XAT 2015 Question Paper

We need to identify opportunities first and then implement them accordingly. Therefore, order is V, I, II.

The answer is option C.

The series is an A.P. with common difference, $$d = -66 - (-64) = -2$$

First term, $$a = -64$$ and last term $$a_n = -100$$

nth term of the series, $$a_n = a + (n - 1) d$$

=> $$-100 = -64 + (n - 1) (-2)$$

=> $$n - 1 = \frac{-36}{-2} = 18$$

=> $$n = 18 + 1 = 19$$

$$\therefore$$ Sum = $$\frac{n}{2} (a + a_n)$$

= $$\frac{19}{2} \times (-64 -100) = \frac{19}{2} \times (-164)$$

= $$19 \times (-82) = -1558$$

Volume of Cylinder = $$\pi r^2 h = \pi \times 7^2 \times 10 = 490\pi$$

Now, The solid metal cylinder is re-cast into two cones in the proportion 3 : 4 i.e. the volumes of cone 1 and cone 2 is

$$210 \pi$$ and $$280 \pi$$ respectively.

So, flat Surface area of cylinder before melting = $$2 \pi r^2 = 2 \pi \times 7^2 = 98 \pi$$

Volume of cone 1 = $$\frac{1}{3} \pi r_1^2 h = 210 \pi$$

=> $$r_1^2 = \frac{210 \times 3}{10} = 63$$

Volume of cone 2 = $$\frac{1}{3} \pi r_2^2 h = 280 \pi$$

=> $$r_2^2 = \frac{280 \times 3}{10} = 84$$

Flat surface area of cones = $$\pi r_1^2 + \pi r_2^2$$

= $$\pi (63 + 84) = 147 \pi$$

$$\therefore$$ Percentage change in surface area = $$\frac{147 \pi - 98 \pi}{98 \pi} \times 100$$

= $$\frac{1}{2} \times 100 = 50 \%$$

Let Manufacturing Cost of the product = $$Rs. 100$$

=> Maximum Retail Price(MRP) = $$100 + \frac{55}{100} \times 100 = Rs. 155$$

Retailer gives 10% discount on MRP

=> Retailer's selling price = $$155 - \frac{10}{100} \times 155 = Rs. 139.5$$

It is given that the retailer earned 23% profit on his purchase price, say $$Rs. x$$

=> $$\frac{123 x}{100} = 139.5$$

=> $$x = \frac{13950}{123} = 113.41$$

Now, the purchase price of retailer = $$x$$ = selling price of Manufacturer

$$\therefore$$ Profit earned by Manufacturer = $$113.41 - 100 = 13.41$$

$$\approx 13 \%$$

The probability that his friend receives the gift in time will be when his friend receives even one gift.

That can be calculated as the probability of his friend receiving at least one gift.

The probability that none of the retailers sends in time = $$(1 - 0.6) \times (1 - 0.8) \times (1 - 0.9) \times (1 - 0.5)$$

= $$0.4 \times 0.2 \times 0.1 \times 0.5 = 0.004$$

$$\therefore$$ Probability of his receiving at least one gift = $$1 - 0.004 = 0.996$$

Area of given figure = 144 sq meter

It is given that BCE becomes square when we will unfold it, so to find the complete area of the figure shown as dotted after unfolding we need to add the area of triangle BCE.

Thus, BC = CE = 6 m

=> Area of $$\triangle$$ BCE = $$\frac{1}{2} \times 6 \times 6 = 18$$ sq meter

$$\therefore$$ Final area of whole figure = 144 + 18 = 162 square meter.

When x  = -3, y = -10
This is satisfied only in option D.

Hence, option D is the correct answer.

Let the quantities of the chemicals X and Y, mixed to produce product M be $$5c$$ and $$4c$$ respectively.

X is prepared by mixing A and B in the ratio = 1 : 3

=> Quantity of B in X = $$\frac{3}{4} \times 5c = \frac{15 c}{4}$$

Y is prepared by mixing B and C in the ratio = 2 : 1

Quantity of B in Y = $$\frac{2}{3} \times 4c = \frac{8 c}{3}$$

Quantity of B in M = $$\frac{15 c}{4} + \frac{8 c}{3} = \frac{77 c}{12}$$

Now, 864 units of M was mixed with water to prepare the final mixture.

=> Total quantity of M = $$9c = 864$$ => $$c = \frac{864}{9} = 96$$

Concentration of raw material B in the final mixture is 50 %

=> Quantity of final mixture = $$\frac{100}{50} \times \frac{77}{12} \times 96 = 1232$$

$$\therefore$$ Quantity of water added to M = $$1232 - 864 = 368$$ units

Perimeter of square ABCD = 200 ft

=> AB = $$\frac{200}{4} = 50$$ ft

=> $$DB = \sqrt{50^2 + 50^2} = 50 \sqrt{2}$$ ft

=> $$BO = r = \frac{50 \sqrt{2}}{2} = 25 \sqrt{2}$$ ft

Width of the road = BX = $$7 \sqrt{2}$$ ft

=> $$BX = R = 25 \sqrt{2} + 7 \sqrt{2} = 32 \sqrt{2}$$

Area of bigger circle = $$\pi R^2 = \pi (32 \sqrt{2})^2 = 2048 \pi$$ sq. ft

Area of smaller circle = $$\pi r^2 = \pi (25 \sqrt{2})^2 = 1250 \pi$$ sq. ft

=> Area of road = $$2048 \pi - 1250 \pi = 798 \times \frac{22}{7} = 2508$$ sq. ft

But we have to calculate cost of construction of 50% road.

Required Construction = $$\frac{2508}{2} = 1254$$ sq. ft

$$\therefore$$ Cost of 1254 ft = $$1254 \times 100 = Rs. 1,25,400$$

In $$\triangle$$ ABF

=> $$tan 30 = \frac{BF}{AB}$$

=> $$\frac{1}{\sqrt{3}} = \frac{10}{AB}$$

=> $$AB = 10 \sqrt{3}$$

Similarly, $$ED = 10 \sqrt{3}$$

Also, $$\angle ECD = \angle BCF = 60$$ (Vertically opposite angles)

In $$\triangle$$ BCF

=> $$tan 60 = \frac{BF}{BC}$$

=> $$\sqrt{3} = \frac{10}{BC}$$

=> $$BC = \frac{10}{\sqrt{3}}$$

=> Height = $$AD = AB + BC + CD = 10 \sqrt{3} + \frac{10}{\sqrt{3}} + 10 = \frac{40 + 10 \sqrt{3}}{\sqrt{3}}$$

$$\therefore area (\triangle AED) = \frac{1}{2} \times AD \times ED$$

= $$\frac{1}{2} \times \frac{40 + 10 \sqrt{3}}{\sqrt{3}} \times 10 \sqrt{3}$$

= $$50 (\sqrt{3} + 4)$$