [found], P(A and B and C) = 0.05
[given] and P(C) = 0.45 [given] => P(C and A' and B') = 0.15
[subtracting the others from P(C)]
We know P(A and B and C') =
0.10 [found], P(A and B and C) = 0.05 [given], P(A and C and B') =
0.20 [found] and P(A) = 0.55 [given] => P(A and B' and C') = 0.20
[subtracting the others from P(A)]
a) P(adult drinks
exactly 1 beverage) = P(A and B' and C') + P(A' and B and C') + P(A'
and B' and C) = 0.20 + 0.05 + 0.15
b) P(adult drinks none of
the beverages) = 1 – P(at least 1 beverage)
= 1 –
[P(A) + P(B) + P(C) – P(A and B) – P(B and C) – P(A
and C) + P(A and B and C)]
= 1 – [0.55 + 0.25 + 0.45 –
0.15 – 0.10 – 0.25 + 0.05]
= 1 – [0.80]
=
0.20
c) P(A | B) = P(A and B) / P(B) = 0.15/0.25
UPDATE
14: Finding the number of sides of a polygon, given a relationship
between its Interior and Exterior angles.
Example.
If the interior angle of a regular convex polygon is 9 times its
exterior angle, how many sides does the polygon have?
Solution.
Since I = 9E and I + E =
180 => 10E = 180 => E = 18degrees
Since in any polygon,
the sum of the exterior angles is 360 degrees, 18n
= 360 => n
= 20sides.
UPDATE
15: Finding the height to which water would rise in a 3–dimensional
solid when poured from another solid.
The
resolution to all such Qs involve equating Volumes.
Example.
If cone [given radius, r, and height, h] was filled
with water, which was then poured into a hexagonal prism of side, s,
to what height would the water rise in the prism?
Solution.
Volume of water in the cone, V1 = 1/3 πr2h1
Surface Area of the hexagon, S2 = 6*Area of 6 equilateral triangles of side, s = 6*(√3/4)*s2
If h2 is the height to which water shall rise in the prism, Volume of water in the prism, V2 = 6*(√3/4)*s2*h2
Known: 1/3 πr2h1 = 6*(√3/4)*s2*h2
Solve for h2 [since all the other dimensions are known!].