Percentage

Sometimes,we are in great hurry because of which we make very simple mistakes and we do not know. Making a mistake is one thing and being unaware of the mistake being done is yet another.
Now I will show you where a large number of students make mistake in questions of percentage.


Problem : If x is 25% more than y. Then how much is y less than x expressed as % of x.

Solution: As soon as the question is read, a majority gives the answer y is 25% less than x. Second category of people those who will see it as a trick and they will take two simple numbers. x=100 and y =75
So that x is 25 more than y. Here they fail to realize the term percentage. so they will also end up saying y is 25 less than x, hence answer is 25%.
But the answer to it must be solved for you now

X is 25% more than y
say y = 100
so x = 125
Talking in Percentage % change (incraese than y) in x = 25 % as per question = (x-y) / y = (125-100)/100 = .25 or 25%
Now similarly % change (decrease than x ) in y = (y-x) / x = (100-125)/125 =  - 25/125 = -1 /5 = -0.2 or -20%
-ve sign indicates the decrease w.r.t. x
Hence the answer must be y is 20% less than x.

Probability Creates a Problem

Questions from probability are really not that difficult as they appear to be. If you have a little command over Permutations and Combinations, then it is just a matter of a minute. But on the contrary, if permutations and combinations are confusing to you, then Probability too can become a nightmare. Though if you are lucky enough you may also see text questions of Favours & Odds, from where you can have a narrow escape.
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Problem : There are seven balls numbered 1 to 7, placed in a each of the two boxes coloured white and black. A ball is drawn out of white box and then from a black box. The process continues till the balls in any draw from both the boxes bear the same number. What is the probability that the balls will exhaust in the box before the process ends.




The solution to the above problem does not need any hi-fi profile. Each one of you with a little understanding can solve this question. If I can frame a question, I hope you can frame a reasonable answer.
If you need help with any topic of maths, you are most welcome to mail me at solveaquestion@gmail.com
I shall be more than happy to help you out...


Have a Nice Concept Foundation.
Bye

IIT JEE EXAM 2011- EXAM DATE 10 April 2011, Last date to Apply = 15 December 2010

Refresh your Concepts

Problem :
A coin is flipped 10 times consecutively. What is the probability that the sixth flip results in a head ?
What is the probability that the 10th flip is a tail ?
Solution :
When a coin is flipped, there is equal possibility for head and tail in each flip for a fair coin. Hence whichever flip it be the probability is 1/2 because each event is independent and mutually exclusive and exhaustive.


Problem:
A basket has three red balls and three white balls. So how many minimum balls must you draw to make a pair of two balls with the same colour ?
Solution : Atleast three balls have to be drawn to ensure that the pair is obtained in any case. How & why not Two? If the first two drawn balls differ in colour, then the third ball is a tie breaking situation between the two coloured balls and whichever colour you get in the third draw, you atleast make a full pair then.
If 10 workers build a wall in 3 days. In how many days will 15 workers build the same wall. ?
Solution :
None. Well, the wall is already built. :) But mathematically working it will be 2 days.

Problem:
A and B play a game and roll a dice. The one who gets a six will win the game. What is the probability of B's winning the game.
Solution:
A can win the game in 1st or 3rd or 5th or 7th try.
P(A) = P(A) + P'(A) P'(B) P(A) + P'(A) P'(B) P'(A) P'(B) P(A) +  . . . . .
= 1/6 + 5/6*5/6*1/6 +5/6*5/6*5/6*5/6*1/6+ . . . .
= 1/6 (1+ 5/6*5/6 + 5/6*5/6*5/6*5/6+.......)          This forms an infinite GP series
=1/6 (   1/(1-(5/6*5/6))   )
= 1/6 (  36/36-25)   )
= 1/6 (36/11)
=6/11

So P(B) = 1-(PA) = 1 - 6/11
=5/11

PROBABILITY OF SOLVING THIS PROBLEM

Probability can be a real fun to solve and scoring topic too but only when right approach is used. A wrong approach can take you far from the answer. Its Probability is much more a case to be thought practically rather than jumping here and there with equations. Probability of solving questions on this topic is inversely proportional to the probability of your losing concentration. So check your concentration.....



Problem :
From a pack of n cards no 1,2,3,4,5.......... n. A draws a card and then puts it back. Now B draws a card. Find the probability that B has drawn the same card as A.
Solution :
A can draw card in n ways.
Similarly, B can also draw card in n ways.
So the total no ways in which A and B can select a card is = n x n.
This becomes the total no of possible cases.

Now A and B can draw the same card nos like in these cases(1,1) or (2,2) or (3,3) or (4,4)....... or (n,n). Hence the total no of favourable cases are n.

So P = No of favourable cases / Total no of cases = n/(nxn) = 1/n



That was quite a simple problem. But the important thing to consider is that how you start to think. If you start with a negative approach of  " its too difficult ..very hard problem... bla bla bla....", it oughts to become the same.  So be positive and explore the ways to achieve the desired results.

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Your suggestions and querries are welcome.
Mail to mathsprobe@gmail.com
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IIT JEE EXAM 2011- EXAM DATE 10 April 2011, Last date to Apply = 15 December 2010