SRCC GBO 2013 Question Paper

$$a = 2^{129} \times 3^{81} \times 5^{128}$$ 

$$b = 2^{127} \times 3^{81} \times 5^{128}$$ 

$$c = 2^{126} \times 3^{82} \times 5^{128}$$ 

$$d = 2^{125} \times 3^{82} \times 5^{129}$$

Eliminating common factors $$(2)^{125}\times(3)^{81}\times(5)^{128}$$

=> $$a=16,b=4,c=6,d=15$$

$$\therefore$$ $$b<c<d<a$$

=> Ans - (B)

Smallest number having $$n$$ digits is $$(10)^{n-1}$$

Smallest possible value of $$a=10$$

Now, $$b$$ has 10 digits, thus smallest possible value of $$b=(10)^9$$

Now, $$c$$ has $$(10)^9$$ digits, thus smallest possible value of $$c=(10)^{10^9-1}$$

=> Ans - (D)

Expression : $$(2)^{36}-1$$

= $$(2^3)^{12}-1=(8)^{12}-1$$

Now, to check whether the number is divisible by 9, we have : $$(-1)^{12}-1=0$$

Thus, $$(2)^{36}-1$$ is divisible by 9, hence sum of its digits is also divisible by 9.

=> $$6+8+a+1+9+4+7+6+7+3+5=56+a$$

Thus, $$56+a=63$$

=> $$a=7$$

=> Ans - (C)

Expression : $$E=\frac{(1.121)^3 - (3.333)^3 + (2.212)^3}{(1.121)(3.333)(2.212)}$$

Let $$x=1.121,y=2.212,z=-3.333$$ and hence $$x+y+z=0$$ ------------(i)

=> $$E=-\frac{(x)^3+(y)^3+(-z)^3}{xy(-z)}$$ -----------(ii)

We know that, $$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$$

=> $$x^3+y^3+z^3=3xyz$$ -------(iii)    [$$\because x+y+z=0$$]

Thus, substituting value from equation (iii) in equation (ii), we get : $$E=-3$$

=> Ans - (C)

Given = $$a : b = 3 : 4, b : c = \frac{8}{9}$$ and $$c : d = \frac{2}{3}$$

Multiplying equation (iii) by 9 and equation (ii) by 2, and (i) by 4

=> $$a:b:c:d=12:16:18:27$$

To find = $$\sqrt[4]{\frac{ad}{b^2}}$$

= $$\sqrt[4]{\frac{12\times27}{16\times16}}$$

= $$\sqrt[4]{\frac{81}{64}}$$

= $$\frac{3}{2\sqrt2}=\frac{3\sqrt2}{4}$$

=> Ans - (D)

Expression : $$\frac{\sqrt{43 - 12\sqrt{7}} - 2}{16 + 6\sqrt7}$$

= $$\frac{\sqrt{(\sqrt7)^2+(6)^2 - 2(\sqrt{7})(6)} - 2}{(\sqrt7)^2+(3)^2 + 2(\sqrt7)(3)}$$

= $$\frac{\sqrt{(6-\sqrt7)^2}-2}{(3+\sqrt7)^2}$$

= $$\frac{4-\sqrt7}{(3+\sqrt7)^2}$$

$$N$$ divides each of 13511, 13903, 14589 leaving remainder $$r$$ in each case.

=> $$13511=aN+r$$ -----------(i)
$$13903=bN+r$$ --------------(ii)
$$14589=cN+r$$ --------------(iii)

Subtracting above equations, we get : $$392=(b-a)N$$
$$686=(c-b)N$$
$$1078=(c-a)N$$

Thus, $$N$$ must be a factor of all three of these numbers. Thus, $$N=$$ H.C.F. (392,686,1078) = 98

$$\therefore$$ sum of digits of $$N$$ = $$9+8=17$$

=> Ans - (D)

Expression : $$\frac{(0.251)^2 - (0.051)^2 - (0.511)^2 - 2(0.051)(0.511)}{(0.251)^2 - (0.051)^2 - 2(0.251)(0.051) - (0.511)^2}$$

Let $$x=0.251,y=0.051,z=0.511$$

= $$\frac{x^2-(y^2+z^2+2yz)}{(x^2-y^2-2xy)-z^2}$$

= $$\frac{x^2-(y+z)^2}{(x-y)^2-z^2}$$

= $$\frac{(x+y+z)(x-y-z)}{(x-y-z)(x-y+z)}$$

= $$\frac{x+y+z}{x-y+z}$$

= $$\frac{0.251+0.051+0.511}{0.251-0.051+0.511}=\frac{0.813}{0.711}=\frac{271}{237}$$

=> Ans - (A)

Let cost of making a table = $$C=X+Y$$ -----------(i)

Now, $$X=k_1A$$, where $$A$$ is area of table

and $$Y=k_2l^2$$, where $$l$$ is length of longer side of table

In making 2 m x 3 m table it costs ₹ 5,000

=> $$C=(6k_1)+(9k_2)=5000$$ ------------(ii)

Similarly, $$6k_1+16k_2=6400$$ -------------(iii)

Subtracting equation (ii) from equation (iii), we get :=> $$7k_2=1400$$

=> $$k_2=\frac{1400}{7}=200$$

Similarly, $$k_1=\frac{3200}{6}$$

Now, for a table having dimension 2.5 m x 2.5 m

=> Cost = $$(6.25\times\frac{3200}{6})+(6.25\times200)$$

= $$6.25\times(\frac{3200}{6}+200)$$

= $$6.25\times\frac{4400}{6}\approx Rs.$$ $$4,583$$

=> Ans - (B)

Let Anu's speed = $$s$$ km/hr

Let length of each train = $$l$$ km and speed of each train = $$v$$ km/hr

Relative speed  of train going in same direction = $$(v-s)$$ km/hr

=> Time taken = $$t_1=\frac{l}{v-s}$$ --------------(i)

Similarly, $$t_2=\frac{l}{v+s}$$ ------------(ii)

Comparing equations (i) and (ii),

=> $$t_1(v-s)=t_2(v+s)$$

=> $$vt_1-st_1=vt_2+st_2$$

=> $$s(t_2+t_1)=v(t_1-t_2)$$

=> $$s=\frac{v(t_1-t_2)}{(t_2+t_1)}$$

=> Ans - (A)