how do u frame the cubes ?______| | ------| 8 | __ | 6 || || 2|| |------------- ------ for max surface area._________| |_2|_| 8 || 6|| || |--------------- for minimum.i got 608 - 536 = 72..whatt m missing?PS : bear with the cumbersome artistry
yea, my answer is wrong.. interpreted the question wrong. let me edit it. :)
x= 1+ หลก2+ หลก3, then the value of 2x^4 - 8x^3 - 5x^2 + 26x - 28?Copied from CGL 2013 Thread!
x-1=rt2 +rt3 square it x^2-2x=4+2rt6..............eq1 again square it x^4 + 4x^2 -4x^3=40+16rt6 Multiply with 2 on both sides 2x^4 + 8x^2 -8x^3=80+32rt6..........eq 2
Multiply eq 1 by 13 then subtract it from eq2, we get 2x^4 - 8x^3 - 5x^2 + 26x=28 + 6rt6 =>2x^4 - 8x^3 - 5x^2 + 26x - 28= 6rt6
how do u frame the cubes ?______| | ------| 8 | __ | 6 || || 2|| |------------- ------ for max surface area._________| |_2|_| 8 || 6|| || |--------------- for minimum.i got 608 - 536 = 72..whatt m missing?PS : bear with the cumbersome artistry
Don't need to calculate full area na...
What I did was, the total area of the cubes is some x. Now let's calculate the reduction from x:
If the three cubes all touch each other, there will be 4 surfaces of 2x2 and 2 of 6x6 eliminated
If the 2-waala cube is in the middle and the other two on either side then just 4 surfaces of 2x2 eliminated.
So difference is 2 surfaces of 6x6 => 72. regards scrabbler
Don't need to calculate full area na...What I did was, the total area of the cubes is some x. Now let's calculate the reduction from x:If the three cubes all touch each other, there will be 4 surfaces of 2x2 and 2 of 6x6 eliminatedIf the 2-waala cube is in the middle and the other two on either side then just 4 surfaces of 2x2 eliminated.So difference is 2 surfaces of 6x6 => 72.regardsscrabbler
Don't need to calculate full area na...What I did was, the total area of the cubes is some x. Now let's calculate the reduction from x:If the three cubes all touch each other, there will be 4 surfaces of 2x2 and 2 of 6x6 eliminatedIf the 2-waala cube is in the middle and the other two on either side then just 4 surfaces of 2x2 eliminated.So difference is 2 surfaces of 6x6 => 72.regardsscrabbler
you are ultimate dude!! #respect maine bhi calculate kia, like @rnishant231did. ๐
let the diameter be 10a+blet oh be xso(10a+b/2 +x)(10a+b/2 -x) = (10b+a/2)^2x = root (11*9(a-b)(a+b))/2as x is a rational number of the form p/q a+b=11a-b=1a=6 b=5diameter 10*6+5=65
Haan that's why I wanted options ๐ Found an approx upper limit and wanted to eliminate. Lazy way out. OK trying for exact answer then :( Edit: @abhishek.2011has already posted an answer so...stopped trying ๐ #sleepy
John has a bag of colored billiard balls which are labeled with the numbers 1,2,3,4,5,6. He reaches into the bag, draws out a ball, records down the number and then places the ball back in the bag. He does this a total of 4 times (hence has 4 numbers). The probability that at least 2 of the numbers are equal is a/b, where a and b are positive, coprime integers. What is the value of a+b?
My approach : solve it using options ..trying option a ie. 60 ..so take 121 ..we need a weight of 61 kg ..this violates the condition as 60 kg is supposed to be the heaviest ..so A option is rejected ..trying option B.. assume heaviest weight is 62 kg.. to make 118 ..other weight has to be 59 kg... similarly to make 120(62+58),,, so our third weight is 58kg... similarly solving we get fourth and fifth weights as 56 and 55 kg respectively ..Hence our weight s are 62,59,58,56 and 55 kg .. This option B satisfies..Hence B is the answer.Team BV--Pratik Gauri
sir aapke sol me ek error hai... 56 +55 =111 which is not there. it shud be 54 56 58 59 62........ yes i have found a bug in ur sol..... i can sleep well now...
given. But Oa is a notch lower than yours..
