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

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

But try the problem first...

Answer: $1976.5$

AMC-10A (2020) Problem 11

Pre College Mathematics

## Try with Hints

First hint

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 ..........

Second Hint

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........

Final Step

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,=\(1976.5\)

## Other useful links

- https://www.cheenta.com/problem-from-probability-amc-8-2004problem-no-21/
- https://www.youtube.com/watch?v=aRevxA0u3D4