XAT 2016 Question Paper

AB + BC + CD = 70 ------------(i)

AB + BD = 45 ------------------(ii)

AC + CD = 30 ------------------(iii)

AC + CB + BD = 65 ------------(iv)

Adding (i) & (iv)

=> AB + BC + CD + AC + CB + BD = 70 + 65

(AB + BD) + 2 BC + (AC + CD) = 135

From (ii) & (iii)

=> 45 + 2 BC + 30 = 135

=> 2 BC = 135 - 75 = 60

=> BC = $$\frac{60}{2} = 30$$ min

10 hour analog clock is used. 

Time in mirror = 3 hours 42 minutes and 20 seconds

=> Actual time = 10 - (3 hrs 42 min 20  sec)

= 6 hrs 17 min 20 sec

$$\therefore$$ Time after 5 minutes = 6 hrs 22 min 20 sec

|a| ≠ |b| ≠|c| and -10 ≤ a, b, c ≤ 10

Expression : $$[abc - (a + b + c)]$$

For maximum value, two of a,b and c should be negative, as all three negative will make abc negative.

Thus, max value will occur if $$a = -10 , b = -9 , c = 8$$

=> Max value = $$[(-10 \times -9 \times 8) - (-10 -9 + 8)]$$

= $$720 + 11 = 731$$

When a square sheet is folded in half, its area is also halved. Now, an isosceles right triangle is formed after folded. For 2nd & 3rd folds, again areas halved each time and isosceles right triangle is formed

Equal sides = 10 cm

Area of last such triangle = $$\frac{1}{2} \times 10 \times 10 = 50 cm^2$$

$$\therefore$$ Area of original square = $$50 \times 2 \times 2 \times 2$$

= $$400 cm^2$$

Let radius of incircle = $$r$$, => Radius of circumcircle = $$2r$$

Difference in area = $$\pi [(2r)^2 - (r)^2] = 2156$$

=> $$3 \times \frac{22}{7} \times r^2 = 2156$$

=> $$r^2 = \frac{2156 \times 7}{66}$$

=> $$r = \sqrt{\frac{686}{3}}$$

Now, height of equilateral triangle = $$3 r = \frac{\sqrt{3}}{2} a$$    (where $$a$$ is side of triangle)

=> $$3 \times \sqrt{\frac{686}{3}} = \frac{\sqrt{3}}{2} a$$

=> $$a = 2 \sqrt{686}$$

$$\therefore$$ Area of triangle = $$\frac{\sqrt{3}}{4} a^2$$

= $$\frac{\sqrt{3}}{4} \times 4 \times 686 = 686 \sqrt{3} cm^2$$

If the object does not change direction at point C then it will be at D above the person. But since it is given that object changed direction and continued flying upwards, thus object would reach point E. 

=> CG is perpendicular bisector to AB => BG = 300 m

In $$\triangle$$ BCG

=> $$cos 30 = \frac{BG}{BC}$$

=> $$\frac{\sqrt{3}}{2} = \frac{300}{BC}$$

=> $$BC = \frac{600}{\sqrt{3}} = 200 \sqrt{3}$$

Now, according to statement I, time can increase only when the angle with ground level will increase.

But, according to statement II, if CD = $$200 \sqrt{3}$$ (= BC), then at constant speed, it will not take any additional time and thus there should not be any increase in angle.

But in first statement direction has been changed whereas in second statement direction has not been changed. 

So both the statements are inconsistent with each other. 

If $$(a + b)^{(a + b)}$$ is divisible by 500, 

$$500 = 2^2 \times 5^3$$

=> Least value of $$a + b = 2 \times 5 = 10$$

For least $$a$$ and $$b$$, let $$a = 1$$

=> $$b = 10 - 1 = 9$$

$$\therefore$$ Min $$(a \times b) = 1 \times 9 = 9$$

Let walking speed = $$a$$

=> Driving speed = $$8a$$ 

Let the distance between temple and office = x

Acc. to ques,

=> $$\frac{x}{a} = \frac{1}{a} + \frac{x + 1}{8 a}$$

=> $$8x = 8 + x + 1$$

=> $$x = \frac{9}{7}$$

Since a,b,c are consecutive integers

=> $$a = b-1$$ and $$c = b+1$$

Expression : $$\frac{a^3+b^3+c^3+3abc}{(a+b+c)^2}$$

= $$\frac{(b - 1)^3 + b^3 + (b + 1)^3 + 3 (b - 1) b (b + 1)}{(b - 1 + b + b + 1)^2}$$

= $$\frac{b^3 + 3b + b^3 + b^3 + 3b + 3b^3 - 3b}{9 b^2}$$

= $$\frac{6 b^3 + 3 b}{9 b^2} = \frac{2 b^2 + 1}{3 b}$$

Putting different values of b from - 10 to 10, we can verify that only - 1 and 1 satisfies to get integer values for the expression.

Ans - (C)

Let P and Q be the centres of the circles with arcs x and y respectively.

Thus, $$\angle APB = 90$$ and $$\angle AQB = 60$$

Also, length of arc $$y = 10 \pi$$ cm

=> $$\frac{\theta}{360} \times 2 \pi r = 10 \pi$$

=> $$\frac{1}{6} \times 2 \times r = 10$$

=> $$r = AQ = 10 \times 3 = 30$$ cm

=> AB = 30 ($$\because \triangle$$ AQB is equilateral triangle)

Also, $$\triangle$$ APB is right isosceles triangle, => $$AP = \frac{30}{\sqrt{2}}$$ 

$$\therefore$$ Arc length = $$x = \frac{90}{360} \times 2 \pi \times \frac{30}{\sqrt{2}}$$

= $$\frac{15 \pi}{\sqrt{2}}$$