CAT 2017 Question Paper Slot 2

The length of the diagonals of a regular hexagon with side s are $$\sqrt{3}s$$.
Here length of AC = $$\sqrt{3}s$$ = $$\sqrt{3}$$ cms
Hence area of the square = $$\sqrt{3}^2$$ = 3 sq cm

See the diagram below

Length of side AD = $$\sqrt{12^2 + 5^2 } = 13$$
Area of the trapezium = 12 * (10 + 20)/2 = 180
Perimeter of the trapezium = 10 + 20 + 13 + 13 = 56
Area of the sides of the pillar = 56 * height  = 56 * 20 = 1120.
Total are of the pillar = 1120 + area of base + area of the top = 1120 + 180 + 180 = 1480

The midpoint of one diagonal lies on the other diagonal.

Midpoint is ((2+6)/2, (5+3)/2) = (4,4)
Hence 4 = 3 * 4 + c => c = -8

$$\angle COD = 120$$ => $$\angle CAD = 120/2 = 60$$ (The angle subtended by the chord DC at the major arc is half the angle subtended at the centre of the circle.)
$$\angle BAC = 30$$
$$\angle BAD = \angle BAC + \angle CAD$$ = 30 + 60 = 90.
$$\angle BCD = 180 - \angle BAD$$ = 180 - 90 = 90

Let the length and breadth of the park be l,b, l > b
Case 1: 2l + b = 400
Area = lb. Area is maximum when 2l * b is maximum, which is maximum when 2l = b (using AM $$ \geq $$ GM inequality) => l = 100, b = 200. Which can't happen since l > b

Case 2: l + 2b = 400
Area = lb. Area is maximum when l *2 b is
maximum, which is maximum when l = 2b (using AM $$ \geq $$ GM
inequality) => l = 200, b = 100.

Hence length of the longer side is 200 ft

Let the length of non-hypotenuse sides of the right angled triangle be $$a$$. Then the hypotenuse h = $$\sqrt{2}a$$
P is equidistant from all the side of the triangle. Hence P is the incenter and the perpendicular distance is the inradius.
In a right angled triangle, inradius = $$\frac{a + b - h}{2}$$
=> $$\frac{a + a - \sqrt{2}a}{2} = 4(\sqrt{2}-1)$$
=> $$ \sqrt{2}a( \sqrt{2} - 1) = 8(\sqrt{2} -1)$$
=> $$ a = 4\sqrt{2}$$
Area of the triangle = $$\frac{1}{2}a^2$$ = 16 sq cm

$$(x -1)x(x+1) = 15600$$
=> $$x^3 - x= 15600 $$
The nearest cube to 15600 is 15625 = $$25^3$$
We can verify that x = 25 satisfies the equation above.
Hence the three numbers are 24, 25, 26. Sum of their squares = 1877

$$1 < \log_{3}5 < 2$$
=> $$ 1 < \log_{5}(2 + x) < 2 $$
=> $$ 5 < 2 + x < 25$$
=> $$ 3 < x < 23$$

$$f[f(g(1)) + g(f(1))]$$ 
= $$f[f(2^1) + g(1^2)]$$
= $$f[f(2) + g(1)]$$
= $$f[2^2 + 2^1]$$
= $$f(6)$$
= $$6^2 = 36$$

Let the roots of the equation $$x^2+(a+3)x-(a+5)=0 $$ be equal to $$p,q$$

Hence, $$p+q = -(a+3)$$ and $$p \times q = -(a+5)$$

Therefore, $$p^2+q^2 = a^2+6a+9+2a+10 = a^2+8a+19 = (a+4)^2+3$$

As $$(a+4)^2$$ is always non negative, the least value of the sum of squares is 3