Consider six distinct natural numbers such that the average of the two smallest numbers is 14, and the average of the two largest numbers is 28. Then, the maximum possible value of the average of these six numbers is:
Question:
Consider six distinct natural numbers such that the average of the two smallest numbers is 14, and the average of the two largest numbers is 28. Then, the maximum possible value of the average of these six numbers is:
Consider six distinct natural numbers such that the average of the two smallest numbers is 14, and the average of the two largest numbers is 28. Then, the maximum possible value of the average of these six numbers is:
Options
Answer: 22.5
Explanation:
Step 1: Define the numbers Let the six numbers in increasing order be: a < b < c < d < e < f Given: Average of two smallest numbers: (a + b)/2 = 14 → a + b = 28 Average of two largest numbers: (e + f)/2 = 28 → e + f = 56 Average of all six numbers = (a + b + c + d + e + f)/6 = (c + d + 84)/6 To maximize the average, we need to maximize c + d. Step 2: Maximize c and d Numbers are distinct natural numbers: a < b < c < d < e < f e + f = 56 and e < f → maximum e = 27, then f = 29 Step 3: Maximize c and d c < d < e = 27 → maximum d = 26, maximum c = 25 a + b = 28 → to keep numbers distinct, maximum b = 24 → then a = 4 So one possible set of numbers: 4, 24, 25, 26, 27, 29 Step 4: Compute average Sum = 4 + 24 + 25 + 26 + 27 + 29 = 135 Average = 135 / 6 = 22.5 Step 5: Verify maximum e + f = 56, distinct natural numbers → maximum c + d = 25 + 26 = 51 → average = 22.5 This is indeed the maximum possible average. Answer: 22.5
Explanation:
Step 1: Define the numbers Let the six numbers in increasing order be: a < b < c < d < e < f Given: Average of two smallest numbers: (a + b)/2 = 14 → a + b = 28 Average of two largest numbers: (e + f)/2 = 28 → e + f = 56 Average of all six numbers = (a + b + c + d + e + f)/6 = (c + d + 84)/6 To maximize the average, we need to maximize c + d. Step 2: Maximize c and d Numbers are distinct natural numbers: a < b < c < d < e < f e + f = 56 and e < f → maximum e = 27, then f = 29 Step 3: Maximize c and d c < d < e = 27 → maximum d = 26, maximum c = 25 a + b = 28 → to keep numbers distinct, maximum b = 24 → then a = 4 So one possible set of numbers: 4, 24, 25, 26, 27, 29 Step 4: Compute average Sum = 4 + 24 + 25 + 26 + 27 + 29 = 135 Average = 135 / 6 = 22.5 Step 5: Verify maximum e + f = 56, distinct natural numbers → maximum c + d = 25 + 26 = 51 → average = 22.5 This is indeed the maximum possible average. Answer: 22.5
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