NICL (GIC) Administrative Officer (AO) Officer Exam 2013 Question paper

A digitizer is a tool used to convert hand-drawn images into a format suitable for computer processing (digital form)

=> Ans - (B)

George Stibits built the world’s first electronic calculator using telephone relays, light bulbs and batteries.

=> Ans - (C)

When $$n$$ is divided by 5, remainder is 2

=> $$n = 5k + 2$$     (k is quotient)

Squaring both sides, we get :

=> $$n^2 = (5k + 2)^2$$

=> $$n^2 = 25k^2 + 20k + 4$$

Now, if we divide $$n^2$$ by 5

=> $$\frac{25k^2 + 20k + 4}{5}$$

$$\because$$ 25 and 20 are divided by 5

=> Remainder = 4

Method II : When ‘n’ is divided by 5 the remainder is 2

Let n = 12

=> $$n^2 = 12^2 = 144$$

Now, when 144 is divided by 5, => Remainder = 4

Hard disks and diskettes (floppy disk) are direct access storage devices.

=> Ans - (A)

Let the number be $$x$$ and $$(24 - x)$$

=> Product = $$x (24 - x) = 143$$

=> $$x^2 - 24x + 143 = 0$$

=> $$(x - 13) (x - 11) = 0$$

=> $$x = 13 , 11$$

Other number = $$11 , 13$$

Thus, the two numbers are = 11 and 13

$$\therefore$$ Sum of their squares = $$(11)^2 + (13)^2$$

= $$121 + 169 = 290$$

The computer operates on binary language (0-1)

ON - 1

OFF - 0

=> Ans - (D)

$$2744 = 8 \times 343$$

=> $$2744 = 2^3 \times 7^3$$

Powers should be even to make it a perfect square

=> Least number which should be multiplied = $$2 \times 7 = 14$$

Let the length, breadth and height of cuboid be $$l , b , h$$ respectively

Area of three consecutive faces :

=> $$l b = 12$$

$$b h = 20$$

$$h l = 15$$

Multiplying the above equations,

=> $$(lbh)^2 = 12 \times 20 \times 15 = 3600$$

=> $$(lbh) = \sqrt{3600} = 60$$

$$\therefore$$ Volume of cuboid = $$60 cm^3$$

Let the number be $$x$$ and $$(45 - x)$$

=> Product = $$x (45 - x) = 500$$

=> $$x^2 - 45x + 500 = 0$$

=> $$(x - 25) (x - 20) = 0$$

=> $$x = 25 , 20$$

Other number = $$20 , 25$$

Thus, the two numbers are = 20 and 25

$$\therefore$$ H.C.F. of 20 and 25 = 5

Let efficiency of P,Q and R to finish the work be $$p , q , r$$ units/day respectively

Let total work to be done = 60 units

P and Q complete the work in 12 days

=> Efficiency of P and Q = $$p + q = \frac{60}{12} = 5$$

Similarly, $$q + r = 4$$ ----------(i)

$$r + p = 3$$

Adding the three equations, we get :

=> $$2 (p + q + r) = 5 + 4 + 3 = 12$$

=> $$p + q + r = \frac{12}{2} = 6$$

From eqn(i), => $$p + 4 = 6$$

=> $$p = 6 - 4 = 2$$ units/day

$$\therefore$$ Time taken by P to finish the work alone = $$\frac{60}{2} = 30$$ days