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 cm
2.

Area of the circle = π (4√3)
2 = 48π cm2.

Required Area = 48(π – √3) cm
2

#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 = (√3s
2)/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 cm
2.

#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, 
= 18.66cm and the hypotenuse, h = 19.32cm.

Area of the isosceles triangle, A
t = ½ ab = 93.3 cm2.

Therefore, Area of the 12–sided polygon, A
p = 12·A= 1119.6 cm2.

Also, radius of the circumscribed circle, r ≈ hypotenuse, 
h = 19.32cm.

Therefore, Area of the circle, A
cc = 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 ≈ cm. 

Then, AD = 
x + 1 cm.

Now, since ΔODC is a right triangle, using Pythagoras' Theorem:

OD
2 + DC2 = OC2.

=> 1
2 + y2 = x2

=> y
2 = x2 – 1    ––––––––––––––––––––––––– (*)

And since ΔADC is a right triangle, using Pythagoras' Theorem:

AD
2 + DC2 = AC2.

=> (x + 1)
2 + y2 = 15/2 = 7.5 –––––––– (**)

Substituting (*) in (**):

(x + 1)
2 + x2 – 1 = 7.5

Simplifying: 2x
2 + 2x – 7.5 = 0

or 4x
2 + 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 TheoremsAttempt 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 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 
* y
2 = 4ax; 
* x
2 / a2 + y2 /  b2 = 1; 
* x
2 / 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



http://upload.wikimedia.org/wikipedia/commons/4/4a/Archimedes_circle_area_proof_-_inscribed_polygons.png



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 –



http://upload.wikimedia.org/wikipedia/commons/thumb/8/81/Stop_sign.png/600px-Stop_sign.png



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!



http://static.howstuffworks.com/gif/wiper-coverage.gif



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· 30–––––––––––––––––– (1) 

Area of the small sector [arm alone] 
= 115/360·π · 10–––––––––––––––––––(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!]

–> 
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 – 
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 + c
= 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: 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!]



http://www.kjmaclean.com/Geometry/Tetrahedron_files/image028.jpg

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 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:


http://www.vias.org/tmdatanaleng/img/cm_prob_tree.png
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:



Tree diagram of two two-outcome events





 


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, 18
n = 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!].