SRCC GBO 2011 Question Paper

x% of y is 1% of z

$$\frac{x}{100} * y = \frac{1}{100} * z$$

xy = z ....... (1)

y% of z is 1% of x 

$$\frac{y}{100} * z = \frac{1}{100} * x$$

yz = x ....... (2)

z% of x is 1% of y

$$\frac{z}{100} * x = \frac{1}{100} * y$$

zx = y ....... (3)

Now,

xy + yz + zx = z + x + y

Put value of z from equation (1) in equaiton (2):

y * xy = x

$$y^2 = 1$$

y = 1 Similarly, 

x = z = 1

Hence, x + y + z = 1 + 1 + 1 = 3

The given expression is $$(10^{4n^2 + 8} + 1)^2$$.

The expression will be of the form $$(10000....1)^2$$

Any number of the form 1000(x times)1 will be of the form 1000(x times)20000(x times)1.

For example $$1001^2$$=1002001

So, the sum of the digits will always we 1+2+1=4

$$=x^4y + xy^4$$

$$=xy(x^3 + y^3)$$

$$=xy(x + y)(x^2 + y^2 - xy)$$

$$=xy(x^2 + y^2 - xy)$$                                  [Since, x + y = 1]

$$=xy[(x + y)^2 - 2xy - xy)]$$

$$=xy[1 - 3xy]$$

Ley xy = a

$$S=a[1 - 3a]$$

$$S=a - 3a^2$$

$$\frac{\text{d}S}{\text{d}a} = 1 - 6a$$

For maximum; 

$$\frac{\text{d}S}{\text{d}a} = 1 - 6a = 0$$

$$a = xy = \frac{1}{6}$$

After putting $$xy = \frac{1}{6}$$ :

$$=xy[1 - 3xy]$$

$$=\frac{1}{6}(1 - 3 * \frac{1}{6})$$

$$=\frac{1}{12}$$

Let first natural number = a

Second natural number = a + 1

15th natural number = a + 14

According to the question:

$$\frac{a + a + 1 + a + 2 + ........... + a + 14}{15} = 13$$

15a + 105 = 195

a = 6

Second largest of these nubers = a + 13 = 19

Let first term $$a_1 = a$$ and common difference = d

$$a_2 = a + d; a_3 = a + 2d$$

Now, 

$$a_1 + a_8 + a_{12} + a_{19} = 224$$

a + a + 7d + a + 11d + a + 18d = 224

4a + 36d = 224

a + 9d = 56 ....... (1)

= $$a_1 + a_2 + a_3 + ........ + a_{19}$$

= a + a + d + a + 2d + .......... + a + 18d

= 19a + 171d

= 19(a + 9d)

= 19 * 56

= 1064

$$V \propto \frac{1}{\sqrt{m}}$$

$$m \propto \frac{1}{t^2}$$

Now, 

$$V \propto \frac{1}{\sqrt{\frac{1}{t^2}}}$$

$$V \propto t$$

Let side of square = 'a'

Diameter of circle = Diagonal of square = $$a\sqrt{2}$$

Radius of circle = $$\frac{a\sqrt{2}}{2} = \frac{a}{\sqrt{2}}$$

Let side of triangle = s

Length of iInradius of triangle = $$\frac{s}{2\sqrt{3}} = \frac{a}{\sqrt{2}}$$

$$s = a\sqrt{6}$$

Area of triangle = $$\frac{\sqrt{3}}{4}s^2 = \frac{\sqrt{3}}{4}(a\sqrt{6})^2 = \frac{3\sqrt{3}a^2}{2}$$

Area of square = $$a^2$$

Required ratio = $$\frac{3\sqrt{3}a^2}{2}: a^2 = 3\sqrt{3}: 2$$

Radius of larger circle = PC + Circum Radius of $$\triangle{ABC}$$            [PC = Radius of smaller circle = 1 cm]

$$\triangle{ABC}$$ is equilateral triangle with side = 2 cm

Circum of an equilateral triangle = $$\frac{a}{\sqrt{3}} = \frac{2}{\sqrt{3}}$$

Radius of large circle = $$1 + \frac{2}{\sqrt{3}} = \frac{\sqrt{3} + 2}{\sqrt{3}} = \frac{3 + 2\sqrt{3}}{3}$$

a + b < c + d ..... (1)

b + c < d + e ...... (2)

c + d < e + a ....... (3)

d + e < a + b ....... (4)

From (1) and (4):

d + e < a + b < c + d => e < c

From (1) and (3):

a + b < c + d < e + a => b < e

From (2) and (4):

b + c < d + e < a + b => c < a

Now, 

b < e < c < a

Let b = 1, e = 2, c = 3 and a = 4

From (1) and (3): 

a + b < c + d => 4 + 1 < 3 + d => d > 2 => d > e

c + d < e + a => 3 + d < 2 + 4 => d < 3 => d < c

Hence, 

b < e < d < c < a

Largest number is a and smallest number is b.

$$2^{x + y} = 10$$ ..... (1)

$$2^{y + z} = 20$$ ....... (2)

$$2^{z + x} = 30$$........ (3)

From (1), (2) and (3):

$$2^{2(x + y + z)} = 10 * 20 * 30 = 6000 = 400 * 15$$

$$2^{x + y + z} = 20\sqrt{15}$$ .... (4)

From (3) and (4):

$$\frac{2^{x + y + z}}{2^{y + z}} = \frac{20\sqrt{15}}{20}$$

$$2^x = \sqrt{15}$$