SRCC GBO 2017 Question Paper

We are given the information that, 
The total capacity of the tank is $$2400m^3$$
The emptying capacity is given to be $$10m^3$$ per minute more than the filling capacity. So, let us consider that the emptying capacity is X+10 and the filling capacity is X. 

We are also given that the filling time is 8 minutes more than emptying time. 

Drawing up the equations, 
$$\frac{2400}{X+10}+8=\frac{2400}{X}$$

$$1=\frac{300}{X}-\frac{300}{X+10}$$

$$1=300\left(\frac{1}{X}-\frac{1}{X+10}\right)$$

$$1=300\left(\frac{10}{X\left(X+10\right)}\right)$$

$$X^2+10X-3000=0$$

$$\left(X+60\right)\left(X-50\right)$$

Since X cannot be negative, X is $$50\ \frac{m^3}{\min}$$

Hence, fillingcapacity is $$50\ \frac{m^3}{\min}$$

A and B walk at a speed of 2 rounds per hour and 3 rounds per hour.

They are walking around a circular track.

They started walking at 8:00 a.m from the same point in opposite directions.

In a circular track, when A and B are walking in opposite directions, the net speed = Speed of A + Speed of B.

We assume, A to be at rest, so speed of B will be = 3+2 = 5 rounds per hour.

So, from 8 to 9:30, in one and half hour, the number of rounds taken by B will be = 5*1.5 = 7.5 round.

Since, 7 rounds are completed, therefore, A and B will cross each other 7 times in this duration.

The given dimensions of the room are 10 m x 7m x 5 m

l x b x h = 10 x 7 x 5

We know the area of walls is = 2(lh + bh) = 2(10*5 + 7*5) 

Hence, the area of the walls = 170 sq m

We have to subtract the area of doors and the windows to get the paintable area

Area of both doors = 2*(1 x 3) = 6 sq m

Area of one window = 2 x 1.5 = 3 sq m

Area of other two windows = 2*(1 x 1.5) = 3 sq m

So, the paintable area = 170 - (6+3+3) = 158 sq m

Given the cost to paint is ₹3 per sq m

So, the total cost is $$158\times3=474$$

The probability of getting 1 or 3 when die is cast is $$\dfrac{2}{6}=\dfrac{1}{3}$$

The probability of selecting a black from the first bag is $$\dfrac{3}{3+4}=\dfrac{3}{7}$$

So, the probability of getting 1 or 3 and selecting a black ball is = $$\dfrac{1}{3}\times\dfrac{3}{7}=\dfrac{3}{21}$$

Similarly, the probability of getting 2, 4, 5 or 6 when die is cast is $$\dfrac{4}{6}=\dfrac{2}{3}$$

The probability of selecting a black from the second bag is $$\dfrac{4}{4+3}=\dfrac{4}{7}$$

So, the probability of getting 2, 4, 5 or 6 and selecting a black ball is = $$\dfrac{2}{3}\times\dfrac{4}{7}=\dfrac{8}{21}$$

Hence, the total probability of selecting a black ball is = $$\dfrac{3}{21}\ +\dfrac{8}{21}=\dfrac{11}{21}$$

It is given that A runs $$1\frac{2}{3}$$ times as fast as B.

Let us assume B runs at a speed of 6 m/s, so A will run at a speed of 6*5/3 = 10 m/s

We know that A gives B a headstart of 80 m.

We need to find the time at which A and B reach at the same time at winning post.

The difference between their speed (A and B ) = 10-6 = 4 m/s

They will reach the winning post at the same time, when A will be able to reach at position of B by covering the headstart distance.

Total distance = 80 m

A's speed which is more than B to cover the distance of 80 m = 4 m/sec

Time it takes to cover the distance = 80/4 = 20 sec

So, the winning post will be reached by A and B together in 20 seconds.

A will cover the distance = 10*20 = 200 meter

Given that the volume of the container is 50, length = 4.5, and breadth = 3.

=> height of the container = 50 / 4.5 * 3 = 3.7 [nearly]

We need to find the surface area of the container [open top]

=> lb + 2(lb + bh) = 69.

Now the weight of lead required = 69 * 27 = 1863 [Option-B].

When the ball is dropped in a cylindrical tub, it will displace the water equal to its volume based on Archimedes' Principle, which states that  the volume of the displaced fluid is the same as the volume of the object submerged.

We know the radius of the cylinder = 12 cm.

The height of water in the cylinder is increased by 6.75 cm.

We need to find the radius of the spherical ball. (Assume r)

Volume of cylinder = $$\pi R^2H$$

The volume of the cylinders increased because of the iron ball by a height of 6.75 cm

Volume increased = $$\pi R^2H$$ = $$\pi\left(12\right)^26.75$$

The volume of spherical iron ball = $$\dfrac{4}{3}\pi r^3$$

Comparing both,

$$\dfrac{4}{3}\pi r^3$$ = $$\pi\left(12\right)^26.75$$

$$r^3=9\times9\times9$$

r = 9 cm.

Let us assume the Selling Price = Rs.100

It is given that the gross profit is 26% of the selling price = 26% of 100 = 26

Therefore, Profit = Rs.26

Cost Price = Selling Price - Profit = 100-26 = Rs. 74

34% of the cost price = 74*0.34 = Rs. 25.16

We need to find What per cent selling price would be 34% of the cost price.

That is (25.16/100)*100 = 25.16%

Let us first understand what the questions ask. We are given that the bill has a certain present value, which seems to be reducing over time. 

We are given that the bill is 1200 rupees after 7 months and 1016 rupees after 2.5 years. That means the calculation for future value considering the current value of the bill using a rate of bill over the years is being done. 

We can write the equation that, Present value equals=$$1200\left(1+\frac{7}{12}\left(\frac{x}{100}\right)\right)$$

We can also write the equation that, Present value equals=$$1016\left(1+2.5\left(\frac{x}{100}\right)\right)$$

Let us equate both these equations by taking x/100 as z

$$1200\left(1+\frac{7}{12}z\right)=1016\left(1+2.5z\right)$$

$$1200+700z=1016+2540z$$

$$184=1840z$$

$$z=0.1$$

$$\frac{x}{100}=0.1$$

$$x=10\%$$

Let's denote their ages as P, Q and R itself.

We are first told that If the ages of P and R are added to twice the age of Q, the total becomes 59

P+R+2Q=59----1

We are then told that the ages of Q and R are added to thrice the age of P, so the total becomes 68

Q+R+3P=68-----2

At last, we are told that the age of P is added to thrice the age of Q and thrice the age of R, so the total becomes 108

P+3Q+3R=108-----3

This is a linear equation with 3 variables, 

Multiplying the second equation with 3 we get, 
3Q+3R+9P=204

Subtracting equation 3 from this equation we get, 
8P=96
P=12

Subtracting equation 1 from 2 we get, 2P-Q=9
Q=15

Substituting this we get, R=17

Hence, P=12.