SRCC GBO 2017

Let height of shorter tower = CD = AB = x

$$\angle ACP = 30^0$$

AP = PB - AB = 150 - x

AC = BD = 60 m

Now, 

$$tan30^0 = \frac{AP}{AC}$$

$$\frac{1}{\sqrt3} = \frac{150 - x}{60}$$

$$150 - x = \frac{60}{\sqrt3} = 20\sqrt3$$

$$x = 150 - 20\sqrt3$$

x = 150 - 34.64

x = 115.36 = 116 m

Let the distance between A and B = 'D' km

Sspeed of the boat in still water =9 km/h

Sspeed of the current =3 km/h

Upstream speed = 9 - 3 = 6 km/h

Downstream speed = 9 + 3 = 12 km/h

Now, 

$$\frac{D}{6} + \frac{D}{12} = 3$$

2D + D = 36

D = 12 km

Total amount of ore in a year = 100x

Wasted ore = 20% off 100x = 20x

Remaining ore = 100x - 20x = 80x

Amount of pure iron = 25% of 80x = 80000

20x = 80000

x = 4000

Qquantity of hematite mined in the year = 100x = 400000 kg

Let speed of car = 'x' km/h

Effective speed when car and pedestrian travelling in the same direction = (x - 2)

Since the pedestrian could see the car up to a distance of 0.6 km. So, total distance distance = 0.6 km

Now, 

$$(x - 2) * \frac{6}{60} = 0.6$$

(x - 2) = 6

x = 8 km/h

Since the work done by A in 5 days is equal to the work done by B in 4 days.

So, time taken by A and B alone to do the work is 5x and 4x respectively. 

Time taken by C alone to do the work = C

Now, 

$$\frac{10}{5x} + \frac{10}{4x} + \frac{3}{C} = 1$$ ........ (1)

$$\frac{3}{5x} + \frac{3}{4x} + \frac{3}{C} = \frac{37}{100}$$ ...... (2)

From (1) and (2):

$$\frac{7}{5x} + \frac{7}{4x} = 1 - \frac{37}{100}$$

$$\frac{28 + 35}{20x} = \frac{63}{100}$$

x = 5 and C = 30

Time taken by A alone to do the work = 5x = 25 days

Time taken by B alone to do the work = 4x = 20 days

Time taken by C alone to do the work = C = 30 days

Hence, the time required by the fastest worker to complete the entire work = 20 days

The volume of three containers are 403 litres, 434 litres and 465 litres. 

The biggest measure that can measure all the different quantities exactly = HCF of 403, 434, 465. 

403 = 13 * 31

434 = 2 * 7 * 31

465 = 3 * 5 * 31

HCF = 31

Hence, required measure = 31 litres

Radius of conical tent = r = 5 m

Height of conical tent = h = 12 m

Lateral height = $$l = \sqrt{r^2 + h^2} = \sqrt{5^2 + 12^2} = 13$$ cm

Aarea of the canvas required in making the tent = Curved Surface area = $$\pi rl = \pi * 5 * 13 = 65 \pi m^2$$ 

BC = 192 m

$$tan\theta = \frac{AP}{AB + BC}$$

$$\frac{5}{12} = \frac{AP}{AB + 192}$$

5AB + 960 = 12AP .......... (1)

$$tan\alpha = \frac{AP}{AB}$$

$$\frac{3}{4} = \frac{AP}{AB}$$

3AB = 4AP

9AB = 12AP .......... (2)

From (1) and (2):

5AB + 960 = 9AB

AB = 240

From equation (2):

9 * 240 = 12AP

AP = 180 m

Let actual height = 100x

Stated height = 125% of 100x = 125x

Correction made by candidate from the stated height = 125x - 100x = 25x

Required percent = $$\frac{25x}{125x} * 100$$ = 20%

Let the number of units produced = x

Materials and labour cost of 'x' units = 40x

Total cost price = 40x + 3000

Total selling price = (40x + 3000) + 1000 = 60x

20x = 4000

x = 200

Hence, he must produce and sell 200 units.