Question

Find the compound interest on Rs.1500 for 2.5 years at 10% per annum.
a..Rs.425.75
b..Rs.305.75
c.Rs.403.58
.d.Rs.255.75

Please tell the correct answer, i will mark brainiest

Answer 1

Solution!!

The concept of compound interest has to be used here. The principal, rate of interest and time is given. We have to find the compound interest.

Principle (P)= Rs 1500

Rate of interest (R) = 10%

Time (T) = 2.5 years = 5/2 years

$\sf \bold{\to A=P\left(1+\dfrac{R}{100}\right)^{T}}$

$\sf \to A=1500\left(1+\dfrac{10}{100}\right)^{\frac{5}{2}}$

$\sf \to A=1500\left(\dfrac{110}{100}\right)^{\frac{5}{2}}$

$\sf \to A=1500\left(\dfrac{11}{10}\right)^{\frac{5}{2}}$

$\sf \to A=1500\times \dfrac{11^{\frac{5}{2}}}{10^{\frac{5}{2}}}$

$\sf \to A=\dfrac{1500\sqrt{11^{5}}}{\sqrt{10^{5}}}\quad (a^{\frac{m}{n}}=\sqrt[n]{a^{m}})$

$\sf \to A=\dfrac{1500\times 11^{2}\sqrt{11}}{10^{2}\sqrt{10}}$

$\sf \to A=\dfrac{1500\times 121\sqrt{11}}{100\sqrt{10}}$

$\sf \to A=\dfrac{15\times 121\sqrt{11}}{\sqrt{10}}$

$\sf \to A=\dfrac{1815\sqrt{11}}{\sqrt{10}}$

$\sf \to A=\dfrac{363\sqrt{110}}{2}$

$\to$A ≈ Rs 1903.58

CI = Amount - Principal

CI = Rs 1903.58 - Rs 1500

CI = Rs 403.58

Compound interest = Rs 403.58

Answer 2

Given : Principal  is  Rs. 1500   , Rate of Interest is Rs. 10%  & Time is 2.5 years .

Need To Find : The Rate of Interest .

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⠀⠀⠀⠀Formula for Finding Amount for Finding Compound Interest  is given by :

$\dag\:\:\boxed{ \sf{ Amount = \bigg[ P \bigg( 1 + \dfrac {R}{100} \bigg) ^T \bigg] }}$

Where,

• P is the Principal , R is the Rate of Interest & T is the Time .

• Given , Time is 2.5 yrs or $\dfrac{5}{2}$ yrs .

⠀⠀⠀⠀⠀⠀$\underline {\bf{\star\:Now \: By \: Substituting \: the \: Given \: Values \::}}$

$\qquad \:\::\implies \sf{ Amount = \bigg[ 1500 \bigg( 1 + \dfrac {10}{100} \bigg) ^{\dfrac{5}{2}} \bigg] }$

$\qquad \:\::\implies \sf{ Amount = \bigg[ 1500 \bigg( 1 + \dfrac {\cancel {10}}{10\cancel{0}} \bigg) ^{\dfrac{5}{2}} \bigg] }$

$\qquad \:\::\implies \sf{ Amount = \bigg[ 1500 \bigg( 1 + \dfrac {1}{10} \bigg) ^{\dfrac{5}{2}} \bigg] }$

$\qquad \:\::\implies \sf{ Amount = \bigg[ 1500 \bigg( \dfrac {11}{10} \bigg) ^{\dfrac{5}{2}} \bigg] }$

$\qquad \:\::\implies \sf{ Amount = \bigg[ 1500 \bigg( \dfrac {11}{10} \bigg) ^{\dfrac{5}{2}} \bigg] }$

$\qquad \:\::\implies \sf{ Amount = \bigg[ 1500 \bigg( \dfrac {11}{10} \bigg) ^{\dfrac{5}{2}} \bigg] }$

$\qquad \:\::\implies \sf{ Amount = \bigg[ 1500 \bigg( \dfrac {11}{10} \bigg) ^{\dfrac{5}{2}} \bigg] }$

$\qquad \:\::\implies \sf{ Amount = \bigg[ 1500 \bigg( 1.1 \bigg) ^{\dfrac{5}{2}} \bigg] }$

$\qquad \:\::\implies \sf{ Amount = \bigg[ 1500 \bigg( 1.1 \bigg) ^{2.5} \bigg] }$

• $1.1^{2.5} \approx 1.27\: [Approx]$

$\qquad \:\::\implies \sf{ Amount = 1500 \times 1.27 }$

$\qquad \:\::\implies \bf{ Amount \approx 1904 }$

As, We know that ,

$\dag\:\:\boxed{ \sf{Compound \:Interest\:= Amount \:- Principal\:}}$

⠀⠀⠀⠀⠀⠀$\underline {\frak{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}$

$\qquad \:\::\implies \sf{ Compound\:Interest\:= 1904 - 1500 }$

$\qquad \:\::\implies \sf{ Compound\:Interest\:\approx 404 \:\:[ Approx]}$ [ As , Our Amount is Approx so Compound Interest is also use to be approx ]

Therefore,

⠀⠀⠀⠀⠀$\therefore {\underline{ \mathrm { Hence\: Compound \:Interest \: \:is\:\bf{Rs.404 \:\: [Approx]}}}}$

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