SRCC GBO 2015 Question Paper

Let us assume A, B, and C have x with the at the end.

In the last round, chocolates went from C to A and B:

=> A and B will have 2x/3 chocolates and the remaining chocolates will be with C = 5x/3 [why y + y/2 = x => y = 2x/3]

Now, B gives chocolates to A and C

=> A will have 4x/9, C will have 10x/9 => B has 13x/9 chocolates

Now, A gave to B and C

=> B will have 26x/27, C will have 20x/27, A will have 7x/27

=> For this to be integer x = 27 => Total chocolates with them = 27 * 3 = 81.

The centre of the circle would act as the centroid of the triangle, from where all the median shall pass. 

The length of the median is $$\ \frac{\ \sqrt{\ 3}}{2}\times\ 10$$, which is $$5\sqrt{\ 3}$$ cm. 

Now, this median shall be divided in 2:1 by the centroid (centre of the circle), and hence the smaller part which is the radius of that circle becomes $$\ \frac{\ 5}{\sqrt{\ 3}}$$cm (the breadth of the rectangle).

Double of this shall be length of the triangle, which is $$\ \frac{\ 10}{\sqrt{\ 3}}$$cm.

Hence, the area of the traingle is $$\ \ \frac{\ 50}{3}$$cm, or 16.67 cm.

L.C.M. of (6,7,8,9,12) = 504

Thus, after every 504 seconds = 8.4 minutes, the bells toll together.

=> Number of times, they will toll together in an hour = $$\frac{60}{8.4}=7.14=$$ 7 times

=> Ans - (D)

From the question, we should arrange younger and older people in an alternate manner.

First, we should arrange the 4 elder people in a circular arrangement in (4 - 1)! = 6 ways.

Now, for the 4 younger people, this is similar to a linear arrangement, as the gaps are distinct positions = 4! = 24 ways.

=> Required number of total arrangements = 6 * 24 = 144 ways.

First 20 prime numbers = 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71

Thus, median is mean of 10th and 11th value = $$\frac{(29+31)}{2}$$

= $$\frac{60}{2}=30$$

=> Ans - (D)

Let the two numbers be $$x$$ and $$y$$

=> Sum = $$x+y=17$$

Squaring both sides, we get : $$(x+y)^2=(17)^2$$

=> $$x^2+y^2+2xy=289$$

Also, it is given that $$x^2+y^2=145$$

=> $$2xy=289-145=144$$

=> $$xy=\frac{144}{2}=72$$

=> Ans - (A)

L.C.M. of (3,5,6,9) = 90

Now, the numbers of the form $$90k$$ will be divisible by 3,5,6,9, where $$k$$ is a natural number

Prime Factorization of 90 = $$3^2\times2\times5$$

Thus, in order to have both even powers, we need to multiply it by 10 = 900

=> Ans - (A)

Expression : $$\left(1 + \frac{x}{144}\right)^{\frac{1}{2}} = 1 + \frac{1}{2}$$

=> $$1+\frac{x}{144}=(\frac{3}{2})^2$$

=> $$\frac{x}{144}=\frac{9}{4}-1$$

=> $$x=\frac{5}{4}\times144=180$$

Let the numbers be $$x$$ and $$y$$

Given : $$(x-y)=3$$ -----------(i) and $$x^2+y^2=369$$ ------------(ii)

Squaring both sides in equation (i), we get : $$x^2+y^2-2xy=9$$

Substituting value from equation (ii), => $$2xy=369-9=360$$

=> $$xy=\frac{360}{2}=180$$

To find : $$(x+y)=z=?$$

Now, we know that $$(x+y)^2=x^2+y^2+2xy$$

=> $$z^2=369+2(180)=729$$

=> $$z=\sqrt{729}=27$$

=> Ans - (C)

Given : $$x = 7 - 4 \sqrt{3}$$ ------------(i)

=> $$\frac{1}{x}=\frac{1}{7-4\sqrt3}$$

Rationalizing the denominator, we get

=> $$\frac{1}{x}=\frac{1}{7-4\sqrt3}\times(\frac{7+4\sqrt3}{7+4\sqrt3})$$

=> $$\frac{1}{x}=\frac{(7+4\sqrt3)}{(7)^2-(4\sqrt3)^2}$$

=> $$\frac{1}{x}=\frac{7+4\sqrt3}{49-48}=7+4\sqrt3$$ ------------(ii)

Adding equations (i) and (ii), we get : $$(x+\frac{1}{x})=14$$

Squaring both sides, $$(x+\frac{1}{x})^2=(14)^2$$

=> $$x^2+\frac{1}{x^2}+2(x)(\frac{1}{x})=196$$

=> $$x^2+\frac{1}{x^2}=196-2=194$$

=> Ans - (D)