IIFT 2017 Question Paper

The given data can be represented as follows.

A1 − C

A2 − D, F (Cricket) A3 − E (Tennis)

From, (g) and (c) it can be concluded that A (tennis) sits in coach A1. From (d) and (e) it can be said that C does not love playing tennis. Hence, C loves playing cricket or volleyball.

From (f) and (d), D does not love playing volleyball.

C loves to play either cricket or volleyball.

Only A and C are seated in coach A1.

The given data can be represented as follows.

A1 − C

A2 − D, F (Cricket) A3 − E (Tennis)

From, (g) and (c) it can be concluded that A (tennis) sits in coach A1. From (d) and (e) it can be said that C does not love playing tennis. Hence, C loves playing cricket or volleyball.

From (f) and (d), D does not love playing volleyball.

C loves to play either cricket or volleyball.

Only A and C are seated in coach A1.

Let us note down the given information.
The sport by star sports I was shown on shown on Friday.
SS1 - Friday - ?
Tennis was shown on Monday.
? - Monday - Tennis
Kabaddi was shown by Sony Six but was not broadcast on Thursday. We know the sports broadcast on Friday and Monday. Therefore, Kabaddi should be broadcast on Tuesday and Football should be broadcast on Thursday or Friday. 
Football is broadcast by Neosports. Therefore, Tennis should be broadcast by Ten sports 2 and Football should be broadcast on Thursday. The final arrangement is as follows:

Ten Sports 2 - Monday - Tennis

Sony Six -  Tuesday - Kabaddi

Neosports - Thursday - Football

Star Sports 1 - Friday - Cricket

Therefore, option B is the right answer.

Let us note down the given information.
The sport by star sports I was shown on shown on Friday.
SS1 - Friday - ?
Tennis was shown on Monday.
? - Monday - Tennis
Kabaddi was shown by Sony Six but was not broadcast on Thursday. We know the sports broadcast on Friday and Monday. Therefore, Kabaddi should be broadcast on Tuesday and Football should be broadcast on Thursday or Friday. 
Football is broadcast by Neosports. Therefore, Tennis should be broadcast by Ten sports 2 and Football should be broadcast on Thursday. The final arrangement is as follows:

Ten Sports 2 - Monday - Tennis

Sony Six -  Tuesday - Kabaddi

Neosports - Thursday - Football

Star Sports 1 - Friday - Cricket

Ten sports 2 broadcasts Tennis on Monday. Therefore, option D is the right answer. 

The sum that was returned is same in both cases that means interest accrued is same in both cases.

Assume that the duration is 't' years for which amount is borrowed in both case and principal amount is 'P'.

                       $$\frac{P*(t-0.5)*10} {100}$$ = $$\frac{P*t*8} {100}$$

                                t=2.5 years

Principal amount that the enterprise borrowed : 

                      $$P(1+\frac{2*10}{100})$$ =62100

                                P=51750

In the given diagram AB=$$\sqrt{10}$$ m

                          

Given that PQRS is a square and the plank is placed symmetrically $$\triangle$$BPA and $$\triangle$$AQC will be isosceles right triangles.

Hence $$\text{AB}^2 = \text{PA}^2 + \text{PB}^2$$

As the plank is symmetrical, PA = PB, 

Hence $$\text{AB}^2 = 2\times\text{PA}^2$$  =>  $$\sqrt{10}^2 = 2\times\text{PA}^2$$

          So PA=PB=$$\sqrt{\frac{10}{2}}$$=$$\sqrt{5}$$ m

          PQ = PA+AQ  

          AQ = PQ-PA = 10-$$\sqrt{5}$$ m

We know that AQ=QC ($$\triangle$$AQC is isosceles right triangle)

     So AC=$$\sqrt{2}$$AQ=$$\sqrt{2}$$*(10-$$\sqrt{5}$$) m

Now we can calculate area of plank 

     Area of ABCD= AB*AC= $$\sqrt{10}$$*$$\sqrt{2}$$(10-$$\sqrt{5}$$)=10($$\sqrt{20}$$-1) sq. mt

Let the duration in which the will be completed be 't' days.

The portion of work done by Nikhil = $$\frac{5}{75}$$= $$\frac{1}{15}$$                    ( Since He worked for 5 days)

Somesh left 2 days early that means Somesh must have worked for 't-2' days

The portion of work done by Somesh = (t-2)*$$\frac{1}{45}$$

Similarly, the portion of work  done by Tarun = t*$$\frac{1}{60}$$

Solving for t

          $$\frac{1}{15}$$+(t-2)*$$\frac{1}{45}$$+t*$$\frac{1}{60}$$ = 1

             $$\frac{4(t-2)+3t}{180}$$   = $$\frac{14}{15}$$

                      t=$$25\frac{1}{7}$$

Therefore, option A is the correct answer. 

Splitting the above mentioned series into two series 

A = $$\log_{e}3+\frac{1}{2!}(\log_{e}3)^{2}+\frac{1}{3!}(\log_{e}3)^{3}+...$$

B = $$5\log_{e}3+\frac{5^{2}}{2!}(\log_{e}3)^{2}+\frac{5^{3}}{3!}(\log_{e}3)^{3}+...$$

We know that $$e^{x}$$ =$$1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+...$$

So  $$e^{x}-1$$ = $$x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+...$$

On solving two series A and B

A = $$\log_{e}3+\frac{1}{2!}(\log_{e}3)^{2}+\frac{1}{3!}(\log_{e}3)^{3}+...$$ =$$e^{\log_{e}3}-1$$ = $$3-1$$ =$$2$$

B = $$5\log_{e}3+\frac{5^{2}}{2!}(\log_{e}3)^{2}+\frac{5^{3}}{3!}(\log_{e}3)^{3}+...$$=$$e^{\log_{e}3^{5}}-1$$=$$3^{5}-1$$=$$242$$

A+B = $$2 + 242$$ = $$244$$

$$f(x) = \frac{1}{1+x}$$

$$f[f(x)]$$ = $$\frac{1}{1+f(x)}$$ = $$\frac{1}{1+\frac{1}{1+x}}$$= $$\frac{1+x}{2+x}$$

$$f[f[f(x)]]$$ = $$\frac{1}{1+f{f(x)}}$$=$$\frac{1}{1+\frac{1+x}{2+x}}$$=$$\frac{2+x}{3+2x}$$

$$f[f[f(x)]]$$=$$\frac{2+5}{3+2*5}$$= $$\frac{7}{13}$$

Let's assume that P gm of chemical X and Q gm of chemical Y to be used to meet the requirement.

So net quantity of salt A in both the chemicals

  A=$$\frac{5}{100}*P$$ + $$\frac{10}{100}*Q \geq 7$$

Similarly net quanity of salt B in both the chemicals 

 B=$$\frac{10}{100}*P$$ + $$\frac{6}{100}*Q \geq 7$$

For minimum cost 

A=$$\frac{5}{100}*P$$ + $$\frac{10}{100}*Q = 7$$

 B=$$\frac{10}{100}*P$$ + $$\frac{6}{100}*Q = 7$$

On solving both equations for P and Q

                            P=40  Q=50

So minimum cost will be = $$40*10.50+50*7.80=810$$