SRCC GBO 2016 Question Paper

Let the capacity of Jar and Drum is 8 and 360 respectively. 

Quantity of water filled in Jar in 1 second = 1

It takes 40 seconds by the hole to empty the Jar completely. 

So. effective time taken to fill the jar = $$\frac{1}{\frac{1}{8} - \frac{1}{40}}$$ = 10 seconds.

It takes 10 seconds to bring the Jar to the drum. So, quantity of water in Jar emptied in 10 seconds = $$8 * \frac{10}{40} = 2$$

Quantity of water poured into Drum by Jar = 6 

Total time taken to pour this 6 unit of water = Time taken to fill the Jar + Time taken to bring the Jar to the Drum = 10 + 10 = 20 seconds

Number of time Jar is used to fill the drum = $$\frac{Capacity of drum}{Quanttiy of water filled by jar in one go} = \frac{360}{6} = 60 times$$

Total time taken to fill the drum = 60 * 20 = 1200 seconds = 20 minutes

One unit of X requires two units of material B, then 25 unit of X will require = 2 * 25 = 50 units of material B

One unit of Y requires five units of material B, then 25 unit of Y will require = 5 * 25 = 125 units of material B

Total requirement of material B = 50 + 125 = 175

Table given shows the operations given: 

Total quantity in container Z = $$\frac{35}{3} + \frac{55}{3} = 30$$

Required percent = $$\frac{\frac{55}{3}}{30} * 100$$ = 61% (Approx)

Let original quantity of pure milk = 100x

Let cost price of '100x' litres of pure milk = 100

Amount of water added = 25% of 100x = 25x

Total quantity of mixture = 100x + 25x = 125x

Total quantity of mixture after evaporation = $$125x * \frac{23}{25} = 115x$$

Selling price of '115x' litres of mixture = Cost price of '115x' litres of pure milk = 115x

Profit amount = 115x - 100x = 15x

Profit % = $$\frac{15x}{100x} * 100$$ = 15%

Let percent of alcohal in mixture M, N and O mixture is 'a', 'ar' and 'a$$r^2$$' respectively.

Now, 

$$\frac{2a + 3ar + 4ar^2}{2 + 3 + 4} = 32$$

$$a(2 + 3r + 4r^2) = 288$$ .......... (1)

$$\frac{3a + 2ar + ar^2}{3 + 2 + 1} = 22$$

$$a(3 + 2r + r^2) = 132$$ .......... (2)

From (1) and (2):

$$\frac{a(2 + 3r + 4r^2)}{a(3 + 2r + r^2)} = \frac{288}{132} \frac{24}{11}$$

$$4r^2 - 3r - 10 = 0$$

r = 2

From equation (1):

$$a(2 + 6 + 16) = 288$$

a = 12

Ppercentage of Alcohol in M = a = 12%

The ratio of milk to water in beaker B after mixing 30 litres of milk from beaker A to beaker = 30: 90 = 1: 3

The remaining quantity of milk in beaker = 90 - 30 = 60 litres

Quantity of milk in beaker A when 12 litres of mixture from beaker B is mixed in beaker = $$60 + 12\frac{1}{4}$$ = 60 + 3 = 63 litres

Quantity of water in beaker A when 12 litres of mixture from beaker B is mixed in beaker = $$12\frac{3}{4}$$ = 9 litres

Total quantity of mixture in beaker A = 63 + 9 = 72 litres

Required percent = $$\frac{9}{72} * 100$$ = 12.5%

Let amount with Kamal, Vishal and Shaan after removing the amount is '11x', '18x' and '24x' respectively. 

Now, 

Total amount = (11x + 10) + (18x + 20) + (24x + 15) = 1105

53x = 1060

x = 20

Amount received by Shaan = (24x + 15) = Rs.495

Let initial number of boys under 15 years and over 15 years are '5x' and '4x' respectively. 

Now, 

$$\frac{5x - 20}{4x + 20} = \frac{7}{8}$$

40x - 160 = 28x + 140

12x = 300

x = 25

Total boys in the exam = 5x + 4x = 9x = 225

Since two sides of triangles are 100 m each.

Now, the third side of triangle must be greater than '0' and less than '200'. 

The Minimum length of the third side when the area is greater than or equal to 4800 = 120 m

The maximum length of the third side when the area is greater than or equal to 4800 = 160 m

Now, Difference between perimeter of two triangle = Difference between length of third side = 160 - 120 = 40 m

Since, total work done $$\propto$$ Volume of pyramids $$\propto side^3$$ 

Total work done to built pyramids = Total men working * Days take

Let days taken by 54 men to build the second pyramid = N

Now, 

$$\frac{36 * 32}{54 * N} = \frac{30^3}{22.5^3} = \frac{4^3}{3^3}$$

$$N = \frac{36 * 32 * 27}{64 * 54}$$

N = 9 days