SRCC GBO 2023 Question Paper

The total number of pencils bought is $$20\times\ 12=240$$ and the price of each pencil is $$\frac{240}{12}=20$$ rs. 
Therefore, the total money invested is 4,800

Now, 2/3 rd of the pencils i.e, $$\frac{2}{3}\times240=160\ $$ pencils are sold at rs 25, earning back $$160\times\ 25=4,000$$
The remaining 80 pencils are sold at rs 22, earning back $$80\times\ 22=1,760$$
Giving us a net revenue of Rs 5,760 and a net profit of Rs 960

The profit percentage is $$\frac{960}{4800}\times\ 100=20\%$$
Therefore, Option A is the correct answer.

We are given that the highest common divisor of x and y is 2, which means that both of them are even. 
Next we are asked, for how many values of x can we have x+y=99
Now, 99 is an odd number, to get an odd number by addition of two numbers, one of those numbers must be odd and the other must be even. 
But we know that both x and y are even as they are both divisible by 2.
Therefore, no value of x and y is possible which satisfy both the conditions. 

Hence, Option C is the correct answer. 

Let's take the original work to be a multiple of the LCM of 24, 30 and 32; that would be 480x.

Now the efficiencies of Ajay, Bharat and Chandu are 20x, 16x and 15x units/day respectively. 

In the process defined in the question, we can add up the efficiencies of Ajay and Chandu, since they always work together, giving us an efficiency of 35x units/day. 
The next day, Bharat does 16x units/day.

In two days, these three do a total of 51x units/two-day.

So, in 9 days(18 days), all three can do a work of 51x*9 = 459x units.

Now, the remaining work is 480x-459x = 21x

The efficiencies of Ajay and Chandu, since they always work together, giving us an efficiency of 35x units/day.

They work together and complete the work in 21x/35x = 0.6 days

Option A is $$18\ \frac{21}{35}$$, this 21/35 is nothing but 0.6
So we can conclude that the answer would be Option A, without calculating the exact value.

Solving the given relation, 
$$\frac{157}{68}=2+\frac{1}{3+\frac{1}{4+\frac{1}{x}}}$$
$$\frac{21}{68}=\frac{1}{3+\frac{1}{4+\frac{1}{x}}}$$
$$3+\frac{1}{4+\frac{1}{x}}=\frac{68}{21}$$
$$\frac{1}{4+\frac{1}{x}}=\frac{5}{21}$$
$$\frac{21}{5}=4+\frac{1}{x}$$
$$\frac{1}{5}=\frac{1}{x}$$

Giving the value of x to be 5, 
Using this we can find the value of polynomial to be 
$$5^3+3\left(5\right)^2-9\left(5\right)+5$$
$$125+75-45+5$$
= 160

Therefore, Option A is the correct answer. 

The distance between the 1st position and 2nd position would be 2 meters. 
The distance between the 2nd position and 3rd position would be 3 meters.
The distance between the 3rd position and the 4th position would be 4 meters. 
we can see a pattern emerging here, 
.....
The distance between the 49th position and the 50th position would be 50 meters. 

So the total distance between the 1st position and the 50th position would be 2+3+4+......+50
We can easily calculate this as the sum of an AP and it turns out to be $$\frac{49}{2}\left[2\left(2\right)+\left(48\times\ 1\right)\right]$$
Which is equal to 1274

Hence, Option A is the correct answer. 

Let's take the original marked price to be MP.
2/3rd of the original 300 pumpkins were sold at the marked price, so the sales from this part is 200 MP.
75% of the remaining 100 pumpkins, which is 75 pumpkins were sold at a 20% discount, the sales from this part would be
 $$75\times\ \frac{4}{5}MP\ =\ 60MP$$
The remaining 25 pumpkins are sold at 40% discount, the sales from this part would be $$25\times\ \frac{3}{5}MP=15MP$$.

This gives us a total sales of $$200MP+60MP+15MP=275MP$$
We are given that this is equal to 2750, which gives us the value of MP to be 10.

Thus, Option B is the correct answer. 

In cases of question asking the probability of an event happening at least one time, it is usually easier to calculating the probability of the event not happening at all and subtracting it from 1.
In this case, we can find the probability of both dices rolling any number but 4, so each dice will have 5 available options. 
The probability of this happening on one dice is $$\frac{5}{6}$$ and the probability of this happening on both the dices simultaneously is $$\frac{5}{6}\times\frac{5}{6}=\frac{25}{36}$$

Now subtracting this from 1 gives us the required probability, $$1-\frac{25}{36}=\frac{11}{36}$$

Hence, Option B is the correct answer. 

Although option C might seem like a good choice,we have to remember that the given function is not algebraic is nature. 
Upon simplifying g(x) we get, 
$$g(x) = \sin (x + \pi) + \sin (2x + \pi) + \cos (3x + \pi)$$
$$g(x) = -\sin (x) - \sin (2x) - \cos (3x)$$
This is because $$\pi\ +x\ $$ lies in the third quadrant, where both sine and cosine functions are negative. 
We can get g(x) upon reflection of f(x) along the x-axis, but it is not possible using the operations given in option A, B and C. 
Therefore, Option D is the correct answer. 

Volume of the semi-sphere = $$\frac{1}{2}\times\ \frac{4}{3}\pi\ r^3$$
Diameter of the Semi-sphere = 2r

Volume of the Cone = $$\frac{1}{3}\times\ \pi\ r^2h$$
Diameter of the base of the cone = 2r

We are given that the diameter of the sphere is 4 inch. 
2r = 4
giving us, r = 2 inches

We are also given that the volume of the cone and volume of the semi-sphere are equal. 
using r =2 gives us,
$$\frac{1}{2}\times\ \frac{4}{3}\pi\ r^3$$ = $$\frac{1}{3}\times\ \pi\ r^2h$$
$$\frac{1}{2}\times\ \frac{4}{3}\pi\ 8$$ = $$\frac{1}{3}\times\ \pi\ 4h$$
this gives us h=4

Now we are to find the volume of the upper half of the toy. 
The total height of the toy is 2 inches of the semi-sphere + 4 inches of the cone = 6 inches
So the upper half of this toy would be the upper 3 inches of the cone. 

We will have to find the radius of base of this section of cone. 
Using the tan of the semi-angle of the cone we get, $$\tan\ \theta\ =\frac{perpendicular}{base}=\frac{2}{4}=\frac{required\ base}{3}$$
This gives the required base radius as $$\frac{3}{2}$$ inches. 

Finding the volume of this section of the cone, $$\frac{1}{3}\pi\ \left(\frac{3}{2}\right)^23\ =\ \frac{9\pi}{4}$$
Hence, Option D is the correct answer. 

Let the efficiency of a single man and single woman by M units/day and W units/day respectively. 

The first statement is : Total work = 18(M + W)                       {Work = Time x efficiency}
The second statement is: Total work = 5(4M  + 3w)

Since, the total work is same in both the equations, we can equate them to get;
$$18M+18W=20M+15W$$
$$3W=2M$$
$$M=\frac{3}{2}W$$

With this relation, we can find the total work in terms of W or M; Using equation 1 to find the total work in terms of W we get, 
$$18\left(\frac{3}{2}W\right)+18W=45W$$
Now the time taken by 2 men and 3 women to do this work will be 
Time x($$2\left(\frac{3}{2}W\right)+3W\ =\ 45W$$
Time = $$\frac{45}{6}=\frac{15}{2}=7\ \frac{1}{2}$$

Therefore, Option B is the correct answer.