A computer is sold either for Rs.19200 cash or for Rs.4800 cash down payment together with five equal monthly installments. If the rate of interest charged is 12% per annum, then the amount of each installment (nearest to a rupee) is:
Amount on which interest will be charged = 19200 - 4800 = 14400
The total amount paid will be equal to the sum of all monthly instalments. Therefore, we have
$$14400*k^{5a} = I (k^{4a} +k^{3a}+k^{2a}+k^{a}+1 )$$ .....(1)
where, k = $$1+\frac{12}{100}$$ & a = $$\frac{1}{12}$$
We know that, $$k^{5a}-1 = (k-1)(k^{4a} +k^{3a}+k^{2a}+k^{a}+1)$$
=>$$k^{4a} +k^{3a}+k^{2a}+k^{a}+1$$ = $$\frac{k^{5a}-1}{k-1}$$ ....(2)
Substituting in equation (1) we get
I = $$14400*k^{5a}*\left[\frac{k-1}{k^{5a}-1}\right]$$ ....(3)
On substituting the values of k and a in equation (3) we get
I $$\approx$$ 2965
Hence, option B.
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