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Maharashtra State BoardSSC (English Medium) 8th Standard

Find the amount and the compound interest. No. Principal (₹) Rate (p.c.p.a.) Duration(Years) 1 2000 5 2 2 5000 8 3 3 4000 7.5 2

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Question

Find the amount and the compound interest.

No. Principal (₹) Rate (p.c.p.a.) Duration
(Years)
1 2000 5 2
2 5000 8 3
3 4000 7.5 2
Sum
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Solution

(1) Here, P = ₹ 2000; R = 5 %; N = 2 years

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

= 2000 `(1 + 5/100)^2`

= 2000`((100+5)/100)^2`

= `2000(105/100)^2`

= `2000(105/100) xx(105/100)`

= 2000 `(21/20)^2`

= 2205 Rupees

Compound interest after 2 years,

I = Amount - Principal

 = 2205 - 2000 

= 205 Rupees

Hence, Amount =  ₹ 2205 and compound interest =  ₹205.

(2) Here, P = ₹ 5000; R = 8 %; N = 3 years

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

= 5000 `(1 + 8/100)^3`

= 5000 `(108/100) xx(108/100) xx(108/100) `

= `(108 xx 108 xx54)/100`

= `629856/100`

= 6298.56 Rupees

∴ Compound interest after 3 years,

I = Amount - Principal

= 6298.56 - 5000

= 1298.56 Rupees

Hence, Amount =  ₹ 6298.56 and compound interest =  ₹1298.56

(3) Here, P = ₹ 4000; R = 7.5 % ; N = 2 years

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

= 4000 `(1 + 7.5/100)^2`

= 4000 `(1 +75/1000)^2`

= 4000 `(1075/1000)xx(1075/1000)`

= `(4xx1075xx1075)/1000`

= `4622500/1000`

A = 4622.5 Rupees

∴ Compound interest after 2 years,

I = Amount - Principal

= 4622.5 - 4000

= 622.5 Rupees

Hence, Amount =  ₹ 4622.5 and compound interest =  ₹ 622.5

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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 1 | Page 90
Balbharati Mathematics Integrated [English] Standard 8 Maharashtra State Board
Chapter 14 Compound Interest
Practice Set 14.1 | Q 1. | Page 45

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