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n = 92.abc, where a, b and c denote the tenths, hundredths and thousandths digits respectively in the decimal representation of n.
When n is rounded to the nearest integer, the result is 92. If c is the average of two consecutive prime numbers such that c > b > a,
what is the difference between the maximum and minimum possible sum of a, b and c?
  • a)
    13
  • b)
    16
  • c)
    17
  • d)
    18
  • e)
    20
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
n = 92.abc, where a, b and c denote the tenths, hundredths and thousan...
Given:
  • n = 92.abc
    • a, b and c denote the tenths, hundredths and thousandths digits respectively
    • Being digits of a number 0 ≤ a, b, c ≤ 9
    • c > b > a
  • Result of rounding n to the nearest integer = 92
  • , where P ,P are consecutive prime numbers
  • To Find: Max(a + b + c) – Min(a + b + c)
Approach:
  1. For finding the value of Max (a + b + c) – Min (a + b + c), we need to find the range of values for a, b and c
  2. We know that the value of n when rounded to the nearest integer will depend on the value of a.
    1. Since we are given that n, when rounded to the nearest integer is 92, we can find the range of values of n.
  3. We know that c is the average of two consecutive prime numbers. As c ≤ 9, we can find the possible values of c.
  4. Also, we will use the relation c > b > a, to find the possible values of b, because we already know the range of possible values of a and c.
Working out:
  1. As n when rounded to the nearest integer = 92, a ≤ 4………….(1)
a. Maximum value of a = 4
b. Minimum value of a = 0
 
2. Since c is an integer ≤ 9 and c is the average of two consecutive prime numbers, possible values of c can be:
3. Now, b is a digit from 0 to 9, however keeping in mind the constraint that b > a, b ≠ 0, because in that case b will not be greater than a.
a. So, Minimum value of b = 1 …….(3)
b. Also, since b < c, b ≠ 9, because in that case b will not be less than c.
c. So, maximum value of b = 8…….(4)

4. Combining (1), (2), (3) and (4), we have
5. Hence, Max (a + b + c) – Min (a+ b + c) = 21 – 5 = 16
Answer B
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Most Upvoted Answer
n = 92.abc, where a, b and c denote the tenths, hundredths and thousan...
To begin, we know that when n is rounded to the nearest integer, the result is 92. This means that the decimal part of n must be less than or equal to 0.5.

Since c is the average of two consecutive prime numbers, it must be an integer. This means that c can only be 1 or 2. However, since c is also the thousandths digit of n, it cannot be 1 because that would imply that n is less than 92. Therefore, c must be 2.

Now, we know that n = 92.2bc, where b is the hundredths digit and c is the thousandths digit. We also know that b + c = 4 because the average of two consecutive prime numbers must be 2.

Since b + c = 4, and c = 2, we can deduce that b = 2. Therefore, n = 92.22.

In conclusion, n = 92.22, a = 2, b = 2, and c = 2.
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n = 92.abc, where a, b and c denote the tenths, hundredths and thousandths digits respectively in the decimal representation of n.When n is rounded to the nearest integer, the result is 92. If c is the average of two consecutive prime numbers such that c > b > a,what is the difference between the maximum and minimum possible sum of a, b and c?a)13b)16c)17d)18e)20Correct answer is option 'B'. Can you explain this answer?
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