REASONING

TOPIC - REASONING
* Selected Questions on Reasoning from various past PO papers appearing very common.

Question : How many such pairs of letters are there in the word CHRONICLE each of which has as many letters in between them as in the english alphabets ?
a) None
b) One
c) Two
d) Three
e) More than three

Solution:
First we must understand what the question intend to convey.
It says that in this word CHRONICLE, you take any two alphabets and then see how many letters are in between them. And if you see in your english alphabets a,b,c,d,e,f...........z and these very two same alphabets, then count how many letters occur in between them. Are they two equal. If yes then this becomes a pair. So how many such pairs can you get ?

There is no fixed formulae to check this it is just that your ability to see that two alphabets are so close or far and compare it in the word there.

BANK PO RESOURCE CONTENTS

TOPICS

* REASONING




PAST QUESTIONS

* INDIAN BANK SELECTED QUESTIONS

Online Mathematics Books - Download Maths Ebook

Download Ebooks for Mathematics.
To Download ---- Right Click the link and then Save Target As
Got something useful for all. I just got to one site where I found much useful content.
You can download the ebook for your use, but do not host it in any form on any website.
I am providing you the link for all the books here at one place in Unit Wise.

1. Algebra
2. Calculus -1
3. Calculus -2
4. Calculus -3
5. Linear Algebra
6. Differential Equations
7. Algebra / Trignometry-1
8. Algebra / Trigonometry -2
9. Complex Number
10. Common Maths Error

The content here will be exhaustive, but it does not mean that you have to imbibe all. Students must be selective to know what actually they should need from here.
Like this Page ?? _______________________________________________________________________________
Download Maths Ebook, Mathematics Ebooks, Ebooks, maths online, Mathematics, Learn Maths online, Download Algebra Ebook, Download calculus Ebook, Download Differential Equations Ebook, Download Trigonometry Ebook, Commmon  Maths Error Ebook, Complex Number Ebook
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Video Learning 4 Kids - 5 ( SIMILAR TRIANGLES )

SIMILARITY OF TRIANGLES

What actually is similarity ?
It just means that the angles are same. So two triangles are similar if all three angles of first triangle are equal to the three angles of the second triangle.
Suppose I say that the two angles A & B of first triangle are equal to the two angles say P & Q of the second triangle, then it is obvious that the third angle of both will be same because the sum of angles in triangle is 180.
That is why it is enough to say Angle-Angle theorem (AA) and not AAA.
We here must also note that the ratio of sides of triangles is same corresponding to the equal angles.
We must also be aware that capital letters are used for angles and small letters are used for the sides.
So if A=P, B=Q & C=R, then we will have a/p = b/q = c/r
Suppose this ratio turns out to be one. Then these become congruent triangles.
Another way to show similarity is SAS, in this the two sides of the two triangles are in same ratio and the angle included between them is equal. This becomes the SAS theorem (Side Angle Side).



Source:Khanacademy.org








Source : khanacademy.org

VIDEO LEARNING 4 KIDS - 4

ANGLES AND TRIANGLES

__________________________

FINDING ANGLES

video (On mathematics )- This video shall explain for how to calculate the angles of the trianlges.

CONGRUENCY OF TRIANGLES

_________________________



VIDEO ON MATHEMATICS

This video shall explain how you can prove traingles are congruent or not !

__________________________

RESOURCE: YOUTUBE + YOURTEACHER.COM
__________________________

VIDEO LEARNING 4 KIDS - 3 ( ALGEBRA & RATIO )

VIDEO LEARNING FOR KIDS


R A T I O- Triangles & Angles






ALGEBRA



Mathematics Plus:

Video Source:
yourteacher.com

Video Learning 4 Kids - 2

VIDEO LEARNING FOR KIDS - SQUARE ROOT

SQUARE ROOT FOR 40



SIMILARLY SQUARE ROOT FOR 90



Mathematics Plus.

Source : www.yourteacher.com

VIDEO LEARNING 4 KIDS - 1 (PYTHAGORAS)

HELLO.
THIS VIDEO IS MAINLY FOR KIDS WHO ARE LEARNING MATHS. IF YOU HAVE KIDS IN YOUR FAMILY THEN YOU CAN SHOW THEM >>>> THEY MIGHT FIND IT INTERESTING TO WATCH IT >> AND AT THE SAME TIME THEY WILL LEARN.....


NOT FOR THE ONES LOOKING FOR HIHGER MATHS


THE BELOW VIDEO SHOWS THE PYTHAGORAS THEOREM









THE BELOW VIDEO SHOWS HOW TO GET AREA OF RECTANGLE - PYTHAGORAS THEROEM USED

Ratio And Proportions - Selected Problems in Ratio and Proportion

Ratio And Proportions

Well if you are planning to go for any interview, sitting for any placement, or any competitive exam for any job application, pin pointing to especially Bank PO exams / Bank Clerical Exams or any exam for recruitment to govt posts, you cannot miss out this topic of Mathematics. Then better have a look at it (Ratio and Proportion). Atelast 3 to 4 Questions in mathematics under algebra will surely be framed from this section because this is neither tough nor lengthy but good and concise questions can be framed from here to test your calculations and understanding. SO DONT MISS OUT THE EASY MARKS IN MATHEMATICS....Just take pain of 10 minutes to digest this capsule on Ratio & Proportion and make yourself little aware / refresh.

Problem in Ratio and Proportion (Mathematics) :
Three positive numbers are in the ratio 2:3:4 and the sum of their square is 1421. Find the numbers?

Solution: Let the numbers in the ratio 2:3:4 be 2x,3x and 4x

Sum of their squares = (2x)² + (3x)² + (4x)² =4x² + 9x² + 16x² = 29x² = 1421

x² = 1421/29 = 49

x = 7

-------------------
Continued Proportion – a, b and c are said to be in continued proportion if a/b =b/c or means b²=ac

Similarly we can also have a/b=b/c=c/d=d/e.......... for more numbers

Problem in Ratio and Proportion (Mathematics):
If a, b, c, d are in continued proportion, the prove that a. b. d. e = c⁴

Solution : Since they are in continued proportion, we can have a/b =b/c =c/d =d/e =say k

so we have

a=bk

b=ck

c=dk or d=c/k

d=ek or e=d/k

We have to prove a. b. c. e = (bk).(ck). (c/k) . (d/k) = (ckk).(ck).(c/k).(c/kk) = c⁴ k³ / k³ = c⁴

Problem in Ratio and Proportion (Mathematics):
If a/b=c/d=e/f

prove that, (a+3c-5e)/(b+3d-5f) = a/b = c/d = e/f

Solution :

Let a/b=c/d=e/f = k(say)

(bk+3dk – 5fk) / (b+3d – 5f) = k (b+3d – 5f)/(b+3d – 5f) = k

Hence Proved

Problem in Ratio and Proportion (Mathematics) :
If p/(b-c) = m/(c-a) = n/(a-b) then prove that ap+bm+cn = 0

Solution: Take each ratio = k

p=(b-c)k

m=(c-a)k

n=(a-b)k

Then ap+bm+cn = a(b-c)k + b(c-a)k + c(a-b)k = abk -ack + bck – abk +ack -cbk = 0

Problem in Ratio and Proportion (Mathematics) :
Ram, Sham and Suresh start business investing in the ratio 1/2 : 1/3: 1/4. The time for which each of them invested their money was in the ratio 8:6:12 respectively. If they get profit of Rs 18000 from the business. Then how much share of profit will each one get ?

Solution: ratio of Investment = 1/2:1/3:1/4=6:4:3 and time for which investment is done = 8:6:12

So in actual their amount was in business in the ratio 6*8 : 4*6 : 3*12 =48:24:36 =4:2:3

So they should also share the profit in this ratio 4:2:3

So Ram gets = 4/9 *18000 =8000

Sham gets = 2/9*18000 = 4000

Suresh gets = 3/9*18000 = 6000

{This is actually based on the concept that a small sum put in for large time and large sum put in for small time, so actually we see for how much period we can have money on same scale. You can compare by money * time }

Problem in Ratio and Proportion (Mathematics) :
p,q and r are three positive numbers and Q=(p+q+r)/2

If (Q-p):(Q-q):(Q-r) = 2:5:7, then find the ratio of p,q and r ?

Solution :

Substitute the value of Q =(p+q+r)/2 everywhere

you will get

(p+q+r-2p)/2 : (p+q+r-2q)/2 : (p+q+r -2r)/2 = 2:5:7

(p+q+r-2p) : (p+q+r-2q) : (p+q+r -2r) = 2:5:7

(q+r-p) : (p-q+r) : (p+q -r) = 2:5:7

or (q+r-p) : (p-q+r) : (p+q -r) = 2k:5k:7k

To solve this you can assume the following equations

q+r-p =2k --------( 1 )

p-q+r =5k --------( 2 )

p+q-r =7k --------( 3 )

Adding (1) & (2)

q+r-p + p-q+r = 7k

2r = 7k

r = 7k/2

Adding (2) & (3)

2p = 12k

p=6k

Adding (1) & (3)

2q = 9k

q=9k/2

So p:q:r = 6k: 9k/2 : 7k/2 = 6 : 9/2 : 7/2 =12:9:7

LOGARITHMS

Logarithm and logarithmic based functions and questions can be framed in every section of mathematics from algebra to calculus and even geometry. But the most confusing and difficulty faced is in the calculus and differentiation and continuity when the question is combined with multiple concepts in the scene. It becomes difficult to handle the question if you do not know in detail about how to move around with logarithms.


Most of you must have got confused with logarithms raised to powers and powers raised to log functions. Yes, it will become more confusing if such things appear in calculus, at that point of time you will have to also keep in mind the continuity, domain, range and base. Yes, another important thing is 'Base'.
Many of you might not even know how the base is changed. Remember if you are not used to such simple simple things, you will get stuck in the middle of any question and to your surprise even you might get stuck at the last stage of your solution because you are helpless to change base, as you dont know how to do it !!

So why not just spend a few minutes, to understand how actually you can mover around playing with indices, properties of log functions and changing bases and raising powers to numbers and even bases of log functions.


Important Points
The logarithmic function is the inverse of exponential function. N = ax → logaN = X
Two of the most important logarithmic base used are 10 and e.
Log to the base e is called Natural Log.
Log of negative number is not defined. You will never find log(-5). Where you find log(-x), it is clear that here x < 0, so that inside parameter of log function becomes positive.

Click on the question to enlarge

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To Download the solutions Click here

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

SIMPLE TEST FOR 10+2 Students

Here is a simple test on mathematics for class 10+2. These 5 questions will test how much hold you have on the concepts & how well prepared you are with mathematics !!!!! Don't think just jump in ! After all Mathematics never killed anyone !!!!

Simple Excercise

For those who are weak in mathematics it is not recommended that you jump to higher concepts question very quickly. If your base is not strong, we cannot expect you to yield results by studying 10 or 20 hours and produce miracle results. It actually is more important first to focus on your foundation and work on it. This starts from some simple simple questions of mathematics which will increase your confidence level, then moving towards the second level which is a little tricky type and more interesting problems in mathematics but not too much difficult to require core mathematics.When finally you build a more stronger base in the mathematics, then at this level it is advisable to go for higher concepts and problems in mathematics.So lets start by checking ourselves at the first two level, that do we indeed to work more before moving up or not.
Question: Try to Simplify
(5√150 + 6√24 - 2√96) : √3

Ans :

(5√150 + 6√24 - 2√96) : √3
=(5√25√2√3 + 6√8√3 - 2√16√2√3) : √3
=(25√2√3 + 12√2√3 - 8√2√3) : √3
=(25√2 + 12√2 - 8√2)√3 : √3
=(25√2 + 12√2 - 8√2) : 1
=29√2 : 1

Question :  Prove that 0/0 ≠ 1

Ans:

Find the area of the unshaded region ?:
Are of the Square = 

10²

Area of the semi circle = ½ ∏R² = ½ ∏5²

Area of unshaded region = 100 - 25∏/2

Solution





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

Equations and Inequations - I

Problem : Find the smallest value of k for which both the roots of the equation  x² - 8 k x+ 16 (k² – k + 1) = 0 will be real roots ?
Solution : For both roots to be real first of all the Discriminant must be greater than zero
Hence,   ∆ ≥ 0

64k²-4 (16(k²-k+1) ) ≥0

64k² -64k² + 4k-4 ≥0

k≥1

So the minimum value is k = 1

Problem : Find the maximum and minimum value of x such that x² - 8 x + 6 < 0

Solution : We can first find where the expression x² - 8 x + 6 = 0 holds true, thereon we can better decide which direction to go.

x² - 8 x + 6 = 0 gives us,

x = -(-8) ± √64-4*6

x = 8 ± √64-16

x = 8 ± √48

x = 8 ± 4√3

So the minimum value of x is 8 – 4√3 and the maximum value is 8 + 4√3.

In between these two points the expression will be less than zero.

+ve ‹‹‹‹ +ve ‹‹‹ (8- 4√3) ›››››› -ve ‹‹‹‹ (8 + 4√3) ›››› +ve ›››› +ve

Problem: Find the values of x satisfying the following,

x²- 5x + 6 > 0, x²- 15x + 16 < 0 and x² – 6x + 16 < 0

Solution: We shall solve each of the inequations seperately and then take the common values in each.

First equation, x²- 5x + 6 > 0

On solving you will get, x >2 & x < 3

Second equation, x²- 15x + 16 < 0

x = [15 ± √225-64 ] /2

x = [15 ± √161] /2

Thus we get,

x > [15 - √161] /2 & x <[15 + √161] /2

Third equation, x² -6x - 16 < 0

On solving we get, x > - 2 & x < 8

The confusing but not difficult part now, taking the common values. We will better do it on a number line to avoid any confusion.

Hence from above it is very clear that the common region is 2 < x < 3

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

Satellite View of India on Diwali

SATELLITE VIEW OF INDIA ON THE OCCASSION OF DIWALI.......

BEFORE DIWALI


India just 2 days Before Diwali

ON DIWALI NIGHT

Viewing India from sky on Deepawali

.......... SIMPLY AWSOME...... WONDERFUL............

Indian Bank PO Question

________________________________________________
The following questions appeared in Indian Bank PO exam. I shall solve these here to show you how easy it is. So if you are planning to get into the Bank Sector at good position it is important that you stop fearing about mathematics and approach in a more tactful & logical way. The data has been manipulated to create different questions.
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Problem : Study the following pie chart & table carefully and answer the questions that follow:

The below figure shows the % breakup of total employees working in various departments of an organisation and the ratio of men to women working in them.

Total number of Employees working = 1800

 

Ration of Men to Women 

Department	Men	Women
Production	11	1
HR	1	3
IT	5	4
Marketing	7	5
Accounts	2	7

Question : What is the number of Men working in the Marketing Department ?

1. 132
2. 174
3. 126
4. 189
5. None of these

Solution : Marketing dept holds 18% of the total 1800 employees. So total number of people in Marketing are = 0.18* 1800 = 324

In the Marketing department the ratio of men is to women is 7 : 5. This means that total number of men in the department = ( 7 / 12 ) x 324 = 189

Number of men = ( men / total men  + women  )  X   Total people in department
This can be weel understood by the fact that out of total 12 people 7 are men and 5 are women. So we use 7/12 of 324 as all the men in the department.

________



Question : What is the ratio of number of men working in Accounts to the number of Women working in IT department ?

Solution :  No of men in the Accounts department = 2 / 9  X   (17% of 1800) = 2/9 x 0.17 x 1800 = 68
No of women in the IT department = 4 / 9  X (23% of 1800) = 4/9 x 0.23 x 1800 = 184

Ratio of Men in Accounts to the Ratio of Women in IT = 68 / 184 = 17 / 46

______

Question : The number of women working in IT department as a percentage of number of women in HR department ?
Solution : Women in IT = 4 /9   X  ( 23% of 1800 ) = 4 / 9 X 0.23  X  1800 = 184

Women in HR deptt = 3/4   X  14% of 1800 = 3/4  X  0.14  X  1800 = 189

Women working in IT as percentage of Women in HR = Women in IT / Women in HR   X  100 %
= 184 / 189 X 100%  =  94.77 %



_______

Question : What is the ratio of  number of women exceeding men in accounts department to the number of men exceeding women in production ?
Solution  : Women exceeding men in accounts = Women - Men

Women in Account =  7 / 9 X 0.17  X 1800 = 238
Men in Account = 2/9 X 0.17 X 1800 = 68
Number of Women Exceeding Men = 238 - 68 =  170


Number of Men in Production = 11/12  X  28%  X  1800 = 11/12 x 0.28 x 1800 = 462
Number o Women in Production = 1 /12  X 28% X 1800 = 1/12  x  0.28  x 1800 = 42
Number of Men Exceeding Women = 462 - 42 =  420

Ratio of Women exceeding men in Acc to the number of Men exceeding Women in Production = 170 /420
= 17 : 42

Alternatively, this can also be done as :
Number of Women exceeding in Acc = ( 7 - 2 ) / 9   X   Number of people in Accounts
= 5/9   X  0.17 X 1800 = 170

Similarly number of Men Exceeding in Production = ( 11 - 1 ) / 12   X  Number of people in Production
=10/12 X  0.28  X 1800 = 420
Required ratio is thus  = 170 / 420 = 17 : 42

______________


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

Salary Day at Various Places -

Hello everybody......This post doesn't relate to mathematics or any question, but just to share the lighter side of salary day. I recieved this mail and thought why not share this too.....Not many of you might realize the importance of salary day today but sooner or later you will... when you start earning.........

SALARY DAY IN RUSSIA



SALARY DAY IN RUSSIA








SALARY DAY IN USA




UK



SALARY DAY IN AUSTRALIA

SALARY DAY IN HONG KONG

SALARY DAY IN TEXAS



SALARY DAY IN INDIA



SALARY DAY IN KERELA



SALARY DAY IN PAKISTAN

SALARY DAY IN AFGHANISTAN

                          

Geometry - Straight Lines & Triangle

Co-Ordinate Geometry

• Centroid : Medians are concurrent (G)
• Incentre: Int angle bisector are concurrent (I)
• Orthocentre : Altitudes are concurrent (H)
• Circumcentre : Perpendicular bisector are concurrent (O)
• H,G,O are collinearG divides HO in the ratio 2:1
• In an isoceles triangle all centres are collinear.

Problems:

1. Find the straight line passing through the point of intersection of straight lines x+y-5 = 0 and 4x-7y+10 = 0 and which is farthest from the point (2,2).
2. Prove that a Δ having all vertices with integer co-ordinates cannot be equilateral.??
   

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Solution: Download PDF

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

Profit Loss & Percentage - 1

A person sold a table at a gain of 15 %. Had he bought for 25% less and sold it for Rs. 80 less, then he would have made a profit of 32%. Find the Cost Price of Table ?

Let the CP be x.
So his SP is at 15% profit means , S.P. = 1.15x
If CP was 25% less, it says if CP was 0.75x, and the SP was Rs 80 less, means if the SP was 1.15x-80 , then the Profit % would have been 32%
Profit% = ( SP - CP ) /  CP = (1.15x-80 - 0.75x) / 0.75x = 32 /100
=>  0.4x - 80 = 0.32 * 0.75x
=>  0.4x - 80 =  0.24x
or 0.4x - 0.24x = 80
      0.16x = 80
Thus we get x= 80 / 0.16 =  500

Hence the Cost Price of the table is Rs 500

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.

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

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)

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 FUNCTIONS
   


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

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

Count on Your Time, your @ssets

Ttime is the most important asset of a student. But this asset does not grow. It only slips out of hand. The wise use of time is very crucial today as there are more candidates to any opportunity than the opportunities itself. The time today most of the students spend on is Facebook. Instead of serving the purpose of networking between people for some or any purpose it has become a network of purposeless activities to kill time and to keep busy for no good reason. The main motive behind such social networking sites my dear friends is not to keep login for whole day and publish your new and indifferent status and feeds, rather it should form a forum for you to serve for a purpose, a means to keep in touch and not keep login. Even the founder of Facebook might not be using his idea too that extent to keep all the day hanging on it. If was it true, he could never have been so capable and developed to develop such a mastermind. The question is not quitting, infact the answer is wisely using.
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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