Math, asked by Shivang6853, 1 day ago

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.

Answers

Answered by robinhood9
48

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%

____________________________________

Answered by GraceS
26

\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%

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