SRCC GBO 2023 Question Paper

We can take the hollow cylinder to me made of two concentric cylinders, with the inner cylinder carved out of the outer cylinder. 
Taking the radius of the outer cylinder to be R, we know that the thickness of the cylinder is 2 cm, therefore, the radius of the inner cylinder will be R-2 cm.
We know that the volume of this hollow cylinder would be $$100\pi\ $$
$$\pi\ R^2h-\pi\ \left(R-2\right)^2h=100\pi\ $$
Putting h=2 and solving the left hand side, we get
$$4R\ -4=5$$
$$R=\frac{9}{4}$$

The inner radius would be $$\frac{9}{4}-2=\frac{1}{4}$$ cm
The capacity of this hollow cylinder would be $$\pi\ \left(\frac{1}{4}\right)^220=\pi\ \frac{20}{16}=\frac{5\pi}{4}\ cm\ $$

Therefore, Option C is the correct answer. 

We should choose a number which is a factor of all of these numbers so that when they are grouped among themselves, there are no leftovers. 
The count in the boxes have HCF 32, so we can distribute the bulbs in groups of 32 so that each box has 32 bulbs and each box has only one type of bulb. 

The bulb with 64 count would require 2 boxes.
The bulb with 192 count would require 6 boxes.
The bulb with 128 count would require 4 boxes.
The bulb with 384 count would require 12 boxes.
The bulb with 256 count would require 8 boxes.
The bulb with 32 count would require 1 boxes.
The bulb with 96 count would require 3 boxes.
The bulb with 288 count would require 9 boxes.

Adding all of them up, we would get that a total of 45 boxes are required. 
Therefore, Option B is the correct answer. 

This question did not require detailed analysis and making the venn-diagram. 
The questions asks the difference between the number of players involved in only swimming and the number of players involved in all three activities. 

We are given both of these exact values in the question. 
Players involved in only swimming: 290
Number of players involved in all three activities: 150

There are 140 less players doing all three activities than players doing only swimming. 
Hence, Option C is the correct answer, 

The number of newly trained people after the first three day period is 3.
Including the initial member who trained them, the total number of trained people would be 1+3

The number of newly trained people after the second trained people is 9.
Including the number of people who are already trained, we get the total number of trained people to be 1+3+9

We can follow through with this reasoning and see that the total number of trained people make the sum of a GP with the common ratio 3 and the number of terms equal to the number of three day periods required. 
We want the total number of trained people to be 1093. 
Using the formula for the sum of an GP with n terms we get, $$\frac{a\left(r^n-1\right)}{r-1}=\frac{1\left(3^n-1\right)}{3-1}=1093$$
$$3^n-1=2186$$
$$3^n=2187$$
from here, we can keep checking the value of n and we find that n=6 satisfies the equation. 

So there are a total of 6 three day periods required, hence, a total of 18 days required for 1093 people to be trained. 

The profit and loss difference in this question is a misdirection. 
The profit on A is equal to the loss on B simply means that the exact values of these are equal. Them being a profit or loss would not make any difference. 

we are given, 
$$10\%\ of\ A\ =\ 20\%\ of\ B\ =\ 15\%\ of\ C$$
= $$2A\ = 4B = 3C$$

From this we can get, A:B = 2:1 and B:C = 3:4
giving us, A:B:C to be 6:3:4

Option A is the correct answer. 

Using the sum and product of roots formula, we get $$\frac{c}{a}=uv$$, and hence, $$c=auv$$
And $$-\frac{b}{a}=u+v$$, and hence, $$-b=a\left(u+v\right)$$

Putting values of c and b in the equation we get, 
$$\sqrt{\ \frac{2\sqrt{\ ac}-b}{c}}=\sqrt{\ \frac{2\sqrt{\ a\left(auv\right)}+a\left(u+v\right)}{auv}}$$
=$$\sqrt{\ \frac{2\sqrt{\ a^2uv}+a\left(u+v\right)}{uav}}=\sqrt{\ \frac{2\sqrt{\ uv}+\left(u+v\right)}{uv}}$$
=$$\sqrt{\ \frac{\left(\sqrt{\ u}+\sqrt{\ v}\right)^2}{uv}}=\sqrt{\ \frac{\left(\sqrt{\ u}+\sqrt{\ v}\right)^2}{\sqrt{\ uv}^2}}=\sqrt{\ \left(\frac{\sqrt{\ u}+\sqrt{\ v}}{\sqrt{\ uv}}\right)^2}$$
=$$\frac{1}{\sqrt{\ u}}+\frac{1}{\sqrt{\ v}}$$

Therefore, Option A is the correct answer. 

This is a series of prime numbers. The prime number next to 71 is 73. 

X can do 60% of the total work in 9 days, this gives us that X can complete the total work in 15 days. 
X and Y can together complete 10% of the work in 1 day, this gives us that X and Y can complete the work in 10 days. 

We can take the total work to be a multiple of the LCM of these values, let it be 30U. 
Now X has an efficiency of 2U units/day.
And X and Y together have an efficiency of 3U units/day. 

This gives us that the efficiency of Y to be 1U units/day.
In order for Y to 5% of the total work, that is 3/2 U units of work, Y would require, $$\frac{3}{2\ }=\ 1\ \frac{1}{2}$$ days. 

Hence, Option A is the correct answer. 

The question had an error in the original questions paper the fraction $$\frac{81}{16}$$ was given as $$\frac{18}{16}$$.

Taking the multiplication factor of this compound interest to be X and the principal amount be P.
We are given that 
$$P\left(X\right)^2=\frac{3}{2}P$$
which would give us, $$X=\ \left(\frac{3}{2}\right)^{\frac{1}{2}}$$

Using this for the second condition we had, 
we can get 
$$P\left(X\right)^N\ =\ \frac{81}{16}P$$
$$\left(\frac{3}{2}\right)^{N\times\ \frac{1}{2}}\ =\ \frac{81}{16}$$
$$\left(\frac{3}{2}\right)^{N\times\ \frac{1}{2}}\ =\ \left(\frac{3}{2}\right)^4$$
Giving us N=8.

Therefore, Option B is the correct answer. 

We can solve this by simply solving for the value under the root, 
$$\sqrt{\ 1600+80\times\ 50+225+30\times\ 35+1225}$$
=$$\sqrt{\ 1600+4000+225+1050+1225}$$
=$$\sqrt{\ 8100}=90$$

Hence, Option B is the correct answer.