SRCC GBO 2013

Let questions answered correctly = $$x$$ and unanswered questions = $$y$$, => incorrect questions = $$(20-x-y)$$

According to ques,

=> $$5x-2(20-x-y)=48$$

=> $$7x+2y=88$$

Since, sum of 2 positive numbers is positive, and 84 is the nearest multiple of 7 (when $$y=2$$)

=> $$7x=84$$ => $$x=12$$

Thus, maximum number of questions answered correctly = 12

=> Ans - (B)

Total students = 100 and Number of girls = 99, Number of boys = 1

Let number of students living off campus = $$x$$, and number of students living on campus = $$(100-x)$$

2% of boys live on campus (since there is only 1 boy)

=> $$\frac{2}{100}\times(100-x)=1$$

=> $$100-x=50$$

=> $$x=50$$

=> Ans - (C)

Let total area of land = $$100$$ sq. units

=> Total area of A's land = $$40$$ sq. units and B's land = $$60$$ sq. units

Also, Total wheat in the land = $$\frac{7}{10}\times100=70$$ sq. units and similarly, total barley = $$30$$ sq. units

A' land : Wheat = $$\frac{4}{5}\times40=32$$ sq. units

Barley = $$40-32=8$$ sq. units

=> Remaining wheat = $$70-32=38$$ sq. units and remaining barley = $$30-8=22$$ sq. units

$$\therefore$$ Ratio of wheat to Barley for B's land = $$\frac{38}{22}=19:11$$

=> Ans - (B)

Let number of apples = $$3x$$ and number of oranges = $$4y$$

=> Number of rotten apples = Number of rotten oranges

=> $$2x=3y$$

=> $$\frac{x}{y}=\frac{3}{2}$$

Let $$x=3$$ and $$y=2$$

=> Fraction of rotten fruits = $$\frac{2x+3y}{3x+4y}$$

= $$\frac{2(3)+3(2)}{3(3)+4(2)}$$

= $$\frac{12}{17}$$

=> Ans - (A)

Principal amount for EMIs = Rs. (19,650 - 3,100) = Rs. 16,550

Rate of interest = 10% and number of EMIs = 3

Amount when compounded annually = $$P(1+\frac{r}{100})^T$$

Let each installment = Rs. $$x$$

=> $$\frac{x}{(1+\frac{10}{100})^1}+\frac{x}{(1+\frac{10}{100})^2}+\frac{x}{(1+\frac{10}{100})^3}=16550$$

=> $$x[(\frac{10}{11})+(\frac{100}{121})+(\frac{1000}{1331})]=16550$$

=> $$x[\frac{1210+1100+1000}{1331}]=16550$$

=> $$x=16550\times\frac{1331}{3310}$$

=> $$x=Rs.$$ $$6,655$$

=> Ans - (C)

The person bought the n article, each at Rs. d.

In that, he sold two articles at Rs. $$\ \frac{\ d}{2}$$, which brings in Rs. d for him. 

Now, he sold the remaining n-2 articles at Rs. (d+8), which brings in Rs. (n-2)(d+8)

His total profit is Rs.72. This means, (n-2)(d+8)+d-nd = 72, or nd+8n-2d-16-nd = 72.

Hence 8n - 2d = 88, or 4n - d = 44

Now, d has to be greater than zero. Hence, the minimum value of n shall be 12 to make d as 4. 

Dimensions of cube = 3 cm x 3 cm x 3 cm and dimensions of square hole = 1 cm x 1 cm

Outer surface area of 1 face = $$(3\times3)-(1\times1)=8$$ $$cm^2$$

=> Outer surface area of the cube (6 faces) = $$6\times8=48$$ $$cm^2$$

Inner surface area of three tunnels = $$3\times2[(2\times1)+(2\times1)]=24$$ $$cm^2$$

$$\therefore$$ Total surface area of solid = $$48+24=72$$ $$cm^2$$

=> Ans - (C)

The square will cover the same area of the triangle as much as covered by the triangle on square, and that at this point the amount of triangle covered is the same as the amount of square covered.

Let area of triangle be $$A$$ and area of square = $$6\times6=36$$ $$cm^2$$

According to ques,

=> $$\frac{60}{100} A=\frac{2}{3}\times36$$

=> $$A=\frac{24}{0.6}=40$$ $$cm^2$$

=> Ans - (C)

Radius of circle = OC = OB = 2 m and OA = 1 m

Area of path = $$2\times$$[Area of $$\triangle$$ OAB+Area of sector BOC] ----------(i)

In right $$\triangle$$ AOB,

=> $$(AB)^2=(OB)^2-(OA)^2$$

=> $$(AB)^2=4-1=3$$

=> $$(AB)=\sqrt3$$ m ------------(ii)

Also, $$tan(\angle AOB)=\frac{AB}{OA}=\sqrt3$$

=> $$\angle OAB=60^\circ$$

Thus, $$\angle BOC=90^\circ-60^\circ=30^\circ$$ --------------(iii)

Now, in equation (i), area of path = $$2\times[(\frac{1}{2}\times OA \times AB)+(\frac{30^\circ}{360^\circ}\pi r^2)]$$

= $$2\times[(\frac{1}{2}\times1\times\sqrt3)+(\frac{1}{12}\times \pi (2)^2)]$$

= $$2\times[\frac{\sqrt3}{2}+\frac{\pi}{3}]$$

= $$\sqrt3+\frac{2\pi}{3}$$ -----------(iv)

$$\therefore$$ Area of plot = Area of shaded region = Area of circle - Area of path

= $$\pi(2)^2-(\sqrt3+\frac{2\pi}{3})$$

= $$4\pi-\sqrt3-\frac{2\pi}{3}$$

= $$\frac{10\pi}{3}-\sqrt3$$ $$m^2$$

=> Ans - (D)

It is given that $$a=8$$ cm and $$b=9$$ cm. Also, $$A_2=\frac{A_1+A_3}{2}$$ ------------(i)

To find : $$c=?$$

Solution : Area of a circle = $$\pi r^2$$

Using equation (i),

=> $$\pi (b)^2=\frac{\pi (a)^2+\pi(c)^2}{2}$$

=> $$a^2+c^2=2b^2$$

=> $$64+c^2=162$$

=> $$c^2=162-64=98$$

=> $$c=\sqrt{98}=7\sqrt2$$ cm

=> Ans - (A)