1. 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:
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By: anil on 05 May 2019 02.34 am
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 = $$14400*k^{5a}*left[frac{k-1}{k^{5a}-1}
ight]$$ ....(3) On substituting the values of k and a in equation (3) we get
I $$approx$$ 2965 Hence, option B.
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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 = $$14400*k^{5a}*left[frac{k-1}{k^{5a}-1} ight]$$ ....(3) On substituting the values of k and a in equation (3) we get
I $$approx$$ 2965 Hence, option B.