XAT 2010 Question Paper

Here we have to find the choice which can compromise on security 

A : Huge public gathering will call for Security

B : If racket players are included in the hit list of terrorist organisations then there is a need of Security

C : Internet based data stealing will definitely call for security 

E : This will call for security measures to be taken 

In Option D , In large stadiums where there are significant number of players they will be provided with Security by default 

So racket players need not focus on this aspect .

Hence D is the correct answer

XADY = XA + AD + DY = 2/2 + (2 * 3.14 * 2)/4 + (30/360) * (2 * 3.14 * 2) = 5.19
XOBY = XO + OB + BY = 2/2 + 2 + (60/360) * (2 * 3.14 * 2) = 5.09
XODY = XO + OD + DY = 2/2 + 2 + (30/360) * (2 * 3.14 * 2) =  4.04
Hence, option D is the correct answer.

Expression : $$x^{2}+ 4xy+ 6y^{2}-4y+ 4$$

= $$(x^2 + 4xy + 4y^2) + (2y^2 - 4y + 4)$$

= $$(x + 2y)^2 + \frac{1}{2} (4y^2 - 8y + 8)$$

= $$(x + 2y)^2 + \frac{1}{2} [(2y)^2 - 2 (2y) (2) + (2)^2 + 4]$$

= $$(x + 2y)^2 + \frac{1}{2} [(2y - 2)^2 + 4]$$

= $$(x + 2y)^2 + \frac{1}{2} (2y - 2)^2 + 2$$

Since, $$x$$ and $$y$$ are real, => Min value of $$(x + 2y)^2 = 0$$

Minimum value of $$(2y - 2)^2 = 0$$

$$\therefore$$ Minimum value of expression = $$0 + 0 + 2 = 2$$

Given that unit digit of $$X$$ and product of all 4 digits of $$X$$ are prime.

The product of two numbers to be prime is possible only when one of the numbers is prime and the other is ‘1’.

The possibilities for the prime unit digits are - 2,3,5,7     (Note that 1 is not a prime number)

Hence the possibility of remaining 3 digits, considering the product of all 4 digits to be prime is ‘111’ only.

Hence all the possible numbers are 1112,1113,1115,1117

$$\therefore$$ Total 4 integers are possible.

Let total employees in Sun Metals = $$100x$$

Number of employees who are general graduates = $$\frac{40}{100} \times 100x = 40x$$

=> Number of employees who are engineers = $$100x - 40x = 60x$$

Now, number of engineers who earn more than Rs. 5 lakh/year = $$\frac{75}{100} \times 60x = 45x$$

Number of employees (both general graduates and engineers) who earn more than Rs. 5 lakhs/year = $$\frac{50}{100} \times 100x = 50x$$

=> Number of general graduates who earn more than Rs. 5 lakhs/year = $$50x - 45x = 5x$$

Thus, number of general graduates who earn less than Rs. 5 lakhs/year = $$40x - 5x = 35x$$

$$\therefore$$ Proportion of the general graduates employed by the organisation earn Rs. 5 lakh or less

= $$\frac{35 x}{40 x} = \frac{7}{8}$$

Given : AB = AC = BC = 3  cm and BD = $$\frac{1}{2}$$ CD

AE is median.

To find : $$AD = ?$$

Solution : BD + CD = 3

=> $$BD + 2BD = 3BD = 3$$

=> $$BD = \frac{3}{3} = 1$$ cm

Also, since AE is media => $$BE = CE = \frac{3}{2}$$ cm

=> $$DE = BE - DE = \frac{3}{2} - 1 = \frac{1}{2}$$ cm

Also, AE = $$\frac{\sqrt{3}}{2} a = \frac{3 \sqrt{3}}{2}$$ cm

In $$\triangle$$ ADE

=> $$(AD)^2 = (AE)^2 + (DE)^2$$

=> $$(AD)^2 = (\frac{3 \sqrt{3}}{2})^2 + (\frac{1}{2})^2$$

=> $$(AD)^2 = \frac{27}{4} + \frac{1}{4} = \frac{28}{4}$$

=> $$AD = \sqrt{7}$$ cm

To find : $$EF = x = ?$$

Solution : In $$\triangle$$ ABC and $$\triangle$$ EFC

$$\angle ACB = \angle ECF$$  (common)

$$\angle ABC = \angle EFC = 90$$

=> $$\triangle ABC \sim \triangle EFC$$

=> $$\frac{x}{2} = \frac{5 - d}{5}$$ ----------Eqn(I)

Similarly, $$\triangle BCD \sim \triangle BFE$$

=> $$\frac{x}{3} = \frac{d}{5}$$ --------Eqn(II)

Adding Eqns (I) & (II), we get :

=> $$\frac{x}{2} + \frac{x}{3} = \frac{5 - d}{5} + \frac{d}{5}$$

=> $$\frac{3x + 2x}{6} = \frac{5}{5}$$

=> $$5x = 6 \times 1 = 6$$ 

=> $$x = \frac{6}{5} = 1.2 m$$

Let the volume of the solution with 30 % acid content lie between $$v_1$$ and $$v_2$$, where we get a 20% acid solution for $$v_1$$

For $$v_2$$, we get a 25 % acid solution as the resultant mixture.

=> $$15 \% (200) + 30 \% (v_1) = 20 \% (200 + v_1)$$

=> $$30 + 0.3 v_1 = 40 + 0.2 v_1$$

=> $$0.1 v_1 = 10$$ => $$v_1 = 10 \times 10 = 100$$ litres

Similarly, $$15 \% (200) + 30 \% (v_2) = 25 \% (200 + v_2)$$

=> $$30 + 0.3 v_2 = 50 + 0.25 v_2$$

=> $$0.05 v_2 = 20$$ => $$v_2 = 20 \times 20 = 400$$ litres

$$\therefore$$ For the acid content in the resultant mixture to lie between 20 % and 25 %, the volume of the 30 % concentration acid solution must lie between 100 litres and 400 litres.

Expected profit when brand ambassador is used = (3.5 - 3.45)[100000 × 0.3 + 80000 × 0.4 + 50000 × 0.3] lakh = 0.05lakh*77000= 3850 lakh

Expected profit when brand ambassador is not used =  (3.5 - 3.45)[80000 × 0.3 + 50000 × 0.4 + 30000 × 0.3] lakh =0.05lakh*53000= 2650 lakh

Difference = (3850 - 2650) = 1200 lakh = 12 crore.

This is the maximum amount of money that the company can afford to pay its brand ambassador.

Hence, option D is the correct answer.

Expected profit when brand ambassador is used = (3.5 - 3.45)[100000 × 0.3 + 80000 × 0.4 + 50000 × 0.3] lakh = 0.05lakh*77000= 3850 lakh

Expected profit when brand ambassador is not used = (3.5 - 3.45)[80000 × 0.3 + 50000 × 0.4 + 30000 × 0.3] lakh =0.05lakh*53000= 2650 lakh

Difference = (3850 - 2650) = 1200 lakh = 12 crore.

When profit is Rs.12.0 crores when Mr. Khan Advertises for the company.

Then profit is (12.0 - 9.0) = Rs.3 crore 

No. of cars sold when Mr. khan advertises = [100000 × 0.3 + 80000 × 0.4 + 50000 × 0.3]= 77000

No. of cars sold when no popular actor is for advertisement= [80000 × 0.3 + 50000 × 0.4 + 30000 × 0.3]=53000

No of extra vehicles sold due to Mr. Khan= 24000

Suppose p is the amount escalated in cost of putting a car on the road,

24000p = 3,00,00,000

p = Rs1250

Had the cost not been increased, the company would have a extra profit of Rs. 1250 due to brand ambassador.

So, this the maximum escalation for which the company can afford signing contract with Mr. Khan.

Hence, option C is the correct answer.