SRCC GBO 2013 Question Paper

To find : $$123456789\times999999999$$

= $$123456789\times(1000000000-1)$$

= $$123456789000000000-123456789$$

= $$12345678876543211$$

Thus, there are 0 9's

=> Ans - (A)

Expression : $$15^{6} \times 28^5 \times 55^7$$

Number of 2's in the expression (in $$28^5$$) = $$(2)^{10}=10$$

Number of 5's = $$(5)^6\times(5)^7=13$$

$$\because$$ number of 2's are less than number of 5's, thus number of zeroes will depend on two, hence number of zeroes are 10

=> Ans - (B)

Since options are not so close, we can think of the 4th root of $$2\times\ 10^{15}$$

This is nothing but 4th root of 2000 * $$10^{12}$$

=> 1000 * 4th root of 2000 [between 6 and 7] => Answer lies between 6000 and 7000 => Option-C

Expression : $$(2002)^{2002}$$

It can be written as : $$(2002)^{4n+2}$$

Thus, the unit's digit will be same as $$(2)^2=4$$

=> Ans - (B)

Let the three original numbers be $$x,y,z$$

According to ques, => $$x+(\frac{y+z}{2})=65$$

Similarly, $$y+(\frac{x+z}{2})=69$$

and $$z+(\frac{y+x}{2})=76$$

Adding all the equations, => $$(x+y+z)+(x+y+z)=65+69+76$$

=> $$x+y+z=\frac{210}{2}=105$$

Dividing both sides by 3, => $$\frac{x+y+z}{3}=\frac{105}{3}=35$$

$$\therefore$$ Required average = 35

=> Ans - (A)

Here, $$c+f=10$$ and $$b+e=9$$ and $$a+d=9$$     [because of carry over 1]

Adding all the equations, => $$a+b+c+d+e+f=10+9+9=28$$

=> Ans - (D)

A number is divided by 9 if sum of its digits is 9

When a number is divided by 9 and leaves no remainder, => sum of digits is 9.

So, when a number is divided by 9 and leaves remainder 4 , then sum of its digits is $$9+4 = 13$$

=> Ans - (A)

Expression : $$\frac{1}{8} + \frac{1}{10} + \frac{1}{11} + \frac{1}{15} + \frac{1}{20} + \frac{1}{41} + \frac{1}{110} + \frac{1}{1640}$$

= $$(\frac{1}{41} + \frac{1}{1640}) + (\frac{1}{10} + \frac{1}{20}) + (\frac{1}{11} + \frac{1}{110}) + (\frac{1}{8} + \frac{1}{15})$$

= $$(\frac{40+1}{1640})+(\frac{2+1}{20})+(\frac{10+1}{110})+(\frac{15+8}{120})$$

= $$\frac{1}{40}+\frac{3}{20}+\frac{1}{10}+\frac{23}{120}$$

= $$\frac{3+18+12+23}{120}=\frac{56}{120}=\frac{7}{15}$$

=> Ans - (D)

When $$(22)^7$$ is divided by 123

= $$\frac{(22)^2\times(22)^2\times(22)^2\times22}{123}$$

Now, when we divide $$(22)^2=484$$ by 123, remainder is $$-8$$

= $$\frac{-8\times-8\times-8\times22}{123}$$

= $$\frac{-20\times22}{123}$$

=> Remainder = -(-52) = 52

=> Ans - (D)

Expression : $$\frac{97}{19} = w + \frac{1}{x + \frac{1}{y}}$$

Breaking the L.H.S. expression = $$5+\frac{2}{19}$$

= $$5+\frac{1}{\frac{19}{2}}$$

= $$5+\frac{1}{9+\frac{1}{2}}$$

Comparing above equation with : $$w+\frac{1}{x+\frac{1}{y}}$$

=> $$w=5, x=9,y=2$$

To find : $$x + 2y - 3w$$

= $$9+2(2)-3(5)=9+4-15=-2$$

=> Ans - (B)