Quant with Kamal Lohia  Part 15

Q1) if a cube is painted and then cut into n number of cubes then
a) how many cubes have all the faces painted
b) how many have two face paintedWhen you are painting a large cube and cut it into n * n * n smaller unit cubes, then there are in all n³ unit cubes. Out of these n³ unit cubes, only the ones which were placed on outer layer are painted (either 1 face, 2 face or 3 face) but the ones which lie inside the outer layer i.e. (n  2)³ are not having any face painted. Is that point clear?
Now to go further, only unit cubes which are placed at vertices of the larger cube are 3face painted because there 3 face are exposed on the outer layer of the larger cube. They number to be 8 as in every cube there are eight vertices only.
The unit cubes which are placed on the edges (except vertices) are exposed on two faces so are 2face painted and they number to 12(n  2).
And the remaining unit cubes on every face (i.e. except the vertices and edges ones) are 1face painted, so they number to 6(n  2)².
Counting back, we can easily deduce that: (n  2)³ + 6(n  2)² + 12(n  2) + 8 = (n  2 + 2)³ = n³.
Q2) Out of 120 boxes of oranges, each box contains at least 120 and at most 144 oranges. The number of boxes containing the same number of oranges is at least
a. 5
b. 103
c. 6
d. cant be determinedThis is simple application of a simple principle known as Pigeonhole Principle or Dirichlet's Box Principle.
It says if there are n + 1 pigeons which are to be kept into n pigeonholes, then there is at least 1 pigeonhole which contains 2 or more pigeons.
Isn't it simple? But it becomes difficult to apply most of the times as one may not be able to identify the pigeons and pigeonholes.
In the given question, we have 120  144 numbered 25 pigeonholes which are to contain 120 boxes i.e. pigeons. Now according to above principle, there is at least one pigeonhole (i.e. number of oranges) which is shared by at least 5 pigeons (i.e. boxes) Or in other words there are at least 5 boxes which contain same number of oranges.Q3) A and B can each run at a uniform speed along a circular track. To cover the whole track, A need 5 seconds less than B. If they start simultaneously from the same place and run in the same direction, they meet 30 seconds later. At how many points will they meet if they run in the opposite directions?
(a) 9
(b) 12
(c) 13
(d) 16
(e) None of theseA takes 10s to complete 1 round and B takes 15s i.e. they will be travelling 3 and 2 rounds in 30s respectively. So if they are moving in opposite direction, then they will at exactly 5 (3 + 2) distinct points.
Find the ratio of speeds of A and B assuming the track length to be d and the two speeds being u and v respectively.
So we have d/v  d/u = 5 and d/(uv) = 30. Also u > v.
Solving two equations, we get u/v = 3/2 and the solution follows.Q4) How many 6digit numbers contain exactly 4 different digits ?
a) 4536
b) 2,94,840
c) 1,91,520
d) noneLet the four different digits be a, b, c, d, then the number can be of two forms i.e. aabbcd or aaabcd.
So to calculate the total numbers; first we need to find the number of ways to select 4 digits (which is C(10, 4)); then number of ways to select the digits to be repeated in each case (which is C(4,2) and C(4,1) respectively); and then the number of ways of arrangement in both the cases (which is 6!/(2!)2 and 6!/3! respectively).
Hence the number of ways to form such 6digit numbers become = C(10, 4)[C(4, 2)6!/(2!)2 + C(4, 1)6!/3!] = 10 * 9 * 8 * 7 * 5 * 13 = 327600 .
But is this the final answer? .....certainly not!
See in the initial selection for the digits, we have used all 10 digits (including 0) for selecting the four distinct digits. That means in the above arrangements there will be some numbers which would be starting with zero. Now thinking alternately, first place of the number can be filled with any 10 digits equally. So removing the cases starting with zero, we are left with (9/10) of previously calculated result.
Thus the answer will be 9 * 9 * 8 * 7 * 5 * 13 = 294840.Q5) In an examination the maximum marks for each of the 3 papers are 50 each.Maximum marks for the fourth paper are 100. Find the number of ways in which the candidate can score 60% marks in the aggregate?
It is simply: 103C3  352C3 = 110551.
Amazed? Now see the explanation part.
Out of total 250 marks, you want to score 150 marks. So number of ways of scoring 150 is equivalent to not scoring the remaining 100 marks. So let's say we have 4 boxes a, b, c, d such that capacity of a, b, c is 50 each and that of d is 100. And all 4 boxes are filled to its capacity and we are to remove 100 balls out of it. Does it sound better?
i.e. we are trying to find the whole number solution of: a + b + c + d = 100 with the restriction that none of a, b, c can be more than 50.OK. Now without restriction, number of whole number solution of above equation is given by 103C3.
We need to subtract the cases when any one of a, b, c is more than 50 (i.e. 51 + x say, where x is a whole number). Remember no more than one of a, b, c can be greater than 50. Also I have taken it as 51 + x because that's how I ensure that one of a, b, c is certainly more than 50. And which one of a, b, c can be determined in 3 ways. Hence the final upfront expression. Hope things are clear nowQ6) Given that 2^2010 is a 606digit number whose first digit is 1, how many elements of the set S = {20, 21, 22, …, 2^2009} have a first digit of 4?
The smallest power of 2 with a given number of digits has a first digit of 1, and there are elements of S with n digits for each positive integer n ≤ 605, so there are 605 elements of S whose first digit is 1. Furthermore, if the first digit of 2^k is 1, then the first digit of 2^(k+1) is either 2 or 3, and the first digit of 2^(k+2) is either 4, 5, 6, or 7. Therefore there are 605 elements of S whose first digit is 2 or 3, 605 elements whose first digit is 4, 5, 6, or 7, and 2010  3(605) = 195 whose first digit is 8 or 9. Finally, note that the first digit of 2^k is 8 or 9 if and only if the first digit of 2^(k  1) is 4, so there are 195 elements of S whose first digit is 4.
Q7) Avatar is hosting a party to celebrate the completion of the ritual, but he needs you help to figure out how many guests he can fit in his home. Avatar has five rooms in his home, and he gives you these five facts:
a. The first room can fit half as many guests as the second room
b. The first room and the last room can together fit 11 guests.
c. The third room can fit three less guests than the fourth room
d. The fourth room can fit three times as many guests as the first room
e. The third room can fit two more guests than the second room.
So if he uses all five rooms how many guests can avatar fit in his roomMinimum number of guests that can be in the home is zero. So I am assuming you want to know the maximum capacity of the home.
Let's say Ist room can accomodate x number of guests.
Then IInd room's capacity = 2x, also Vth = 11  x
IVth = 3x. Now IIIrd = 3x  3 = 2x + 2, => x = 5.
So Ist + IInd + IIIrd + IVth + Vth = (x) + (2x) + (2x + 2) + (3x) + (11  x) = 7x + 13 = 7(5) + 13 = 48Q8) A hollow cube of size 5 cm is taken, with a thickness of 1 cm. It is made of smaller cubes of size 1 cm. If one face of the outer surface of the cube are painted, totally how many faces of the smaller cubes remain unpainted?
a) 900
b) 488
c) 563
d) 800There are 5^3  3^3 = 98 smaller cubes being used with a total of 6 * 98 = 588 faces and out of which 5^2 = 25 faces are painted which leaves 588  25 = 563 unpainted faces of the smaller cubes
Q9) Rohit drew a grid of 529 cells, arranged in 23 rows and 23 columns. n filled each cell with a no. the no. with which he filled each cell were such that the no. of each row taken from left to right formed an AP and the no. of each column taken from top to bottom formed an AP. the 7th and 17th no of 5th row were 47 and 63, while the 7th and 17th no of 15th rows were 53 and 77. what is the sum of all the no. in the grid.
a. 32798
b. 65596
c. 52900
d. noneAs every row and every column is in AP. So sum of all the columns follow an AP and likewise sum of rows. Given sum of 5th row as 23 * (47+63)/2 = 23 * 55 and sum of 15th row as 23 * (53+77)/2 = 23 * 65.
So sum of 12th row is = 23 * 62 and sum of sum of all rows
i.e. sum of all the numbers in grid = 23 * 23 * 62 = 32798Q10) The integers 1, 2, 3, .... , 40 are written on a blackboard. The following operation is then repeated 39 times. In each operation two numbers say 'a' , 'b' currently on the blackboard are erased and a new number 'ab + a + b' is written. What will be the number left on the board at the end ?
a) 40^3  40 + 1
b) 40^3  40  1
c) No unique value
d) None of these
e) 41!  1If two numbers are a, b, then the new number you are putting in the place of these two numbers is = ab + a + b = (a + 1)(b + 1)  1 = c (say)
Now when you take c and d next, then the replacement number becomes = cd + c + d = (c + 1)(d + 1)  1 = (a + 1)(b + 1)(d + 1)  1.
So final number remaining will be (1 + 1)(2 + 1)(3 + 1)....(39 + 1)(40 + 1)  1 = 41!  1.