The average weight of a class is 20 kg. If the strength of the class is increased by 25%, then the average weight of the class gets increased by 1%. What is the average weight of the additional students ?<\/p>\n
[amp_mcq option1=”20.50 kg” option2=”21.00 kg” option3=”21.50 kg” option4=”22.00 kg” correct=”option2″]<\/p>\n
The strength of the class is increased by 25%.
\nNew strength = $N + 0.25N = 1.25N$.
\nThe number of additional students is $N_{add} = 0.25N$.<\/p>\n
The average weight of the class gets increased by 1%.
\nInitial average weight = 20 kg.
\nIncrease in average weight = 1% of 20 kg = $0.01 \\times 20 = 0.2$ kg.
\nNew average weight = $20 + 0.2 = 20.2$ kg.<\/p>\n
The new total weight of the class is $W_{new} = (\\text{New strength}) \\times (\\text{New average weight})$
\n$W_{new} = (1.25N) \\times 20.2$ kg.<\/p>\n
The total weight of the class after adding students is the sum of the initial total weight and the total weight of the additional students.
\nLet the average weight of the additional students be $W_{avg\\_add}$.
\nTotal weight of additional students = $N_{add} \\times W_{avg\\_add} = (0.25N) \\times W_{avg\\_add}$.<\/p>\n
So, $W_{new} = W_{initial} + \\text{Total weight of additional students}$.
\n$(1.25N) \\times 20.2 = (N \\times 20) + (0.25N \\times W_{avg\\_add})$.<\/p>\n
We can divide the entire equation by N (assuming $N > 0$):
\n$1.25 \\times 20.2 = 20 + 0.25 \\times W_{avg\\_add}$.<\/p>\n
Calculate the left side:
\n$1.25 \\times 20.2 = (5\/4) \\times 20.2 = 5 \\times (20.2 \/ 4) = 5 \\times 5.05 = 25.25$.<\/p>\n
The equation becomes:
\n$25.25 = 20 + 0.25 \\times W_{avg\\_add}$.<\/p>\n
Subtract 20 from both sides:
\n$25.25 – 20 = 0.25 \\times W_{avg\\_add}$
\n$5.25 = 0.25 \\times W_{avg\\_add}$.<\/p>\n
Solve for $W_{avg\\_add}$:
\n$W_{avg\\_add} = 5.25 \/ 0.25 = 5.25 \/ (1\/4) = 5.25 \\times 4$.
\n$W_{avg\\_add} = (5 + 0.25) \\times 4 = 5 \\times 4 + 0.25 \\times 4 = 20 + 1 = 21$.<\/p>\n