SRCC GBO 2023

Taking X and Y as one pair, we can have 8 people on 8 seats. These 8 elements can be seated in 8! ways and X and Y can be seated in 2! ways. Therefore, the total number of ways X and Y can be seated together are 2!8!
Total number of ways 9 people can be seated is 9!

The probability that X and Y will be seated together is $$\frac{2!8!}{9!}=2\times\ \frac{8!}{9\times\ 8!}=\frac{2}{9}$$
Therefore, Option C is the correct answer.

The simplest way to solve this question would be through the options. 

Option A: $$3x^{2} — 7x + k = 0$$ and $$—7x^{2} + kx + 3 = 0$$ on putting k= 3 becomes, 
$$3x^{2} — 7x + 3 = 0$$ and $$—7x^{2} + 3x + 3 = 0$$
Where we can not calculate the roots by factorisation. 

Option B: $$3x^{2} — 7x + k = 0$$ and $$—7x^{2} + kx + 3 = 0$$ on putting k= 6 becomes, 
$$3x^{2} — 7x + 6 = 0$$ and $$—7x^{2} + 6x + 3 = 0$$
Where again,we can not calculate the roots by factorisation.

Option C: $$3x^{2} — 7x + k = 0$$ and $$—7x^{2} + kx + 3 = 0$$ on putting k= 4 becomes, 
$$3x^{2} — 7x + 4 = 0$$ and $$—7x^{2} + 4x + 3 = 0$$
Where the first equation has roots 1 and 4/3 and the second equation has roots 1 and 3/7

Option D: $$3x^{2} — 7x + k = 0$$ and $$—7x^{2} + kx + 3 = 0$$ on putting k = 2 becomes, 
$$3x^{2} — 7x + 2 = 0$$ and $$—7x^{2} + 2x + 3 = 0$$
Where the first equation has roots 2 and 1/3 but for the second equation, we can not calculate the roots by factorisation.


We can see that option C gives us one common root, hence, that would be our answer. 

We can find the remainder on dividing by 9 and then find the negative remainder in order to find the reminder we would have gotten on dividing by -9. 
$$\left[\frac{8-5^{17}\times\ 2^{20}}{9}\right]_R$$
$$\left[\frac{8}{9}\right]_R-\left[\frac{5^{17}\times\ 2^{17}\times\ 2^3}{9}\right]_R$$
$$-1-\left[\frac{10^{17}}{9}\right]_R\times\ \left[\frac{8}{9}\right]_R$$
$$-1-1^{17}\times\ \left(-1\right)$$
$$-1+1\ =\ 0$$
Upon reversing the sign to calculate the remainder on dividing it by -9, we get 0 again. 
Hence, the remainder is 0.

When $$5x^4-3x^3+2x^2-1\ $$ is divided by $$x^2+4$$, the quotient is $$5x^2-3x-18$$ and the remainder is $$12x+71$$
Here we are asked to take the quotient as f(x);
f(x) = $$5x^2-3x-18$$

When $$2x^3-x+1$$ is divided by $$x^2+x+1$$ the quotient is $$2x-2$$ and the remainder is $$-x+3$$
Here we are asked to take the remainder as g(x);
g(x) = $$-x+3$$

We are to find the remainder when f(x) is divided by g(x)
Upon dividing we would get the quotient as -5x-12 and the remainder is 18. 

Hence, Option B is thee correct answer. 

Statement 1: 
Remainder of 3864335 when divided by 300 is a short calculation and gives us the remainder as 35. 
Remainder of 3864335 when divided by 382 is lengthy but after dividing the remainder would be 23.
Sum of these remainders is 58, this is twice of the remainder we get on dividing the unknown number by 32; 
This remainder would be 29
We now have to check the remainder when 1021 is divided by 32, 
1024 is $$32^2$$, so $$\left[\frac{1024}{32}\right]_R=-3=\ 29$$
And hence, we can confirm that statement 1 is correct. 

Statement 2: According to Wilson's theorem, 
$$\left[\frac{\left(p-2\right)!}{p}\right]_R=1$$ where P is a prime number. 
Taking P =47 we get, $$\left[\frac{45!}{47}\right]_R=1$$ and subtracting 1 from this would mean that 45!-1 is divisible by 47. 
Therefore, Statement II is incorrect. 

We need to first convert $$20_{\left(base-3\right)}$$ and $$202_{\left(base-3\right)}$$ into decimal system so as to get the values of a and b in the decimal system.
20 in 10 base system would be : $$\left(2\times\ 3^1\right)+\left(0\times\ 3^0\right)$$ = 6
202 in 10 base system would be : $$\left(2\times\ 3^2\right)+\left(0\times\ 3^1\right)+\left(2\times\ 3^0\right)=20$$

so a+b = 6 and $$a^2+b^2=20$$
From this we get, ab = 8
Trying out different values we can get the values of a and b to be 4 and 2 
converting these from decimal to 3 base system we get 2 as 2 only and 4 as $$\left(1\times\ 3^1\right)+\left(1\times\ 3^0\right)$$ = $$11_{\left(base-3\right)}$$

Therefore, Option A is the correct answer. 

David faces east and walks 70 steps in the east direction. 
After this, without turning his back, David walks 30 steps in reverse direction. Landing him 40 steps east of the starting point, facing eastwards. 
After this he reverses his back, and is now facing west towards his starting point. He then walk 20 steps in the direction he is facing i.e, 20 steps towards the starting point. Making his final position 20 steps east of the starting point. 
1 step measures 2 feet, so 20 steps would measure 40 meters. 
Therefore, David is 40 meters east of the starting point. 

Hence, Option C is the correct answer. 

Lateral surface area of a cube is $$4a^2$$, where a is the side length.
We are given that the lateral surface area is 2304, solving this for a, we get
$$a^2=576$$
$$a=24$$

Now we need to divide this cubic room which is of 24 m side length, in small 4m wide rooms which have the same length and height. 
For this we will be placing plywood parallel to the height of the room and a distance of 4 meters.

The room will be divided in 6 small rooms of 4 meter width and for this we would require 5 plywood sheets. 
The number of segments is 1 +  the number of cuts made. 

{Since, the length of the room is close is the the width section we want to divide the room in, we can count the number of plywood sheets we would need by drawing a rough sketch. But it is recommended to know such common logical reasoning solutions.}

Therefore, the correct answer would be Option D. 

Suppose we have two lines $$a_1x+b_1y+c_1=0$$ and $$a_2x+b_2y+c_2=0$$
The equation of pair of these two equations would be obtained by multiplying these two equations, 
Giving us, $$a_1a_2x^2+b_1b_2y^2+x\left(a_1c_2+a_2c_1\right)+y\left(b_1c_2+b_2c_1\right)+xy\left(a_1b_2+a_2b_1\right)+c_1c_2=0$$
Comparing this with the equation we have, we can start finding the values, 
$$a_1=a_2=1$$
From coefficient of x, we get $$c_1+c_2=-1$$
From the constant in the equation we get, $$c_1c_2=-2$$
The value of $$c_{1\ }and\ c_2$$ in some order must be -2 and 1 

From coefficient of xy, we get $$b_2=-b_1$$
And through coefficient of $$y^2$$, we get that $$b_{1\ }or\ b_2=\pm\ 2$$
Since the coefficients of x are equal in both the equations, b becomes the only differentiating point. 
The equations will therefore now be, 
$$x+2y+c_1=0$$ and $$x-2y+c_2=0$$
Where we can have two case:
Case 1: $$c_1=-2\ \&\ c_2=1$$
The required value would be : -6

Case 2: $$c_2=-2\ \&\ c_1=1$$
The required value would be : 6

The product of possible values therefore is -36

Upon finding the prime factors of 2835, we get $$7\times\ 5\times\ 3^4$$
Since we know that the HCF of x and y is 9, the $$3^4$$ must be split equally between x and y. 
Now the possible pair values of x and y are (315, 9) or (45,63) {This we get by distributing the remaining 5 and 7 between x and y}
of these pairs, we cannot we cannot find the exact values of x and y. 
So we will have to find the possible combinations for 2x+y and find those values. 
x=9 and y=315 gives us 333
x=63 and y=45 gives us 171
x=45 and y=63 gives us 153
x=315 and y=9 gives us 639

We can see that of all of these values, 333 and 153 are in the option C. 
Hence, Option C is the correct answer.