SRCC GBO 2009

For the first case where the square (side being a cm) is inscribed in a circle, the square's diagonal is equal to the circle's diameter (2r).

Hence, 2r = a$$\sqrt{\ 2}$$, Hence a= r$$\sqrt{\ 2}$$

For the second case, where the square (side is b cm) is inscribed in the semi circle with the same radius, the radius, half of the square side, and a complete side will form a right angle triangle (from the mid point on the base of the semi circle). 

Hence, $$r^2=\ \ \frac{\ r^2}{4}\ +\ b^2$$, or b= $$\ \frac{\ 2r}{\sqrt{\ 5}}$$

Hence the area of square with side a to the area of square with side b is 5:2

Total area of the cardboard is $$20\times\ 15=300m^2$$ Area of the four squares is $$2\times\ 2\times\ 4=16\ m^2$$

Remaining area is 300-16=284 

The perimeter of the smaller ring is $$4\pi\ $$ . The perimeter of the larger ring is $$20\pi\ $$

As we can clearly see, it will take 5 rotations of smaller ring to completely cover the larger circle.

Area of the equilateral triangle park is $$16\sqrt{\ 3}=\frac{\sqrt{\ 3}}{4}a^2$$   where "a" is the length of the side of the triangle

$$a^2=64\ or\ a=8\ cm$$ Now cost of fencing triangular and square park is same, hence their perimeter will also be same. Perimeter of the square park is 24cm

Hence the length of each side of square park is 6 cm. $$\sqrt{\ 6^2+6^2}=\sqrt{\ 72}=6\sqrt{\ 2}$$

Length of each diagonal is $$6\sqrt{\ 2}cm$$

The length of glass tape required will be the sum of lengths of all the edges. Hence total length required will be  4(30+25+20)=300 cms of total tape will be required

Radius of circle at centre O = 5 m and the triangle formed is an equilateral triangle.

Let side of triangle be $$s=?$$

O is the centroid of the triangle dividing $$a:h=2:1$$

=> $$h=2.5$$ m

=> $$tan(30^\circ)=\frac{h}{\frac{s}{2}}$$

=> $$\frac{1}{\sqrt3}=\frac{5}{s}$$

=> $$s=5\sqrt3$$ m

=> Ans - (B)

Let length, breadth and height of Meeta's lunch box be 10 cm each

=> Length and Breadth of Rita's lunch box = 11 cm and height = 8 cm

Volume of a cuboid = $$lbh$$

=> Required ratio = $$\frac{11\times11\times8}{10\times10\times10}$$

= $$121:125$$

=> Ans - (D)

Number of visitors who paid for the ticket = $$\frac{90}{100}\times35000=31500$$

Let number of children be $$x$$ and number of adults = $$(31500-x)$$

According to ques,

=> $$40(31500-x)+20x=9,50,000$$

=> $$12,60,000-20x=9,50,000$$

=> $$20x=3,10,000$$

=> $$x=\frac{310000}{20}=15,500$$

=> Ans - (B)

Let radius of new sphere = $$r$$ cm

Volume of a sphere = $$\frac{4}{3}\pi r^3$$

=> $$\frac{4}{3}\pi r^3=(\frac{4}{3}\pi\times3^3)+(\frac{4}{3}\pi\times4^3)+(\frac{4}{3}\pi\times5^3)$$

=> $$r^3=27+64+125=216$$

=> $$r=\sqrt[3]{216}=6$$ cm

=> Ans - (A)

Mean of 30 observations = 150

For correcting the error, we need to subtract 135 and add 165, 

=> New average = $$\frac{(150\times30)+165-135}{30}$$

= $$\frac{150\times30+30}{30}$$

= $$150+1=151$$

=> Ans - (B)