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In a class of 5 students, average weight of the 4 lightest students is 40 kgs, Average weight of the 4 heaviest students is 45 kgs. What is the difference between the maximum and minimum possible average weight overall?
- a)3.2 kgs
- b)3 kgs
- c)2.8 kgs
- d)4 kgs
Correct answer is option 'B'. Can you explain this answer?
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Most Upvoted Answer
In a class of 5 students, average weight of the 4 lightest students i...
Given:
- Class of 5 students
- Average weight of the 4 lightest students is 40 kgs
- Average weight of the 4 heaviest students is 45 kgs
To find:
- Difference between the maximum and minimum possible average weight overall
Approach:
- Let's assume the weights of the 5 students are a, b, c, d, and e
- We need to find the maximum and minimum values of the average weight, which means we need to find the maximum and minimum values of the sum of weights of the students
- We can use the given information to form equations and solve them to get the values of a, b, c, d, and e
- Once we have the values of the weights, we can find the maximum and minimum values of the average weight
Calculation:
- Let the weights of the 5 students be a, b, c, d, and e
- Average weight of the 4 lightest students is 40 kgs, which means the total weight of the 4 lightest students is 4*40 = 160 kgs
- Therefore, the weight of the heaviest student is (a+b+c+d+e) - 160 kgs
- Average weight of the 4 heaviest students is 45 kgs, which means the total weight of the 4 heaviest students is 4*45 = 180 kgs
- Therefore, the weight of the lightest student is (a+b+c+d+e) - 180 kgs
Equations:
- (a+b+c+d) = 160 (total weight of 4 lightest students)
- (a+b+c+e) = 160 (total weight of 4 lightest students)
- (a+b+d+e) = 160 (total weight of 4 lightest students)
- (a+c+d+e) = 160 (total weight of 4 lightest students)
- (a+b+c+d+e) - 160 = 45 (total weight of 4 heaviest students)
- (a+b+c+d+e) - 180 = 40 (weight of the lightest student)
Solving the equations:
- Adding the first 4 equations, we get: 4(a+b+c+d) + e = 640
- Adding the last 2 equations, we get: 2(a+b+c+d+e) = 425
- Substituting the value of (a+b+c+d+e) from the second equation in the first equation, we get: 4(425/2 - e) + e = 640
- Simplifying the equation, we get: e = 75
- Substituting the value of e in the second equation, we get: a+b+c+d = 175
- Substituting the values of a+b+c+d and e in the last 2 equations, we get: a+b+c+d = 175 and a+b+c+d+e = 425
- Solving these equations, we get: a = 25, b = 50, c = 50, d = 50, e = 75
Maximum and minimum values of the average weight:
- Maximum value: If the heaviest student is not included in the average, then the average weight would be (25+50+
- Class of 5 students
- Average weight of the 4 lightest students is 40 kgs
- Average weight of the 4 heaviest students is 45 kgs
To find:
- Difference between the maximum and minimum possible average weight overall
Approach:
- Let's assume the weights of the 5 students are a, b, c, d, and e
- We need to find the maximum and minimum values of the average weight, which means we need to find the maximum and minimum values of the sum of weights of the students
- We can use the given information to form equations and solve them to get the values of a, b, c, d, and e
- Once we have the values of the weights, we can find the maximum and minimum values of the average weight
Calculation:
- Let the weights of the 5 students be a, b, c, d, and e
- Average weight of the 4 lightest students is 40 kgs, which means the total weight of the 4 lightest students is 4*40 = 160 kgs
- Therefore, the weight of the heaviest student is (a+b+c+d+e) - 160 kgs
- Average weight of the 4 heaviest students is 45 kgs, which means the total weight of the 4 heaviest students is 4*45 = 180 kgs
- Therefore, the weight of the lightest student is (a+b+c+d+e) - 180 kgs
Equations:
- (a+b+c+d) = 160 (total weight of 4 lightest students)
- (a+b+c+e) = 160 (total weight of 4 lightest students)
- (a+b+d+e) = 160 (total weight of 4 lightest students)
- (a+c+d+e) = 160 (total weight of 4 lightest students)
- (a+b+c+d+e) - 160 = 45 (total weight of 4 heaviest students)
- (a+b+c+d+e) - 180 = 40 (weight of the lightest student)
Solving the equations:
- Adding the first 4 equations, we get: 4(a+b+c+d) + e = 640
- Adding the last 2 equations, we get: 2(a+b+c+d+e) = 425
- Substituting the value of (a+b+c+d+e) from the second equation in the first equation, we get: 4(425/2 - e) + e = 640
- Simplifying the equation, we get: e = 75
- Substituting the value of e in the second equation, we get: a+b+c+d = 175
- Substituting the values of a+b+c+d and e in the last 2 equations, we get: a+b+c+d = 175 and a+b+c+d+e = 425
- Solving these equations, we get: a = 25, b = 50, c = 50, d = 50, e = 75
Maximum and minimum values of the average weight:
- Maximum value: If the heaviest student is not included in the average, then the average weight would be (25+50+
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Community Answer
In a class of 5 students, average weight of the 4 lightest students i...
Let's say that the students are named a, b, c, d and e, in increasing order of weights. The average of a, b, c and d is 40 kg, whereas the average of b, c, d and e is 45 kg.
The sum of a, b, c and d is 160 kg, and the sum of b, c, d and e is 180 kg.
There are two ways of looking at this.
a) 160 + e
b) 180 + a
Or, e is 20 more than a.
The total weight is 160 + e. So, the highest value of e will correspond to the highest possible average. The highest possible value of e occurs when it is 20 higher than the highest possible value for a, which is 40 (all the first 4 scores are equal to 40).
So, the highest possible average is 160 + 605/5 = 44
This will be the case when the weights are 40 kgs, 40kgs, 40 kgs, 40 kgs and 60 kgs.
Conversely, the least possible value for the average occurs when a is the least. This happens when e is the least too (since a is 20 less than e).
The least possible value for e is 45 = 180/4
So, the least possible value for a would be 25.
The least possible average = 180 + 25/5 = 41
This will be the case when the weights are 25 kgs, 45kgs, 45 kgs, 45 kgs and 45 kgs. So, the difference between maximum possible and minimum possible average = 3 kgs.
Hence, the correct option is (b).
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