IIFT 2016 Question Paper

Number of ways in which the candidate can secure the cut-offs in all 6 sections = $$6C_{6}$$ = 1

Number of ways of securing marks in 6 sections 

= $$6C_{0} + 6C_{1} + 6C_{2} + 6C_{3} + 6C_{4} + 6C_{5} + 6C_{6}$$

= $$2^{6}$$ = 64

Thus, the number of ways in which he fails to secure the cut offs in at least one section = 64 − 1 = 63 

Given that, S = 4+44+444+...

$$\Rightarrow$$ $$S = \frac{4}{9}(9+99+999+...)$$

$$\Rightarrow$$ $$S = \frac{4}{9}(10-1+10^2-1+10^3-1+...+10^n-1)$$

$$\Rightarrow$$ $$S = \frac{4}{9}(10+10^2+10^3+...+10^n-n)$$

$$\Rightarrow$$ $$S = \frac{4}{9}(\frac{10(10^n - 1)}{10-1}-n)$$

$$\Rightarrow$$ $$S = \frac{40}{81}(10^n - 1) - \frac{4n}{9}$$

Hence, option C is the correct answer. 

Alternate method: 

Solving for n = 2

Sum of series = 4+44 = 48

Substituting n = 2 in options

 (A) $$\frac{40}{81}(8^{2}-1)-\frac{5*2}{9}$$ = $$\frac{280-10}{9}$$ = 30

 (B) $$\frac{40}{81}(8^{2}-1)-\frac{4*2}{9}$$ = $$\frac{280-8}{9}$$ =  $$\frac{272}{9}$$

 (C) $$\frac{40}{81}(10^{2}-1)-\frac{4*2}{9}$$ = $$\frac{440-8}{9}$$ = 48

 (D) $$\frac{40}{81}(10^{2}-1)-\frac{5*2}{9}$$ = $$\frac{440-10}{9}$$ = $$\frac{430}{9}$$

The two lines are   4y - 3x = 2                                 ...(1) 

                      and 6x - 8y= 15                                    

                            4y - 3x= -7.5                               ...(2)

As we can see $$\frac{a_{1}}{a_{2}}$$ = $$\frac{b_{1}}{b_{2}}$$ $$\neq$$ $$\frac{c_{1}}{c_{2}}$$; these two lines are parallel to each other.Hence the distance between these two parallel lines will be the side of the square i.e. 

                              d  = $$\frac{|c_{1} - c_{2}|}{\sqrt{a^{2}+b^{2}}}$$      (here a = -3,  b = 4 c_{1} = 2  c_{2} = -7.5 )

                              d =  $$\frac{2-(-7.5)}{\sqrt{(-3)^{2}+(4)^{2}}}$$ = $$\frac{9.5}{5}$$ = 1.9

The distance between the parallel lines will be equal to the length of side of the square. 

 $$\therefore$$ Area of square = (1.9)^{2} = 3.61 Sq. units