Question

help me number 9 help me plss be fast​

Answer 1

Answer:

Qᴜᴇsᴛɪᴏɴ :

↝ Find to the nearest rupee the compound interest on ₹1800 for 3 years at 5% rate of interest, compound annually.

$\begin{gathered}\end{gathered}$

Gɪᴠᴇɴ :

• ➛ Principle = Rs.1800
• ➛ Time = 3 years
• ➛ Rate of Interest = 5%

$\begin{gathered}\end{gathered}$

Tᴏ Fɪɴᴅ :

• ➛ Amount
• ➛ Compound

$\begin{gathered}\end{gathered}$

Cᴏɴᴄᴇᴘᴛ :

↝ First we will find the Amount using the formula to calculate compound interest. Where, Principle is Rs.1800, ↝ Rate is 5% and Time is 3 years. Then we will use the formula to find the compound interest.

$\begin{gathered}\end{gathered}$

Usɪɴɢ Fᴏʀᴍᴜʟᴀs :

$\longrightarrow\small{\underline{\boxed{\pmb{\sf{A= P\bigg(1 + \dfrac{ {R}}{100} \bigg)^{T}}}}}}$

$\longrightarrow\small{\underline{\boxed{\pmb{\sf{{C.I=A- P}}}}}}$

☼ Where :-

• ➛ A = Amount
• ➛ P = Principle
• ➛ R = Rate
• ➛ T = Time
• ➛ C.I = Compound Interest

$\begin{gathered}\end{gathered}$

Sᴏʟᴜᴛɪᴏɴ :

☼ Finding amount by substituting the values in the formula :-

$\longrightarrow\small{\sf{Amount= P\bigg(1 + \dfrac{ {R}}{100} \bigg)^{T}}}$

$\longrightarrow\small{\sf{Amount= 1800\bigg(1 + \dfrac{{5}}{100} \bigg)^{3}}}$

${\longrightarrow{\small{\sf{Amount= 1800\bigg( \dfrac{(1 \times 100) + (5 \times 1)}{100} \bigg)^{3}}}}}$

${\longrightarrow{\small{\sf{Amount= 1800\bigg( \dfrac{100 + 5}{100} \bigg)^{3}}}}}$

${\longrightarrow{\small{\sf{Amount= 1800\bigg( \dfrac{105}{100} \bigg)^{3}}}}}$

${\longrightarrow{\small{\sf{Amount= 1800\bigg( \cancel{\dfrac{105}{100}} \bigg)^{3}}}}}$

${\longrightarrow{\small{\sf{Amount= 1800\bigg( {\dfrac{21}{20}} \bigg)^{3}}}}}$

${\longrightarrow{\small{\sf{Amount= 1800\bigg(\dfrac{21}{20} \times \dfrac{21}{20} \times \dfrac{21}{20}\bigg)}}}}$

${\longrightarrow{\small{\sf{Amount= 1800\bigg(\dfrac{9261}{8000}\bigg)}}}}$

${\longrightarrow{\small{\sf{Amount= 1800 \times \dfrac{9261}{8000}}}}}$

${\longrightarrow{\small{\sf{Amount= 18\cancel{00} \times \dfrac{9261}{80 \cancel{00}}}}}}$

${\longrightarrow{\small{\sf{Amount= 18\times \dfrac{9261}{80}}}}}$

${\longrightarrow{\small{\sf{Amount= \dfrac{18 \times 9261}{80}}}}}$

${\longrightarrow{\small{\sf{Amount= \dfrac{166698}{80}}}}}$

${\longrightarrow{\small{\sf{Amount= Rs.2083.725}}}}$

${\longrightarrow{\small{\underline{\boxed{\sf{Amount= Rs.2083.725}}}}}}$

∴ The amount is Rs.2083.725 or Rs.2083.73.

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☼ Finding compound interest by substituting the values in the formula :-

$\longrightarrow\small{\sf{Compound \: Interest=A- P}}$

${\longrightarrow{\small{\sf{Compound \: Interest=2083.725- 1800}}}}$

${\longrightarrow{\small{\sf{Compound \: Interest=Rs.283.725}}}}$

${\longrightarrow{\small{\underline{\boxed{\sf{Compound\:Interest=Rs.283.725}}}}}}$

∴ The compound interest is Rs.283.725 or Rs.283.73.

$\begin{gathered}\end{gathered}$

Lᴇᴀʀɴ Mᴏʀᴇ :

$\dashrightarrow\small{\underline{\boxed{\sf{\green{Amount = Principle + Interest}}}}}$

$\dashrightarrow\small{\underline{\boxed{\sf{\green{ Principle=Amount - Interest }}}}}$

$\dashrightarrow\small{\underline{\boxed{\sf{\green{ Simple \: Interest = \dfrac{P \times R \times T}{100}}}}}}$

$\dashrightarrow{\small{\underline{\boxed{\sf{\green{Amount={P{\bigg(1 + \dfrac{R}{100}{\bigg)}^{T}}}}}}}}}$

$\dashrightarrow\small{\underline{\boxed{\sf{\green{Principle = \dfrac{Amount\times 100 }{100 + (Time \times Rate)}}}}}}$

$\dashrightarrow\small{\underline{\boxed{\sf{\green{Principle = \dfrac{Interest \times 100 }{Time \times Rate}}}}}}$

$\dashrightarrow\small{\underline{\boxed{\sf{\green{Rate = \dfrac{Simple \: Interest \times 100}{Principle \times Time}}}}}}$

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