XAT 2014 Question Paper

Expression : $$S=\frac{\alpha\times\omega}{\tau+\rho\times\omega}$$

=> $$\frac{1}{S} = \frac{\tau+\rho\times\omega}{\alpha\times\omega}$$

=> $$\frac{1}{S} = \frac{\tau}{\alpha \omega} + \frac{\rho}{\omega}$$

Since, $$\tau, \rho$$ and $$\alpha$$ are constant,

=> $$\frac{1}{S} = \frac{k_1}{\omega} + k_2$$

Thus, $$S \propto \omega$$

$$\therefore$$ When $$\omega$$ increases, S increases.

Grade A $$\geq$$ 90 and Grade B = 87 to 89

If Ramesh scores 70 instead of 97, => Change of marks = 97 - 70 = 27

It creates a change from grade A to B, this means an overall change in average by

= Minimum marks for grade A - Minimum marks for Grade B = 90 - 87 = 3

$$\therefore$$ Number of subjects = $$\frac{27}{3} = 9$$

Expression :  $$ax^{3} + bx^{2 }+ cx + d$$

When it intersects x-axis at x = 1, => Point = (1,0)

=> $$a(1)^3 + b(1)^2 + c(1) + d = 0$$

=> $$a + b + c + d = 0$$ --------Eqn(I)

Similarly at (-1,0)

=> $$a(-1)^3 + b(-1)^2 + c(-1) + d = 0$$

=> $$-a + b -c + d = 0$$ => $$(a + c) = (b + d)$$

Substituting it in eqn(I), we get : 

=> $$2 (b + d) = 0$$ => $$b + d = 0$$ ---------Eqn(II)

When it intersects y-axis at  = 2, => Point = (0,2)

=> $$a(0)^3 + b(0)^2 + c(0) + d = 2$$

=> $$d = 2$$

Substituting it in Eqn(II), => $$b = -2$$

Number of factors of $$10^{29} = 2^{29} \times 5^{29}$$

= $$30 \times 30 = 900$$

Factors of $$10^{29}$$ which are multiple of $$10^{23}$$

= $$10^6 = 2^6 \times 5^6$$

= $$7 \times 7 = 49$$

=> Required probability = $$\frac{49}{900} = \frac{a^2}{b^2}$$

=> $$\frac{a}{b} = \frac{7}{30}$$

$$\therefore b - a = 30 - 7 = 23$$

Radius = 3 units, => Diameter = PQ = 6 units

y-coordinates of P and Q are same as PQ is parallel to x-axis

x-coordinates of P and Q will have a difference of 6 units.

Equating y-coordinate of P and Q

=> $$a^x = 2 a^{x + 6}$$

=> $$\frac{1}{2} = \frac{a^{x + 6}}{a^x}$$

=> $$a^{x + 6 - x} = \frac{1}{2}$$

=> $$a = \frac{1}{\sqrt[6]{2}}$$

Let us draw the figure according to the available information, 

In $$\triangle$$ AMP and $$\triangle$$ CMR

$$\angle$$ MAP = $$\angle$$ MCR

$$\angle$$ AMP = $$\angle$$ CMR

Therefore, we can say that $$\triangle$$ AMP $$\sim$$ $$\triangle$$ CMR

Hence, we can say that $$\dfrac{AP}{CR}=\dfrac{MP}{MR}$$

$$\Rightarrow$$ $$\dfrac{AB/3}{AB/2}=\dfrac{MP}{MR}$$

$$\Rightarrow$$ $$MR=\dfrac{3}{2}*MP$$

Therefore, we can say that $$\Rightarrow$$ $$MR=\dfrac{3}{5}*RP$$   ... (1)

Similarly, in $$\triangle$$ ANQ and $$\triangle$$ CNR

$$\angle$$ NAQ = $$\angle$$ NCR

$$\angle$$ ANQ = $$\angle$$ CNR

Therefore, we can say that $$\triangle$$ ANQ $$\sim$$ $$\triangle$$ CNR

Hence, we can say that $$\dfrac{AQ}{CR}=\dfrac{NQ}{NR}$$

$$\Rightarrow$$ $$\dfrac{2AB/3}{AB/2}=\dfrac{NQ}{NR}$$

$$\Rightarrow$$ $$NR=\dfrac{3}{4}*NQ$$

Therefore, we can say that $$\Rightarrow$$ $$NR=\dfrac{3}{7}*RQ$$  ... (2)

In $$\triangle$$ RMN and $$\triangle$$ RPQ 

$$\dfrac{\text{Area of triangle RMN}}{\text{Area of triangle RPQ}} = \dfrac{0.5*RM*RN*sinMRN}{0.5*RP*RQ*sinPRQ}$$

$$\text{Area of triangle RMN}=\dfrac{3}{5}*\dfrac{3}{7}*\text{Area of triangle RPQ}$$

We know that, Area of triangle RPQ = 1/6*Area of rectangle ABCD = 1/6*90 = 15 sq. units

$$\Rightarrow$$ $$\text{Area of triangle RMN}=\dfrac{3}{5}*\dfrac{3}{7}*15$$ = $$\dfrac{27}{7}$$ sq. units

Hence, the area of the quadrilateral PQNM = 15 - $$\dfrac{27}{7}$$ = $$\dfrac{78}{7}=11\dfrac{1}{7}$$ sq. units. Therefore, option D is the correct answer. 

In Base B, $$297_B = 2B^2 + 9B + 7$$

and $$792_B = 7B^2 + 9B + 2$$

It is given that $$297_{B}$$ is a factor of $$792_{B}$$

=> $$\frac{7B^2 + 9B + 2}{2B^2 + 9B + 7}$$ must be an integer    

=> $$\frac{(2B^2 + 9B + 7) + (5B^2 - 5)}{2B^2 + 9B + 7}$$

=> $$\frac{5B^2 - 5}{2B^2 + 9B + 7} + 1 = k$$

=> $$5B^2 - 5 = (2B^2 + 9B + 7) k$$      (where $$k$$ is factor)

Put $$k = 1$$

=> $$5B^2 - 5 = 2B^2 + 9B + 7$$

=> $$B^2 - 3B - 4 = 0$$

=> $$(B - 4) (B + 1) = 0$$

=> $$B = 4 , -1$$

Since, B is a base,so B must be greater than 9. Hence, it is not possible

Put $$k = 2$$

=> $$5B^2 - 5 = 4B^2 + 18B + 14$$

=> $$B^2 - 18B - 19 = 0$$

=> $$(B - 19) (B + 1) = 0$$

=> $$B = 19 , -1$$

$$\therefore B = 19$$

Let $$x, y, z$$ be the number of students getting 6, 8 and 20 marks respectively.

=> $$6x + 8y + 20z = 504$$ -------------Eqn(I)

Since, the mode is 8, => Most populated category = $$y$$

=> $$y = x + 2z$$

=> $$x = y - 2z$$ -------------Eqn(II)

Substituting it in eqn(I), we get : 

=> $$(6y - 12z) + 8y + 20z = 504$$

=> $$14y + 8z = 504$$

By hit and trial, since $$x,y,z$$ are integers, we get : $$y = 32$$ and $$z = 7$$

Putting it in Eqn(II), => $$x = 32 - 2 \times 7 = 18$$

$$\therefore$$ Total students = $$x + y + z = 18 + 32 + 7 = 57$$

Expression : $$\sum_{n=1}^{13}\frac{1}{n} = \frac{x}{13!}$$

=> $$\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + ...... + \frac{1}{13} = \frac{x}{13!}$$

=> $$x = \frac{13!}{1} + \frac{13!}{2} + \frac{13!}{3} + ......... + \frac{13!}{13}$$

Now, if $$x$$ is divided by 11

=> $$\frac{13! + \frac{13!}{2} + \frac{13!}{3} + ......... + \frac{13!}{13}}{11}$$

All terms are divisible by 11 except $$\frac{13!}{11}$$

$$\therefore$$ Remainder if x is divided by 11 = Remainder of $$\frac{13!}{11 \times 11}$$

= $$(10! \times 12 \times 13) \% 11$$

= $$(10 \times 1 \times 2) \% 11 = 20 \% 11 = 9$$

For every 2.6 m that one walks along the slanting part of the pool, there is a height of 1 m that is gained.

=> $$\frac{AC}{BC} = \frac{2.6}{1}$$

=> $$AC = 2.6 \times BC$$

Also, dimensions of cuboidal part = $$48 \times 20 \times 1$$

 In $$\triangle$$ ABC

=> $$(AC)^2 = (AB)^2 + (BC)^2$$

=> $$(2.6 \times BC)^2 = (48)^2 + (BC)^2$$

=> $$6.76 (BC)^2 - (BC)^2 = 2304$$

=> $$(BC)^2 = \frac{2304}{5.76} = 400$$

=> $$BC = \sqrt{400} = 20$$ m

$$\therefore$$ Volume of water in the pool = Volume of cuboid + Volume of triangle

= $$(l \times b \times h) + (\frac{1}{2} \times AB \times BC) \times height$$

= $$(48 \times 20 \times 1) + (\frac{1}{2} \times 48 \times 20 \times 20)$$

= $$960 + 9600 = 10560 m^3$$