Question

A deposit of $6,500 was invested and has grown to $8,300 in 5 years. The growth in value was via simple annual interest, calculated on the basis of the initial investment. Calculate the interest rate.

Answer 1

Formula used-

$\: \\ si \: = \frac{p \times r \times n}{100}$

Solution-

∴A deposit of 6500 was invested and expanded to 8300 , this means the simple interest is

= 8300 - 6500

= $ 1800

∴ Principal = $6500

∴No. of years = 5 years

∴Simple interest = $1800

$= > 1800 = \frac{6500 \times r \times 5}{100} \\ \\ = > 1800 \times 100 = 6500 \times r \times 5 \\ \\ = > 180000 = 32500 \times r \\ \\ = > r = \frac{180000}{32500} \\ \\ = > r = 5.5\%$

ANS- Rate of interest = 5.5%

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Answer 2

$\sf\huge\bold{Answer:}$

Given :

• Invested deposit = $6500
• Grown deposit = $8300
• Time = 5years
• Growth in value was via Simple Interest (SI)

To find :

• Interest rate (r)

Solution :

Formula used :

$\boxed{\tt \: \red{simple \: interest = \frac{principal \times rate \times time}{100} } }$

• Simple interest = Grown deposit-Invested deposit

Simple interest = $8300-$6500

Simple interest = $1800

• Principal value = Invested deposit

Principal value = $6500

• Time = 5

$\tt \: SI = \frac{6500 \times r \times 5}{100}$

$\tt \: SI = \frac{ \cancel{6500} \times r \times 5}{ \cancel{100}}$

$\tt\ 1800 = 65 \times r \times 5$

$\tt\ 1800 = 65 \times 5 \times r$

$\tt\ r = \frac{1800}{65 \times 5}$

$\tt\ r = \frac{ \cancel{1800}}{ 65 \times \cancel { \: \: \: 5}}$

$\tt\ r = \frac{360}{65}$

$\tt\ r = \cancel\frac{360}{65}$

$\tt\ r = 5.5$

$\huge \boxed{\tt\ r = 5.5\%}$

Rate of interest is 5.5%