{
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 {
		   "cell_type": "markdown",
	   "metadata": {},
	   "source": [
       "# Chapter 2: Simple and compound interest"
	   ]
	},
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 2.10: Calculate_simple_interest.sce"
		   ]
		  },
  {
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"//Exa9\n",
"clc;\n",
"clear;\n",
"close;\n",
"//given data is :\n",
"CI=102;//in rupees\n",
"r=4;//in % per annum\n",
"n=2;//in years\n",
"//Let principal amount is P\n",
"//Amount will be: A=P+102\n",
"//formula : A=P(1+r/100)^n=P+102;\n",
"P=102/((1+r/100)^n-1);\n",
"SI=(P*r*n)/100;\n",
"disp('Simple interest is : '+string(SI)+' Rupees.')"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 2.1: Calculate_compound_interest.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"//Exa1\n",
"clc;\n",
"clear;\n",
"close;\n",
"//given data is :\n",
"P=10000;//in rupees\n",
"n=3;//in years\n",
"r=10;//% per annum\n",
"A=P*(1+r/100)^n;\n",
"CI=A-P;//in rupees\n",
"disp('Compound interest is : '+string(CI)+' Rupees.')\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
""
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 2.2: Find_compound_interest.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"//Exa2\n",
"clc;\n",
"clear;\n",
"close;\n",
"//For first year\n",
"P1=500;//in rupees\n",
"n=3;//in years\n",
"r=10;//% per annum\n",
"T=1//in year\n",
"I1st=(P1*r*T)/100;\n",
"A1=P1+I1st;\n",
"//For second year\n",
"P2=A1;\n",
"I2nd=(P2*r*T)/100;\n",
"A2=P2+I2nd;\n",
"//For third year\n",
"P3=A2;\n",
"I3rd=(P3*r*T)/100;\n",
"A3=P3+I3rd;\n",
"//compound interest or 3 years\n",
"CI=A3-P1;\n",
"disp('Compound interest is : '+string(CI)+' Rupees.')\n",
"\n",
"\n",
"\n",
""
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 2.3: Find_the_ammount_and_compounded_interest.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"//Exa2\n",
"clc;\n",
"clear;\n",
"close;\n",
"//given data is :\n",
"P=5000;//in rupees\n",
"n=3/2;//in years\n",
"r=10/2;//% per annum paid half yearly\n",
"m=2;//freq of compounding\n",
"A=P*(1+r/100)^(m*n);\n",
"CI=A-P;//in rupees\n",
"disp('Compound interest is : '+string(CI)+' Rupees.')\n",
"\n",
"\n",
"\n",
""
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 2.4: Find_the_time.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"//Exa4\n",
"clc;\n",
"clear;\n",
"close;\n",
"//given data is :\n",
"n=3;// in years \n",
"disp('Let P=x then A=2*x');\n",
"disp('Let r% be the rate of interest');\n",
"//formula : A=P(1+r/100)^n;\n",
"//putting values\n",
"disp('2*x=x(1+r/100)^3');\n",
"disp('or');\n",
"disp('2=(1+r/100)^3')\n",
"//on solving this eqn\n",
"r=((2^(1/3))-1)*100;//in %\n",
"disp(r,'rate is comtuted :')\n",
"disp('suppose in n years the amount x will become 16*x, then by formula')\n",
"//16=(1+r/100)^n;\n",
"n=log(16)/log(1+r/100);\n",
"disp('Time is : '+string(n)+' years');"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 2.5_a: Find_compound_interest_reckoned_quarterly.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"//Exa5(a)\n",
"clc;\n",
"clear;\n",
"close;\n",
"//given data is :\n",
"P=4000;//in rupees\n",
"N=9;// months\n",
"R=6;// in % per annum\n",
"//if interest is reckoned quarterly\n",
"r=R/4;// in % per quarter,as there are 4 quarters in a year\n",
"n=(N/12)*4;//in quarters\n",
"Amount1=P*(1+r/100)^n;\n",
"CI1=Amount1-P;\n",
"disp(CI1,'Compound interest while reckoned quarterly :')\n",
"//Ans in the book is not correct"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 2.5_b: Find_compound_interest_reckoned_half_yearly.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
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	   },
	   "outputs": [],
"source": [
"//Exa5(b)\n",
"clc;\n",
"clear;\n",
"close;\n",
"//given data is :\n",
"P=4000;//in rupees\n",
"N=9;// months\n",
"R=6;// in % per annum\n",
"//if interest is reckoned half yearly\n",
"r=R/2;// in % per half yearly,as there are 2 half years in a year\n",
"n=(N/12)*2;//in half years\n",
"Amount2=P*(1+r/100)^n;\n",
"CI2=Amount2-P;\n",
"disp(CI2,'Compound interest while reckoned half yearly :')\n",
"//Ans in the book is not correct"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 2.5_c: Find_compound_interest_reckoned_yearly.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"//Exa5(c)\n",
"clc;\n",
"clear;\n",
"close;\n",
"//given data is :\n",
"P=4000;//in rupees\n",
"N=9;// months\n",
"R=6;// in % per annum\n",
"//if interest is reckoned yearly\n",
"r=R;// in % per annum\n",
"n=(N/12);//in years\n",
"Amount3=P*(1+r/100)^n;\n",
"CI3=Amount3-P;\n",
"disp(CI3,'Compound interest while reckoned yearly :')\n",
"//Ans in the book is not correct"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 2.6: Find_compound_interest.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
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"source": [
"//Exa6\n",
"clc;\n",
"clear;\n",
"close;\n",
"//given data is :\n",
"P=10000;//in rupees\n",
"N=3;// years\n",
"r1=4;// in % per annum for 1st year\n",
"r2=5;// in % per annum for 2nd year\n",
"r3=10;// in % per annum for 3rd year\n",
"A=P*(1+r1/100)*(1+r2/100)*(1+r3/100);\n",
"CI=A-P;\n",
"disp('Compound interest is : '+string(CI)+' Rupees.')\n",
"\n",
"\n",
"\n",
"\n",
""
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 2.7: Find_the_amount.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"//Exa7\n",
"clc;\n",
"clear;\n",
"close;\n",
"//given data is :\n",
"CI=496.50;//compound interest in rupees\n",
"n=3;//in years\n",
"r=10;//rate in % per annum\n",
"disp('CI is given by : ');\n",
"disp('CI=P(1+r/100)^n-P');\n",
"//solving this eqn\n",
"P=CI/((1+r/100)^n-1);\n",
"disp('Principal amount is : '+string(P)+' Rupees.')\n",
""
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 2.8: Find_the_time.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"//Exa8\n",
"clc;\n",
"clear;\n",
"close;\n",
"//given data is :\n",
"P=2000;//in rupees\n",
"A=2662;//in rupees\n",
"r=10;//% per annum\n",
"//formula : A=P(1+r/100)^n;\n",
"//solving for n\n",
" n=log(A/P)/log(1+r/100);\n",
" disp('The time in which Rs.2000 will rise to Rs. 2662 is  : '+string(n)+' years.')"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 2.9: Find_the_principal_amount.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
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"source": [
"//Exa9\n",
"clc;\n",
"clear;\n",
"close;\n",
"//given data is :\n",
"r=5;//% per annum\n",
"n=2;//in years\n",
"//let amount=P\n",
"//CI=P(1+r/100)^n-P;\n",
"//SI=(P*r*n)/100;\n",
"//CI-SI=15 Rupees;given\n",
"disp('solving eqns for CI and SI, we get : ')\n",
"disp('CI=0.1025*P');\n",
"disp('SI=0.10*P');\n",
"P=15/(0.1025-0.10);//using CI-SI\n",
"disp('Principal amount is : '+string(P)+' Rupees.')\n",
"\n",
"\n",
"\n",
""
   ]
   }
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