Calculate the compound interest for the second year on * 12,000 invested for 3 years at 10% per year. Also, find the sum due at th

Question

Calculate the compound interest for the second
year on * 12,000 invested for 3 years at 10%
per year. Also, find the sum due at the end of
the third year.​

in progress 0
Delilah 1 month 2021-08-19T06:18:59+00:00 2 Answers 0 views 0

Answers ( )

    0
    2021-08-19T06:20:00+00:00

    {\huge{\huge{\underline{\underline{\mathrm{\red{QuesTion:-}}}}}}}

    Calculate the compound interest for the second

    year on * 12,000 invested for 3 years at 10%

    per year. Also, find the sum due at the end of

    the third year.

    {\huge{\huge{\underline{\underline{\mathrm{\green{SoluTion:-}}}}}}}

    \underline {\purple{\boxed{ \sf{Given:-}}}}

    ✍︎Principal=12000

    ✍︎Rate=10%

    ✍︎Time=3 years

    Lets understand

    ☞︎︎︎The money given to the borrower is called Principal.

    ☞︎︎︎the extra money given to the lender by borrower is called interest.

    ☞︎︎︎the money given to borrower for a specified time is called Time.

    lets learn the formula for calculacting amount for CI(compound interest).

    \mathrm{ a = p(1 + \frac{r}{100}) {}^{n}  }

     ‎

    \underline {\green{\boxed{ \sf{Here:-}}}}

     ‎

     ‎

    A=Amount

    P=Principal

    r=rate

    n=time

     ‎

     ‎

    lets solve the formula

    \mathrm{ a = 12000( 1 + \frac{10}{100}) {}^{3}  }

     ‎

     ‎

    100 will be simplified by 10

    so

     ‎

     ‎

    \mathrm{ a = 12000( 1 + \frac{1}{10}) {}^{3}  }

     ‎

     ‎

    now lets evaluate the formula

     ‎

     ‎

    \mathrm{ a = 12000(  \frac{11}{10}) {}^{3}  }

     ‎

     ‎

    \mathrm{ a = 12000 \times   \frac{11}{10} \times  \frac{11}{10} \times  \frac{11}{10}  }

     ‎

     ‎

    all zeroes of 10s will be divided by 12,000

     ‎

     ‎

    \mathrm{ a = 12 \times   \frac{11}{1} \times  \frac{11}{1} \times  \frac{11}{1}  }

     ‎

     ‎

    there is no value of 1 in the formula so

    we can write formula as

     ‎

     ‎

    \mathrm{ a = 12 \times   \frac{11}{} \times  \frac{11}{} \times  \frac{11}{}  }

     ‎

     ‎

    now lets evavulate

     ‎

     ‎

    \mathrm{ a = 12 \times   \frac{11}{} \times  \frac{11}{} \times  \frac{11}{}=15972  }

     ‎

     ‎

    \underline {\purple{\boxed{ \sf{Amount=15972}}}}

     ‎

     ‎

    lets find out CI(compound interest)

    CI=A-P

     ‎

     ‎

    CI=15972-12000

    =3972

     ‎

     ‎

    The amount is 15972 so the sum due at the end of third year will be 15,972.

    ____________________________________

    0
    2021-08-19T06:20:25+00:00

    Given :

    • Principal = Rs. 12,000
    • Time = 3 years
    • Rate of interest = 10% p.a.

    To find :

    • Compound interest for second year !
    • The sum at the end of third year !

    Solution :

    To find the compound interest for second year :

    :\implies \sf{A = P\bigg(1 + \dfrac{R}{100}\bigg)^{n}} \\ \\ \\

    Where :-

    • P = Principal
    • R = Rate of interest
    • n = Time period
    • A = Amount\\ \\ \\

    :\implies \sf{A = 12000 \times \bigg(1 + \dfrac{10}{100}\bigg)^{3}} \\ \\ \\

    :\implies \sf{A = 12000 \times \bigg(\dfrac{100 + 10}{100}\bigg)^{3}} \\ \\ \\

    :\implies \sf{A = 12000 \times \bigg(\dfrac{110}{100}\bigg)^{3}} \\ \\ \\

    :\implies \sf{A = 12000 \times \dfrac{110}{10} \times \dfrac{110}{10} \times \dfrac{110}{10}} \\ \\ \\

    :\implies \sf{A = 15972} \\ \\ \\

    \underline{\therefore \sf{Amount\:(A) = Rs. 15972}} \\ \\ \\

    Thus, the amount for second year is Rs 15972.

    And we know that Compound Interest is the difference of the amount and the principal.

    → CI = A – P

    → CI = 15792 – 12000

    → CI = 3792

    Hence, the compound interest for second year is Rs.3792

    To Find the sum at the end of third :

    We know that the amount for the second year is equal to the principal for the third year, hence the sum of money at the end of third year is 15972.

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