XAT 2014

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.