find simple interest and compound interest both on rupees 30000 at 8 p.c.p.a for 2 years​

2 thoughts on “find simple interest and compound interest both on rupees 30000 at 8 p.c.p.a for 2 years​”

1. Step-by-step explanation:

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2. Answer :-

   • The simple interest is Rs 4800.
   • The compound interest is Rs 4992.

   Step-by-step explanation :-

   • In this question, the principal, rate and time have been given to us. We have to find the simple interest and compound interest.

   —————-

   Let’s find the simple interest first!

   We know that :-

   [tex] \underline{ \boxed{ \sf SI = \dfrac{Principal \times Rate \times Time}{100}}}[/tex]

   Here,

   • Principal = Rs 30000.
   • Rate = 8 p.c.p.a.
   • Time = 2 years.

   Hence,

   [tex]\tt SI = \dfrac{30000 \times 8 \times 2}{100}[/tex]

   Cutting off the zeroes,

   [tex] \tt SI = \dfrac{300 \times 8 \times 2}{1} [/tex]

   Now let’s multiply the remaining numbers.

   [tex] \tt SI = 300 \times 8 \times 2[/tex]

   Multiplying the remaining numbers,

   [tex]\overline{\boxed{\tt SI = Rs \: 4800}}[/tex]

   • The simple interest is Rs 4800.

   —————-

   Now let’s find the compound interest!

   First let’s find the amount.

   We know that :-

   [tex] \underline{\boxed{\sf Amount = Principal \Bigg(1 + \dfrac{Rate}{100} \Bigg)^{Time}}}[/tex]

   Here,

   • Principal = Rs 30000.
   • Rate = 8 p.c.p.a.
   • Time = 2 years.

   Hence,

   [tex] \rm Amount = 30000 \bigg(1 + \dfrac{8}{100} \bigg)^{2} [/tex]

   Making 1 a fraction by taking 1 as the denominator,

   [tex] \rm Amount = 30000 \bigg( \dfrac{1}{1} + \dfrac{8}{100} \bigg)^{2} [/tex]

   The LCM of 1 and 100 is 100, so adding the fractions using their denominators,

   [tex] \rm Amount = 30000 \bigg( \dfrac{1 \times 100 + 8 \times 1}{100 } \bigg)^{2} [/tex]

   On simplifying,

   [tex] \rm Amount = 30000 \bigg( \dfrac{100 + 8}{100} \bigg)^{2} [/tex]

   Adding 8 to 100,

   [tex] \rm Amount = 30000 \bigg( \dfrac{108}{100} \bigg) ^{2} [/tex]

   The power here is 2, so removing the brackets and multiplying 108/100 with itself 2 times,

   [tex] \rm Amount = 30000 \times \dfrac{108}{100} \times \dfrac{108}{100} [/tex]

   Let’s multiply 108/100 with itself 2 times first.

   [tex] \rm Amount = 30000 \times \dfrac{108 \times 108}{100 \times 100} [/tex]

   On multiplying,

   [tex] \rm Amount = 30000 \times \dfrac{11664}{10000} [/tex]

   Cutting off the zeroes,

   [tex] \rm Amount = 3 \times \dfrac{11664}{1} [/tex]

   Now let’s multiply the remaining numbers.

   [tex] \rm Amount = 3 \times 11664[/tex]

   Multiplying 3 with 11664,

   [tex] \overline{\boxed{ \rm Amount = Rs \: 34992}}[/tex]

   —————-

   Now, as we know the amount, let’s find the compound interest!

   We know that :-

   [tex] \underline{ \boxed{\sf CI = Amount – Principal}}[/tex]

   Here,

   • Amount = Rs 34992.
   • Principal = Rs 30000.

   Hence,

   [tex] \boxed{ \bf CI = 34992 – 30000}[/tex]

   Subtracting 30000 from 34992,

   [tex] \overline{\boxed{ \bf CI = Rs \: 4992}}[/tex]

   • The compound interest is Rs 4992.

   —————-

   Abbreviations used :-

   [tex] \sf SI = Simple \: Interest. [/tex]

   [tex] \sf CI = Compound \: Interest. [/tex]

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