Question

The average of 4 distinct prime numbers a, b, c, d is 35, where a<b<c<d. a and d are equidistant from 36 and b and c are equidistant from 34 and a, b are equidistant from 30 and c and d are equidistant from 40. The difference between a and d is:

• A. 30
• B. 14
• C. 21
• D. Cannot be determined

Answer 1

The Correct Option is B

To solve this problem, let's break down the information given and use the properties of prime numbers:

1. The average of the four prime numbers \(a, b, c, d\) is 35. Therefore, their sum is: \(a + b + c + d = 4 \times 35 = 140.\)

2. It is given that \(a\) and \(d\) are equidistant from 36. Therefore: \(36 - a = d - 36.\) Solving, \(d = 72 - a.\)

3. Prime numbers \(b\) and \(c\) are equidistant from 34: \(34 - b = c - 34.\) Solving, \(c = 68 - b.\)

4. Prime numbers \(a\) and \(b\) are equidistant from 30: \(b - 30 = 30 - a.\) Solving, \(b = 60 - a.\)

5. Prime numbers \(c\) and \(d\) are equidistant from 40: \(d - 40 = 40 - c.\) Solving, \(d + c = 80.\)

Now, substitute the expressions for \(b, c,\) and \(d\) in terms of \(a\) (from points 2-4) into \(a + b + c + d = 140\):

\(a + (60 - a) + (68 - (60 - a)) + (72 - a) = 140\)

Upon simplification, the equation reduces to:

\(68 - b + 72 - a = 140\)

Solving the equation for \(a\) using \(d = 72 - a\) and \(d = 40 - a\), we find the values:

Suppose \(a = 29\):

Then \(b = 60 - 29 = 31,\) \(c = 68 - 31 = 37,\) and \(d = 72 - 29 = 43.\)

Verify if these numbers satisfy \(c + d = 80:\)

\(37 + 43 = 80\)

Therefore, with the correct set of prime numbers being \([29, 31, 37, 43]\), the difference between \(a\) and \(d\) is:

\(d - a = 43 - 29 = 14\)

Thus, the correct answer is 14.