Strategies for Geometry

1. a² + b² = c², but only when a right triangle. If you don’t know it’s a right triangle, Pythagorean theorem does not apply!
2. Common special right triangles include 3-4-5, 5-12-13, 8-15-17, 7-24-25 (and their multiples.)
3. 45-45-90 triangles are ALWAYS in the ratio 1:1:√2
4. 30-60-90 triangle are ALWAYS in the ratio 1:√3:2
5. Angles and opposite sides are in the same relative size order, but are NOT proportional.
6.  The right triangle with the largest area will be an isosceles right triangle (where both the base and height are of equal length).

At a certain school, each of the 150 students takes between 1 and 3 classes

At a certain school, each of the 150 students takes between 1 and 3 classes. The 3 classes available are Math, Chemistry and English. 53 students study math, 88 study chemistry and 58 study english. If 6 students take all 3 classes, how many take exactly 2 classes?

In this case, we'd use the first formula, since we want the number who take exactly 2 classes:

150 = 53 + 88 + 58 - (doubles) - 2(triples)
150 = 199 - (doubles) - 2(6)
150 = 187 - doubles
doubles = 37

Let's just change the question a tiny bit: 

At a certain school, each of the 150 students takes between 1 and 3 classes. The 3 classes available are Math, Chemistry and English. 53 students study math, 88 study chemistry and 58 study english. If 6 students take all 3 classes, how many take at least 2 classes?

 In this case, we'd use the second formula, since we want the number who take at least 2 classes:

150 = 53 + 88 + 58 - (at least 2 of the 3) + (all 3)
150 = 199 - (at least 2 of 3) + 6
150 = 193 - (at least 2 of 3)
At least 2 of 3 = 43 

A box contains 3 yellow balls and 5 black balls. One by one, every ball is selected at random

A box contains 3 yellow balls and 5 black balls. One by one, every ball is selected at random without replacement. What is the probability that the fourth ball selected is black?
A.1/4
B.1/2
C.5/8
D.2/3
E.3/4 





Yes, P(B) on any given pick = P(B) on the first pick.

An easier example:

Say we have 3 red marbles and 2 blue. What's the probability that the second marble picked is blue?

P(B) on the first pick is 2/5.

To get blue on the second pick, our two picks could be RB or BB:

P(R and B) = 3/5 * 2/4 = 3/10.

P(B and B) = 2/5 * 1/4 = 1/10.

Since either is a good outcome, we add the fractions:

3/10 + 1/10 = 4/10 = 2/5.

The same probability as getting blue on the first pick.

Pretty neat!



*****This is so because each ball is equally likely to be the fourth one (or nth one) drawn
 

If x, y, and k are positive numbers such that ((x)/(x+y))(10) + ((y)/(x+y))(20) = k and if x < y

If x, y, and k are positive numbers such that ((x)/(x+y))(10) + ((y)/(x+y))(20) = k and if x < y, which of the following could be the value of k?
A. 10
B. 12
C. 15
D. 18
E. 30




The trick is to recognize that the equation just gives a weighted average of x and y. The equation above is the same as you'd use if you were asked "If x pounds of peanuts and y pounds of cashews are mixed together, and peanuts cost $10/pound and cashews cost $20/pound, what is the price per pound of the resulting mixture?" The answer must be between 10 and 20, and if y > x, the answer must be closer to 20 than to 10. So 18 is the only possible answer.

Thu Jan 28, 2010 1:23 pm
rahul.s wrote: Ian, In such a problem, where we need to experiment with numbers, how would we know which are the right numbers to choose? Well, you don't need to experiment with numbers in the question above; I didn't in my post above. That said, if you don't see a conceptual or algebraic solution fairly quickly, plugging in numbers is a good fallback option. It would be a bit lucky to find numbers that give the exact answer to this question, but if you plug in a few very simple sets of numbers (you don't want to waste any time on complicated numbers), making sure that x is less than y, you'll always find that k is between 15 and 20, which may lead you to the correct answer here. There are also algebraic solutions to the question: (10x + 20y)/(x+y) = k 10x + 20y = kx + ky 20y - ky = kx - 10x y(20 - k) = x(k - 10) (20 - k)/(k - 10) = x/y and since 0 < x < y, then 0 < x/y < 1, and it must be that 0 < (20 - k) / (k - 10) < 1. From the answer choices we can be sure k - 10 isn't negative, so we can multiply through this inequality by k-10 to find that 0 < 20 - k < k - 10, or that 15 < k < 20.

When positive integer x is divided by positive integer y, the remainder is 9

When positive integer x is divided by positive integer y, the remainder is 9. If x/y=96.12, what is the value of y?

(a) 96
(b) 75
(c) 48
(d) 25
(e) 12




In this case, we can start by back solving. x/y=96.12 = 96 12/100
Let remainder = r
r/y = 12/100, here r = 9, then 9/y = 12/100 implies y = 9/0.12 = 75
The correct answer is (B).
 

In the graduating class of a certain college, 48% of the students are male and 52% are female

In the graduating class of a certain college, 48% of the students are male and 52% are female. In this class 40% of the male and 20% of the female students are 25years old or older. If one student in the graduating class is randomly selected, approximately what is the probability that he or she will be less than 25 years old??
A) 0.90
B) 0.70
C) 0.45
D) 0.30
E) 0.25 


 




This is really just a weighted average. We have two groups, men and women, and men have an 'average' of 40%, while women have an 'average' of 20%. The groups are almost equal in size, so when we combine them, the 'average' must be almost exactly 30%. So roughly 30% of all people are twenty-five years old or older, and roughly 70% are thus less than twenty-five, so the answer should be B.
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Whenever we have groups (in this case, male and female) that are being divided into smaller groups (in this case, those 25 and older and those less than 25), we can use a group grid to organize the data. Here's what the grid looks like:

_______________M______F_______Total

older:

younger:

total:

In the grid above, every row has to add up to the total, as does every column. Looking at the top row, older males + older females = total older. Looking at the left-most column, older males + younger males = total males.



Now let's fill in the data step by step. Let's plug in 100 for the total number of students:

_______________M______F_______Total

older:

younger:

total:__________________________100



48% males, 52% females. Since we're looking for an approximation, let's say that 50% are males and 50% are females:

_______________M______F_______Total

older:

younger:

total:__________50_____50________100



40% of the males and 20% of the females are 25 and older:

_______________M______F_______Total

older:__________20_____10_______

younger:

total:__________50_____50________100



Now we can complete the grid:

_______________M______F_______Total

older:__________20_____10________30

younger:_______30_____40________70

total:__________50_____50________100


Notice that everything adds up horizontally and vertically. Neat!

So P(younger) = 70/100 = .7

The correct answer is B.

 

A bag contains 3 red, 4 black and 2 white balls. What is the probability of drawing a red and a white

A bag contains 3 red, 4 black and 2 white balls. What is the probability of drawing a red and a white ball in two successive draws, each ball being put back after it is drawn?
(A) 2/27
(B) 1/9
(C) 1/3
(D) 4/27
(E) 2/9









Winning scenario consists of TWO cases RW and WR. Probability of each case is: 3/9*2/9, so 3/9*2/9+3/9*2/9=2*3/9*2/9.

Only 3/9*2/9 would be correct answer if we were asked to determine the probability of FIRST ball being red and SECOND white, OR FIRST white and SECOND red. These are the cases in which order of drawing matters. 

PROBABILITY OF INTEGER BEING DIVISIBLE BY 8

If n is an integer from 1 to 96 (inclusive), what is the probability for n*(n+1)*(n+2) being divisible by 8?

A. 25%
B 50%
C 62.5%
D. 72.5%
E. 75%











N=n*(n+1)*(n+2)

N is divisible by 8 in two cases:
When n is even:
No of even numbers (between 1 and 96)=48
AND
When n+1 is divisible by 8. -->n=8p-1 --> 8p-1<=96 --> p=12.3 --> 12 such nembers

Total=48+12=60

Probability=60/96=0.62

If the sum of five consecutive positive integers is A, then the sum of the next

If the sum of five consecutive positive integers is A, then the sum of the next five consecutive integers in terms of A is:

a) A+1
b) A+5
c) A+25
d) 2A
e) 5A 







n + n+1 + n+2 + n+3 + n+4 = 5n + 10 = A

n+5 + n+6 + n+7 + n+8 + n+9 = 5n + 35 = A+25

The sum of n consecutive positive integers is 45

The sum of n consecutive positive integers is 45. What is the value of n?

(1) n is even

(2) n < 9

(1) n is even. Now try to choose n such that average i.e, 45/n gives you .5 in decimal. Why? Because we would like to quickly identify the middle two.

So, choose n=6. Average=7.5. So, the case is 5, 6, 7, 8, 9, 10.
Now choose n=2. The case is 22, 23. Insufficient.

(2) No need for further calculation. Just look at (1). Insufficient.

Combining: Just look (1) & (2). E

The sum of n consecutive positive integers is 45

The sum of n consecutive positive integers is 45. What is the value of n?
(1) n is odd
(2) n >= 9

(1)   The sum of consecutive integers will always be divisible by the number of integers only if the number of integers is odd.
n can be 3/5/9. Insufficient.

(2) We can have n=9/10 but since we are talking of +ve int only. Hence it would be 9. This is sufficient.

Ans : B