IIFT 2010 Question Paper

If the quadratic equation is of type $$Ax^+Bx+C=0$$ then the roots of the quadratic equation are given by

$$x = \dfrac{-B\pm\sqrt{B^2 - 4AC}}{2A}$$

Comparing $$Ax^+Bx+C=0$$ with $$bx^{2}-2ax+a=0$$, A = b,  B = -2a,  C = a

Hence, the roots = $$\dfrac{2a\pm\sqrt{4a^2 - 4ba}}{2b}$$

x = $$\dfrac{a\pm\sqrt{a^2 - ba}}{b}$$

Let $$x_{1}$$ = $$\dfrac{a-\sqrt{a^2 - ba}}{b}$$,  $$x_{2}$$ = $$\dfrac{a+\sqrt{a^2 - ba}}{b}$$

Rationalizing  $$x_{1}$$ = $$\dfrac{a-\sqrt{a^2 - ba}}{b}$$

$$\Rightarrow$$ $$x_{1}$$ = $$\dfrac{a-\sqrt{a^2 - ba}}{b}*\dfrac{a+\sqrt{a^2 - ba}}{a+\sqrt{a^2 - ba}}$$

$$\Rightarrow$$ $$x_{1}$$ = $$\dfrac{a^2-(a^2 - ba)}{b*(a+\sqrt{a^2 - ba})}$$

$$\Rightarrow$$ $$x_{1}$$ = $$\dfrac{ab}{b*(a+\sqrt{a^2 - ba})}$$

$$\Rightarrow$$ $$x_{1}$$ = $$\dfrac{a}{a+\sqrt{a^2 - ba}}$$

$$\Rightarrow$$ $$x_{1}$$ = $$\dfrac{a}{a+\sqrt{a}*\sqrt{a - b}}$$

$$\Rightarrow$$ $$x_{1}$$ = $$\dfrac{\sqrt{a}}{\sqrt{a}+\sqrt{a - b}}$$

Similarly, $$x_{2}$$ = $$\dfrac{\sqrt{a}}{\sqrt{a}-\sqrt{a - b}}$$

Therefore, x = $$\dfrac{\sqrt{a}}{\sqrt{a}\pm\sqrt{a - b}}$$. Hence, option C is the correct answer. 

let G be number of days in which Dr. gupta can complete the evaluation of answer sheet , let Sh be number of days in which Dr. Sharma can complete the evaluation of answer sheet , let Si be number of days in which Dr. Singh can complete the evaluation of answer sheet

Given Dr. Gupta is 40% more efficient than Dr. Sharma therefore number of days required by Dr. Sharma to complete the evaluation of answer sheet = 1.4*G

Dr. Gupta takes 10 days less than Dr. Sharma to complete the evaluation work i.e. 1.4G-G = 10 therefore G = 25 and hence Sh = 35

Also given that Dr. Sharma is 20% more efficient than Dr. Singh therefore number of days required by Dr. Singh to complete the evaluation of answer sheet = $$\frac{35}{1.2}$$ = 42 i.e. Si = 42

Dr. Gupta starts the evaluation work and works for 10 days therefore he completes $$\frac{10}{25}$$ of work then Dr. Sharma takes over. Dr. Sharma evaluates for next 15 days therefore he completes $$\frac{15}{35}$$ of work. Fraction of work left is 1 - $$\frac{10}{25}$$ - $$\frac{15}{35}$$ = $$\frac{6}{35}$$ now time required by Dr. Singh to complete this amount of work is $$\frac{6*42}{35}$$ = 7.2 days

Therefore our answer is option 'A'

$$[\sqrt{1}]+[\sqrt{2}]+[\sqrt{3}]+[\sqrt{4}]+....+[\sqrt{360}]$$ = 1 + 1 + 1 + 2 + 2 + 2 + 2 + 2 + 3 + 3 + 3.. 7 times + 4 + 4 +......9 times +.........+ 18 + 18 + ......37 times

= 1*3 + 2*5 + 3*7 + 4*9+............+18*37 

= $$\sum_1^{18}$$ n(2n+1)

=  $$\sum_1^{18}$$ $$2n^2+n$$

= $$\frac{2n*(n+1)*(2n+1)}{6}$$ + $$\frac{n*(n+1)}{2}$$

= $$\frac{2*18*(19)*(37)}{6}$$ + $$\frac{18*19}{2}$$

= 4389

Therefore our answer is option 'A' 

$$\sqrt{\frac{a}{b}}$$, If $$\log_{4}\log_{4}4^{a-b}=2\log_{4}(\sqrt{a}-\sqrt{b})+\log_{4}{4}$$

i.e. $$\log_{4}\log_{4}4^{a-b}=\log_{4}((\sqrt{a}-\sqrt{b})^2)*4$$

i.e. $$\log_{4}4^{a-b}=((\sqrt{a}-\sqrt{b})^2)*4$$

i.e. (a-b)*$$\log_{4}4=((\sqrt{a}-\sqrt{b})^2)*4$$

i.e. a-b = 4a+4b-8$$\sqrt{ab}$$

i.e. 3a + 5b - 8$$\sqrt{ab}$$ = 0

i.e. $$3\sqrt\frac{a}{b}^2$$ - 8$$\sqrt\frac{a}{b}$$+5 = 0

put $$\sqrt\frac{a}{b}$$ = t

therefore 3$$t^2$$ - 8t + 5 = 0 

solving we get t = 1 or t = $$\frac{5}{3}$$

i.e. $$\sqrt\frac{a}{b}$$ = 1 or $$\frac{5}{3}$$

but if $$\sqrt\frac{a}{b}$$ = 1 then a=b then $$\log_{4}(\sqrt{a}-\sqrt{b})$$ will become indefinite

Therefore  $$\sqrt\frac{a}{b}$$ = $$\frac{5}{3}$$

Therefore our answer is option 'C' 

Let the capacity of the tank be 60 litres.
Capacity of the first pipe = 12 l/hr
Capacity of the second pipe = 6 l/hr
Capacity of the third pipe = 4 l/hr
In 2 hrs, first pipe fills (9 + 12) l = 21 l
In 2 hrs, second pipe fills (4.5 + 6) = 10.5 l
In 2 hrs, third pipe fills (16/3) l
In 2 hrs, tank filled = (21 + 10.5 + 5.33) l = 36.83 l
Tank left to be filled = (60 - 36.83) l = 23.17 l
Time required = (23.17/22) hr = 1.05 hrs
Total time = 3.05 hrs
Hence, option C is the correct answer.

Using AM $$\geq$$ GM, We can say that 

$$\dfrac{3^{sinx}+3^{cosx}}{2}$$ $$\geq$$ $$\sqrt{3^{sinx}*3^{cosx}}$$

$$\Rightarrow$$ $$3^{sinx}+3^{cosx}$$ $$\geq$$ $$2*3^\frac{sinx+cosx}{2}$$  ... (1)

We know that -$$\sqrt{A^2+B^2}$$ $$\leq$$ Asinx+Bcosx $$\leq$$ $$\sqrt{A^2+B^2}$$

Therefore,   -$$\sqrt{1^2+1^2}$$ $$\leq$$ sinx+cosx $$\leq$$ $$\sqrt{1^2+1^2}$$

$$\Rightarrow$$ -$$\sqrt{2}$$ $$\leq$$ sinx+cosx $$\leq$$ $$\sqrt{2}$$

Hence, we can say that the minimum value of sinx+cosx = -$$\sqrt{2}$$  ... (2)

From equation (1) and (2) we can say that,

$$\Rightarrow$$ $$3^{sinx}+3^{cosx}$$ $$\geq$$ $$2*3^\frac{-\sqrt{2}}{2}$$

$$\Rightarrow$$ $$3^{sinx}+3^{cosx}$$ $$\geq$$ $$2*3^\frac{-1}{\sqrt{2}}$$

Therefore, option B is the correct answer. 

Let the number of professors, associate professors and assistant professors be x, y and z respectively and their average ages be a, b and c respectively.
xa + yb + zc = 2160   -------------------(1)

Average age =36

$$\therefore$$ x+y+z=60

xa + yb = 39 (x + y)
11(yb + zc) = 360(y + z)
3(xa + zc) = 110(x + z)
x(a + 1) + y(b + 6) + z(c + 7) = 2160 + 5*(60) -------(2)            [$$\because$$ Average increases by 5]

Eq 2- Eq 1

We get x+6y+7z=300

Lets solve the options one by one
Option A: x = 16 , y = 24, z = 20

16+6*24+7*20=300 which satisfies the equations.

So either A or C can be the answer.

Now check for the values of a, b, c

a=45, b=35, c=30

ax+by+cz = 45*16+35*24+30*20

=2160

Hence, option A is the correct answer.

Let $$\log_{5}{2}$$ = y

Let us assume  $$\log_{5}{2}$$ is a rational number.

$$\log_{5}{2}$$ = p/q, where p and q are co primes.

5^(p/q)=2 => 5^p=2^q.

5^p=5*5*5*5*5*5*5..................p times

2^p=2*2*2*2*2*2*2..................q times

No value of p and q can satisfy the equation. Hence y is an irrational number.

Let us assume that 'a' is the length of the side of equal sides of the isosceles triangle. 

We can see that $$a+\sqrt{2}*a+a=2$$

$$\Rightarrow$$ a = $$\dfrac{2}{2+\sqrt{2}}$$

Therefore, the perimeter of the regular octagon = $$8*a*\sqrt{2}$$

$$\Rightarrow$$ $$\dfrac{8\sqrt{2}*2}{2+\sqrt{2}}$$

$$\Rightarrow$$ $$\dfrac{16}{1+\sqrt{2}}$$ ... (1)

Area of regular octagon = $$2x^2(1+\sqrt{2})$$ where 'x' is the length of each side. 

Here, x = $$a\sqrt{2}$$ = $$\dfrac{2\sqrt{2}}{2+\sqrt{2}}$$ = $$\dfrac{2}{1+\sqrt{2}}$$

Therefore, area of the regular octagon = $$2*(\dfrac{2}{1+\sqrt{2}})^2(1+\sqrt{2})$$ = $$\dfrac{8}{1+\sqrt{2}}$$ =$$\dfrac{8}{1+\sqrt{2}}*\dfrac{1+\sqrt{2}}{1+\sqrt{2}}$$ = $$\dfrac{8*(1+\sqrt{2})}{3+2\sqrt{2}}$$

Hence, we can say that none of the given answer is correct. 

7! can be broken into prime factors.

7! = 1*2*3*4*5*6*7 = $$2^4*3^2*5*7$$

Hence, the smallest perfect square which is divisible by 7! will $$2^4*3^2*5^2*7^2$$ = 5040*5*7 =  176400