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Maharashtra State BoardSSC (English Medium) 8th Standard

To start a business, Shalaka has taken a loan of ₹8000 at a rate of 10 12 p.c.p.a. After two years, how much compound interest will she have to pay?

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Question

To start a business, Shalaka has taken a loan of ₹8000 at a rate of 10 `1/2` p.c.p.a. After two years, how much compound interest will she have to pay?

Sum
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Solution

Here, P = ₹8000, R = `10 1/2 = 21/2` p.c.p.a, N = 2 Years

A = `"P"(1 + "R"/100)^"N"`

= `8000(1 + (21/2)/100)^2`

= `8000(1 + 21/200)^2`

= `8000((200 + 21)/200)^2`

= `8000(221/200)^2`

= `8000 xx 221/200 xx 221/200`

= `(2 xx 221 xx 221)/10`

= `97682/10`

∴ A = 9768.2

∵ A = P + I

9768.2 = 8000 + I

9768.2 − 8000 = I

I = ₹ 1768.2

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Chapter 14: Compound interest - Practice Set 14.1 [Page 90]

APPEARS IN

Balbharati Mathematics [English] Standard 8 Maharashtra State Board
Chapter 14 Compound interest
Practice Set 14.1 | Q 3 | Page 90
Balbharati Mathematics Integrated [English] Standard 8 Maharashtra State Board
Chapter 14 Compound Interest
Practice Set 14.1 | Q 3. | Page 45

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