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Find the Compound Interest to the Nearest Rupee on Rs. 10,800 for 2 1/2 Years at 10% per Annum. - Mathematics

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Question

Find the compound interest to the nearest rupee on Rs. 10,800 for `2 1/2` years at 10% per annum.

Sum
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Solution

Given : P = Rs. 10,800 ; Time = `2 1/2` years and Rate = 10% p.a.

For 2 years
A = P`( 1 + r/100)^n` = 10,800( 1 + 10/100 )^2 = Rs. 13,068

For `1/2` year
∴ A = P`( 1 + r/[ 2 xx 100 ])^( n xx 2)  = 13068( 1 + 10/[ 2 xx 100 ])^( 1/2 xx 2)`

= 13068 x `21/20` = Rs. 13721.40 = Rs. 13721 ( nearest rupee)
Rs.13,721 - Rs.10,800 = Rs.2,921

shaalaa.com
Concept of Compound Interest - Use of Compound Interest in Computing Amount Over a Period of 2 Or 3-years
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Chapter 3: Compound Interest (Using Formula) - Exercise 3 (E) [Page 54]

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Selina Concise Mathematics [English] Class 9 ICSE
Chapter 3 Compound Interest (Using Formula)
Exercise 3 (E) | Q 2 | Page 54
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