hi
wallid,
in
linear interpolation, you assume the quantities are changing with respect to each other at a
constant rate - meaning that if you were to plot them on a graph of, say,
percentage vs. payment, then the graph would be that of an oblique
line.
define the function for payment as P(%), where '%' = the rate-of-interest variable.
ie. so the equation of the graph would be:
P(%) = m.% + b ; where 'm' and 'b' are constants.
you have two sets of corresponding values in your possession already, namely:
1) at % = 6 , P = 1810 ; and
2) at % = 6.5 , P = 1763
substituting this into the expression for P yields two simultaneous equations:
i) 1810 = 6m + b ; and,
ii) 1763 = 6.5m + b
solving these equations gives you:
m = -94 ; and
b = 2374
---> equation for P becomes:
P(%) = -94(%) + 2374
now you want to know for what value of %, will give you a P = 1776 ; to do this, simply substitute
P = 1776 into the new equation:
ie. 1776 = -94(%) + 2374
---> % = [approx.]
6.36
Therefore, for a payment of $1776, the
rate of interest is approximately 6.36% .
hope that helps
P.S. if you need to do this
graphically, then draw up a set of appropriately scaled axes, and then plot the two point you already know onto the plane - then, join those two points with a
straight line, and then simply 'read off' the line for the corresponding value of rate of interest that gives you
P = 1776.
also, linear interpolation is not part of the 4u syllabus btw...
