# Quant Boosters - Kamal Lohia - Set 5

• 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 now

• Q26) Given that 2^2010 is a 606-digit 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.

• Q27) 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 room

• Minimum 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 = 48

• Q28) 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) 800

• There 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

• Q29) 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. none

• As 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 = 32798

• Q30) 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! - 1

• If 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.

48

63

63

58

1

61