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Find the Compound Interest, Correct to the Nearest Rupee, on Rs. 2,400 for 2 1/2 Years at 5 per Cent per Annum. - Mathematics

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Question

Find the compound interest, correct to the nearest rupee, on Rs. 2,400 for `2 1/2` years at 5 per cent per annum.

Sum
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Solution

For 1st years

P = Rs. 2400

R = 5%

T = 1 year

I = `[ 2400 xx 5 xx 1]/100` = 120

A = 2400 + 120 = Rs. 2520

For 2nd year

P = Rs. 2520

R = 5%

T = 1 year

I = `[ 2520 xx 5 xx 1]/100` = Rs. 126.

A = 2520 + 126 = Rs. 2646

For final `1/2` year,

P = Rs. 2646

R = 5%

T = `1/2` year

I = `[2646 xx 5 xx 1]/[100 xx 2]` = Rs. 66.15

Amount after `2 1/2` years = 2646 + 66.15

= Rs. 2712.15

Compound interest = 2712.15 - 2400 

= Rs. 312.15

= Rs. 312

shaalaa.com
Concept of Compound Interest - Compound Interest as a Repeated Simple Interest Computation with a Growing Principal
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Chapter 2: Compound Interest (Without using formula) - Exercise 2 (A) [Page 28]

APPEARS IN

Selina Concise Mathematics [English] Class 9 ICSE
Chapter 2 Compound Interest (Without using formula)
Exercise 2 (A) | Q 4 | Page 28

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