XAT 2014 Question Paper - QUANT

The terms $$x, 17, 3x - y^{2} - 2$$ and $$3x + y^{2} - 30$$ are in A.P.

=> Common difference = $$d = 17 - x$$ ----------Eqn(I)

$$d = 3x - y^2 - 19$$ ----------Eqn(II)

$$d = 2y^2 - 28$$ ----------Eqn(III)

From eqn(I) & (II), => $$17 - x = 3x - y^2 - 19$$

=> $$4x - y^2 = 36$$ -------Eqn(IV)

From eqn(II) & (III), => $$3x - y^2 - 19 = 2y^2 - 28$$

=> $$x - y^2 = -3$$ ---------Eqn(V)

Solving eqn(IV) & (V), we get : 

$$x = 13 , y^2 = 16$$

=> Terms are = $$13,17,21,25$$

$$\therefore$$ Sum = $$13+17+21+25 = 76$$, which is divided by 2.   (among the given options)

In a triangle , sum of two sides is greater than the third side and difference of two sides is less than third side.

In $$\triangle$$ PQS

=> $$QS < 9 + 5$$ => $$QS < 14$$ ------Eqn(I)

In $$\triangle$$ QRS

=> $$QS > 17 - 5$$ => $$QS > 12$$ ------Eqn(II)

From eqn(I) & (II)

=> $$12 < QS < 14$$

Expression : $$|x + 7| + |x - 8| = 16$$

Case 1 : $$x < -7$$

=> $$-(x + 7) - (x - 8) = 16$$

=> $$-2x + 1 = 16$$

=> $$x = \frac{-15}{2} = - 7.5$$

Case 2 : $$-7 \leq x < 8$$

=> $$(x + 7) - (x - 8) = 16$$

=> $$15 = 16$$, which is not possible.

Case 3 : $$x \geq 8$$

=> $$(x + 7) + (x - 8) = 16$$

=> $$2x - 1 = 16$$

=> $$x = \frac{17}{2} = 8.5$$

$$\therefore$$ Sum of all possible values of $$x = 8.5 - 7.5 = 1$$

Given : Length of ladder = PA = P'B = 30 m and PQ = 26 m

$$\angle$$ PAQ = 2 $$\angle$$ P'BQ

To find : AB = ?

Solution : In $$\triangle$$ PAQ

=> $$(QA)^2 = (PA)^2 - (PQ)^2$$

=> $$(QA)^2 = 30^2 - 26^2 = 900 - 676$$

=> $$QA = \sqrt{224} \approx 15$$

Also, $$cos 2 \theta = \frac{QA}{PA}$$

=> $$2 \theta = cos^{-1}(\frac{15}{30})$$

=> $$2 \theta = 60$$ => $$\theta = \frac{60}{2} = 30$$

In $$\triangle$$ P'QB

=> $$cos 30 = \frac{QB}{P'B}$$

=> $$QB = \frac{\sqrt{3}}{2} \times 30$$

=> $$QB = 15 \times 1.732 = 25.98$$

$$\therefore$$ AB = QB - QA = 25.98 - 15 = 10.98 m

Let the no. chosen by Amitabh be x

In the first step, the result got by Amitabh is 2x and the result got by Shashi is 2x+50.

In the second step, the result got by Amitabh is 4x+100 and the result got by Shashi is 4x+150.

In the third step, the result got by Amitabh is 8x+300 and the result got by Shashi is 8x+350.

In the fourth step, the result got by Amitabh is 16x+700 and the result got by Shashi is 16x+750.

In the fifth step, the result got by Amitabh is 32x+1500 and the result got by Shashi is 32x+1550.

In step 5, since both of their results are greater than 1000, so step 4 will be the last step.

Since Amitabh wins the game the result of Shashi must be greater than 1000.

Now, $$16x+750\ge\ 1000$$

or $$x\ge\ 16$$

Hence the minimum value of x should be 16 and the its sum of digits is 7.

Natural numbers = $$x , y , (x+y) , (x-y)$$

Statement I : As all the numbers are prime, therefore, either x or y has to be 2 because otherwise (x+y) cannot be prime.

Case 1 : If x = 2, then (x-y) cannot be prime

Case 2 : If y = 2, numbers = $$(x-2) , x , (x+2)$$

These numbers are prime, hence all possibility = 3,5,7

$$\therefore$$ Sum = 2+3+5+7 = 17

Using statement II, we cannot find the required sum, as no specific value of mean is given.

Thus, statement I alone is sufficient.

ABC is right angled triangle. D and E are mid points of AB and BC respectively.

As it is not given that which angle is 90°. So we need statement (III) to find the value of AC.

Whereas using statement (I) & (II) alone, we cannot find the value of AC.

But using all the three statements. We can find value of $$(AB)^2 + (BC)^2$$.

Hence, the value of $$AC = \sqrt{(AB)^2 + (BC)^2}$$ using Pythagoras Theorem.

=> Ans - (E) : All statements are required.

Given : $$OP_1 = 3 , O'P_2 = 8 , OX = 5$$ units

To find : $$P_1P_2 = ?$$

Solution : In $$\triangle OP_1X$$

=> $$(P_1X)^2 = (OP_1)^2 - (OX)^2$$

=> $$(P_1X)^2 = 5^2 - 3^2 = 25 - 9$$

=> $$P_1X = \sqrt{16} = 4$$

In $$\triangle OP_1X$$ and $$\triangle O'P_2X$$

$$\angle OXP_1 = O'XP_2$$   (Vertically opposite angles)

$$\angle OP_1X = O'P_2X = 90$$

=> $$\triangle OP_1X \sim \triangle O'P_2X$$

=> $$\frac{XP_1}{XP_2} = \frac{OP_1}{O'P_2}$$

=> $$XP_2 = 4 \times \frac{8}{3} = 10.66$$

$$\therefore P_1P_2 = P_1X + XP_2 = 4 + 10.66 = 14.66$$ units

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

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

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

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

Expression : $$\sum_{i = 2}^{100} \frac{1}{log_{i}100!}$$

= $$\frac{1}{log_2 100!} + \frac{1}{log_3 100!} + \frac{1}{log_4 100!} +$$ ..... $$+ \frac{1}{log_{100} 100!}$$

We know that $$\frac{1}{log_b a} = log_a b$$

= $$log_{100!} 2 + log_{100!} 3 + log_{100!} 4 +$$ ..... $$+ log_{100!} 100$$

Also, $$log_a b + log_a c = log_a (b \times c)$$

= $$log_{100!} (2 \times 3 \times 4 \times 5 \times ..... \times 100)$$

= $$log_{100!} 100! = 1$$

Area of $$S_1 = 8$$ sq. units

=> Side of $$S_1 = PS = \sqrt{8} = 2 \sqrt{2}$$ units

Similarly, Side of $$S_2 = CD = \sqrt{9} = 3$$ units

=> $$a + b = 3$$

In $$\triangle$$ PDS

=> $$b^2 + a^2 = 8$$

=> $$b^2 + (3 - b)^2 = 8$$

=> $$b^2 + 9 + b^2 - 6b = 8$$

=> $$2b^2 - 6b + 1 = 0$$

=> $$b = \frac{6 \pm \sqrt{36 - 8}}{4} = \frac{6 \pm \sqrt{28}}{4}$$

=> $$b = \frac{3 + \sqrt{7}}{2}$$     $$(\because b > a)$$

=> $$a = 3 - \frac{3 + \sqrt{7}}{2} = \frac{3 - \sqrt{7}}{2}$$

$$\therefore \frac{b}{a} = \frac{\frac{3 + \sqrt{7}}{2}}{\frac{3 - \sqrt{7}}{2}}$$

= $$\frac{3 + \sqrt{7}}{3 - \sqrt{7}} \approx 15.9$$

Radius of cone = $$BP = 13$$ cm

In $$\triangle$$ ABC

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

=> $$(AB)^2 = 17^2 - 15^2 = 289 - 225$$

=> $$AB = \sqrt{64} = 8$$ cm

=> AP = OC = $$BP - AB = 13 - 8 = 5$$ cm

Let time elapsed before water stops coming out of the pipe = $$T$$ min

Volume of frustum = Volume discharged

=> $$\frac{1}{3} \pi h (R^2 + r^2 + R r) = \pi  (0.5)^2 \times f \times T$$    (where f is the flowrate given by 10m/s = 1000cm/s)

(Total volume = Cross sectional area of pipe $$\times$$ flowrate $$\times$$ Total time)

=> $$\frac{1}{3} \times 15 \times (13^2 + 5^2 + 13 * 5) = 0.25 \times 1000 \times T$$

=> $$5 \times 259 = 250 \times T$$

=> $$T = \frac{259}{50} = 5.18$$ mi

Probability of both marbles being red = Probability of red from 1st * probability of red from 2nd

= $$\frac{5}{16}$$

$$\because$$ Each bag has marbles of both colors and probability cannot be greater than 1

=> $$\frac{5}{16} = \frac{5}{8} \times \frac{1}{2}$$ 

where, probability of red marbles from 1st bag = $$\frac{5}{8}$$

=> Probability of blue marbles from 1st bag = $$1 - \frac{5}{8} = \frac{3}{8}$$

Similarly, Probability of red marbles from 1st bag = $$\frac{1}{2}$$

=> Probability of blue marbles from 2nd bag = $$1 - \frac{1}{2} = \frac{1}{2}$$

$$\therefore$$ Probability of both blue marbles = $$\frac{3}{8} \times \frac{1}{2}$$

= $$\frac{3}{16}$$

From the above table, it is clear that the percentage increase in the number of Indians going abroad was greater than the percentage increase in the number of domestic tourists

for the years 2005 and 2007

C is the correct answer.

from the above table, it is clear that the value of rupee wrt to the dollar is cheapest in the year 2002

Hence B is the correct answer.

If we try to plot a pie chart using the values of R for all the years, 2011 will subtend maximum value since its R value is maximum among all the years.

Angle = $$\frac{2633.39}{26415.49}\times\ 360^0$$

=35.88 = $$36^{0}$$ approximately.

C is the correct answer.

After the USA, we have to find a country with the highest spending on the Military as a % of GDP.

From the chart for Military Expenditure, it can be easily seen that the bar graph for India is higher than any other country except the USA.

Hence, option C is the answer.

This change can be observed by viewing the chart.

Where all other countries saw a gradual drop in their “industry as a % of GDP”, for Malaysia in 2008-09, the fall was sudden and had the maximum value among the options.

Hence, option D is the answer.

From the chart for “services, value-added as a % of GDP” we can see that every other country, except India, has a constant value or decrease from 2000-10, whereas India shows a slow but gradual increase.

Hence, option B is the answer.

We have to mark the option which displays an ascending order of democratic regions for the poor.

From the charts, we can see that the lowest value is displayed by C & E Europe (<5).

Option D has C & E Europe as the first name, so it could be the answer. Checking the remaining regions in option D, we get Africa (approx. 10), South America (14), Western Europe (19.99) and Scandinavia (20).

Thus, the order should be C&E Europe, Africa, South America, Western Europe and Scandinavia.

Hence, option D is the answer.

We have to find the region with the highest disparity in democratic participation between rich and poor. We can check this by calculating the difference between the highest and lowest values from the chart.

North America = 20 - 18 = 2

C&E Europe = 20 - 5 = 15

Africa = 11 - 10 = 1

South America = 20 - 14 = 6

Western Europe = 19.95 - 19.85 = 0.1

Thus, C&E Europe has the highest disparity between the rich and the poor.

Hence, option B is the answer.

From the graph, we can say that the maximum GDP of African and South American region has the same value. i.e 20

So E is the correct answer.