IIFT 2016 Question Paper

Let SP of each soap be $$X$$.

Let's calculate Cost Price (CP) of each soap separately.

$$CP_{1} = \frac{100}{100-13}*X = \frac{100X}{87}$$

$$CP_{2} = \frac{100}{100+23}*X = \frac{100X}{123}$$

$$CP_{2} = \frac{100}{100-26}*X = \frac{100X}{74}$$

Total cost price of 3 soaps= $$\frac{100X}{87}+\frac{100X}{123}+\frac{100X}{74}=3.31X$$

Total  selling price of 3 soaps= $$3X$$

Cost price is higher than the selling price by =$$\frac{3.31X-3X}{3X}*100 \approx10.5$$ percent 

Therefore, option A is the correct answer. 

Let's assume my roommate completes 'x' amount of work in a day, I will complete '2x' amount of work.

Entire assignment is completed in  30 days hence total work = 30*(x+2x) = 90x

I will complete the entire assignment alone in = $$\frac{90x}{2x} = 45$$ days

My roommate will complete the entire assignment alone in = $$\frac{90x}{x} = 90$$ days

It's given that PQRSTU is a regular hexagon and O is the center of the hexagon. 

If we fold $$\triangle$$TSR ,$$\triangle$$PQR ,$$\triangle$$TUP along lines TR, PR, PT respectively then vertices S,Q,U will overlap each other exactly at center of the hexagon.

Hence we can say that Area of hexagon PQRSTU =  2($$\triangle$$PRT)

So Area of $$\triangle$$PRT = 0.5(Area of hexagon PQRSTU)

                                                                

It's given that 120 students study Spanish :         A + B = 120                                        ...(1)

              Also 100 students study Mandarin :       B + C = 100                                        ...(2)

At least 80 students will not opt for any of these two languages i.e.      

                                                     Spanish + Mandarin $$\leq$$ (290-80)

                                                     Spanish + Mandarin $$\leq$$ 210

                                                        A + B + C $$\leq$$ 210

Case 2: When B is minimum and A is maximum. 

Solving for boundary condition  A + B + C = 210                                                             ...(3)

Solving equations (1)+(2)-(3) 

                                                                   B = 10 

                                            i.e. A = 120 - B = 120 - 10 =110 {Maximum possible value of A}

Case 2: When B is maximum and A is minimum. 

Since B+C = 100 so, Maximum value B can attain is 100. Hence, the minimum value of A = 120 - B = 120 - 100 = 20 {Minimum possible value of A}

So we can say that the number of students who study Spanish but not Mandarin will be A [20 ,110].

Volume of liquid = Volume of cylinder - Volume of rectangular solid

Volume of cylinder = $$\pi*r^{2}*h$$

                               =$$\pi*7^{2}*15$$ = $$735\pi$$

Volume of rectangular solid = Area of square base * height 

In square ABCD,  AC = $$\sqrt{2}$$*AB        so AB = $$\frac{14 }{\sqrt{2}}$$ = $$7\sqrt{2}$$

Volume of rectangular solid = Area of square base * height  = $$(AB)^{2}$$*height = $$(7\sqrt{2})^{2}*12$$

                                                                                                        =  $$98*12$$ = $$1176$$

So Volume of liquid = Volume of cylinder - Volume of rectangular solid

                                                                                     = $$735\pi$$ - $$1176$$ = $$147(5π-8)$$

In this case first we have to select the top three salesmen out of 15 and then arrange them as first, second and third.

∴ Number of arrangements = 15C3 × 3!  = 2730

Sum of integers from 54 to 196 = (sum of integers from 1 to 196) - (sum of integers from 1 to 53) 

Sum of first N natural numbers = (N)(N+1)/2

Sum from 1 to 196 = (196)(197)/2 = 98 × 197 = 19306

Sum from 1 to 53 = (53)(54)/2 = 53 × 27 = 1431

Sum of integers from 54 to 196 = 19306 - 1431 = 17875 

Let's first estimate the sample space. In a committee of n members, there can be the following cases:

(0 males, n females), (1 male, n-1 females), (2 males, n-2 females) ...(n males, 0 females).

Thus, the sample space has n+1 cases.

n(S) = n+1 cases

Barring 2 cases from the sample space, event P occurs for all other cases. 

=> n(P) = n+1-2 = n-1 cases

=> p(P) = (n-1)/(n+1)

At most 1 female can be expressed as the sum of the cases exactly 0 females and exactly 1 female.

=> n(Q) = 2 cases

=> p(Q) = 2/(n+1)

The set $$P \cap Q$$ occurs when there is exactly one female member.

Hence, n($$P \cap Q$$) = 1

p($$P \cap Q$$) = 1/(n+1)

When two events are independent, the probability of both of them occurring is equal to the product of their individual probabilities.

Thus,

p($$P \cap Q$$) = p(P) * p(Q)

1/(n+1) = (n-1)/(n+1) * 2/(n+1)

=> n+1 = 2n -2

=> n= 3

                    $$R_{1}$$= 20 , $$R_{2}$$= 21 , $$R_{3}$$= 23 ...

                    Sum = 20+$$\underbrace{21+23+...upto 16 terms}$$

                    So apart from first term the above mentioned sum represents an Arithmetic Progression that has 16 terms.

                    Sum of A.P. = $$\frac{16}{2}[2(21)+(16-1)2]$$  = 576

                    So total number of parking spaces = 20+576 = 596

Given that 

a = $$\sqrt{2}, b = \sqrt[3]{3},$$ and c = $$\sqrt[4]{4}$$ 

a = $$(2)^{\frac{1}{2}}$$, b = $$(3)^{\frac{1}{3}}$$ and c = $$(4)^{\frac{1}{4}}$$

Taking log both sides   $$\log_{}{a}$$ = $$\log_{}{(2)^{1/2}}$$ , $$\log_{}{b} = \log_{}{(3)^{1/3}}$$ , $$\log_{}{c} = \log_{}{(4)^{1/4}}$$

we know that ($$\log_{}{2}$$ = 0.3010 , $$\log_{}{3}$$ = 0.4771 , $$\log_{}{4}$$ = $$2\log_{}{2}$$ =0.6020)

Substituting these values 

$$\log_{}{a}$$ = $$\frac{1}{2}$$*0.3010 , $$\log_{}{b}$$ =$$\frac{1}{3}$$*0.4771 , $$\log_{}{c}$$= $$\frac{1}{4}$$*0.6020

$$\log_{}{a}$$ = 0.1505 , $$\log_{}{b}$$ = 0.1590 , $$\log_{}{c}$$= 0.1505

We know that if $$A_{1}$$ > $$A_{2}$$ > $$A_{3}$$ > 1   Then  $$log_{}{A_{1}}$$ > $$log_{}{A_{2}}$$ > $$log_{}{A_{3}}$$

 Here clearly  $$log_{}{b}$$ > $$log_{}{a}$$ = $$log_{}{c}$$  hence we can say that b = $$\sqrt[3]{3}$$ is the highest number among all.

Alternate method: 

a = $$\sqrt{2}, b = \sqrt[3]{3},$$ and c = $$\sqrt[4]{4}$$

$$a^{12} = 2^6, b^{12} = 3^4, c^{12} = 4^3$$

$$a^{12} = 64, b^{12} = 81, c^{12} = 64$$

We can see that $$b^{12}$$ > $$c^{12}$$ = $$a^{12}$$

Also, a, b, c > 1. Hence, we can say that b > a = c. 

***  Point to remember ------ $$\sqrt[N]{N}$$ is highest for N=3 (N being natural number) ***