Direction: Study the following information carefully and answer the questions given below.
There is a 7×5 matrix which can produce frequency signals that helps in the illumination of some bulbs. The rows of the matrix are denoted by %, #, @, &, $, € and ¥ from top to bottom respectively and the columns are denoted by I, II, III, IV, and V from left to right respectively:
• % row (topmost row) contains the numbers, which are the consecutive multiples of ‘3’ starting from ‘6’.
• # row contains the numbers, which are the consecutive numbers following the given equation, n (n + 1)/2, starting from n = 4.
• @ row contains the numbers, which are the consecutive multiples of ‘12’ starting from ‘24’.
• &row contains the numbers, which are the consecutive multiples of ‘14’ starting from ’14’.
• $ row contains the numbers, which are the consecutive prime numbers starting from ‘13’.
• € row contains the numbers, which are the consecutive numbers following the given equation, (4n2+ 3), starting from n = 2.
• ¥ row contains the numbers, which are the consecutive multiples of ‘4’ starting from ‘20’.
The matrix helps in producing frequency which is a string of numbers. There are five bulbs White, Purple, Black, Cyan and Orange. Based on the outcome of the string frequency one of the bulbs blinks.
Condition for blink:-
I. If the outcome of the string is between 25 and 75, then Black bulb blinks.
II. If the outcome of the string is between 75 and 125, then Cyan bulb blinks.
III. If the outcome of the string is between 125 and 150, then Orange bulb blinks.
IV. If the outcome of the string is between 150 and 200, then Purple bulb blinks.
V. If the outcome of the string is between 200 and 275, then White bulb blinks.
VI. If none of the above conditions follow, then no bulb blinks. For the outcome of the string:-
I. If all the numbers of the string are an even number, then the resultant frequency can be obtained by multiplying the tenth place digits of all the two-digit numbers.
II. If a second number is a prime number and the second last number is the multiple of ‘3’, then the resultant frequency can be obtained by the approximate average of all the numbers.
III. If a prime number is followed by a perfect square, then the resultant frequency can be obtained by multiplying the one’s place digits of all the numbers.
IV. If all the numbers in the string are an odd number, then the resultant frequency can be obtained by double the difference between the highest and second lowest number.
V. If none of the above conditions follows, then the resultant frequency can be obtained by the difference between the highest and lowest number.
VI. If more than one string is given, then the resultant frequency can be obtained by the sum of the results of individual strings.
28) If X = &II $IV %V €II ¥III, then which of the following bulb blinks?
A. Yellow
B. Black
C. Pink
D. Red
E. None of these
Sol:
X = 28 23 18 39 28
From condition II: (28+23+18+39+28) / 5 = 27.2
Thus, Black bulb blinks.
29) If Y = $III €IV €III %IV and Z = $IV %II @V #III, then which of the following bulb blinks?
A.Black
B.Orange
C.Purple
D.White
E.No bulb blinks
Sol:
Y = 19 103 67 15
From condition IV: 2(103-19) = 168
Z = 23 9 72 21
From condition III: (3*9*2*1) = 54
From condition VI: 168+54 = 222
Thus, White bulb blinks.
30) If W = ¥IV &III #IV @II and T = $II €III &IV @V, then which of the following bulb blinks?
A.Cyan
B.Black
C.Orange
D.Purple
E.No bulb blinks
Sol:
W = 32 42 28 36
From condition I: (3*4*2*3) = 72
T = 17 67 56 72
From condition V: (72-17) = 55
From condition VI: (72+55) = 127
Thus, Orange bulb blinks.