Find the compound interest, if the amount of a certain principal after

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» $\underline{ \underline{ \Large \pmb{\mathit{ {Given:}} }} }$ • 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 $\longrightarrow \sf {4036.80 = {P \Bigg \lgroup 1 + \dfrac{16}{100} \Bigg \rgroup}^{2}}\\$ $\longrightarrow \sf {4036.80 = {P \Bigg \lgroup \dfrac{100 +16}{100} \Bigg \rgroup}^{2}}\\$ $\longrightarrow \sf {4036.80 = {P \Bigg \lgroup \dfrac{<klux>116</klux>}{100} \Bigg \rgroup}^{2}}\\$ $\longrightarrow \sf {4036.80 = P \times \dfrac{116}{100} \times \dfrac{116}{100} }\\$ $\longrightarrow \sf {4036.80 \times \dfrac{116}{100} \times \dfrac{100}{116} = P}\\$ $\longrightarrow \sf { \dfrac{403680}{\cancel{100}} \times \dfrac{\cancel{100}}{116} \times \dfrac{100}{116} =P } \\$ $\longrightarrow \sf {403680 \times \dfrac{1}{116} \times \dfrac{100}{116} = P } \\$ $\longrightarrow \sf { \dfrac{403680 \times 1 \times 100}{116 \times 116} = P } \\$ $\longrightarrow \sf { \dfrac{\cancel{403680} \times 1 \times 100}{\cancel{116} \times 116} = P }\\$ $\longrightarrow \sf { \dfrac{3480 \times 1 \times 100}{116} = P } \\$ ( 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.

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»

$\underline{ \underline{ \Large \pmb{\mathit{ {Given:}} }} }$

• 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

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

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

$\longrightarrow \sf {4036.80 = {P \Bigg \lgroup \dfrac{<klux>116</klux>}{100} \Bigg \rgroup}^{2}}\\$

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

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

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

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

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

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

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

( 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.

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

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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​

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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