1.

There is 3x 5 matrix which can produce signals which in turn help in the illumination some bulbs. The row of the matrix are denoted by alpha,beta and gamma from bottom to top and the columns are denoted by the alphabets P,Q,R,S and T from left to right. alpha row contains number which are consecutive multiple of 7, starting from 28 (from left to right ) gamma row contains number which are consecutive multiple of 13 starting from 13 (from left to right ) beta row contains consecutive multiples of 11 starting from 11 (from left to right) The matrix helps in producing signals which can be either a single string of number A-or two -line string A andB. there are 4 lights W,X,Y and Z. based on the outcome of the string mentioned above one of the light blinks Conditions for blink 1. If the outcome is below 85, then W will blink 2. If outcome range is 85-110, then X 3. If outcome range is 111-210, then Y blinks 4. If outcome is greater than 210, then Zblinks For outcome of the string 1. If the string has all even numbers , then outcome of the string is obtained by adding all the numbers 2. If an odd number is followed by an even number then the one's places of all the two -digit numbers are deleted and, tenth place are multiplied to get the outcome 3. If the string contains 2 prime number, then the tenth's place is deleted from each of the two -digit number and remaining number are multiplied. 4. if no above logic is followed , then simple outcome is addtion of the numbers If A=alpha Pgamma Q alpha S beta R , then which bulb blink ?

Answer»

<P>Z
X
W
Y

Solution :From the above given CONDITIONS the MATRIX will be

`A =ALPHA P gamma Q alpha S BETA R `
so, A=2826 49 33
As it is clear that conditon (4) is applicable in the above ques so the outcome will be outcome =28+26+49+33=136
As the outcome is 111-210, so clearly Y will blink


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