SRCC GBO 2010

It is clear from the question that the centers of all three figures coincide. 

Let x be the length side of the equilateral triangle, hence the area (A) is ($$\sqrt{\ 3}$$/4)$$x^2$$

From this, the radius of the inscribed circle is= ($$\sqrt{\ 3}$$/2)x * (1/3) = x/2$$\sqrt{\ 3}$$

So, the diameter is x/$$\sqrt{\ 3}$$

Now, the diameter of the circle coincides with the diagonal of the square.

Hence, assuming the sides of square to be y, we have y$$\sqrt{\ 2}$$= x/$$\sqrt{\ 3}$$

y= x/$$\sqrt{\ 6}$$

From this, the area of square (B) is $$y^2$$, which is $$x^2$$/6

Now, the ratio of A and B is = $$3\sqrt{\ 3}$$/2

The given shape will essentially have one cuboid with dimensions 26*10*1.6 and a right-angle triangular shape with a height of 26 m, a base of 10 m, and a width of 2.8 m.

Now, the volume of the cuboid is 26*10*1.6 $$m^3$$, which is 416$$m^3$$.

The volume of the triangular shape is (Area)*(width), which is 1/2*26*10*2.8= 364$$m^3$$.

Adding the two, we get a total volume of 780$$m^3$$.

Let's name the vertices for better understanding. 

Since we are given that EC is 7 cm and CD is 8 cm, the length DE would be 1 cm. 
Joining E and A, we would get two right angles triangles, ADE and AEG

DE is 1 cm, and AD is 12 cm, this would give the length of side AE to be $$\sqrt{\ 145}$$ cm
Further, AE is $$\sqrt{\ 145}$$ cm and AG is 8 cm, giving the length of GE to be 9 cm using the Pythagorean theorem.

In order to find the area of the shaded region, we need to find the area of the rectangle ABCD and subtract the area of triangles ADE and AGE

Area of ABCD - Aread of ADE - Area of AGE
= $$\left(8\times\ 12\right)-\left(\frac{1}{2}\times\ 12\times\ 1\right)-\left(\frac{1}{2}\times\ 8\times\ 9\right)$$ 
= $$96-6-36$$
=54

Therefore, Option C is the correct answer. 

Let l,b,h be the dimensions of the rectangular box. 

Sum of 12 edges is 112cm, which means 112 = 4l+4B+4h

Hence, l+b+h= 28

Now, the surface area of the box:

384= 2lb+2bh+2lh

This means, lb+bh+lh= 192

Also: due to the rectangular box being inscribed in the sphere, the diagonal of the box will be same as the diameter of the sphere.

Hence, $$\ \ \sqrt{\ l^2+b^2+h^2}$$=$$\ \sqrt{\ (l+b+h)^2-2\left(lh+bh+lh\right)}$$= 2r

2r= 20

r=10 cm

Surplus for the years are:

2003: 4.1 million tons

2004: 9.3 million tons

2005: 9.4 million tons

2006: 1.7 million tons

2007: -25.8 million tons

2008: -2.5 million tons

Hence, it is clear that the surplus is highest in 2005.

The average supply of steel from 2003 to 2008= (49.2$$\ +\ $$52.8$$\ +\ $$50.1$$\ +\ $$31.9$$\ +\ $$30.8$$\ +\ $$60)$$\div$$6, which is equal to 45.8.

From simple observation, we can see that in 2007, the demand of steel went up from 30.2 to 56.6, which is way greater than the previous year. Apart from 2007, the only year where the demand increased was 2008, but the percentage increase is lesser than the what we saw in the previous year. 

Surplus = Supply - Demand

2003 surplus = 49.2-45.1= 4.1

2004 surplus = 52.8 - 43.5 = 9.3, Hence, the increase = 9.3-4.1 = 5.2

2005 surplus = 50.1 - 40.7 = 9.4, Hence, the increase = 9.4-9.3 = 0.1

2006 surplus = 31.9 - 30.2 = 1.7, increase = 1.7-9.4 = - 7.7

Hence, it is highest in 2004.

By observing the graphs and remembering that the total cost is a sum of labour cost and raw material cost, we can see that S had the highest raw material cost (total cost-labour cost), which is 40 cr.