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ATQ,
A number is divisible by 11 if difference between the sum of digits at odd places and the sum of the digits at even places is either zero or divisible by 11. (We can count from either side). So, sum of digits at odd places (Counting from right to left) = 7 + 8 +a = 15 + a And, sum of digits at even places = 9 + 1 + 5 = 15 So, required difference = 15 + a – 15 = a So, ‘a’ must be either ‘0’ or a multiple of ‘11’. But since only ‘0’ is in the options. So, ‘a’ must be 0
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