Directions (1-5): When a letter/number/symbol arrangement machine is given an input line of letters/numbers/symbol, it arranges them following a certain rule. Following is an illustration of Input and rearrangement. Read the following information and answer the questions that follow
Input: A O $ @ 5 H I % Q 2 4 E A K 9 > P U E # S ^ *
Step I: A $ @ 5 H I % Q 2 4 E A K 9 > P U E # S * P ^
Step II: A $ @ H I % Q 2 4 E A K 9 > P U E # * P ^ 4 T
Step III: A $ @ H I Q 2 4 E A K 9 > P U E * P ^ 4 T % #
Step IV: A $ @ H I Q 2 E A K 9 > P U * P ^ 4 T % # 3 F
Since there is no end to the number of steps of the above input one can make as many steps as intended. As per the rules followed in the above input, work out the steps for the following input:
Input: H # β 4 O E ! A $ € I K 5 9 @ S A U 8 E Q 1
The number of steps to be worked out is more than four.
Q1. How many vowels appear in step V of the given input?
(a) None
(b) One
(c) Two
(d) Three
(e) More than three
Q2. In which of the following given steps, the last element is not a number?
(i) Step II
(ii) Step III
(iii) Step V
(iv) Step IV
(a) Only (ii)
(b) (i), (iii) and (iv)
(c) Only (iii)
(d) Both (ii) and (iv)
(e) All the given
Q3. Which element is third to the right of symbol ‘@’ in step IV?
(a) A
(b) R
(c) K
(d) #
(e) None of these
Q4. How many elements are between ‘β’ and ‘@’ in step II?
(a) Nine
(b) Eight
(c) Ten
(d) Seven
(e) None of these
Q5. Which of the following element is 4th to the left of 12th element from the right end in step IV?
(a) €
(b) !
(c) β
(d) E
(e) None of these
Solutions
Solutions (1-5):
Sol. The logic followed here is –
The elements from the left and right ends are picked which appear at even numbered positions from both ends, starting from 2nd position for step I. Then, the following operations is applied –
If the element is a number- it is decreased by 1.
If the element is a symbol – remains the same.
If the element is a letter – changed to just next letter as per alphabetical series.
The element picked from the right end is placed at rightmost end and the element picked from the left end is placed second from the right end.
These ways, in step I second element from both ends is picked. In step II, 4th from both ends is picked, and so on.
For the given input we get
Input: H # β 4 O E ! A $ € I K 5 9 @ S A U 8 E Q 1
Step I: H β 4 O E ! A $ € I K 5 9 @ S A U 8 E 1 # R
Step II: H β 4 E ! A $ € I K 5 9 @ S A U 8 1 # R P F
Step III: H β 4 E ! $ € I K 5 9 @ S A U 1 # R P F B 7
Step IV: H β 4 E ! $ € K 5 9 @ S A 1 # R P F B 7 J V
Step V: H β 4 E ! $ € K 5 @ S 1 # R P F B 7 J V 8 B
And, so on. Step V is not the last step
S1. Ans. (b)
S2. Ans. (b)
S3. Ans. (e)
S4. Ans. (c)
S5. Ans. (a)
. .