Find the compound interest to the nearest rupee on Rs- 10-800 for 2 -1-2 years at 10- per annum-

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Standard IX

Mathematics

Finding CI When Rate Compounded Annually

Find the comp...

Question

Find the compound interest to the nearest rupee on Rs. 10,800 for $2 - \frac{1}{2}$ years at 10% per annum.

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Solution

Answer: Amount =18000(1+10100)2

=18000×1110×1110=21780

Interest after 2 years =21780−18000= Rs 3780

Interest after half year =21780×10100×12= Rs 1089

Total interest after 2.5 years =3780+1089= Rs 4869

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Calculate the amount and compound interest on
(a) $R s . 10, 800$ for $3$ years at $12 \frac{1}{2} %$ per annum compounded annually.
(b) $R s . 18, 000$ for $2 \frac{1}{2}$ years at $10 %$ per annum compounded annually.

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Find the compound interest to the nearest rupee on Rs. 10,800 for 2−12 years at 10% per annum.

Answer: Amount =18000(1+10100)2 =18000×1110×1110=21780 Interest after 2 years =21780−18000= Rs 3780 Interest after half year =21780×10100×12= Rs 1089 Total interest after 2.5 years =3780+1089= Rs 4869

Find the compound interest to the nearest rupee on Rs. 10,800 for $2 - \frac{1}{2}$ years at 10% per annum.

Answer: Amount =18000(1+10100)2

=18000×1110×1110=21780

Interest after 2 years =21780−18000= Rs 3780

Interest after half year =21780×10100×12= Rs 1089

Total interest after 2.5 years =3780+1089= Rs 4869