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01 Apr 2026, 19:00
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77% (02:27) correct
23%
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The average weight of students in a class of 20 was 80 pounds. Five new students joined and the weight increased to 84 pounds. If weights of the five new students are in integers, weight of no two students is same, and the lightest new student weighs 84 pounds, what is the highest possible weight of the heaviest new student?
A. 100
B. 116
C. 158
D. 164
E. 172
A. 100
B. 116
C. 158
D. 164
E. 172
01 Apr 2026, 22:31
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This is pretty easy one.
Sum = 20*80 =1600
New sum = 25*84 = 2100
Difference btw them is 500
This 500 has be split btw the 5 members such that we need to find the highest possible weight of the heaviest student. For this we need to minimize all the other 4 members weight to their lowest possible. We have also been given that the lowest weight of the 5 guys is equal to 84. Therefore, the set can be formed as : 84,85,86,87,x
Now, 84+85+86+87+x = 500
x = 158
Option C
Sum = 20*80 =1600
New sum = 25*84 = 2100
Difference btw them is 500
This 500 has be split btw the 5 members such that we need to find the highest possible weight of the heaviest student. For this we need to minimize all the other 4 members weight to their lowest possible. We have also been given that the lowest weight of the 5 guys is equal to 84. Therefore, the set can be formed as : 84,85,86,87,x
Now, 84+85+86+87+x = 500
x = 158
Option C
02 Apr 2026, 02:22
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Given avg weight of 20 students is 80 p
So total weight to 20 students = 80*20=1,600 p
Given the 5 students got added and lightest one is 84 and the new avg of 25 student is 84
So new total weight = 84*25=2,100 p
Extra weight added = 2100-1600=500 p
Since the weight are integers and distinct and we want to find the heaviest weight possible for the heavy kid
We will assume the 4 remaining kids weigh = 84+85+86+87=342 p
So weight of heavy kid = 500-342=158 p
Answer C
So total weight to 20 students = 80*20=1,600 p
Given the 5 students got added and lightest one is 84 and the new avg of 25 student is 84
So new total weight = 84*25=2,100 p
Extra weight added = 2100-1600=500 p
Since the weight are integers and distinct and we want to find the heaviest weight possible for the heavy kid
We will assume the 4 remaining kids weigh = 84+85+86+87=342 p
So weight of heavy kid = 500-342=158 p
Answer C
