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<H2>Math circle level problem on average</H2> <H3>Problem 1</H3>The average of four three digit numbers is 500. Two of the numbers are 150 and 230. What is the largest difference between the other two numbers? <B>Solution</B> <pre> Let the four numbers be "a", "b", "c" and "d". We are given that {{{(a+b+c+d)/4}}} = 500, which implies a + b + c + d = 4*500 = 2000. (1) We also are given that two of them are 150 and 230: let a = 150 and b = 230. Then from (1) c + d = 2000 - a - b = 2000 - 150 - 230 = 1620. Thus we know about "c" and "d" that c + d = 1620 and c and d are 3-digit numbers. +---------------------------------------------------------------+ | The fact that "c" and "d" are 3-digit integer numbers | | means that they are | | greater than or equal to 100 and lesser than or equal to 999. | +---------------------------------------------------------------+ So again, we know that c + d = 1620 and "c" and "d" are integer numbers greater than or equal to 100 and lesser than or equal to 999. When the difference |c - d| is maximum ? First answer, which lies on the surface, is this: when "c" is as great as it is possible; then d is minimal. Second answer, which lies on the surface, is this: when "d" is as small as it is possible; then c is maximal. So, we should check, which of these two possibilities will really happen. (1) "c"s greatest possible value is 999; then "d"'s value is 1620-999 = 621 and it is in the allowable range between 100 and 999; then the difference |c-d| is |999-621| = 378. (2) "d"s least possible value is 100; then "c"'s value is 1620-100 = 1520 and it is OUT of the allowable range between 100 and 999. Therefore, the answer to the problem's question is THIS the greatest possible difference |c-d| is 378, when c = 999 and d = 621. </pre> It can be derived in more formal fashion, but I consciously appeal to your intuition and common sense. To make it even MORE VISIBLE, you may think this way: <pre> the sum c + d = 1620 is frozen; hence, the average of numbers "c" and "d", {{{(c+d)/2}}} = {{{1620/2}}} = 810 is frozen, too, and 100 <= c, d <= 999. When the difference |c-d| is maximum ? - It is maximum when the distance between "c" and "d" on the number line is maximum, with their frozen average. But it is obvious that it will happen when "c"-value will reach its maximum, which is 999. </pre> My other lessons on <B>Average, Mean and Median</B> in this site are - <A HREF=https://www.algebra.com/algebra/homework/Average/Geometric-Mean.lesson>WHAT IS Geometric mean</A> - <A HREF=https://www.algebra.com/algebra/homework/Average/Difference-between-Amean-and-Gmean.lesson>Difference between Arithmetic mean and Geometric mean</A> - <A HREF=https://www.algebra.com/algebra/homework/Average/What-is-Median.lesson>WHAT IS Median</A> - <A HREF=https://www.algebra.com/algebra/homework/Average/Solved-problems-on-average-scores-weight-height-and-temperature.lesson>Solved problems on average scores, weight, height and temperature</A> - <A HREF=https://www.algebra.com/algebra/homework/Average/Solved-problems-on-average-scores.lesson>Solved problems on average scores</A> - <A HREF=https://www.algebra.com/algebra/homework/Average/Solved-problems-on-average-age.lesson>Solved problems on average age</A> - <A HREF=https://www.algebra.com/algebra/homework/Average/Miscellaneous-problems-on-average-values.lesson>Miscellaneous problems on average values</A> - <A HREF=https://www.algebra.com/algebra/homework/Average/Entertainment-problems-on-average.lesson>Entertainment problems on average</A> - <A HREF=https://www.algebra.com/algebra/homework/Average/The-mean-and-the-total-value-problem-for-the-International-Fools-day-of-April-1.lesson>The mean and the total value problem for the International Fools' day of April, 1</A> - <A HREF=https://www.algebra.com/algebra/homework/Average/OVERVIEW-of-lessons-on-Average-Mean-and-Median.lesson>OVERVIEW of lessons on Average</A>