@hiteshkhurana82 said:How many distinct right triangles with integer side lengths are there one of whose sides is of length 60?
4 .. ?
@sahil.oberoi
Let T be the total fixed garbage an let a be the garbage added every day..
Thus, 15*100 = 15a + T..(i)
And, 60*30 = 30a + T..(ii)
Solving, (i) and (ii), we have T=1200 and a=20
Thus, 10*x= 200 + 1200, or, 10x=200 + 1200, or, 10x=1400, or, x=140 Million??
@pyashraj said:@sahil.oberoiLet T be the total fixed garbage an let a be the garbage added every day..Thus, 15*100 = 15a + T..(i)And, 60*30 = 30a + T..(ii)Solving, (i) and (ii), we have T=1200 and a=200Thus, 10*x= 200 + 1200, or, 10x=200 + 1200, or, 10x=1400, or, x=140 Million??
Sahi hai...I committed a blunder still managed to match a option!!!!!
@krum said:a - rate of decompositionb - initial quantity of dumpc - garbage added per day100*a*15=b+15c60*a*30=b+30c300a=15c==> c=20a,b=1200asay x bacteria neededx*a*10=b+10c==> x=(1200+200)/10=140
Sahi hai...I committed a blunder still managed to match a option!!!!!
@YouMadFellow said:Nope, OA = (2^(n/2)) * cos(n*pi/4)
Yeah, but even cos can be represented in terms of log na??
100 million bacteria can completely decompose a garbage dump in 15 days and 60 million can do so in 30 days. If same quantity of garbage is added to the initial quantity of the dump everyday , how many bacteria will be required to completely decompose the dump in 10 days?
cc: @sahil.oberoi
Let us say 1 Million bacteria can clear 1 unit of garbage in 1 day.
Initially there were 'x' units of garbage and everyday 'y' units of garbage gets added.
Case 1: 100 M bacteria can decompose in 15 days.
Total garbage present = Total garbage decomposed
=> x + 15y = 100*15 ...(1)
Case 2: 60 M bacteria can decompose in 30 days.
=> x + 30y = 100*18 ...(2)
Subtract equ(1) from equ(2) to get
15y = 300
=> y = 20 {This means that 20 units of garbage gets added every day}
=> x = 1200 {We got this by putting the value of 'y' in equ(1) or equ(2). This means that there was 1200 units of garbage present initially}
Case 3: We have 10 days to decompose the bacteria
Total garbage = 1200 + 20*10 = 1400 units
Bacteria required = 1400 / 10 = 140 Million
@YouMadFellow said:Nope, OA = (2^(n/2)) * cos(n*pi/4)
(1+i)^(n-1) --- x + iy??
Which can be represented in terms of cos/sin or log?? Am I right or wrong?
@ravihanda said:
@ravihanda said:100 million bacteria can completely decompose a garbage dump in 15 days and 60 million can do so in 30 days. If same quantity of garbage is added to the initial quantity of the dump everyday , how many bacteria will be required to completely decompose the dump in 10 days?cc: @sahil.oberoiLet us say 1 Million bacteria can clear 1 unit of garbage in 1 day.Initially there were 'x' units of garbage and everyday 'y' units of garbage gets added.Case 1: 100 M bacteria can decompose in 15 days.Total garbage present = Total garbage decomposed=> x + 15y = 100*15 ...(1)Case 2: 60 M bacteria can decompose in 30 days.=> x + 30y = 100*18 ...(2)Subtract equ(1) from equ(2) to get 15y = 300=> y = 20 {This means that 20 units of garbage gets added every day}=> x = 1200 {We got this by putting the value of 'y' in equ(1) or equ(2). This means that there was 1200 units of garbage present initially}Case 3: We have 10 days to decompose the bacteriaTotal garbage = 1200 + 20*10 = 1400 unitsBacteria required = 1400 / 10 = 140 Million
Thanks sir for a detailed solution...I just want your advice regarding some issues...I am working so i need some tips and proper methodology for going about things....
Can I have your contact....I have read your articles....
Hopefully you will help me out with what to do and more importantly what not to do
Can I have your contact....I have read your articles....
Hopefully you will help me out with what to do and more importantly what not to do
@ScareCrow28 said:(1+i)^(n-1) ---Circle banega?? Which can be represented in terms of cos/sin or log?? Am I right or wrong?
Well, why are you going for log series !? Log series would anyway be just an approximation.
(1/2)*[ (1 + i)^n + (1 - i)^n ] => this can be found out using (1 + x)^n
Now, (1 + i)^n = [root(2)]^n * ( cos(n*pi/4) + i* sin(n*pi/4) )
using the same for (1-i)^n and adding, we get the final expression
@sahil.oberoi Just keep working hard. There isn't much to it. My mail id is ravihanda[at]gmail[dot]com
@hiteshkhurana82 said:approach?
Getting much more than 4 
Messed up the equtions
Messed up the equtionsSee, a^2 + 60^2 = b^2 (One possibility)
So, (b-a)*(a+b) = 60^2 = 2^4 * 3^2 * 5^2
Here, both (b+a) and (b-a) should be even or both should be odd
But, both cannot be odd as product is even.
So, Take out 2^2 out of product
(b-a)(b+a) = 2^2*3^2*5^2
Total 27 ways
But, (b-a)
Total 13 ways
Other equation : 60^2 = a^2 + b^2 ..
Some possibilities :(
@YouMadFellow said:Well, why are you going for log series !? Log series would anyway be just an approximation.(1/2)*[ (1 + i)^n + (1 - i)^n ] => this can be found out using (1 + x)^nNow, (1 + i)^n = [root(2)]^n * ( cos(n*pi/4) + i* sin(n*pi/4) ) using the same for (1-i)^n and adding, we get the final expression
This is what I got 😛 But thought log series would look better than cos or sin! Anyways, At least I got the method right
@ScareCrow28 said:Getting much more than 4 Messed up the equtionsSee, a^2 + 60^2 = b^2 (One possibility)So, (b-a)*(a+b) = 60^2 = 2^4 * 3^2 * 5^2 Here, both (b+a) and (b-a) should be even or both should be oddBut, both cannot be odd as product is even. So, Take out 2^2 out of product(b-a)(b+a) = 2^2*3^2*5^2Total 27 waysBut, (b-a) Total 13 waysOther equation : 60^2 = a^2 + b^2 ..Some possibilities
60^2=48^2+36^2
The lower corner of a page in a book is folded over so as to reach the inner edge of the page.What is the fraction of the width folded over when the area of the folded part is minimum.
1)1/3
2)2/3
3)3/4
4)None of these
1)1/3
2)2/3
3)3/4
4)None of these
On a straight line of length 10 cm, two points A, B are selected at random. What is the probability that AB >4