diff --git a/Tutorial_02/Numpy-Solved.ipynb b/Tutorial_02/Numpy-Solved.ipynb
index 9830969..99d2f96 100644
--- a/Tutorial_02/Numpy-Solved.ipynb
+++ b/Tutorial_02/Numpy-Solved.ipynb
@@ -599,6 +599,7 @@
    ]
   },
   {
+   "attachments": {},
    "cell_type": "markdown",
    "metadata": {},
    "source": [
@@ -612,14 +613,14 @@
     "$$m-M=5(\\log(d)-1)$$\n",
     "where $M$ is the absolute magnitude and $d$ the distance of the star from us in parsecs(pc).\n",
     "\n",
-    "Now let $m_\\odot$ denote the apparent magnetude of the sun and $d_\\odot$ the distance of the sun from us. We can then write the following formula assuming $m$ to be the apparent magnitude of the star considered and $d$ its distance:\n",
+    "Now let $m_\\odot$ denote the apparent magnitude of the sun and $d_\\odot$ the distance of the sun from us. We can then write the following formula assuming $m$ to be the apparent magnitude of the star considered and $d$ its distance:\n",
     "$$m = -2.5log\\left(\\frac{L/4\\pi d^2}{F_0}\\right)$$\n",
     "$$m_\\odot = -2.5\\log\\left(\\frac{L_\\odot/4\\pi d_\\odot^2}{F_0}\\right)$$\n",
     "Subtracting the two we get, \n",
-    "$$m_\\odot-m = -2.5\\log\\left(\\frac{L}{L_\\odot}\\right) + 5\\log\\left(\\frac{d}{d_\\odot}\\right)$$ \n",
+    "$$m_\\odot-m = 2.5\\log\\left(\\frac{L}{L_\\odot}\\right) - 5\\log\\left(\\frac{d}{d_\\odot}\\right)$$ \n",
     "\n",
     "Now using\n",
-    "$$m_\\odot - 4.83 = 5(\\log(d) - 1)$$\n",
+    "$$m_\\odot - 4.83 = 5(\\log(d_\\odot) - 1)$$\n",
     "and substituting in the above equation we get after some rearrangement:\n",
     "$$5 \\log(d) = m + 2.5\\log\\left(\\frac{L}{L_\\odot}\\right) + 0.17$$\n",
     "$$\\implies \\log(d) = \\left[m + 2.5\\log\\left(\\frac{L}{L_\\odot}\\right) + 0.17\\right]/5$$\n",