SRCC GBO 2016

Total number of sweets distributed = 720

Let total children = 100x

Number of sweets received by each child = 20% of 100x = 20x

Total number of sweets = 100x * 20x = 720

$$x^2 = 0.36$$

x = 0.6

Total number of sweets did each child receive = 20x = 12

Let eligible voters = 100x

Eligible voters between 18 and 25 years of age = 15% of 100x = 15x

Eligible voters between 18 and 25 years of age who actually voted = 75% of 15x = 11.25x

Required percent = $$\frac{11.25x}{100x} * 100 = 11.25$$%

LCM of sides of cuboid = 2.6

So, side of each cube = 2.6 m

Least number of cubes formed = $$\frac{5.2 * 13 * 39}{2.6 * 2.6 * 2.6} = 150$$

Total surface area of each = $$6 * 2.6^2 = 40.56 m^2$$

Sum of total surface area of all the cubes = 150 * 40.56 = 6084 $$m^2$$

Let us say that the cross section areas of A, B, and C are 3a, 2a, and 4a.

Now, assume that the water flows at the rate of x, y, and z per min for each of the pipes respectively.

The amount of water flowing through the pipes (per min) shall be: 3ax, 2ay, 4az

Now, the time taken by each of them to fill the tank is 20 min, 50 min, and 25 min.

This means, 3ax$$\times\ $$20 = 4az$$\times\ $$25, or 100:60, or 5:3

According to the question: 

Total work = (3M + 4W + 6C) * 7

Now, 

A woman does double the work a man does, So, the ratio of efficiency of a man to that of a woman = 1: 2

M: W = 1: 2

2M = W .......... (1)

A child does half the work a man does. So, the ratio of efficiency of a man to that of child = 2: 1

M: C = 2: 1

M = 2C ....... (2)

From (1) and (2):

4C = W ....... (3)

Let the required number of women = 'x'

Now, 

(3M + 4W + 6C) * 7 = (xW) * 7

From (1) and (3):

1.5W + 4W + 1.5W = xW

7W = xW

x = 7

Let total time taken to complete the work = 'x' days

C work for 'x' days.

A left 8 days before the completion of the work. So, A works for 'x - 8' days

B left 12 days before the completion of the work. So, B works for 'x - 12' days

Now, 

$$\frac{x - 8}{36} + \frac{x - 12}{54} + \frac{x}{72} = 1$$

$$\frac{6(x - 8) + 4(x - 12) + 3x}{216} = 1$$

6x - 48 + 4x - 48 + 3x = 216

x = 24

Hence, C works for total 24 days. 

Ratio of efficiency of $$1^{st}$$ boy to $$2^{nd}$$ boy = 1: 2

Ratio of efficiency of $$2^{nd}$$ boy to $$3^{rd}$$ boy = 1: 2

And so on...

So ratio of efficiency of $$1^{st}$$, $$2^{nd}$$, $$3^{rd}$$, $$4^{th}$$ and $$5^{th}$$tboy = 1: 2: 4: 8: 16

Let unit of work done by $$1^{st}$$, $$2^{nd}$$, $$3^{rd}$$, $$4^{th}$$ and $$5^{th}$$boy is 1, 2, 4, 8 and 16 respectively. 

Since, job is completed by all the five boys together in 5 days. Hence,total work = 5 * (1 + 2 + 4 + 8 + 16) = 155

Time taken by $$2^{nd}$$ and $$4^{th}$$ boys together to do the work = $$\frac{155}{2 + 8} = 15.5$$ days

Time taken by $$5^{th}$$ boy alone to do the work = $$\frac{155}{16}$$ days

Required diffrence = 15.5 - $$\frac{155}{16} = \frac{93}{16} = 5.8125$$ days = 6 days (Approx)

Let time taken by Atul, Shiva and Sapan together to finish the work = 'x' hours

Time taken by Atul alone to finish the work = (x + 6) years

Time taken by Shiva alone to finish the work = (x + 1) years

Time taken by Sapan alone to finish the work = 2x years

Now, 

$$\frac{1}{x + 6} + \frac{1}{x + 1} + \frac{1}{2x} = \frac{1}{x}$$

$$x = \frac{2}{3}$$ hours

Pipes A and B together can fill the tank in 12 hours and C alone can empty the tank in 60 hours. 

So, part of tank emptied by pipe C in 12 hours = $$\frac{12}{60} = \frac{1}{5}$$

Now, this $$\frac{1}{5}^{th}$$ part of the tank will be filled by pipes A and B together in $$\frac{1}{5} * 12 = 2.4$$ hours

Time taken by $$T_1 and T_2$$ together to fill the tank = $$\frac{1}{\frac{1}{50} + \frac{1}{60}} \frac{300}{11}$$ minutes

Time taken by $$T_1 and T_2$$ together to fill half the tank = $$\frac{300}{2 * 11} = \frac{150}{11}$$ minutes

Time takne by $$T_1$$ alone to fill half the tank = $$\frac{50}{2} = 25$$ minutes

Total time taken to fill the tank completely = 25 + $$\frac{150}{11}$$ minutes