SRCC GBO 2011

Since 'x' and 'y' are positive integers. Let x = y = 1

x + y = 1 + 1 = 2

$$\sqrt{3xy} = \sqrt{3 * 1 * 1} = \sqrt{3} = 1.732$$

$$\frac{x^2 + y^2}{x + y} = \frac{1^2 + 1^2}{1 + 1} = 1$$

$$\frac{x^4 + x^2y^2 + y^4}{x^3 + y^3} = \frac{1^4 + 1^2 * 1^2 + 1^4}{1^3 + 1^3} = 1.5$$

Hence, x + y is largest. 

The equation shall be: 25a+16b+20c = 213 and a+b+c = 12

Since the final total is 213, the only possibility of a sum ending with 3 is when the resultant of 16 ends with 8 (48 or 128) and the resultant of 25 ends with 5, thereby giving 3 as the unit digit of the sum.

Case 1: When 16b is 48

This means b = 3

Hence, 25a + 20c = 165

Now, a+c = 9 (as b = 3)

These two equations are inconsistent since one of the values is coming out a negative integer. 

Case 2: When 16b = 128

This means b = 8

Hence, 25a + 20c = 85

And a+c=4 (as b=8)

Hence, a=1 and c=3

By this, we know that the number of months with 25 problems is 1.

Let y $$=\sin^6 x + \cos^6 x$$

y $$=(\sin^2 x)^3 + (\cos^2 x)^3$$

y $$=(\sin^2 x + \cos^2 x)^3 - 3\sin^2 x \cos^2 x(\sin^2 x + \cos^2 x)$$

y $$=1 - 3\sin^2 x \cos^2 x$$

y $$=1 - 3(\sin x \cos x)^2$$

y $$=1 - \frac{3}{4}(2\sin x \cos x)^2$$

y $$=1 - \frac{3}{4}(sin2 x)^2$$

When sin2x is maximum, y will be minumum and maximum value of sin2x is 1. 

y (min) = $$1 - \frac{3}{4} * 1 = \frac{1}{4}$$

When sin2x is minimum, y will be maximum and minimum value of sin2x is 0.

y (max) = $$1 - \frac{3}{4} * 0 = 1$$

Required ratio = 1: $$\frac{1}{4}$$ = 4: 1

The best way to proceed with this question is to assume m as a very small integer. For this question, let us say that m=2

Hence, a = 11 and b = 105

Hence,  $$\sqrt{ab + 1}$$ is $$\sqrt{1156}$$, which is 34.

Clearly, the number of 3s is m-1, aa m-1 = 1.

This is clearly in line with option A.

Simplifying the expression: $$\ \frac{\ \log\ x}{\log\ 3}\times\ \ \frac{\ \log\ 2x}{\log\ x}\times\ \ \frac{\ \log\ y}{\log\ 2x}$$, which is $$\ \log_3y$$

Now, the RHS of the equation is equal to 2. 

Hence, $$\ \log_3y$$ = 2, or y is 

Let the length of each small rectangular strips is 'l' and its breadth is 'b'.

Breadth of piece of paper = Length of each small strips = l

Length of piece of paper = Sum of breadth of 'n' small strips = nb

Now, 

$$\frac{l}{b} = \frac{nb}{l}$$

$$l^{2} = nb^{2}$$

$$l = \sqrt{n}b$$

$$l: b = \sqrt{n}: 1$$

Since, $$\triangle{PQR}$$ and $$\triangle{QRS}$$ are similar equilateral triangles. 

Side of $$\triangle{PQR}$$ = Diameter of any circle = 4 cm

$$Area of \triangle{PQR} = \frac{\sqrt{3}}{4}PR^{2} = \frac{1}{2} * QR * \frac{PS}{2}$$

$$\frac{\sqrt{3}}{4}4^{2} = \frac{1}{2} * 4 * \frac{PS}{2}$$

$$PS = 4\sqrt{3}$$

Lenght of BC = PS + 2 + 2 = $$4\sqrt{3} + 4 = (4{\sqrt{3} + 1}) cm$$

Let length, breadth and height of cuboid is 'l', 'b' and 'h' respectively.

According to the question: 

l + b + h = 19 ........ (1)

Diagonal of cuboid = $$\sqrt{l^{2} + b^{2} + h^{2}} = 5\sqrt{5}$$

$$l^{2} + b^{2} + h^{2} = 125$$ ...... (2)

We know that: 

$$(l + b + h)^{2} = l^{2} + b^{2} + h^{2} + 2(lb + bh + hl)$$

From eqaution (1) and (2):

$$19^{2} = 125 + 2(lb + bh + hl)$$

$$2(lb + bh + hl) = 361 - 125 = 236$$

Surface area of cuboid = $$2(lb + bh + hl) = 236 cm^{2}$$

Base radius of cone = AQ = BQ = 5 cm

AB = 2AQ = 10 cm

Height of cone = PQ = 12 cm

Let QY = x and OQ = 2x

MN = XY = 2x

Now, 

$$\frac{MN}{AB} = \frac{PO}{PQ}$$

$$\frac{2x}{10} = \frac{12 - 2x}{12}$$

6x = 30 - 5x

$$x = \frac{30}{11}$$

Ddiameter of the cylinder = XY = 2x = $$\frac{60}{11} cm$$