[Wonder Boy]

Whenever you encounter such a problem try to simplify it as far as possible and then accumulate the terms which can cancel out or give some known results

Lets try to simplify the expression  (x²-x+1)/(x-1).

So, (x²-x+1)/(x-1) = (x²-2x+1 +x )/(x-1) = [(x-1)² + x]/(x-1) = (x-1) + x/(x-1) = (x-1) + (x-1+1)/(x-1)

= (x-1)  + 1 + 1/(x-1)

Now we know that -2 <= y +1/y <= 2

–> -2 <= (x-1)  + 1/(x-1) <= 2

–> -1 <= (x-1) + 1 + 1/(x-1) <= 3

Hence the value of expression cant lie between (infinite, -1) U (3, infinite)

[Wonder Boy]

Let the roots of quadratic equation be a and 2a .

so sum of roots = -I/(I-m) = 3a

---

(1)

also product of roots = 1/(I-m) = 2a² —-(2)

—> -I(2a²) = 3a –>I= -3/2a

also I-m = 1/2a²

Since both roots are real. Hence D= B²– 4AC >=0

–> I² – 4(I-m)(1) >= 0

–> (-3/2a)² – 4 (1/2a²) >=0

9/4a² -2a² >=0

a² /4 >= 0 –> a²>=0

 

[Wonder Boy]

To solve  this  type of problem, where reciprocals are involved we will use the concept of partial fraction.

so here we can see each term is of the form

¹/[_{n.(n+1).(n+2)]}

Let ¹/_{[n.(n+1).(n+2)]}  =  ^A/_n + ^B/_(n+1) + ^C/_(n+2)

–> ¹/_{[n.(n+1).(n+2)]} = ^{[ A(n+1)(n+2) + B(n)(n+2) +C(n)(n+1)]}/_{[n.(n+1).(n+2)]} 

–> 1 = [ A(n+1)(n+2) + B(n)(n+2) +C(n)(n+1)]

putting n=-1, we get

B = -1

putting n = -2 , we get

C = 1/2

putting n= 0, we get

A= 1/2

So , ¹/_{[n.(n+1).(n+2)]} = ¹/_2n –¹/_(n+1) + ¹/_2(n+2)

1/(1*2*3) +1/(3*4*5) + 1/(5*6*7) + …….. = 1/2.1-1/2+1/2.3   +  1/2.3-1/4+1/2.5  +  1/2.5-1/6+1/2.7 +……

= 1/2(1+1/3 + 1/5+1/7 + …..) – (1/2 +1/4 +1/6 +…… ) + 1/2(1/3 +1/5 +1/7 +…..)

= 1/2(1+1/3 + 1/5+1/7 + …..) – (1/2 +1/4 +1/6 +…… ) -1/2

= (1-1/2+1/3-1/4+ ….)  -1/2

= 0.69314 -0.5 = 0.19314

 

 

 

[Wonder Boy]

This problem can be easily solved by using binomial theorem.

According to binomial theorem one can expand any power of (x+y) into the sum of the form.

In short form we can write it as

 

so here we will try to expand the given numbers.

9 = 1 + 8

–> 9¹²³ = (1+8)¹²³

= ¹²³C₀ 1¹²³ 8⁰ + ¹²³C₁ 1¹²² 8¹ +………. +¹²³C₁₂₃ 1⁰ 8¹²³

Since leaving first two terms other terms contains power of 8² = 64 so all the other terms are divisible by 64.

Now we have to check the divisibility of only 1st 2 terms i.e. ¹²³C₀ 1¹²³ 8⁰ + ¹²³C₁ 1¹²² 8¹

Again 

7 = -1 + 8

–> 7¹²³ = (-1+8)¹²³

= ¹²³C₀ (-1)¹²³ 8⁰ + ¹²³C₁ (-1)¹²² 8¹ +………. +¹²³C₁₂₃ (-1)⁰ 8¹²³

Since leaving first two terms other terms contains power of 8² = 64 so all the other terms are divisible by 64.

Now we have to check the divisibility of only 1st 2 terms i.e. ¹²³C₀ (-1)¹²³ 8⁰ + ¹²³C₁ (-1)¹²² 8¹

 

So we have

7¹²³ + 9 ¹²³ =  ¹²³C₀ 1¹²³ 8⁰ + ¹²³C₁ 1¹²² 8¹ + 8²x some number  + ¹²³C₀ (-1)¹²³ 8⁰ + ¹²³C₁ (-1)¹²² 8¹ + 8²x some number

= ¹²³C₀ 1¹²³ 8⁰ + ¹²³C₁ 1¹²² 8¹ +¹²³C₀ (-1)¹²³ 8⁰ + ¹²³C₁ (-1)¹²² 8¹ + 8²x some number

= 1 + 123 x 8 + (-1) + 123 x 8  + 8²x some number  = 2x123x8 + 1 + (-1) + 8²x some number

= 16 x 123 +  8²x some number

 

Hence remainder of 7¹²³ + 9 ¹²³ is divided by 64 = remainder of 16×123 /64 = 16x[remainder of 123/4] = 16 x3 = 48 .

 

 

[Wonder Boy]

[sol1] = 50%

[sol2] = 80%

now formula for mixture problem is new conc c3 = [c1 . n1 + c2. n2]/[n1+n2]

where c1 & c2 are concentrations of the 2 solutions and n1 & n2 are the amounts(in lit.) of the solutions with conc 50% & 80% respectively. They are mixed to give sol. of conc. c3

 

so 0.62 = [.50xn1 + .80 x n2] /[n1 + n2]

–> 0.62 n1 + 0.62 n2 = .5 n1 + .8 n2

–> .12 n1 = .18 n2 –> 2 n1 = 3 n2 –> n1/n2 = 3/2 ………(i)

Also let n3=(n1 + n2) litres of 62 % acid sol. is mixed with 6 lit of water (pure water is used so concentration of it is 0% as it doesnt contain any sulphuric acid).

 

using the forumla again  .50 = [.62 x n3 + 0 x 6] /[n3 + 6]

.5 n3 + 3 = .62 n3

–> .12 n3 = 3 –> n3 = 300/12 –> n3 = 25 lit.

–> n1 + n2 = 25

 

putting n1= (3/2).n2  from (i)

(3/2).n2  + n2 = 25

–> 5/2 . n2 = 25 –> n2 = 10 lit.

So 10 lit. of 80% solution has been used in  entire process.

[Wonder Boy]

How to be at Top.

Bullshit might get you to the top, but it won’t keep you there.

[Wonder Boy]

Get to know what it is getting More than 100%…

We have all been to those meetings where someone wants “more than 100%.”

Well here’s how you do that. Here’s how you can achieve 103%. First of all,here’s a little math that might prove helpful in
the future. How does one achieve 100% in LIFE? Begin by noting the following.

IF :

A = 1, B = 2, C = 3, D = 4, E = 5, F = 6, G = 7, H = 8, I = 9, J = 10, K = 11, L = 12, M = 13, N = 14, O = 15, P = 16, Q = 17, R = 18, S = 19, T = 20, U = 21, V = 22, W = 23, X = 24, Y = 25, Z = 26

Then:

H A R D W O R K = 8+1+18+4+23+15+18+11 = Only 98%

Similarly,

K N O W L E D G E = 11+14+15+23+12+5+4+7+5 = Only 96%

But interesting (and as you’d expect),

A T T I T U D E = 1+20+20+9+20+21+4+5 = 100%…..

This is how you achieve 100% in LIFE.

But EVEN MORE IMPORTANT TO NOTE (or REALIZE), is

B U L L S H I T = 2+21+12+12+19+8+9+20 = 103%

So now you know what all those high-priced consultants, upper management,

and motivational speakers really mean when they want to exceed 100%!

Here’s a word on how communication travels across the management….

[Wonder Boy]

Mnemonic : Abase > to bring someone down back to A BASE level.

[Wonder Boy]

Mnemonic : Ab for Abraham (john abraham)and Ash for Aishwarya roy .when they come close, Abhishek get abashed.

[Author]

Posts