1.

Find the compound interest, if the amount of a certain principal after 2 years is ₹ 4036.80 at a rate of 16 p.cp.a * 1 point₹ 3000₹ 1035.80₹ 1036.80₹ 1020.00​

Answer»

align="absmiddle" alt="\underline{ \underline{ \Large \pmb{\mathit{ {Given:}} }} }" class="latex-formula" id="TexFormula1" src="https://tex.z-dn.net/?F=%5Cunderline%7B%20%5Cunderline%7B%20%20%5CLarge%20%5Cpmb%7B%5Cmathit%7B%20%7BGiven%3A%7D%7D%20%7D%7D%20%7D%20" title="\underline{ \underline{ \Large \pmb{\mathit{ {Given:}} }} }">

• Amount (A) = ₹ 4036.80

• Rate (R) = 16 % p.a

• TIME (n YEARS) = 2 years

\underline{ \underline{ \Large \pmb{\mathit{ {To \: calculate:}} }} }

• Compound Interest (C.I)

\underline{ \underline{ \Large \pmb{\mathit{ {Calculation:}} }} }

As we know that,

\bigstar \: \boxed{\sf {C.I = Amount - Principal}} \\

We aren't given the Principal, let's calculate the Principal first.

As we know that,

\bigstar \: \boxed{\sf {A = {P \Bigg \lgroup 1 + \dfrac{R}{100} \Bigg \rgroup}^{n} }} \\

Where,

• P = Principal

• A = Amount

• n = number of years/time

• R = Rate

\longrightarrow \sf {4036.80 = {P \Bigg \lgroup 1 + \dfrac{16}{100} \Bigg \rgroup}^{2}}\\

\longrightarrow \sf {4036.80 = {P \Bigg \lgroup \dfrac{100 +16}{100} \Bigg \rgroup}^{2}}\\

\longrightarrow \sf {4036.80 = {P \Bigg \lgroup \dfrac{116}{100} \Bigg \rgroup}^{2}}\\

\longrightarrow \sf {4036.80 = P \times \dfrac{116}{100} \times \dfrac{116}{100} }\\

\longrightarrow \sf {4036.80 \times \dfrac{116}{100} \times \dfrac{100}{116} = P}\\

\longrightarrow \sf { \dfrac{403680}{\cancel{100}} \times \dfrac{\cancel{100}}{116} \times \dfrac{100}{116} =P } \\

\longrightarrow \sf {403680 \times \dfrac{1}{116} \times \dfrac{100}{116} = P } \\

\longrightarrow \sf { \dfrac{403680 \times 1 \times 100}{116 \times 116} = P } \\

\longrightarrow \sf { \dfrac{\cancel{403680} \times 1 \times 100}{\cancel{116} \times 116} = P }\\

\longrightarrow \sf { \dfrac{3480 \times 1 \times 100}{116} = P } \\

( Cancel 3480 in numerator and 116 in denominator by 4)

\longrightarrow \sf { \dfrac{ \cancel{870} \times 1 \times 100}{\cancel{29}} = P } \\

\longrightarrow \sf { 30 \times 1 \times 100= P }\\

\longrightarrow \sf { 30 \times 100= P } \\

\longrightarrow \boxed{ \sf { Rs. \: 3000= P }}\\

Therefore, principal us Rs. 3000. Now, substitute the values in the formula of C.I to FIND the compound interest.

\bigstar \: \boxed{\sf {C.I = Amount - Principal}} \\

\longrightarrow \sf {C.I =Rs. \: 4036.80 - 3000} \\

\longrightarrow \boxed { \pmb {\rm \red {C.I =Rs. \: 1036.80 } }}\\

Therefore, C.I is ₹ 1036.80 and OPTION C is correct.


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