NICL (GIC) Administrative Officer Finance 2013 Exam Question paper

Average temperature of Gwalior, Agra and Lucknow is 37° C

=> Sum of temperatures of Gwalior, Agra and Lucknow = $$G + A + L = 37 \times 3 = 111$$ -----------(i)

Similarly, $$G + A + D = 38 \times 3 = 114$$ ----------(ii)

It is given that temperature of Delhi, $$D = 39$$

Putting it in eqn(ii), => $$G + A = 114 - 39$$

=> $$G + A = 75$$

Now, putting above value in equation (i), we get :

=> $$L = 111 - 75 = 36$$

$$\therefore$$ Ratio of temperature of Delhi to that of Lucknow = $$\frac{D}{L}$$

= $$\frac{39}{36} = \frac{13}{12}$$

= $$13 : 12$$

Two digit numbers which remain same after interchanging the digits = 11,22,33,....,99

The above expression is an A.P with first term, $$a = 11$$ and common difference, $$d = 11$$

Number of terms = 9

Sum of A.P. = $$\frac{n}{2} (a + a_n)$$

= $$\frac{9}{2} (11 + 99) = \frac{9}{2} \times 110$$

= $$9 \times 55$$

$$\therefore$$ Required average = $$\frac{9 \times 55}{9} = 55$$

Expression : 4899 x 999

Since, the last digit of each number is 9, => last digit of the resultant number = (9*9 = 81) = 1

Only, option (A) matches above description.

=> 4899 x 999 = 4894101

=> Ans - (A)

1000 position or 1024 bit is represented by K

=> Ans - (A)

Given : 90% of A = 30% of B

=> $$\frac{90}{100} \times A = \frac{30}{100} \times B$$

=> $$9A = 3B$$

=> $$\frac{B}{A} = \frac{9}{3} = 3$$

Also, B = x % of A

=> $$B = \frac{x}{100} \times A$$

=> $$\frac{x}{100} = \frac{B}{A} = 3$$

=> $$x = 100 \times 3 = 300$$

The word computer is a combination of Central Processing unit and the internal memory.

=> Ans - (B)

L.C.M. = 120 and H.C.F. = 10

Let the numbers be $$10a , 10b$$     (where a,b are co prime)

=> $$10a \times 10b = 120 \times 10$$

=> $$a \times b = 12$$

Since a,b are co primes, the only possibility is $$a = 3 , b = 4$$     (or vice-versa)

=> Numbers = 30 and 40

$$\therefore$$ Sum = 30 + 40 = 70

External radius, $$R = 2$$ cm and Internal radius, $$r = 1$$ cm

=> Volume of external sphere = $$\frac{4}{3} \pi R^3$$

= $$\frac{4}{3}\pi (2)^3 = \frac{32 \pi}{3}$$

Volume of internal sphere = $$\frac{4}{3} \pi r^3$$

= $$\frac{4}{3}\pi (1)^3 = \frac{4 \pi}{3}$$

$$\therefore$$ Volume of silver = $$\frac{32 \pi}{3} - \frac{4 \pi}{3}$$

= $$\frac{28 \pi}{3} cm^3$$

Expression : $$(123)^{2.5} \div (123)^{?} = 123$$

=> $$\frac{123^{2.5}}{123^x} = 123$$

=> $$123^x = \frac{123^{2.5}}{123}$$

=> $$123^x = 123^{2.5 - 1} = 123^{1.5}$$

=> $$x = 1.5$$

An internal storage consisting of arrays of small rings of magnetic material is called magnetic core.

=> Ans - (B)