हिंदी

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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प्रश्न

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

योग
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उत्तर

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
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 3: Compound Interest (Stage 2) [Applications] - Exercise 3 (E) [पृष्ठ ५४]

APPEARS IN

सेलिना Concise Mathematics [English] Class 9 ICSE
अध्याय 3 Compound Interest (Stage 2) [Applications]
Exercise 3 (E) | Q 2 | पृष्ठ ५४

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