@rnishant231 said:isn't it fine if we consider a cross section of 3D figure and solve it...like in ths case , sabse middle wala cross section with max area ?
@scrabbler said:Yes, but which cross section? juining the midpoints of the sides of the square? Or the vertices? If you take the midpoints, the vertices will protrude outside the cone.regardsscrabbler
@bodhi_vriksha said:use similar triangles ..10-x/x=10/8 .. from this we get upper triangle's height Team BV--Pratik Gauri
@mailtoankit said:sir, please explain, how can side of cube be diameter of upper cone? i think diagonal of the sides of cube will be the diameter of the upper cone...correct me if i m wrong...
@bodhi_vriksha said:The idea of my this post is to walk you through an observation and pattern.Suppose x is the sum of digits of a 3 digit number and N(x) denotes the number of ways of getting x as the sum. Then, N(x) =N(28-x), where x -> [1,27] For this reason maximum possible sum corresponds to N(14)= 45 +9 + 7 +5 +3 +1If the three digit number is abc, then for x = 9 to 19 we have 9 values of a. However, for remaining values of x, there are x or 28 窶度 (which ever is single digit) values of a.So the total sum possibilities of any 3 digit number can be generated as follows:For x= 1 to 9 and 19 to 27 -> Sum = Summation (Summation(n))X=10 ->Summation(9) + 9 = 45 +9X=11 -> 45 +9 + 7X=12 -> 45 +9 + 7 +5 =66 = N(16) {This is the same value that we got in our previous problem on N(16). Remember?}X=13 -> 45 +9 + 7 +5 +3X=14 -> 45 +9 + 7 +5 +3 +1-> Maximum SumX=15 -> 45 +9 + 7 +5 +3 +1 -1 = 45+9+7+5+3 = N(13)X=16 -> 45 +9 + 7 +5 +3 +1 -1 -3=45+9+7+5 = N(12)X=17 -> 45 +9 + 7 +5 +3 +1 -1 -3-5= 45+9+7= N(11)X=18-> 45 +9 + 7 +5 +3 +1 -1 -3-5-7= 45+9 = N(10)X=19-> 45 +9 + 7 +5 +3 +1 -1 -3-5-7-9= 45= N(9)For the purposes of the problem at hand, Summation(Summation(16)) = Summation (Summation (9)) + 45*7 + 9*7 +7*6+5*5+3*4+1*3-1*2-3*1=>Summation(Summation(16))= 165 +420 +40-5=620Team BV - Vineet
@rnishant231 said:as we have a 3D fig with cone , cube and sphere(all 3d objects) , if we assume cutting the objects along the vertices of the cone in vertical direction , we get a 2d fig with corresponding triangle , square and circle..hope m able to make sense
Draw the figure and see, I am not in a position to do so right now...sorry...@scrabbler said:Bhai cone ka ek hi vertex hota hai Draw the figure and see, I am not in a position to do so right now...sorry...regardsscrabbler
i already took my words back 
in my prev post..Q 8
A motorist used 10% of his fuel to cover 20% of his journey. he covered another 40% of his total journey under similar condition . for the rest of journey the conditions were different. Find the maximum % by which his fuel efficiency (distance covered per unit of fuel) can drop, so that he can still cover the remaining journey without a refill.
@vbhvgupta said:Q 8
@pratskool said:in IV it is 2000, in III it is 1600, for II it has to be 1280
@vbhvgupta said:did u solved it using options.
@vbhvgupta said:did u solved it using options.
@vbhvgupta said:A motorist used 10% of his fuel to cover 20% of his journey. he covered another 40% of his total journey under similar condition . for the rest of journey the conditions were different. Find the maximum % by which his fuel efficiency (distance covered per unit of fuel) can drop, so that he can still cover the remaining journey without a refill.
at the begining of 2002 sunil had 4 dozen goat with him. he increased his stock by X%. at the end of year he sold by Y%. at the begining of 2003 he again increased his stock by X%. at the end of year he sold by Y%. At the end of 2003 he had 5 dozen goats with him after his sales.
Q28 Q29