@bada.bhai said:but book also says 2:3 is correct
both are right . they havent said ratio of wat to wat . its just ratio of 'these metals' .
@kingsleyx said:Here's an easy one . : You walk upwards on an escalator, with a speed of 1 step per second. After 50 steps you are at the end. You turn around and run downwards with a speed of 5 steps per second. After 125 steps you are back at the beginning of the escalator.
and then i go home ? 
Here's an easy one . : You walk upwards on an escalator, with a speed of 1 step per second. After 50 steps you are at the end. You turn around and run downwards with a speed of 5 steps per second. After 125 steps you are back at the beginning of the escalator.How many steps do you need if the escalator stands still ??
@rubikmath Thanx !! :)
@bada.bhai said:but book also says 2:3 is correct
2 mahine regular yahan raho book mei se answer dekhne ki jarurat nahi padegi 
LCM of 2 nos(say a and b) is 300. How many sets of A and B are possible..??
LCM of 2 nos(say a and b) is 300. How many sets of A and B are possible..??
@kingsleyx said:Here's an easy one . : You walk upwards on an escalator, with a speed of 1 step per second. After 50 steps you are at the end. You turn around and run downwards with a speed of 5 steps per second. After 125 steps you are back at the beginning of the escalator.
50 + 50x = 125 - 25x
=> x = 1 = speed of escalator
number of steps = 100
=> x = 1 = speed of escalator
number of steps = 100
@kingsleyx said:Here's an easy one . : You walk upwards on an escalator, with a speed of 1 step per second. After 50 steps you are at the end. You turn around and run downwards with a speed of 5 steps per second. After 125 steps you are back at the beginning of the escalator.How many steps do you need if the escalator stands still ??@rubikmath Thanx !!
100?
@jain4444 said:2 mahine regular yahan raho book mei se answer dekhne ki jarurat nahi padegi LCM of 2 nos(say a and b) is 300. How many sets of A and B are possible..??
38 !! ?? There is a formula for this.!!
@chillfactor
r=2,c=3
hence number of possible paths = (3+2)!/(3!2!) + {(3-1)+(2-1)+1}!/(2!1!1!) + {(3-2)+(2-2)+2}/(1!0!2!)
10 + 12 + 3 = 25
Generalized formula
if there are r rows and c columns and diagonal movement is allowed. then total paths using which we can move from one end of rectangle to the other diagonal end can be given by:
Sum[ (r+c-d)!/ {(r-d)!(c-d)!d!} ] where 0
I had derived this sometime back when I was presented with general version of this problem. Will post the detailed explanation if needed.
ATDH.
@scrabbler ...hello! its been long time.
@jain4444 said:2 mahine regular yahan raho book mei se answer dekhne ki jarurat nahi padegi LCM of 2 nos(say a and b) is 300. How many sets of A and B are possible..??
300 = 2^2 * 3 * 5^2
A and B are of the form 2^x * 3^y * 5^z
Power of 2:
x for each of A or B can be 0,1,2, but atleast one should be 2 -> 3^2 - 2^2 ways = 5
Same way, for Power of 3 -> 2^2 - 1^2 = 3 ways
Power of 5 = 5 ways
So total pairs = 5*5*3 = 75
Note: If we are looking for unordered sets, then barring the (300,300) pair, all others would have a symmetric pair. So total = (75 - 1)/2 + 1 = 38 unordered sets.
@kingsleyx said:38 !! ?? There is a formula for this.!!
@Torque024 said:what's the formula, I have done like this 300=2^2*3*25 A 2^2 *ways A 2^2*ways B 2*ways or B 2^0*ways For ways we have 2 options of either 3 or 25 total will be 2*2+2*2 = 8 Arrangement possible= 8*2=16 A,B can be 300,1 ; two ways Total=18; what am I doing wrong? @@jain4444
I guess everyone has slept before IIFT. I'm the only one drinking
@Ibanez said:I guess everyone has slept before IIFT. I'm the only one drinking
:nono:BTW all the Best 😛 :D
If log x/(b-c)= log y/(c-a)= log z/(a-b),
mark all the correct options
a. xyz = 1
b. x^a*y^b*z^c = 1
c. x^(b+c) *y^(c+a)* z^(a+b) = 1
d. x^(b + c)*y(c+a)*z^(a+b) = 0
@pratyush9811 said:@Ibanez SIRji aap to waise hi phodne waale ho
@vijay_chandola said:BTW all the Best
All the best guys! Very bored right now and not sleepy 
Ok ill post a ques then,
Consider points A(1,-3) and B(5,2). Let P be a point on y=x. Find co-ordinates of P for which |AP+PB| is minimum
1. (10/9,10/9)
2. (11/7,11/7)
3. (11/9,11/9)
4. (13/7,13/7)
@vijay_chandola said:BTW all the Best If log x/(b-c)= log y/(c-a)= log z/(a-b),mark all the correct optionsa. xyz = 1 b. x^a*y^b*z^c = 1 c. x^(b+c) *y^(c+a)* z^(a+b) = 1 d. x^(b + c)*y(c+a)*z^(a+b) = 0
a,b?
@Ibanez said:Ok ill post a ques then,Consider points A(1,-3) and B(5,2). Let P be a point on y=x. Find co-ordinates of P for which |AP+PB| is minimum1. (10/9,10/9)2. (11/7,11/7)3. (11/9,11/9)4. (13/7,13/7)
13/7 ?? 
