Answer the questions based on the information given below. Five different schools of a district participated in multiple tournaments during a year. Each school won some gold, silver and bronze medals. The pie chart given below shows the percentage distribution of number of gold medals won by each school out of total gold medals won by all schools together. (x-4)% 24% A B C (x+5)% D 16.50% E 18.50% The table chart given below shows the difference between number of gold and number of silver medals, won by each school and percentage of number of bronze medals won by each school as percentage of number of silver medals won by that school. School Difference between number of gold Percentage of number of bronze and number of silver medals won medals won by each school as percentage of number of silver medals won A 116 84% B 186 78% C 104 (3.5x+1)% D 120 75% E 144 4x% Note: Total number of medals won by school 'E' were 86 more than that by school 'D'. Find the average number of medals won by schools 'C' and 'D' together.

Explanation

4 Detailed solution: ATQ; 24 + 16.5 + 18.5 + x + 5 + x - 4 = 100 Or, 2x = 40 So, x = 20 So, percentage of gold medals won by school 'D' out of total gold medals won by all 5 schools together = 25% And, that by school 'E' = 20 - 4 = 16% Similarly, Percentage of number of bronze medals won as percentage of number of silver medals won by school 'C' = 3.5 X 20 + 1 = 71% Percentage of number of bronze medals won as percentage of number of silver medals won by school 'E' = 4 X 20 = 80% Let total number of gold medals won by all five schools together be '400y'. So, number of gold medals won by school 'D' = 400y X 0.25 = '100y' Number of silver medals won by school 'D' = 100y + 120 (Case 'P') or 100y - 120 (Case 'Q') Number of gold medals won by school 'E' = 400y X 0.16 = 64y For case 'P': Number of bronze medals won by school 'D' = (100y + 120) X 0.75 = 75y + 90 Case P (i): Number of silver medals won by school 'E' = 64y + 144 So, number of bronze medals won by school 'E' = (64y + 144) X 0.8 = 51.2y + 115.2 ATQ; 100y + 75y + 90 + 100y + 120 = 64y + 64y + 144 + 51.2y + 115.2 – 86 Or, 95.8y = -36.8 (Since 'y' is a positive integer, we may discard this case). Case P (ii): Number of silver medals won by school 'E' = 64y – 144 So, number of bronze medals won by school 'E' = (64y - 144) X 0.8 = 51.2y - 115.2 ATQ; 100y + 75y + 90 + 100y + 120 = 64y + 64y - 144 + 51.2y - 115.2 – 86 Or, 95.8y = - 555.2 (Since 'y' is a positive integer, we may discard this case) Case Q: Number of bronze medals won by school 'D' = (100y - 120) X 0.75 = 75y – 90 Case Q (i): Number of silver medals won by school 'E' = 64y + 144 So, number of bronze medals won by school 'E' = (64y + 144) X 0.8 = 51.2y + 115.2 ATQ; 100y + 75y - 90 + 100y - 120 = 64y + 64y + 144 + 51.2y + 115.2 – 86 Or, 95.8y = 383.2 So, y = 4 Case Q (ii): Number of silver medals won by school 'E' = 64y – 144 So, number of bronze medals won by school 'E' = (64y - 144) X 0.8 = 51.2y - 115.2 ATQ; 100y + 75y - 90 + 100y - 120 = 64y + 64y - 144 + 51.2y - 115.2 – 86 Or, 95.8y = 135.2 So, y ~ 1.41 (Since 'y' is not an integer, we may discard this case) So, y = 4 So, total number of gold medals won by all five schools together = 4 X 400 = 1600 For school 'A': Total number of gold medals won by the school = 1600 X 0.24 = 384 Number of silver medals won by the school = 384 + 116 = 500 or 384 - 116 = 268 When number of silver medals won = 268, then number of bronze medals won = 268 X 0.84 = 225.12 (Since number of medals won must be an integer, this case can be discarded) So, number of silver medals won by school 'A' = 500 And, number of bronze medals won by the school = 500 X 0.84 = 420 Similarly, School Number of gold Number of Number of Total number of medals won silver medals bronze medal medals won won won A 384 500 420 1304 B 264 450 351 1065 C 296 400 284 980 D 400 280 210 890 E 256 400 320 976 Total 1600 2030 1585 5215 Required average = {(980 + 890)/2} = 935