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TOUGH & TRICKY SET Of PROBLEMS

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Re: TOUGH & TRICKY SET Of PROBLMS [#permalink] New post 26 Oct 2009, 19:05
Awesome questions and explanations. I could solve 80%, dist from circle to line was the best and I had no clue...
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Re: TOUGH & TRICKY SET Of PROBLMS [#permalink] New post 26 Oct 2009, 19:17
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9. PROBABILITY OF INTEGER BEING DIVISIBLE BY 8:
If n is an integer from 1 to 96, 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%

Ans C 62.5%

Given, n is an integer from 1 to 96.
Case 1:
If n is even then n*(n+1)*(n+2) is divisible by 8.
reason: n is even then n is divisible by 2 and n+2 is divisible by 4. Hence the product should be divisible by 8.
ex: 2*3*4.

1 to 96 there are 48 even integers. Hence 48 possibilities divisible by 8.

If n is odd, then n*(n+1)*(n+2) is divisible by 8 when the even integer n+1 is divisible by 8.
There are 12 possible cases here. Write now multiple of 8s you will get till 96, 12 possibilities.

Add 48+12 and divide by total 96 = .625. Hence, 62.5% is correct answer.
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Re: TOUGH & TRICKY SET Of PROBLMS [#permalink] New post 31 Oct 2009, 05:43
Bunuel wrote:
SOLUTION:
1. THE SUM OF EVEN INTEGERS:
The sum of the even numbers between 1 and k is 79*80, where k is an odd number, then k=?

(A) 79
(B) 80
(C) 81
(D) 157
(E) 159

The solution given in the file was 79, which is not correct. Also the problem was solved with AP formula thus was long and used the theory rarely tested in GMAT. Here is my solution and my notes about AP. Some may be useful:

The number of terms in this set would be: n=(k-1)/2 (as k is odd)
Last term: k-1
Average would be first term+last term/2=(2+k-1)/2=(k+1)/2
Also average: sum/number of terms=79*80/((k-1)/2)=158*80/(k-1)
(k+1)/2=158*80/(k-1) --> (k-1)(k+1)=158*160 --> k=159

Answer E.


How is n = (k-1)/2??

If the last term is K, and the 1st term is 1, the no. of terms should be k-1+1
Hence, here the no. of terms should be =>(k-1+1)/2 => k/2 since there should be half as many even numbers...
Am I missing something?
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Re: TOUGH & TRICKY SET Of PROBLMS [#permalink] New post 31 Oct 2009, 10:45
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vkb16 wrote:
Bunuel wrote:
SOLUTION:
1. THE SUM OF EVEN INTEGERS:
The sum of the even numbers between 1 and k is 79*80, where k is an odd number, then k=?

(A) 79
(B) 80
(C) 81
(D) 157
(E) 159

The solution given in the file was 79, which is not correct. Also the problem was solved with AP formula thus was long and used the theory rarely tested in GMAT. Here is my solution and my notes about AP. Some may be useful:

The number of terms in this set would be: n=(k-1)/2 (as k is odd)
Last term: k-1
Average would be first term+last term/2=(2+k-1)/2=(k+1)/2
Also average: sum/number of terms=79*80/((k-1)/2)=158*80/(k-1)
(k+1)/2=158*80/(k-1) --> (k-1)(k+1)=158*160 --> k=159

Answer E.


How is n = (k-1)/2??

If the last term is K, and the 1st term is 1, the no. of terms should be k-1+1
Hence, here the no. of terms should be =>(k-1+1)/2 => k/2 since there should be half as many even numbers...
Am I missing something?


The problem is that the formula you proposed wont't give you the integer value as k is an odd number. Consider the list from 1 to 9, clearly there are 4 even numbers in the list(2,4,6,8):
if we use your formula k/2=9/2=4.5 --> incorrect;
if we use the formula proposed in solution (k-1)/2=(9-1)/2=4 --> correct.

Hope it's clear.
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Re: TOUGH & TRICKY SET Of PROBLMS [#permalink] New post 18 Dec 2009, 10:53
A very good (ofcourse tough and tricky) set of problems. But when i was prcticing the tests from book that time i felt they are quite easier. Are these the actual type of questions that come in GMAT?
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Re: TOUGH & TRICKY SET Of PROBLMS [#permalink] New post 22 Dec 2009, 20:28
9. PROBABILITY OF INTEGER BEING DIVISIBLE BY 8:
If n is an integer from 1 to 96, 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%

Ans C 62.5%
--------------------------------------------------------------
I solve it in an opposite side of view.
How about n is odd and n,n+2 are both odds, which means only if n+1 is indivisible by 8 can satisfy the product of n*(n+1)*(n+2) is indivisible.
Let's take a look at n+1 which is even.
There are 48 even integers from 1 to 96. 96/2 = 48, and among these 48 even integers, 96/8=12 of them are divisible by 8. So 48-12=36 even integers are indivisible. 1 - 36/96 = 0.625
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Re: TOUGH & TRICKY SET Of PROBLMS [#permalink] New post 23 Dec 2009, 14:04
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7. THE DISTANCE BETWEEN THE CIRCLE AND THE LINE:
What is the least possible distance between a point on the circle x^2 + y^2 = 1 and a point on the line y = 3/4*x - 3?

A) 1.4
B) sqrt (2)
C) 1.7
D) sqrt (3)
E) 2.0

I find useful to roughly draw such questions. You can instantly recognize that line y=3/4*x-3 and axises form a right triangle with sides 3 and 4, on Y-axis and X-axis respectively. The hypotenuse lies on the line y=3/4*x-3 and equals 5 (Pythagorean Triple). The closest distance from any point to line is perpendicular from that point to line. Perpendicular to the hypotenuse (from origin in this case) will always divide the triangle into two triangles with the same properties as the original triangle. So, lets perpendicular will x, then

\frac{x}{3}=\frac{4}{5} and x=3*\frac{4}{5}=\frac{12}{5}=2,4

To find distance from point on circle to line we should substract lenth of radius from lenth of perpendicular: 2,4-1=1,4

So, the answere is A.

Unfortunately, I cannot attach illustration. But if you draw this problem (even roughly), you will find it really easy to solve.
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Re: TOUGH & TRICKY SET Of PROBLMS [#permalink] New post 23 Dec 2009, 14:35
Expert's post
Vyacheslav wrote:
7. THE DISTANCE BETWEEN THE CIRCLE AND THE LINE:
What is the least possible distance between a point on the circle x^2 + y^2 = 1 and a point on the line y = 3/4*x - 3?

A) 1.4
B) sqrt (2)
C) 1.7
D) sqrt (3)
E) 2.0

I find useful to roughly draw such questions. You can instantly recognize that line y=3/4*x-3 and axises form a right triangle with sides 3 and 4, on Y-axis and X-axis respectively. The hypotenuse lies on the line y=3/4*x-3 and equals 5 (Pythagorean Triple). The closest distance from any point to line is perpendicular from that point to line. Perpendicular to the hypotenuse (from origin in this case) will always divide the triangle into two triangles with the same properties as the original triangle. So, lets perpendicular will x, then

\frac{x}{3}=\frac{4}{5} and x=3*\frac{4}{5}=\frac{12}{5}=2,4

To find distance from point on circle to line we should substract lenth of radius from lenth of perpendicular: 2,4-1=1,4

So, the answere is A.

Unfortunately, I cannot attach illustration. But if you draw this problem (even roughly), you will find it really easy to solve.


Excellent! +1.

For those who don't know the distance formula (which is in fact very rarely tested) this is the easiest and most elegant solution.
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Re: TOUGH & TRICKY SET Of PROBLMS [#permalink] New post 30 Dec 2009, 03:25
there is an errata on solution of 4. the answer requested is the value of C not CD.

Thus the answer is 9, option D

Megan :idea:

Bunuel wrote:
As asked I'm combining all the problems from Tough & tricky problems in single thread. Here are the first ten questions. Next set and solutions to these ten will follow in couple of hours.

1. THE SUM OF EVEN INTEGERS:
The sum of the even numbers between 1 and k is 79*80, where k is an odd number, then k=?

(A) 79
(B) 80
(C) 81
(D) 157
(E) 159

2. THE PRICE OF BUSHEL:
The price of a bushel of corn is currently $3.20, and the price of a peck of wheat is $5.80. The price of corn is increasing at a constant rate of 5x cents per day while the price of wheat is decreasing at a constant rate of 2^1/2*x-x cents per day. What is the approximate price when a bushel of corn costs the same amount as a peck of wheat?

(A) $4.50
(B) $5.10
(C) $5.30
(D) $5.50
(E) $5.60

3. LEAP YEAR:
How many randomly assembled people are needed to have a better than 50% probability that at least 1 of them was born in a leap year?

A. 1
B. 2
C. 3
D. 4
E. 5

4. ADDITION PROBLEM:
AB + CD = AAA, where AB and CD are two-digit numbers and AAA is a three digit number; A, B, C, and D are distinct positive integers. In the addition problem above, what is the value of C?

(A) 1

(B) 3

(C) 7

(D) 9

(E) Cannot be determined

5. RACE:
A and B ran, at their respective constant rates, a race of 480 m. In the first heat, A gives B a head start of 48 m and beats him by 1/10th of a minute. In the second heat, A gives B a head start of 144 m and is beaten by 1/30th of a minute. What is B’s speed in m/s?
(A) 12
(B) 14
(C) 16
(D) 18
(E) 20

6. PROBABILITY OF DRAWING:
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

7. THE DISTANCE BETWEEN THE CIRCLE AND THE LINE:
What is the least possible distance between a point on the circle x^2 + y^2 = 1 and a point on the line y = 3/4*x - 3?

A) 1.4
B) sqrt (2)
C) 1.7
D) sqrt (3)
E) 2.0

8. THE AVERAGE TEMPERATURE:
The average of temperatures at noontime from Monday to Friday is 50; the lowest one is 45, what is the possible maximum range of the temperatures?

A. 20
B. 25
C. 40
D. 45
E. 75

9. 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%

10. SUM OF INTEGERS:
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 inquiry
B. A+5
C A+25
D 2A
E. 5A

THE OA WITH SOLUTIONS WILL BE PROVIDED.
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Re: TOUGH & TRICKY SET Of PROBLMS [#permalink] New post 30 Dec 2009, 03:42
The problem number 9 was very tough.

I would appreciate if someone can provide an alternative explaination to it. I don't quite understand the one that is already given.
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Re: TOUGH & TRICKY SET Of PROBLMS [#permalink] New post 09 Jan 2010, 04:10
Bunuel wrote:
SOLUTION:
6. PROBABILITY OF DRAWING:
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

This is with replacement case (and was solved incorrectly by some of you):

P=2*3/9*2/9=4/27

We are multiplying by 2 as there are two possible wining scenarios RW and WR.

Answer: D.


I've some problem to understand this ; as the two events are mutually exclusive , independent.
So why not the answer is 3/9*2/9 = 2/27
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Re: TOUGH & TRICKY SET Of PROBLMS [#permalink] New post 09 Jan 2010, 04:49
Expert's post
angel2009 wrote:
Bunuel wrote:
SOLUTION:
6. PROBABILITY OF DRAWING:
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

This is with replacement case (and was solved incorrectly by some of you):

P=2*3/9*2/9=4/27

We are multiplying by 2 as there are two possible wining scenarios RW and WR.

Answer: D.


I've some problem to understand this ; as the two events are mutually exclusive , independent.
So why not the answer is 3/9*2/9 = 2/27


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.
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Re: TOUGH & TRICKY SET Of PROBLMS [#permalink] New post 19 Jan 2010, 05:43
Bunuel wrote:
SOLUTION:
2. THE PRICE OF BUSHEL:
The price of a bushel of corn is currently $3.20, and the price of a peck of wheat is $5.80. The price of corn is increasing at a constant rate of 5x cents per day while the price of wheat is decreasing at a constant rate of 2^1/2*x-x cents per day. What is the approximate price when a bushel of corn costs the same amount as a peck of wheat?

(A) $4.50
(B) $5.10
(C) $5.30
(D) $5.50
(E) $5.60

Note that we are not asked in how many days prices will cost the same.
Let y # of days when these two bushels will have the same price.
First 2^1/2*x-x=0.41x
3.2+5xy=5.8-0.41xy solving for xy=0.48
The cost of a bushel of corn=3.2+5*0.48=5.6

Answer: E.


How do you make this Bunuel , I dont get it :(

Quote:
First 2^1/2*x-x=0.41x
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Re: TOUGH & TRICKY SET Of PROBLMS [#permalink] New post 19 Jan 2010, 06:19
Expert's post
GMATMadeeasy wrote:
Bunuel wrote:
SOLUTION:
2. THE PRICE OF BUSHEL:
The price of a bushel of corn is currently $3.20, and the price of a peck of wheat is $5.80. The price of corn is increasing at a constant rate of 5x cents per day while the price of wheat is decreasing at a constant rate of 2^1/2*x-x cents per day. What is the approximate price when a bushel of corn costs the same amount as a peck of wheat?

(A) $4.50
(B) $5.10
(C) $5.30
(D) $5.50
(E) $5.60

Note that we are not asked in how many days prices will cost the same.
Let y # of days when these two bushels will have the same price.
First 2^1/2*x-x=0.41x
3.2+5xy=5.8-0.41xy solving for xy=0.48
The cost of a bushel of corn=3.2+5*0.48=5.6

Answer: E.


How do you make this Bunuel , I dont get it :(

Quote:
First 2^1/2*x-x=0.41x


Yes it's kind of confusing, partly because I didn't use the formula formatting. Edited the original solution. So here it is:


Note that we are not asked in how many days prices will cost the same.

Let y be the # of days when these two bushels will have the same price.

First let's simplify the formula given for the rate of decrease of the price of wheat: \sqrt{2}*x-x=1.41x-x=0.41x, this means that the price of wheat decreases by 0.41x cents per day, in y days it'll decrease by 0.41xy cents;

As price of corn increases 5x cents per day, in y days it'll will increase by 5xy cents;

Set the equation: 3.2+5xy=5.8-0.41xy, solve for xy --> xy=0.48;

The cost of a bushel of corn in y days (the # of days when these two bushels will have the same price) will be 3.2+5xy=3.2+5*0.48=5.6.

Answer: E.

Hope it's clear. Tell me if it needs more clarification.
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PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: TOUGH & TRICKY SET Of PROBLMS [#permalink] New post 19 Jan 2010, 06:43
Bunuel wrote:
SOLUTION:
6. PROBABILITY OF DRAWING:
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

This is with replacement case (and was solved incorrectly by some of you):

Quote:
P=2*3/9*2/9=4/27


We are multiplying by 2 as there are two possible wining scenarios RW and WR.

Answer: D.


The probability of drawing a red and a white ball has to be( 2/9)*(3/9) and then the solution follows. How come it is (2*3/9) / (2/9), i dont get it again .

Will continue again with rest of the questions.

P.S. I have never liked questions who have just been tough but no brainerbut the one in the list here are some of the best questions to train on many tricks and fundamentals I believe. More I see the stuff on quant , more I admire the work of the people engaged with the site. Bravo.. and cheers from cold Paris but grâce à GMAT for keeping me hot :)

Last edited by GMATMadeeasy on 19 Jan 2010, 07:44, edited 1 time in total.
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Re: TOUGH & TRICKY SET Of PROBLMS [#permalink] New post 19 Jan 2010, 07:42
Bunuel wrote:
GMATMadeeasy wrote:
Bunuel wrote:
SOLUTION:
2. THE PRICE OF BUSHEL:
The price of a bushel of corn is currently $3.20, and the price of a peck of wheat is $5.80. The price of corn is increasing at a constant rate of 5x cents per day while the price of wheat is decreasing at a constant rate of 2^1/2*x-x cents per day. What is the approximate price when a bushel of corn costs the same amount as a peck of wheat?

(A) $4.50
(B) $5.10
(C) $5.30
(D) $5.50
(E) $5.60

Note that we are not asked in how many days prices will cost the same.
Let y # of days when these two bushels will have the same price.
First 2^1/2*x-x=0.41x
3.2+5xy=5.8-0.41xy solving for xy=0.48
The cost of a bushel of corn=3.2+5*0.48=5.6

Answer: E.


How do you make this Bunuel , I dont get it :(

Quote:
First 2^1/2*x-x=0.41x


Yes it's kind of confusing, partly because I didn't use the formula formatting. Edited the original solution. So here it is:


Note that we are not asked in how many days prices will cost the same.

Let y be the # of days when these two bushels will have the same price.

First let's simplify the formula given for the rate of decrease of the price of wheat: \sqrt{2}*x-x=1.41x-x=0.41x, this means that the price of wheat decreases by 0.41x cents per day, in y days it'll decrease by 0.41xy cents;

As price of corn increases 5x cents per day, in y days it'll will increase by 5xy cents;

Set the equation: 3.2+5xy=5.8-0.41xy, solve for xy --> xy=0.48;

The cost of a bushel of corn in y days (the # of days when these two bushels will have the same price) will be 3.2+5xy=3.2+5*0.48=5.6.

Answer: E.

Hope it's clear. Tell me if it needs more clarification.


My problem was in understanding how to get .41x . It is crystal clear now.
Thank you so much.
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Re: TOUGH & TRICKY SET Of PROBLMS [#permalink] New post 19 Jan 2010, 10:52
Expert's post
GMATMadeeasy wrote:
Bunuel wrote:
SOLUTION:
6. PROBABILITY OF DRAWING:
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

This is with replacement case (and was solved incorrectly by some of you):

Quote:
P=2*3/9*2/9=4/27


We are multiplying by 2 as there are two possible wining scenarios RW and WR.

Answer: D.


The probability of drawing a red and a white ball has to be( 2/9)*(3/9) and then the solution follows. How come it is (2*3/9) / (2/9), i dont get it again .

Will continue again with rest of the questions.

P.S. I have never liked questions who have just been tough but no brainerbut the one in the list here are some of the best questions to train on many tricks and fundamentals I believe. More I see the stuff on quant , more I admire the work of the people engaged with the site. Bravo.. and cheers from cold Paris but grâce à GMAT for keeping me hot :)


We want out of two balls one to be red (R) and another white (W).

There are two ways this could happen: we can draw FIRST ball red and SECOND white OR FIRST ball white and SECOND red.

Probability of each case is: \frac{3}{9}*\frac{2}{9}, so P=\frac{3}{9}*\frac{2}{9}+\frac{2}{9}*\frac{3}{9}=2*\frac{3}{9}*\frac{2}{9}.

One more thing worth of mentioning: \frac{3}{9}*\frac{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.

Hope it's clear.
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Re: TOUGH & TRICKY SET Of PROBLMS [#permalink] New post 19 Jan 2010, 12:08
SOLUTION:
6. PROBABILITY OF DRAWING:
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

This is with replacement case (and was solved incorrectly by some of you):

Quote:
P=2*3/9*2/9=4/27


We are multiplying by 2 as there are two possible wining scenarios RW and WR.

Answer: D.[/quote]

The probability of drawing a red and a white ball has to be( 2/9)*(3/9) and then the solution follows. How come it is (2*3/9) / (2/9), i dont get it again .

Will continue again with rest of the questions.

P.S. I have never liked questions who have just been tough but no brainerbut the one in the list here are some of the best questions to train on many tricks and fundamentals I believe. More I see the stuff on quant , more I admire the work of the people engaged with the site. Bravo.. and cheers from cold Paris but grâce à GMAT for keeping me hot :)[/quote]

We want out of two balls one to be red (R) and another white (W).

There are two ways this could happen: we can draw FIRST ball red and SECOND white OR FIRST ball white and SECOND red.

Probability of each case is: \frac{3}{9}*\frac{2}{9}, so P=\frac{3}{9}*\frac{2}{9}+\frac{2}{9}*\frac{3}{9}=2*\frac{3}{9}*\frac{2}{9}.

One more thing worth of mentioning: \frac{3}{9}*\frac{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.

Hope it's clear
Perfect. That is how I solved it.

Next attack : Inequality : Will be going through your links to flex "inequality muscles" . Can't thank enough for the efforts you guys have put.
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Re: TOUGH & TRICKY SET Of PROBLMS [#permalink] New post 15 Feb 2010, 17:42
This is fantastic, Bunuel! Thank you so much!
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Re: TOUGH & TRICKY SET Of PROBLMS [#permalink] New post 04 Mar 2010, 10:06
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@ Bunuel

Q7 isn't that tough. we need a better approach.
Here is what I did.
Draw a circle with radius 1 and center as origin. Now draw the line y=(3/4)x-3. This has y intercept -3 and x-intercept =4. Now draw a line(blue line) from (0,0) perpendicular to the line. What we need is the length of red line. See the attached Diagram
Attachment:
q7.JPG
q7.JPG [ 12.01 KiB | Viewed 13841 times ]

Area of the right triangle with 3 as base is same as that with 5 as base
(1/2)*3*4 = (1/2)*5*(blue line) ==> blue line = 2.4 units
Hence red line = 2.4 - 1(radius of circle)
= 1.4

Hope this helps.
Re: TOUGH & TRICKY SET Of PROBLMS   [#permalink] 04 Mar 2010, 10:06
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