Math@Home: Lift – number of floors travelled

This is just for fun.....

One day I stood at the lift in 5^th floor. The lift-indicator showed that the lift is at 10^th floor. I wanted to go to ground floor and so I pressed the ‘down-arrow’.

Of course I always have a doubt. Which button I need to press? Especially the doubt arises when the lift is in the down floors and I need to go in the down direction. I think in this way: shall I press up-arrow such that the lift comes up to me and takes me or shall I press down-arrow as I need to go in the down direction. The whole dilemma is about which thing I need to pursue first? Pulling the lift ‘up’ to me? or the ultimate direction of my going ‘down’? The dilemma comes again when I need to go in the up-direction and the lift also stays at an upper floor.

Come back to our discussion. Luckily this time I pressed the right button. Lift arrived at 5^th floor and stopped. After I got inside, it started and took me to the ground floor.

Then I asked myself, total how many floors the lift has travelled?

My thought process went like this: First, the lift travelled from 10^th to 5^th floor. ie., 10-5 = 5 floors. Then after taking me in, it travelled from 5^th to ground, ie., 5-0 = 5 floors. So, it travelled 5+5 = 10 floors.

When I told this story to my wife, she laughed at me and said “why do you calculate in a round-about-way? lift travelled from 10^th floor to ground floor. So it travelled 10-0 = 10 floors”

What has influenced me to go in a so called round-about-way?

Probably the word ‘total’. In order to find the ‘total’ number of floors, I might have brought two numbers in to picture and ‘totalled’ them.

Is this the influence of so called ‘mathematical approaches’ learnt for years?

My point here is: Applying a little commonsense and logic to math funda really blossoms the beauty of math.

Numbers in Coalition Governments: How Math works?

‘Mamata withdraws support to UPA government’ this is the front page column in today’s news paper ‘The Hindu’. The part of the news attracted me more is a table named ‘How the arithmetic works’ right at the centre of the article.

UPA – Trinamool	246
UPA – Trinamool + SP + BSP + RJD	293
UPA – Trinamool - SP + BSP + RJD	271
UPA – Trinamool - SP - BSP + RJD	251*
UPA – Trinamool + SP + BSP + RJD +JD(S)	296
UPA – Trinamool - SP - BSP + RJD +JD(S)	253

 *original value given in the article is 250. It is changed for the ease of calculations

Here I do not want to contemplate on the fate of UPA in case Mamata Deedi withdraws support or Lalluji & Malluji extend it or what so ever. What I see is this: six relationships are given between member-strengths of six parties. Can’t we find the member-strengths of each party? Let us give a try..

First let us write down these in the form of equations:

U-T = 246                                --(1)

U-T+S+B+R = 293                    --(2)

U-T-S+B+R = 271                     --(3)

U-T-S-B+R = 251                      --(4)

U-T+S+B+R+J = 296                 --(5)

U-T-S-B+R+J = 253                  --(6)

Now let us solve:

(2)-(1) gives => S+B+R = 47    --(7)

(3)-(1) gives => -S+B+R = 25   --(8)

(4)-(1) gives => -S-B+R = 5      --(9)

(7)-(8) gives => 2S = 22 => S = 11      

(8)-(9) gives => 2B = 20 => B = 10     

(7)+(9) gives => 2R = 52 => R = 26    

(5) - (2) gives => J = 3

By now, we have got strengths of four parties SP, BSP, RJD and JD(S). But can we do it for UPA and Trinamool?

We can’t. With the given set of data it is not possible. Even though six equations in six unknowns are given, we cannot solve for all the six variables. Why so?

Here U and T are represented together as (U-T) in all the equations. It is represented as a single entity and the value of that entity (U-T) is only known. Even though we are provided with n number of such equations we can’t find the individual value of either U or T. And one more thing, only one of the equations 5 or 6 is enough to find the value of J and the other equation is useless (‘redundant’ in mathematical sense). So effectively there are less than six equations.

(To prove this mathematically, math-savvies can try out one of the standard methods: Cramer’s or Matrix or Rank method. Here Rank method is suitable as in the other methods it involves dealing with determinant of 6X6 matrix, which is difficult)
Click on the compressed sheet down here for the mathematical proof by Rank method:

Prerequisites to crack IIT-JEE Math section

The following topics mostly cover the mathematics section of the IIT-JEE:

Complex Numbers, Matrices & Determinants, , Inverse Trigonometric Functions, Trigonometric Equations, Quadratic Equations, Differential Equations, Circles, Probability, Statistics, P&C, Progressions, Functions.

In my opinion, one should master the following concepts for cracking IIT-JEE mathematics paper. These funda are important in a way that these are tested in different forms and are used in different topics.

Inverse trigonometric functions: Generally we neglect the ranges of sin^-1, cos^-1 etc.,

Range of sin^-1 => [-π/2, π/2]

Range of cos^-1 => [0, π]

Range of tan^-1 => (-π/2, π/2)

Range of cot^-1 =>(0, π)

Formulae for expansions of fractional powers. ex: (1+x)^1/2

Equations of circle, ellipse, parabola, hyperbola etc., in complex plane

Mixing of concepts (for ex: complex numbers with conic sections, trigonometry with conic sections)

How these graphs look like, just an idea: k^x, f(x)^k, |x-k| etc.,

Complex number theory approach is a powerful tool. This can be used to resolve geometry problems on locus, trigonometric problems etc.

Ex: Try out complex number approach for the following one:

Q) Find the locus of centre of circle which touches two circles externally?

Answer) Hyperbola

"AM ≥ GM" and "AM ≥ HM" conditions are very much useful in solving Maxima-minima problems in 2 or 3 variables. Try out some problems in this angle.

Grip on special functions such as [x], {x} and their properties.

Expansions of special functions like e^x, log(1+x), log(1-x) etc.

                         

General equation in ‘x’- relationship between the coefficients and roots:

xⁿ+p₁x^n-1+ p₂x^n-2+ p₃x^n-3+..... + p_n = 0

If α₁, α₂,.... α_n are the roots of this equation, then the following relations hold good:

S₁ = sum of roots taken one at a time = α₁+ α₂+.... +α_n = -p₁

S₂ = sum of roots taken two at a time = α₁α₂+ α₁α₃+..... = p₂

S₃ = sum of roots taken three at a time = α₁α₂α₃+ α₁α₂α₄+..... = -p₃

This logic extends up to n iterations.

Important Algebra concept frequently required:

a³+b³+c³-3abc = 0

=> (a+b+c)(a²+b²+c²-ab-bc-ca) = 0

=> a+b+c = 0 (or) a²+b²+c²-ab-bc-ca = 0

a²+b²+c²-ab-bc-ca = 0 => 2(a²+b²+c²-ab-bc-ca) = 0

=> (a-b)²+(b-c)²+(c-a)² = 0 => a-b = b-c = c-a = 0 => a=b=c

If a³+b³+c³-3abc = 0, then the possibilities are

(1)    a+b+c = 0 (2) a²+b²+c²= ab+bc+ca (3) a=b=c

Important Complex numbers concept frequently required:

e^ix+e^iy+e^iz = 0

a = e^ix ;b= e^iy;c=e^iz

a+b+c = 0 => cosx+cosy+cosz+i(sinx+siny+sinz) = 0

=> cosx+cosy+cosz = sinx+siny+sinz = 0

=> cosx+cosy+cosz-i(sinx+siny+sinz) = 0

=> (cosx-isinx)+ (cosy-isiny)+ (cosz-isinz) = 0

=> 1/a + 1/b + 1/c = 0 => ab+bc+ca = 0

=> cos(x+y)+cos(y+z)+cos(x+z) = cos(x+y)+cos(y+z)+cos(x+z)

a+b+c = 0 => (a+b+c)² = 0

=> a²+b²+c²+2(ab+bc+ca)=0

=> a²+b²+c²=0 => cos2x+cos2y+cos2z = sin2x+sin2y+sin2z = 0

a+b+c = 0 => a³+b³+c³ = 3abc

=> cos3x+cos3y+cos3z = 3cos(x+y+z)

sin3x+sin3y+sin3z = 3sin(x+y+z)

A useful wikipedia link for inverse trigonometry formulae, where the formulae are given in tables: Inversetrig formulae

The story does not end here and will be updated from time to time.... 

***A Land Mark - Page views reached Fifth Digit***

"MathByVemuri" blog crossed a land mark. Its page views count crossed 9999 mark on 12thSep (of-course it excludes my own page views). I felt this occasion as an important land mark and am glad to share with you all. 

This blog on math was kicked-off on 4th Oct'2011 and the journey so far is thrilling. Thanks to all the followers for the wonderful support all the way. Thanks a lot to the people supported me and encouraged me to continue the journey. Thanks for those who gave valuable suggestions in promoting the blog. 

You can reach me at mathbyvemuri@gmail.com for all your inputs.

The following are some statistics about the blog and I am glad to share this info:

Each parameter is considered up to top ten counts only.

Page views by Countries:

Page views by Posts:

Page views by Browsers:

Page views by Referring Sites: