Mathematics Average and Weight Problems

Finding the Seventh Number

• The average of the first 7 numbers is 5, meaning the sum of these numbers is 7 * 5 = 35.
• The average of the first 6 numbers is 4, meaning the sum of these numbers is 6 * 4 = 24.
• The seventh number is the difference between the sum of the first 7 numbers and the sum of the first 6 numbers: 35 - 24 = 11.

Finding the New Average

• The average of 11 numbers is 24, meaning the sum of these numbers is 11 * 24 = 264.
• Adding 108 to the sum makes the new sum 264 + 108 = 372.
• The new average is the new sum divided by the total number of values: 372 / 12 = 31.

Maximum Possible Average Weight

• The initial total weight of the class is 40 students * 40 kgs/student = 1600 kgs.
• Let 'm' be the number of new students and 'n' be their average weight.
• The new total weight is 1600 kgs + (m * n) kgs.
• The new average weight is (1600 + m * n) / (40 + m).
• Since m + n = 50, we want to maximize 'n' to maximize the average weight.
• To maximize 'n', we set 'm' to its minimum value of 1.
• Therefore, 'n' = 50 - 1 = 49.
• The maximum possible average weight is (1600 + 1 * 49) / (40 + 1) = 40.49.

Average of Averages

• The average of the 7 numbers is 'Z', meaning the sum of these numbers is 7 * Z.
• Taking any 4 numbers and calculating their average will always result in a value less than or equal to 'Z'.
• The average of all possible averages of groups of 4 numbers will also be less than or equal to 'Z'.
• Therefore, the average of all possible averages is not equal to the average of the 7 numbers.

Maximum Possible Score of the Topper

• The total score of all 10 students is 10 * 60 = 600.
• The total score of the bottom 5 students is 5 * (60 - 5) = 275.
• The total score of the top 5 students is 600 - 275 = 325.
• Since none of the students failed, the minimum possible score for each of the bottom 5 students is 40.
• Therefore, the maximum possible score for the topper is (325 - 40 - 40 - 40 - 40 - 40) + 40 = 95 (adding 40 back because the minimum score for the topper is 40).

Mixing Wheat

• Let 'x' be the quantity of wheat at 3.20 Rs/kg and 'y' be the quantity of wheat at 2.90 Rs/kg.
• The total cost of the mixture is 3.20x + 2.90y.
• The total quantity of the mixture is x + y.
• The price of the mixture is (3.20x + 2.90y) / (x + y) = 3 Rs/kg.
• Solving this equation, we get the ratio x : y = 1 : 2.
• This means 1 kg of wheat at 3.20 Rs/kg should be mixed with 2 kg of wheat at 2.90 Rs/kg to achieve a mixture worth 3 Rs/kg.

Mixing Tea

• Let 'x' be the quantity of tea costing 192 Rs/kg and 'y' be the quantity of tea costing 150 Rs/kg.
• The total cost of the mixture is 192x + 150y.
• The profit is 20%, meaning the selling price is 120% of the cost price: 1.20 * (192x + 150y) = 194.40 * (x + y).
• Solving this equation, we get the ratio x : y = 1 : 3.
• This means 1 kg of tea costing 192 Rs/kg should be mixed with 3 kg of tea costing 150 Rs/kg to achieve a mixture that gives a 20% profit when sold at 194.40 Rs/kg.

This quiz explores concepts of averages and calculations related to weights and sums. You will practice finding averages based on given data and determining the effects of new values on overall calculations. Ideal for students looking to sharpen their math skills in practical scenarios.