Download Non-Overlapping Sausage Ends

TestPrep Toolkit Primer – Math
Concept Lesson I
Overview
Problem Solving Procedure/Approach
Word problems
1) Recognize Word Problem by 5 Answer Choices – will be ‘Numbers’ generally. Problems will
be ‘spelled out’, be fairly lengthy and will have sentences describing elements (variables) as
well as quantifiables – such as pure numbers
2) Record on scratch paper all Numbers and Quantifiable words (twice, half)
3) Write Question, so you identify precise value & units seeking
4) Identify how many variables: 1, 2 , or 3
- Objective: Number of Variables should = Number of Equations to be formed
- Anchor Variables at single specific point in time (due to moving targets i.e. age changes
every year) before working with altered equation(s)
5) Impute: Additional Necessary Facts/Formulae
6) Now formulate the appropriate number of necessary equations
7) System of solving 2 Simultaneous equations
- Strategy: 1) Man in Middle, 2) Addititive, 3) Substitution
Man in the Middle
I.
C+D=G
II.
E+F=G
Deduction: C + D = G= E + F
{Eliminate} Man in the Middle Method: C + D = E + F
Additive
3X – Y = 14
Y + X = 21
Priority: See if you can Add or Subtract to eliminate a variable
3X – Y = 14
X + Y = 21
4X
= 35
Substitution
3X + Y = 14
5X + 4Y = 28
Priority: Choose Equation with the “Easiest” numbers to isolate your variable
3X + Y = 14 Identify Easiest equation to Isolate
5X + 4Y = 28
Y = 14 – 3X Isolated Variable Deduction
5X + 4Y = 28 Unmanipulated Equation
5X + 4[14 – 3X] = 28 Substitute Isolated Variable Deduction in other equation and Solve
Back Solve Approach
1) Start by Testing Answer Choice B
- 20% chance that B is correct. If the B answer choice it too large, answer is A
- 40% Chance of getting Correct Answer on First Trial
2) If First Trial is Inconclusive Test Answer Choice D
3) At Most 2 Calculations
4) During Ramp-up, Also Plug in/Test Final Answer choice in Hypothetical Equation for Sanity
Check
Average/Aggregate Approach
1) Aggregate ( ∑ ) = Average * # of Values
2) Aggregate = ( ∑ ) Field Set Characteristics
3) Average * # of Values = ( ∑ ) Field Set
Boys Club Example
12 , 14 , 16 , 18 are the ages of four members of a boy’s club
Additional Fifth Member – Average age is now 16 Years
Question – How Old is # 5
To Compute, Need to deal in Aggregates, and Dynamic Change
Average
# of People
Aggregate
Avg (1) = 15
4
60
Avg (2) = 16
5
80
Dynamic Change
20
Dynamic Event: 5th Member is 20
3 Puppy Example
2 Dachshunds (Identical Twins) & 1 Rottweiler
Rottweiler increases weight 15%
Average
Avg (Day 1) = 12 lbs
Avg (Day 30) = 13 lbs
# of People
Aggreagate
3
3
36
39
Dynamic Change
3
3 = 15
X
100
2
Period
Before
After
Dachshund #1
8
8
Dachshund #2
8
8
Rottweiler
20
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Age Problems
The combined age of Bill and his dad is 60. Six years from now, Bill's dad will be twice Bill's age.
How old was Bill's dad when Bill was born?
1) B + D = 60 (Lock in Age Right Now) Write: B is Bill’s age NOW
Derivative D = 60 - B
1st draft
B
=
D
2)
Add Compensation 2(B + 6) = (D + 6)
2B + 6 = D + 6
2B + 12 = 60 – B + 6
3B = 54
B = 18
D = 42
NOTE: This subject has been covered in more detail in the Basic Word Problem Module.
Symmetrical Integer Series (Evenly Spaced Integers)
Suppose you were asked to compute the sum of 15 consecutive integers, starting at X and ending at Y.
Yes, you could add them up by counting on your fingers, or place the numbers in a column and add
them all up, like a guy in a deli adding up your six purchases, on the grocery sack.
There is a more sophisticated approach – one that is faster, and – amazingly – takes the same amount
of time whether you add up 15, 35, or 101 consecutive numbers. To understand the concept, look at the
following, easier-to-illustrate example: Add up the consecutive integers between 97 and 103.
1.Determine the ‘bookend’ integers (namely 98 and 102)
2.Use the ‘deli counter’ principle to determine the NUMBER of integers (in this case, 5)
3. Determine the middle one (provided an odd number of integers) (98 + 102)/2
4. Bracket the offset integers around the middle integer – see how they nicely form brackets of 200
(‘bracket sums’). The middle integer is the median
5. Compute how many brackets there are : (5-1)/2 = 2
6. Add numbers up : Bracket sum * (number of brackets) + the single median
200 * 2 + 100 = 500
98
99
100
101
102
200
3
200
∑=
500
Avg = 100
100
Avg = 100
∑ =1100
Let’s elaborate on this approach. We earlier noted that 100, the number in the middle, was the median
number. Since this is a symmetrical series, the median is also the average number. By looking at the
diagram above, instead of looking at the bracket amounts, we can visualize that each bracket ‘pair’ –
adding up to 200 as they do, in fact imply that each integer, regardless of where it sits in the sequence,
carries an average value of 100. Since the only unbracketed member (namely 100, itself) also has –
obviously – a value of 100, we discover, much to our delight that a second way to compute the sum of
the evenly spaced integers is to simply compute their aggregate. Their aggregate is – as we already
know - their average times the number of integers. In the simple case above, their sum, or aggregate,
is 5 * 100 or 500. Does this men that in ANY series of integers, be they consecutive, or consecutive
even integers, or consecutive odd integers, all we need to do, is find the middle value ascertain the
number of integers, and multiply the two??
The answer is an emphatic YES.
What if there is an EVEN number of integers. Rats – it is not as easy, but we can still carry on with the
principle. If after using the deli line principle you find that the number of integers, including the two
bookends, constitute an even number, then you know that their average is the average of the two
middle numbers. To illustrate. Suppose the series are the integers 5, 6, 7, 8, 9, and 10
Here the average (5 + 10)/2 comes to 7.5 - that looks ominous. How can the sum of perfectly wellrounded integers become associated with the value ‘One Half’ ?? Not to worry – when we multiply the
average (7.5) by the (even) number of integers, the product will always round up to a full number – in
this case 45. Saved by the bell !!
Q. Variables a, b, c, d, e, f, g, h, and i are consecutive even integers whose sum is 792. What is the
value of a?
88
∑=792
A B
C
D
Avg = 88
E
F
G
H
I
4
80
82
84
86
88
90
92
94
96
792=9*Avg
Avg =88
Deli Line (Inclusive)
3…………376
Process:
1) Subtract Smaller number from Larger number
2) Add 1 to the result to reach total inclusive set
376
- 3
373
+ 1
374
Movie Line (Between)
8…………15
Process:
15
1) Subtract Smaller number from Larger number
-8
2) Subtract 1 from the result to reach total set in between 7
-1
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Q. P is the sum of the all the positive even integers between 0 and 100
Q is the sum of the 19 largest negative integers What is P – Q?
P
1) Draw Sample {2,4,6,8…92,94,96,98}
2) Compute the number of integers. This is not so easy – it is obviously a lot easier if we are
dealing with consecutive integers. However help is on the way.
3) If on your scratchpaper you write out, by using the ellipsis in the middle, the first and the last
portions of the sequence, and underneath, construct its countepart, by dividing each integer by
two, the bottom line will consist of consecutive integers. Their number is the same as the top
line, but their value has simply been cut in half. This doesn’t really matter – your only purpose
is to ascertain how many integers there are in the sequence. We have transcribed a sequence of
integers with a ‘gap’ of 2, into a sequence of consecutive integers
4) The bottom line therefore will look like this (1,2,3,4…46,47,48,49)
# of values is 98/2 = 49
On the bottom line, which integer is the middle, median or average?? (1 + 49)/2 or 25
Since the top and bottom lines have a 1:2 relationship, the middle value on the top line therefore is 2 *
25, or 50
Middle/Center = 50
4) Plug in derived values in Aggregate Formula
Aggregate ( ∑ ) = Average * # of Values
2450
= 50
*
49
Q
1) Draw Sample {-19, -18, -17….-3, -2, -1}
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2) # of values is 19 ??? (How to derive formula or framework to derive) Reverse order of
integers, and multiply eaxh one by –1
set now looks like {1 2 3 …. 19} – use this ONLY to figure out # elements
5) Determine Middle, Median??? (How to provide formula or framework to derive)
6) (1 + 19)/2 = 10
7) Middle/Center = -10
4) Plug in derived values in Aggregate Formula
Aggregate ( ∑ ) = Average * # of Values
-190
= -10
*
19
P – Q = 2450 – (-190)
Answer: 2640
Non-Overlapping Sausage Ends
Q. What is ∑A - ∑B
Set A {1,2,3…31} Consecutive numbers
Set B {4…32} Consecutive numbers
Approach 1 (SAT)
Aggregate A – Aggregate B
A
# of Elements (31)
Avg (16)
∑A = 31 * 16 = 496
B
# of Elements (29)
The Median or middle integer = (15th) In other words, we must have 14 integers, followed by the
median/average integer, followed by 14 more, integers. We arrive at the median by counting UP to the
15th from the beginning, or counting DOWN 15 integers from the end.. In each case, we arrive at the
integer 18 (check yourself – the integers 4 thru 17, by the deli rule, number 14. Same thing with the
integers 19 thru 32.
∑B = 29 * 18 = 522
∑A(496) - ∑B(522) = - 26
Approach 1 (GMAT)
Set A 1,2 3…..29, 30,31
Set B
4,5 …30,31,32
By lining these integers up underneath each other, and making sure that the same integers line up
directly underneath each other, we see the nonoverlapping sausage ends
The top sequence has 1, 2 and 3 sticking out to the left, and the bottom line has 32 sticking out to the
right.
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Since the objective is to DEDUCT sequewnce from set B from Set A, instead of computing the
aggregates first and THEN subtracting one aggregate from each other, we perform the subtraction in
the microcosm way, first,a and then simply see what’s left. The sequence 4 through 31 repeats on both
top and bottom row, so all those integers neutralize each other – pooof, they become nothing, and we
are left with + 1 + 2 + 3 and – 32. They sum up to negative 26.
Distance / Resource Allocation
1. Julian drives from home to work in the morning managing to maintain an average speed of 45
Miles Per Hour. After work, he uses the same route to drive home, and his evening commute
averages 30 Miles Per Hour. If Julian spends one hour a day on his commute, what is the
distance between his home and his office?
Formula: DISTANCE = AVG SPEED * TIME
Approach A: Assume a hypothetical distance
Assume 15 miles each way and compute the commuting time under this assumption.@ 45 MPH
D = S* T
15 = 45 * T
Going to Work T= 1/3(H) or 20 mins
D = S* T
15 = 30 * T
Returning Home T= 1/2(H) or 30 mins
Assuming Distance was 15 miles each way, Total time elapsed was 20 + 30 = 50 Minutes
In fact the actual commuting time was one full hour, so we UNDER estimated the hypothetical
distance. Let’s compute the actual distance:
15 * 50 (Where x is Actual Distance)
X 60
X=18 Miles each way, or 36 miles roundtrip.
Avg Distance (The slower time takes up a larger weighting of the trip)
30* 3 + 45* 2 = 18 + 18 = 18
5
5
2
Approach B: Resource allocation, and ratios.
Recognizing that there is an INVERSE relationship between the speed he drives, versus the time it
takes for each trip, we can simply find out the allocation ratio. We have a resource that needs to be
allocated two ways. The resource is the given 1 hour it takes him to commute each day, and we
allocate it in two portions – the out trip, and the home trip. The allocating factors are 45 and 30 (his
relative speeds so, the ratio 45:30 can be simplified as 3:2
This is the pizza pie approach – we add up the ratio factor (3 + 2 = 5) so we slice the element being
partitioned (his one hour commute time) into slices, each amounting to 1/5 of an hour, and we allocate
THREE sllices to his slower evening commute, and we allocate TWO slices to his faster morning
commute.
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Ipso Facto, his morning commute takes 2/5 hours or 24 minutes, and his slower evening commute
takes 3/5 hours or 36 minutes. Given those time parameters, let us compute his commuting distance.
2/5 * 45 MPR produces a distance of 18 miles
3/5 * 30 MPR likewise produces a distance of 18 Miles. Guess that’s got to be the answer.
2. Work Rate Problem
Paul can cut a lawn in 5 hours; Jason can cut the same lawn in 3 hours. If Paul and Jason work
together, how long will it take them to cut the lawn?
P cuts lawn in 5 hrs
J cuts lawn in 3 hrs
General Principle of Work Rate
Formula: 1 + 1 = 1 = _1_
A
B F
T
1+1 =1
P J
T (T is total time working together)
1+1 =1
P J T (T is total time working together)
1+1 =1
5 3 7
= _8_ = _1_
15
T
T = _15_ or
8
17/8
hours
Related Problem.
If the lawn is 12,000 square feet, how much more lawn (in square feet) does Jason cut than Paul,
assuming they work on different parts of the lawn using two different lawn movers?
Calculation Approach
Paul cuts @ rate 12,000 sq feet/ hr
If Alone
5
= 2400 sq feet/ hr
Jason cuts @ rate 12,000 sq feet/ hr
If Alone
3
= 4000 sq feet/ hr
*Note Reverse Relationship Jason produces the most- therefore works fewer hours if alone
Jason 4000 X 15 = 7,500
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Paul 2400 X 15 = 4,500
8
12,000
Ratio Approach
*Note Reverse Relationship
12,000 X 3
12,000 X
8
_5_
8
8
Time = Requirement/Production rate
Paul: Time = 12000/2400
Jason: Time = 12000/4000
Q. 3 men paint 6 houses in 72 days. How long will it take 4 men to build 5 houses?
“Unit” Problem
6 houses = 72(days) * 3(men) = 216 Man-Days
Unit
1 (Unit) House = 216 (Man-Days) = 36 Man-Days
6 (Houses)
5 houses = 5 X 36 = 180 Man-Days
4 people would have to provide 180 Man-Days total
Therefore, Each Man works 45 Days
For the ‘Unit’ productivity problems, the distinguishing factor is that all men are presumed to have
equal productivity. (As opposed to the ‘work rate’ problems where A can do a certain job in x time, but
B requires y units of time to complete the same job)
You should set up a matrix
ENGINE
UNIT of PRODUCTION TIME UNIT
3 men
6 houses
72 days
1 man
6 houses
72x3 = 216 days We MULTIPLY - Harder
1 man
1 house
216/6 = 36 days We divide - Easier
Now, do the REQUIREMENTS computation
We have 4 men, and need 5 houses
1 man
1 house
36 days (unit rate)
1 man
5 houses
36x5
4 men
5 houses
36x5/4 = 45 days
3. Ratio Setups
1. For all ration problems always ADD Ratios together
- Below 3 + 4 + 6 + 7 + 10 = 30
2. Visualize Pizza in [X Slices]
3. Divide Grand Total by Ratio Aggregate
-Below: 14,200 = 440 Units
30
1 Unit Equivalent = 440
4. Populate each slice with [ X unit Pepperonis]
5. Let each “Candidate grab # of slices equal to Ratio * units
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The Mayor of White Plains
Candidates A:B:C:D:E
Votes, Ratio 3: 4:6: 7:10
Total Votes: 14,200
Q. What was winning margin (Who won?)
TIC / TOC / TAC
3 Tics = 4 Tocs
5 Tocs = 7 Tacs
Q. What is TIC / TAC Ratio?
Cross Multiplication Matrix
TIC
TOC
TACS
3
4
5
7
15 = 5* 3=LCD/4 = 5 LCD 28 = 4* 7=LCD/5 = 4
15
20
28
15 TICS = 28 TACS
Keys = Do a Reverse Cross Multiplication
TICS = 28
TACS
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Geometry
1. The Triangle
Always ask yourself the question: Is this a righ triangle??
1) Look for Explicit or Implied Right Angle
2) Determine if you have the length of 1 or 2 sides
1 Side Length Approach
Assumption dealing with a 30-60-90 or 45-45-90 Triangle
Look for 30,60,90 or 45 angle – there HAS to be one of theose – if not you would not be able to
answer any questions about the triangle’s area
2 Side Length Approach
Assumption dealing with Pythagorean Theorem or Magic Triangle
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Pythagorean Theorem
A2 + B2 = C2
Magic Triangles Question is this a magic triangle
Question is this a magic triangle?
3-4-5 6-8-10 or 9-12-15
5-12-13 10-24-26 or 15-12-25? 15-12-25? Which one is this, does not seem to be a multiple??
Multiple of 5 & 13 need to be on Hypotenuse? So if they give you a triangle with 2 sides shown, as 15
and 12, is it or isn’t it a magic triangle? If 15 is the longest of the short legs, it could not be a MT. In
any scaled version of a 3-4-5 triangle, the multiple of 5 must be the hypotenuse. In any scaled version
of a 5-12-13 triangle, the multiple-of –5 length short leg needs to be the smallest (the other short leg
will be a multiple of 12, and be substantially longer)
Look for one of
lengths below
Rohrshah
3 15 12 10 9
8 4 18 12
16 26 6 20
25 5 13 24
Special Triangles.
The 30-60-90 or the 45-45-90 triangles are the ONLT ones where you can determine the area, even if
the problem only gives you ONE side of the triangle.
If the triangle in question does NOT have a right angle, then it is simply a ‘trivial triangle’ and all they
can ask, are questions about relative length of sides etc.
ONE EXCEPTION. If the triangle is an equilateral triangle, it is one of the Uniform Geometric
Objects, and its area, perimeter can be determined if wes imply know the length of its side, s.
Paul’s Triangle in a Triangle
See separate page writeup called ‘Special Triangles’
Paul Crutches in Geometry
See separate writeup
-Problem appears non-solveable
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5 Crutches of Geometry
Problem
Solution
Triangle is unsolveable
Draw a perpendicular
Quadrilateral is unsolveable
Draw a diagonal. Quadrilateral
now becomes TWO triangles
Circle is unsolveable
1. Find R
2. Identify additional R’s
3. Draw additional R’s as required
Problems that utilize these tools:
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