Categories
AMC 8 Math Olympiad USA Math Olympiad

Median of numbers | AMC-10A, 2020 | Problem 11

Try this beautiful problem from Geometry based on Median of numbers from AMC 10A, 2020. You may use sequential hints to solve the problem.

Try this beautiful problem from Geometry based on Median of numbers from AMC 10A, 2020.

Median of numbers – AMC-10A, 2020- Problem 11


What is the median of the following list of $4040$ numbers$?$

\(1,2,3,…….2020,1^2,2^2,3^2………..{2020}^2\)

  • $1989.5$
  • $ 1976.5$
  • $1972.5$

Key Concepts


Median

Algebra

square numbers

Check the Answer


Answer: $1976.5$

AMC-10A (2020) Problem 11

Pre College Mathematics

Try with Hints


To find the median we need to know how many terms are there and the position of the numbers .here two types of numbers, first nonsquare i.e (1,2,3……2020) and squares numbers i.e \((1^2,2^2,3^2……2020^2)\).so We want to know the \(2020\)th term and the \(2021\)st term to get the median.

Can you now finish the problem ……….

Now less than 2020 the square number is \({44}^2\)=1936 and if we take \({45}^2\)=2025 which is greater than 2020.therefore we take the term that \(1,2,3…2020\) trms + 44 terms=\(2064\) terms.

can you finish the problem……..

since $44^{2}$ is $44+45=89$ less than $45^{2}=2025$ and 84 less than 2020 we will only need to consider the perfect square terms going down from the 2064 th term, 2020, after going down $84$ terms. Since the $2020$th and $2021$st terms are only $44$ and $43$ terms away from the $2064$th term, we can simply subtract $44$ from $2020$ and $43$ from $2020$ to get the two terms, which are $1976$ and $1977$. Averaging the two,=\(1976.5\)

Subscribe to Cheenta at Youtube


Leave a Reply

Your email address will not be published. Required fields are marked *