JUNE 2013 - PAPER III

10. Match the following :
a. RAID 0       i. Bit interleaved parity
b. RAID 1       ii. Non redundant stripping
c. RAID 2       iii. Mirrored disks
d. RAID 3       iv. Error correcting codes
Codes :
        a  b  c  d
   (A)  iv  i   ii  iii
   (B)   iii   iv   i   ii
   (C)   iii   i   iv   ii
   (D)   iii   ii   iv   i

Ans:- None of the options

Explanation:-
The following are the one liners explaining different RAIDs and their specification.
RAID 0 - It has no redundancy.It is JBOD(Just a Bunch of Disks)
RAID 1 - Mirroring or shadowing
RAID 2 - Applying memory-style error correcting codes to disks
RAID 3 - Bit interleaved parity.
So, a - ii, b-iii, c-iv, d-i. These are the correct matches. There is no answer which matches this. So, the answer is none of the options are right.

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11. The golden ratio ϕ and its conjugate –ϕ both satisfy the equation
(A) x³ – x – 1 = 0
(B) x³ + x – 1 = 0
(C) x² – x – 1 = 0
(D) x² + x – 1 = 0

Ans:-C

Explanation:
The best explanation for golden ratio is in some well known maths websites. I am providing their links down.
https://www.mathsisfun.com/numbers/golden-ratio.html

http://mathworld.wolfram.com/GoldenRatio.html

Golden ratio is represented by the symbol ϕ(Phi), and its conjugate is –ϕ(phi, also called as silver ratio). Both are satisfied by the equation, x² – x – 1 = 0, as given in the explanation for Golden ratio. Since this equation is a quadratic equation, we are going to solve it the usual way, by calculating its roots.
Roots=-b + or - (sqrt(b²-4ac))/2a. Given the equation, x² – x – 1 = 0, value of a=1,b=-1 and c=-1. So, the equation for calculating root will become

1 + or - sqrt(1+4) / 2

(1 + or - sqrt(5))/2

If we calculate the roots we will get the value 1.61 and -0.61 which are actually the values of Golden ratio and its conjugate. So, the Golden ratio and its conjugate both satisfy the equation x² – x – 1 = 0. The answer is C.

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13. In any n-element heap, the number of nodes of height h is
(A) less than equal to [ n/2^h]
(B) greater than [ n/2^h]
(C) greater than [ n/2^h+1]
(D) less than equal to [ n/2^h+1]

Ans:-D

Explanation:-
A heap of size n has at most [ n/2^h+1] nodes with height h. So, it can be less than equal to [ n/2^h+1

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