diff --git a/src/chapters/6/sections/means_medians_modes_moments/index.tex b/src/chapters/6/sections/means_medians_modes_moments/index.tex
index 1f290f4f..7280c601 100644
--- a/src/chapters/6/sections/means_medians_modes_moments/index.tex
+++ b/src/chapters/6/sections/means_medians_modes_moments/index.tex
@@ -4,3 +4,5 @@ \subsection{problem 1}
 \input{problems/1}
 \subsection{problem 2}
 \input{problems/2}
+\subsection{problem 3}
+\input{problems/3}
diff --git a/src/chapters/6/sections/means_medians_modes_moments/problems/3.tex b/src/chapters/6/sections/means_medians_modes_moments/problems/3.tex
new file mode 100644
index 00000000..2de38f3b
--- /dev/null
+++ b/src/chapters/6/sections/means_medians_modes_moments/problems/3.tex
@@ -0,0 +1,18 @@
+Let k be the median, and F be the CDF of X, such that
+\[F(k) = 1/2\]
+since X is continuous from 1 onwards, and 0 otherwise,
+\begin{align*}
+F(k) &= \int_{-\infty}^{k}{\frac{a}{x^{a+1}}} \\
+&= \int_{1}^{k}{\frac{a}{x^{a+1}}} \\
+&= 1 - k^{-a}
+\end{align*}
+It follows that
+\begin{align*}
+1-k^{-a} &= \frac{1}{2} \\
+k &= \sqrt[a]{2}
+\end{align*}
+The median is therefore \(\sqrt[a]{2}\). \\ 
+
+Let $f$ be the PDF of X, then,
+\[f'(x) = -a(a+1)x^{-a-2}\]
+since $f'(x) \le 0$ for all $x \in [1,\infty)$, $f(x)$ is strictly decreasing between 1 and \(\infty\) and 0 otherwise, $f(x)$ has maximum at 1, so 1 is the mode.