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.
Few Simple Things
There are few simple things which make our calculations little faster without efforts. Students applying these simple simple methods in various situation sometimes come across very quick results.
Let us discuss a few
Multiplication by 11
It is the simplest of all next to multiplication by zero.
For eg.
960403 x 11 = 10564433
How this is done in single line.
Start from right and approach left while adding the current digit to the preceeding digit and at the extremes the two numbers come as they are with carry anywhere if applicable is to be added.
960403 x 11
Step1 ) ________3 last digit as it is
Step 2) _______33 Added 0 to previous digit 3 of original no.
Step 3) ______433 Added 4 to the previous digit 0
Step 4) _____4433 Added 0 to thee previous digit 4
Step 5) ____64433 Added 6 to previous digit 0
Step 6) ___564433 Addded 9 to previous digit 6. this becomes 15. Write 5 and take 1 as carry for next step
Step 7) 10564433 Added the carry 1 to 9. After this there is no more digit to the left so the process completes and the answer is 10564433
Squaring of number ending in 5
To square a number which ends in 5 is very easy. For eg to square a number 25,35,15,85 or any other
The last two digits will always be 25 because both 5 will multiply to give 25.
What will come before 25 ? Take the digits ahead of 5 in the original number and multiply it with one more.
That means for a number say 85
The last digits will be ____25
And before that will come 8 x (8+1) = 8 x 9 =72
So the answer is 725
You can check this with 25 .
Last digit ____25
Before will come 2 x (2+1) = 2 x 3 =6
So the answer is 625
This will work with bigger numbers also
For eg 115
Last digit ______25
Before will come 11 x (11+1) = 11 x 12 = 132
So the answer is 13225
Let us discuss a few
Multiplication by 11
It is the simplest of all next to multiplication by zero.
For eg.
960403 x 11 = 10564433
How this is done in single line.
Start from right and approach left while adding the current digit to the preceeding digit and at the extremes the two numbers come as they are with carry anywhere if applicable is to be added.
960403 x 11
Step1 ) ________3 last digit as it is
Step 2) _______33 Added 0 to previous digit 3 of original no.
Step 3) ______433 Added 4 to the previous digit 0
Step 4) _____4433 Added 0 to thee previous digit 4
Step 5) ____64433 Added 6 to previous digit 0
Step 6) ___564433 Addded 9 to previous digit 6. this becomes 15. Write 5 and take 1 as carry for next step
Step 7) 10564433 Added the carry 1 to 9. After this there is no more digit to the left so the process completes and the answer is 10564433
Squaring of number ending in 5
To square a number which ends in 5 is very easy. For eg to square a number 25,35,15,85 or any other
The last two digits will always be 25 because both 5 will multiply to give 25.
What will come before 25 ? Take the digits ahead of 5 in the original number and multiply it with one more.
That means for a number say 85
The last digits will be ____25
And before that will come 8 x (8+1) = 8 x 9 =72
So the answer is 725
You can check this with 25 .
Last digit ____25
Before will come 2 x (2+1) = 2 x 3 =6
So the answer is 625
This will work with bigger numbers also
For eg 115
Last digit ______25
Before will come 11 x (11+1) = 11 x 12 = 132
So the answer is 13225
FUNCTIONS
EVEN AND ODD FUNCTIONS
If f(x)=f(-x), then f(x) is even function
and if f(x) =-f(-x) then f(x) is odd function
For example Sin(x) is odd function while cos(x) is even function. This means that sin(x)=-sin(-x) and cos(x)=cos(-x)
With this basic understanding, let us see this question.
Problem : If function f satisfies the relation f(x+1)+f(1-x)=2f(x) for all real values of x and f(0)< or >0, then prove that f(x) is even.
Solution:
f(x+1)+f(1-x)=2f(x)
Replacing x with -x
f(1-x)+f(1+x)=2f(-x)
Comparing both we get, 2 f(x) = 2 f(-x)
Hence f(x) = f(-x)
Hence proved that, f(x) is an even function.
COMPOSITE FUNCTIONS
Let f : A à B, and g : B à C, be two functions, then gof : A à C
This is known as the product function or composite of f and g, given by gof(x)=g{f(x)} for all real values of x.
The symbols f : A à B is read as Function F maps A to B.
It is to be noted that fog = f{g (x)} and (f ± g) x = f(x) ± g(x) and (f/g)x=f(x)/g(x)
A function f(x) is said to be periodic function of x, if there exists a positive real number T such that f(x+T) = f(x). Then the smallest value of T is called the Period of the function.
For example Sin x is periodic because
IIT JEE EXAM 2011- EXAM DATE 10 April 2011, Last date to Apply = 15 December 2010
If f(x)=f(-x), then f(x) is even function
and if f(x) =-f(-x) then f(x) is odd function
For example Sin(x) is odd function while cos(x) is even function. This means that sin(x)=-sin(-x) and cos(x)=cos(-x)
With this basic understanding, let us see this question.
Problem : If function f satisfies the relation f(x+1)+f(1-x)=2f(x) for all real values of x and f(0)< or >0, then prove that f(x) is even.
Solution:
f(x+1)+f(1-x)=2f(x)
Replacing x with -x
f(1-x)+f(1+x)=2f(-x)
Comparing both we get, 2 f(x) = 2 f(-x)
Hence f(x) = f(-x)
Hence proved that, f(x) is an even function.
COMPOSITE FUNCTIONS
Let f : A à B, and g : B à C, be two functions, then gof : A à C
This is known as the product function or composite of f and g, given by gof(x)=g{f(x)} for all real values of x.
The symbols f : A à B is read as Function F maps A to B.
It is to be noted that fog = f{g (x)} and (f ± g) x = f(x) ± g(x) and (f/g)x=f(x)/g(x)
Example:
If F : RàR and G : RàR, be two mappings such that f(x) = sin x and g(x)=x2
Then show that fog ≠ gof
Solution : Let x € R,
So (fog) x = f {g(x)} = sin(x2) -----(1)
While (gof) x = g {f(x)} = g {sinx} = (sinx)2 = sin2x -----(2)
From (1) & (2), (fog) x ≠ (gof) x
PERIODIC FUNCTIONSA function f(x) is said to be periodic function of x, if there exists a positive real number T such that f(x+T) = f(x). Then the smallest value of T is called the Period of the function.
For example Sin x is periodic because
Sin(2π+x)= Sin(x) and also sin(4π+x)=Sin(x) = Sin(6π+x) and so on…. But 2π being the smallest value is the period of the function.
Example: Show tha cos(√x) is non-periodic.
Solution: Let Cos(√x) be periodic with T as the period.
So, Cos(√x) = Cos{ √ (T+x) }
à √ (T+x) = 2nπ ± √x
Putting x = 0, we get, √T= 2nπ -----(1)
Putting x=T, we get √2T = 2nπ ± √T ----(2)
From (1) and (2) we get, √2T = √T ± √T
Or √T x √2 = √T (1 ± 1)
Or √2 = 1 ± 1, which is not possible, Hence Cos√x is not periodic function.
IIT JEE EXAM 2011- EXAM DATE 10 April 2011, Last date to Apply = 15 December 2010
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.
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
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
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
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
Count on Your Time, your @ssets
Talking of time I just recalled another puzzling question about time. Consider this:
Problem :
Mohan lives 24 km from the railway station. He has one hour to catch the train. He leaves home at the speed of 18 km/hr. Travels half the distance and then increases his speed to 30km/hr. He with this speed for the next quarter of the distance and then plans to speed up again. Now what should be his speed in order to catch the train.
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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
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
The Divisibility Test...
This question will test basic general understanding. This question can be solved by students over 8th class onwards very easily.
Consider two numbers a & b which when divided by 6, leave remainder 3 and 2 respectively. So what will be the remainder when each one of the following is divided by 6:
(ii) a-b
(iii) a*b
(iv) b-a
(v) b*b
Solution :
The answers are (i) 5 (ii) 1 (iii) 0 (iv) 5 (v) 4
How ?
Just take these two numbers a and b as 6x+3 and 6y+2 so that they leave the remainder 3 & 2 respectively when divided by 6.
Now you can see that a+b= 6(x+y)+5
a-b = 6(x+y)+1
a*b=(6x+3)*(6y+2)=36xy+18y+12x+6 is divisible by 6
b-a = 6(y-x)-1 = 6(y-x-1)+ 6-1=6(y-x-1)+5
b*b= (6y+2)*(6y+2) = 36yy+12y+12y+4
I hope now you get how we arrived at the solution.
Counting On Matches
Problem:
There are 17 teams to play in a Game. Matches are played between two teams and No match ends is draw. The loser is out of the game and the winner stays in the game to play ahead. The matches continue to be played among the teams until one team emerges as the winner. So can you tell me how many matches will be played in all to decide the winner of the games.
There are 17 teams to play in a Game. Matches are played between two teams and No match ends is draw. The loser is out of the game and the winner stays in the game to play ahead. The matches continue to be played among the teams until one team emerges as the winner. So can you tell me how many matches will be played in all to decide the winner of the games.
Solution:
It’s so simple, only if you think it practically and at the same time becomes much difficult the moment you try making any sort of equations etc.
Suppose you and 15 of your friends come and play with me. So you all are 16 and me is the 17th person. Right.
Since No match ends in a draw, so for me to emerge as the sole Winner, I will have to play with each one of you and defeat you to return you back to pavilion. So tell me how many people I will send back if I win. Sixteen. To defeat sixteen people, then how many matches have I played ? Sixteen.
Agreed!!!
That’s it. So simple. Bubye have a nice concept !
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