The mean of 14 numbers is 72. If remove the least and the largest number, the mean of remain numbers is 76 If the least number is 12 less than the largest number, find the largest number. geqslant leqslant t equiv a(b)/(c) (a)/(b) sqrt ((a)/(b)) (sqrt (a))/(sqrt (b)) sqrt (n) sqrt [n](n) sqrt [n](m) a^b (sqrt (a))/(b) (a)/(sqrt (b))

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The mean of 14 numbers is 72. If remove the least and the largest number, the mean of remain numbers is 76 If the least number is 12 less than the largest number, find the largest number. geqslant leqslant t equiv a(b)/(c) (a)/(b) sqrt ((a)/(b)) (sqrt (a))/(sqrt (b)) sqrt (n) sqrt [n](n) sqrt [n](m) a^b (sqrt (a))/(b) (a)/(sqrt (b))

Answer 1

Question 1: The mean of 14 numbers is 72. If remove the least and the largest number, the mean of remain numbers is 76. If the least number is 12 less than the largest number, find the largest number.Solution:Let's denote the smallest number as x and the largest number as y. the problem, we have two conditions:1. The sum of all 14 numbers is 72*14, because the mean is the sum divided by the quantity.2. If we remove the smallest and the largest number, the mean of the remaining 12 numbers is 76. So, the sum of these 12 numbers is 76*12.3. The smallest number is 12 less than the largest number, so we have x = y - 12.From conditions 1 and 2, we can form the equation:72*14 - x - y = 76*12Substitute x from condition 3 into the equation:72*14 - (y - 12) - y = 76*12Solve this equation for y to find the largest number.

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