Question

Find the rate of interest at which Rs8000 will amount to Rs10648 after 3 years, interest being compounded annually.


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Answer 1

☘⫷❥ᴀ᭄n §₩ΣR⫸☘

Given,

Principal= Rs 8000

Amount= Rs 10648

Time = 3yrs.

Required to find = Rate of interest.

By the problem,

8000( 1+ r/100)³= 10648

or, (1+r/100)³= 10648/8000

or, ( 1+r/100)³ = 1331/1000

or, (1+r/100)³ = (11/10)³

or, (1+r/100)= 11/10

or, 100+r/100= 11/10

or, 100+r= 110

or, r = 110-100 = 10

Therefore, Rate of interest= 10%

Hope it helps you

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Answer 2

Step-by-step explanation:

Principal (P) = Rs.8000

Amount (A) = Rs.10648

Time (n) = 3 years

Compound Interest = Amount - Principal

= 10648 - 8000

= 2648

The formula for finding Amount is

$\rm Amount = P \bigg(1 + \dfrac{r}{100}\bigg)^{n}$

$\rm \dashrightarrow 10648 = 8000 \times \bigg(1 + \dfrac{r}{100}\bigg)^{3}$

$\rm \dashrightarrow \dfrac{ 10648}{ 8000} = \bigg(1 + \dfrac{r}{100}\bigg)^{3}$

$\rm \dashrightarrow \dfrac{1331}{ 1000} = \bigg(1 + \dfrac{r}{100}\bigg)^{3}$

$\rm \dashrightarrow \dfrac{1331}{ 1000} = \bigg( \dfrac{100 + r}{100}\bigg)^{3}$

$\rm \dashrightarrow \sqrt[ 3]{ \dfrac{1331}{ 1000} } = \dfrac{100 + r}{100}$

$\rm \dashrightarrow \dfrac{11}{ 10} = \dfrac{100 + r}{100}$

$\rm \dashrightarrow \dfrac{11}{ 10} \times 100= 100 + r$

$\rm \dashrightarrow 110= 100 + r$

$\rm \dashrightarrow r = 110 - 100$

$\rm \dashrightarrow r = 10$

$\underline{\boxed{\rm \therefore Rate \: of \: interest = 10\%}}$