3 Quant Questions

Q1. Find the shortest path that a spider, at point A, the midpoint of an edge of a unit cube, walks on the faces of the cube to catch a fly at point B, a vertex. Q2. Given the sets of consecutive integers { 1 } , {2, 3}, {4, 5, 6}, . . , where each set contains one more element than the preceding one, and where the first element of each set is one more than the last element of the preceding set. Let Sn be the sum of the elements in the nth set. For example, S1 = 1 and S2 = 5. Then S21 equals Q3.Three spheres of radius one are placed on a table with each touching the other two, one additional sphere of the same size is stacked on top, as sketched on the left. What is the radius of the largest sphere that can be placed in the hollow spot between the four spheres? * * * *

Answer for Question 2

Let T be the total number of elements till the nth set. Hence, T = n * (n+1) / 2. So, the number of elements till (n-1)th set = T - n. Sum of all elements till nth set (A) = T * (T+1) / 2. (since first element of first set is 1) Similarly, sum of all elements till (n-1)th set (B) = (T-n) * (T-n+1) / 2. Therefore, Sn = A - B. For S21: T = 21*22/2 = 231 T-n = 231-21 = 210 A = 231*232/2 = 26796 B = 210*211/2 = 22155 Therefore, Sn = 26796-22155 = 4641(ans)

Thanks Sumit

Thanks Sumit and what about the other two I think for the first question it shd be straight line if a be the side of the cube then from the mid point A to the mid of another side =√ {(a/2)2 + (a/2)2} = a/√2 and the other part shd be =√ {(a)2 + (a/2)2} = (√5 a)/2 So the total length = sum of above two = a/√2 + (√5 a)/2 DID IT MAKE ANY SENSE ??

Answer for the first one

Guys n Gals first try to visulise the prob here. Imagine 3-D Room like stucture. Then Divide each of the two adjacent squares into 2 equal rectangles. Then the minimum path for the spider will be = 2* Diagonl of the Rectangle = sqrt(5a) Which is less than considering that the spider will approach the opposite corner of one square and then vertically go downward. In this case, total path length = a + a * sqrt(2) What almost every1 think at first to be the shortest path....

1 and 3

So the total length = sum of above two = a/?2 + (?5 a)/2
I dont think this is correct though your approach was in right direction.
Then the minimum path for the spider will be = 2* Diagonl of the Rectangle = sqrt(5a)
not correct √ {a2 + [a + (a/2)]2} = (√13 a)/2 The last question : The radius of the small sphere in the middle is (3/√6) - 1

How do u approached for the

How do u approached for the 3rd Question?

guys i think ans for third

guys i think ans for third question is sqrt(2)-1. first take 4 circles join them such that each circle touches other 2 circles and find the radius of the circle that can fit in the space as said in the question. correct me if i am wrong......

I wanna know Ur solution.

Spread a box as sheet... Then to move to the biggest diagonally opposite corner the spider has to move in a rectangle of length 2a & breadth is a. Then how can the minimum path length can be less than the diagonal i.e. sqrt(5a)

I wanna know Ur solution.

Spread a box as sheet... Then to move to the biggest diagonally opposite corner the spider has to move in a rectangle of length 2a & breadth is a. Then how can the minimum path length can be less than the diagonal i.e. sqrt(5a) Interested to know about Ur approach...