SRCC GBO 2012

Let the whole numbers be $$a,b,c,d$$

According to ques, => $$a+b+c=180$$ -------(i)
$$a+b+d=197$$ ------------(ii)
$$a+c+d=208$$ ------------(iii)
$$b+c+d=222$$ -----------(iv)

Adding above equations, we get : $$3(a+b+c+d)=180+197+208+222$$

=> $$(a+b+c+d)=\frac{807}{3}=269$$

Substituting value from equation (i) in above equation,

=> $$d=269-180=89$$

$$\therefore$$ Largest number = 89

=> Ans - (A)

Radius of small pipes = 0.5 cm and big pipes = 3 cm

=> Number of circular pipes required = $$\frac{\pi R^2}{\pi r^2}$$

= $$\frac{3^2}{0.5^2}=\frac{9}{0.25}$$

= $$9\times\frac{4}{1}=36$$

=> Ans - (D)

Dimensions of city after rainfall = 10km x 10km x 1 cm

=> Volume = $$10^7\times10^7\times10$$ mm

= $$10^{15}$$ mm

Volume of raindrop = $$10$$ $$mm^3$$

$$\therefore$$ Number of raindrops = $$\frac{10^{15}}{10}=10^{14}$$

=> Ans - (D)

Let price of each pencil = Rs. $$x$$ and price of each pen = Rs. $$2x$$

Let number of pencils ordered = $$y$$ and number of pens ordered = $$4$$

=> Original cost price = Rs. $$(xy+8x)$$

After interchanging the price, => New price of bill = Rs. $$(4x+2xy)$$

According to ques,

=> $$\frac{4x+2xy}{xy+8x}=1.5$$

=> $$4x+2xy=1.5xy+12x$$

=> $$4+2y=1.5y+12$$

=> $$0.5y=12-4=8$$

=> $$y=\frac{8}{0.5}=16$$

$$\therefore$$ Ratio of pens to pencils = $$\frac{4}{16}=1:4$$

=> Ans - (B)

Let side of square inscribed in the semi circle be $$a$$ cm and radius of semi circle = $$r$$ cm

In right $$\triangle$$, => $$r^2=a^2+(\frac{a}{2})^2$$

=> $$r^2=a^2+\frac{a^2}{4}$$

=> $$r^2=\frac{5a^2}{4}$$

=> $$a^2=\frac{4r^2}{5}$$ --------------(i)

Similarly, let side of another square be $$b$$ cm and radius of same circle = $$r$$ cm

=> $$b^2+b^2=(2r)^2$$

=> $$b^2=2r^2$$ --------------(ii)

$$\therefore$$ Ratio of area of first square to area of second square = $$\frac{a^2}{b^2}$$

Substituting values from equation (i) and (ii), => $$\frac{\frac{4r^2}{5}}{2r^2}$$

= $$2:5$$

=> Ans - (C)

Let original price = Rs. $$x$$

Price after $$p\%$$ increase = $$x+\frac{p}{100}\times x=Rs.$$ $$(x+\frac{px}{100})$$

Price after $$p\%$$ decrease = $$(x+\frac{px}{100})-(\frac{p}{100})(x+\frac{px}{100})$$

According to ques, => $$x+\frac{px}{100}-\frac{px}{100}-\frac{p^2x}{10000}=100$$

=> $$x(1-\frac{p^2}{10000})=100$$

=> $$x=Rs.$$ $$\frac{100}{1-\frac{p^2}{100^2}}$$

=> Ans - (A)

Let the common perimeter be $$c$$

So, each side of square = $$\frac{c}{4}$$

=> Circum radius of square = $$r_s=\frac{s}{\sqrt2}=\frac{c}{4\sqrt2}$$

Thus, area of circle circumscribed about the square will be

=> $$A=\pi (r_s)^2=\pi\frac{c^2}{32}$$ -------------(i)

Now, each side of triangle = $$\frac{c}{3}$$

=> Height of triangle = $$\frac{\sqrt3}{2}\times\frac{c}{3}$$

=> Circum radius of triangle = $$r_t=\frac{2}{3}\times(\frac{\sqrt3}{2}\times\frac{c}{3})=\frac{c}{3\sqrt3}$$

So area of the circle circumscribed about the triangle 

=> $$B=\pi (r_t)^2=\pi\frac{c^2}{27}$$ -----------(ii)

$$\therefore$$ Required ratio $$A:B=27:32$$

=> Ans - (C)

Let length, breadth and height be $$l,b,h$$ cm respectively.

=> Surface area of cuboid = $$(lb+bh+hl)=281$$ -----------(i)

Sum of all edges = $$(l+b+h)=24$$

Squaring both sides, => $$(l^2+b^2+h^2)+2(lb+bh+hl)=576$$

Substituting value from equation (i), we get :

=> $$(l^2+b^2+h^2)+2(281)=576$$

=> $$(l^2+b^2+h^2)=576-562=14$$

$$\therefore$$ Diagonal of cuboid = $$\sqrt{l^2+b^2+h^2}=\sqrt{14}$$

=> Ans - (D)

Let radius of cylinder = $$x$$ cm and its height = $$2x$$ cm

Height of cone = $$h=12$$ cm and radius of cone = $$r=5$$ cm

Since, the axes coincide, thus in the two similar triangles, we have : $$\frac{y}{h}=\frac{x}{r}$$

=> $$\frac{12-2x}{12}=\frac{x}{5}$$

=> $$60-10x=12x$$

=> $$12x+10x=22x=60$$

=> $$x=\frac{60}{22}=2\frac{8}{11}$$ cm

=> Ans - (A)