IIFT 2017 Question Paper

Let's assume 'n' be the number of units that Suntex needs to manufacture and sell in order for their revenue to be equal to their total costs for each production run.

Cost incurred for manufacturing these 'n' units = $$50400+\frac{50}{100}*12n$$

                                                  =$$50400+6n$$

Revenue generated by selling these units = $$12n$$

For break-even   Revenue generated = Cost incurred

                                      $$50400+6n$$ = $$12n$$

                                            $$6n = 50400$$

                                             $$n = 8400$$

Given that $$x_{n}=5x_{n-1} - \frac{3}{4}x_{n-2}$$

        $$\therefore$$  $$x_{2}=5x_{1} - \frac{3}{4}x_{0}$$

            $$x_{2}=5*2 - \frac{3}{4}*4$$ = $$10-3$$ = 7

Similarly  $$x_{3}=5x_{2} - \frac{3}{4}x_{1}$$ 

               $$x_{3}=5*7 - \frac{3}{4}*2$$ = 35 - $$\frac{3}{2}$$ = $$\frac{67}{2}$$

Let's assume that revenue reported from ACN-I and ACN-II sales in 2015 be X and Y respectively.

So revenue generated from ACN-I sales in 2016 = $$X(1- \frac{12}{100})$$ =$$0.88X$$

Similarly revenue generated from ACN-II sales in 2016 = $$Y(1+ \frac{9}{100})$$ =$$1.09Y$$

                   $$0.88X$$ + $$1.09Y$$ = $$(X+Y)(1+\frac{3}{100})$$

                    $$0.88X$$ + $$1.09Y$$ =     $$1.03X$$ + $$1.03Y$$

                                $$6Y = 15X$$

                                   $$\frac{X}{Y}$$ = $$\frac{2}{5}$$

$$10^{67}- 87$$ = $$9999....99913$$ (total 67 digits)

Sum of digits =$$65*9 + 1 + 3$$ = 589

In the given diagram AE is the mall building and DE is pole. The observer is at C point which is 400 mts from the ground. 

It is given that $$\angle$$DCE = $$\angle$$ECA = $$\theta$$

In $$\triangle$$ACD and $$\triangle$$ECD

tan(2$$\theta$$) = $$\frac{400}{x}$$

tan($$\theta$$) = $$\frac{75}{x}$$

We know that tan(2$$\theta$$) = $$\frac{2tan(\theta)}{1-tan^{2}(\theta)}$$

                  $$\therefore$$ $$\frac{400}{x}$$ = $$\frac{2*\frac{75}{x}}{1-(\frac{75}{x})^{2}}$$

                          $$x^{2}$$ = 9000 = $$30\sqrt{10}$$

Given that  probabilities of his delivering the product late to the destination by using  bike, scooter, tempo and truck are 3/5, 2/5, 1/5 and 4/5 respectively.

Hence the probabilities of his delivering the product on time by using these modes of transport are 2/5, 3/5, 4/5 and 1/5 respectively.

Hence the required probability  

                         P = $$\frac{\frac{1}{9}*\frac{3}{5}}{\frac{2}{9}*\frac{2}{5}+\frac{1}{9}*\frac{3}{5}+\frac{4}{9}*\frac{4}{5}+\frac{2}{9}*\frac{1}{5}}$$

                         P = $$\frac{3}{25}$$

Distance covered by the pest control person = Circumference of the circular hall = $$2*\frac{22}{7}*49$$ = 308 m

So machine's speed = $$\frac{308}{148}$$ m/min = 2.08 m/min

The distance that is to be covered =  Perimeter of hexagon = 6 * 54 = 324 m

Time consumed  = $$\frac{Distance}{Speed}$$

                           = $$\frac{324}{2.08}$$ 

                           =155.7 min

So additional time taken by the machine is = 155.7-148 = 7.7 $$\approx$$ 7.69 minutes.

With the help of given information we can make Venn diagram:

                                                   

                           It's given that   e=2d= 4(12)=48 and d=48/2= 24

                             So                    e= 12+f  so f=e-12=48-12=36

                            total number of students in GSC= 120 =a+e+d+12

                                             a=120-48-24-12 =36

                             Similarly in case of CG,   CG=124=c+e+f+12

                                               c=124-12-48-36=28

                                   total number of students a+b+c+d+e+f+g+12=360

                                          36+b+28+24+48+36+72+12=360

                                                b=104 

The distance covered by Ramesh in 90 minutes = 1.5*3 = 4.5 kms

Similarly the distance covered by Sohan in 90 minutes = 1.5*4 = 6 kms

The distance between Sohan and Ramesh after 90 minutes will be :

                               Cos(A) = $$\frac{b^{2}+c^{2}-a^{2}}{2bc}$$

                                     Cos(120) = $$\frac{4.5^{2}+6^{2}-d^{2}}{2*4.5*6}$$

                                                $$d^{2}$$ = $$4.5^{2}+6^{2}-2*4.5*6*Cos(120)$$

                                                        $$d^{2}$$ = 20.25+36+27 = 83.25

                                                              d =  $$\sqrt{83.25}$$

                                                             $$d \approx 9.12$$

  In $$\triangle$$OAB      

          OC is drawn perpendicular to AB 

           By symmetry AC=CB=AB/2 =24/2= 12 

                                OA=13

                    $$OC^{2}=OA^{2}-AC^{2}$$

                    $$OC^{2}=13^{2}-12^{2}$$=25

                                 OC=5

So area of shaded region = Area of sector OAPB - Area of $$\triangle$$OAB

                                      =$$\frac{x}{360}*\pi*(OA)^{2}$$-$$\frac{1}{2}*(AB)*(OC)$$

                                      =$$\frac{x}{360}*\pi*(13)^{2}$$-$$\frac{1}{2}*(24)*(5)$$

                                      =$$\frac{169\pi x}{360}-60cm^{2}$$