XAT 2010 Question Paper

As calculated in the last question, the company would have an additional profit or Rs. 1250/unit.

But,  the cost is going up by Rs. 1000/unit

Therefore, profit/unit = Rs. 1250 - Rs. 1000 = Rs. 250

∴    Addition/increase in profit

Rs. $$3 * \dfrac{250}{1250}$$ = Rs. 60 lakh

Hence, option B is the correct answer.

Let us draw the diagram first,

Let L$$_{1}$$: 2x+ 3y -1 = 0

L$$_{2}$$: x+ 2y -3 = 0

L$$_{3}$$: 5x-6y- 1 = 0

With respect to L$$_{1}$$, we can see that the point $$(a, a^{2})$$ lies within the triangle and (0, 0) are opposite side. Therefore,

 L$$_{(a, a^2)}$$* L$$_{(0, 0)}$$ < 0

$$\Rightarrow$$ $$(2a+ 3a^2-1)(-1)<0$$

$$\Rightarrow$$ $$(3a^2+2a-1)>0$$

$$\Rightarrow$$ a < -1 or a > $$\frac{1}{3}$$   ... (1)

With respect to L$$_{2}$$, we can see that the point $$(a, a^{2})$$ lies within the triangle and (0, 0) are on the same side. Therefore,

 L$$_{(a, a^2)}$$* L$$_{(0, 0)}$$ > 0

$$\Rightarrow$$ $$(a+ 2a^2-3)(-3)>0$$

$$\Rightarrow$$ $$(2a^2+a-3)<0$$

$$\Rightarrow$$ $$\frac{-3}{2}$$ < a < 1  ... (2)

With respect to L$$_{3}$$, we can see that the point $$(a, a^{2})$$ lies within the triangle and (0, 0) are on the same side. Therefore,

 L$$_{(a, a^2)}$$* L$$_{(0, 0)}$$ > 0

$$\Rightarrow$$ $$(5a- 6a^2-1)(-1)>0$$

$$\Rightarrow$$ $$(6a^2-5a+1)>0$$

$$\Rightarrow$$ a < $$\frac{1}{3}$$ or a > $$\frac{1}{2}$$   ... (3)

From equation (1), (2) and (3) we can say that 

a $$\epsilon$$ (-3/2,-1) ⋃ (1/2, 1). Hence, option C is the correct answer.

Let $$p \times \frac{0.34 mt}{mt}$$m packets of milk be prepared in unit time at the normal speed.

Now, at normal speed in $$t$$ time, the number of packets of milk that would be produced = $$mt$$

=> Number of packets of milk produced at fast speed = $$(\frac{110}{100} \times m) + (\frac{60}{100} \times t) = 0.66 mt$$

The target for the supervisor = $$mt$$ packets

Number of packets produced at normal speed = $$mt - 0.66 mt = 0.34 mt$$

Let the probability of a packet being damaged when produced at normal speed = $$p$$

=> Probability that a packet is damaged when produced at fast speed = $$2p$$

The probability that a packet selected at random will be damaged = 0.112

=> $$(p \times \frac{0.34 mt}{mt}) + (2p \times \frac{0.66 mt}{mt}) = 0.112$$

=> $$0.34p + 1.32p = 1.66 p = 0.112$$

=> $$p = \frac{0.112}{1.66} = 0.067$$

$$\therefore$$ Probability that a packet will not be damaged at normal speed = $$1 - 0.067 = 0.93$$

$$x_1 , x_2 , x_3,......., x_n$$ are in A.P. Let the first term be $$a$$ and common difference be $$d$$

Also, $$x_2 = a + d = 2$$

and $$x_{24} = a + 23d = 68$$

Solving above equations, we get : $$a = -1$$ and $$d = 3$$

=> $$x_8 = a + 7d = -1 + 7(3) = 20$$

To find y coordinate, we will use $$y = px + q$$

$$\because A_2 (2, -2)$$ and $$A_{24} (68, 31)$$

Substituting in above equation, => $$-2 = 2p + q$$

and $$31 = 68p + q$$

Solving above equations, we get : $$p = \frac{1}{2}$$ and $$q = -3$$

$$\therefore y_8 = p x_8 + q = \frac{1}{2} (20) + (-3) = 7$$ 

$$x_1 , x_2 , x_3,......., x_n$$ are in A.P. Let the first term be $$a$$ and common difference be $$d$$

Also, $$x_2 = a + d = 2$$

and $$x_{24} = a + 23d = 68$$

Solving above equations, we get : $$a = -1$$ and $$d = 3$$

=> x coordinates = {$$x_1 , x_2 , x_3, x_4 , x_5 ,.......$$} = {$${-1 , 2 , 5 , 8 , 11 , 14 ,......}$$}        -------------(i)

To find y coordinates, we will use $$y = px + q$$

$$\because A_2 (2, -2)$$ and $$A_{24} (68, 31)$$

Substituting in above equation, => $$-2 = 2p + q$$

and $$31 = 68p + q$$

Solving above equations, we get : $$p = \frac{1}{2}$$ and $$q = -3$$

=> $$y_1 = px_1 + q = \frac{1}{2}(-1) + (-3) = -3.5$$

Similarly, y coordinates = {$$y_1 , y_2 , y_3, y_4 , y_5 ,.......$$} = {$${-3.5 , -2 , \frac{-1}{2} , 1 ,......}$$}         -------(ii)

From (i) and (ii),

=> $$A_1 = (-1 , -3.5)$$

$$A_2 = (2 , -2)$$

$$A_3 = (5 , \frac{-1}{2})$$

$$A_4 = (8 , 1)$$

For the points to lie in the first quadrant, the coordinates of both $$(x_k , y_k)$$ must be positive.

Since, $$d$$ is positive and $$y$$ is a linear relation in $$x$$, the corresponding coordinates of $$A_k$$ i.e. $$A_5$$ onwards will be increasing.

Thus, only for $$A_1 , A_2$$ and $$A_3$$ we do not have both coordinates positive.

$$\therefore$$ Only 3 points do not lie in the first quadrant.

Expression : $$f(x + y) = f(x).f(y)$$

and $$f(1) = 2$$

Putting, $$x=y=1$$, => $$f(1 + 1) = f(1).f(1)$$

=> $$f(2) = 2 \times 2 = 4$$

If $$x = 2 , y = 1$$ => $$f(3) = f(2).f(1)$$

=> $$f(3) = 4 \times 2 = 8$$

Similarly, $$f(4) = f(3).f(1) = 8 \times 2 = 16$$

The pattern followed : $$f(n) = 2^n$$

Now,  $$\sum_{x=1}^nf(x) = 1022$$

= $$f(1) + f(2) + f(3) + ............+ f(n) = 1022$$

=> $$2^1 + 2^2 + 2^3 + ......... + 2^n = 1022$$

The series is a G.P. with first term, $$a = 2$$ and common ratio, $$r = 2$$

=> Sum of G.P. = $$\frac{a (r^n - 1)}{r - 1}$$

=> $$\frac{2 (2^n - 1)}{2 - 1} = 1022$$

=> $$2^n - 1 = \frac{1022}{2} = 511$$

=> $$2^n = 511 + 1 = 512 = 2^9$$

Comparing both sides, we get : $$n = 9$$

Amarendra took 25 minutes to cover 15 km. In the same time Dharmendra travel 2500m less, i,e, 12.5 km

=> Speed of Amarendra = $$\frac{15}{25} = 0.6$$ km/min

Speed of Dharmendra = $$\frac{12.5}{25} = 0.5$$ km/min

=> Time taken by Dharmendra to reach the station = $$\frac{15}{0.5} = 30$$ minutes

Next day, Dharmendra started 7 minutes early, so he will reach the station = 30 - 25 - 7 = -2 minutes

=> 2 minutes or 120 seconds before Amarendra.

Length of side of $$S_{n + 1}$$ = Length of diagonal of $$S_n$$

=> Length of side of $$S_{n + 1}$$ = $$\sqrt{2}$$ (Length of side of $$S_{n}$$)

=> $$\frac{\textrm{Length of side of }S_{n + 1}}{\textrm{Length of side of }S_n} = \sqrt{2}$$

=> Sides of $$S_1 , S_2 , S_3 , S_4,........, S_n$$ form a G.P. with common ratio, $$r = \sqrt{2}$$

It is given that, $$S_3 = ar^2 = 4$$

=> $$a (\sqrt{2})^2 = 4$$

=> $$a = \frac{4}{2} = 2$$

$$\therefore$$ $$n^{th}$$ term of G.P. = $$a (r^{n - 1})$$

= $$2 (\sqrt{2})^{n - 1}$$

=$$2^[{\frac{n+1}{2}}]$$

Angle covered by the hour hand in 12 hours = 360°

In 1 hour = $$\frac{360}{12} = 30^{\circ}$$

and in 1 minute = $$\frac{30}{60} = \frac{1}{2}^{\circ}$$

Similarly, angle covered by minute hand in 1 hour = 360°

In 1 minute = $$\frac{360}{60} = 6^{\circ}$$

=> Every minute, the angle between the two hands changes by = $$6 - \frac{1}{2} = \frac{11}{2}^{\circ}$$

$$\therefore$$ From 7:45 A.M. to 7:47 A.M.,i.e. in 2 minutes the angle between the two hands will change by

= $$2 \times \frac{11}{2} = 11^{\circ}$$

In the figure, $$\triangle AQR \sim \triangle APS$$

=> $$\frac{AQ}{AP} = \frac{QR}{PS} = \frac{AR}{AS} = k$$ --------Eqn(I)

Statement I : PQ = 3 cm , RS = 4 cm , $$\angle$$ QPS = 60°

In right $$\triangle$$ PQM

=> $$sin 60^{\circ} = \frac{QM}{QP}$$

=> $$\frac{\sqrt{3}}{2} = \frac{QM}{3}$$

=> $$QM = \frac{3 \sqrt{3}}{2} = RN$$

Similarly, $$sin (\angle RSN) = \frac{3 \sqrt{3}}{8}$$

=> $$\angle RSN = sin^{-1} (\frac{3 \sqrt{3}}{8})$$

$$\therefore$$ In $$\triangle$$ APS

=> $$\angle PAS = 180^{\circ} - \angle APS - \angle PSA$$

=> $$\angle PAS = 120^{\circ} - sin^{-1} (\frac{3 \sqrt{3}}{8})$$

Thus, statement I alone is sufficient. 

Statement II : PS = 10, QR = 5

From eqn(I), $$k = \frac{1}{2}$$

But, we do not know anything regarding the measures of the remaining sides or any of the angles.

So, statement II is not sufficient.