Solve for A- A-P 1-R100n P-4000 R-7-5 n-2

A = P ( 1 + R100)^n Given: P = 4000 R = 7.5 n = 2 ⇒ A = 4000 ( 1 + 7.5100)^2 ⇒ A = 4000 ( 107.5100)^2 ⇒ A = 0.4 × 11556.25 = ...

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Standard XII

Mathematics

Property 7

Solve for A: ...

Question

Solve for A:
$A = P {(1 + \frac{R}{100})}^{n}$
$P = 4000$
$R = 7.5$
$n = 2$

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Solution

$A = P {(1 + \frac{R}{100})}^{n}$
Given:-
$P = 4000$
$R = 7.5$
$n = 2$
$\Rightarrow A = 4000 {(1 + \frac{7.5}{100})}^{2}$
$\Rightarrow A = 4000 {(\frac{107.5}{100})}^{2}$
$\Rightarrow A = 0.4 \times 11556.25 = 4622.5$

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Solve for A:A=P(1+R100)nP=4000R=7.5n=2

A=P(1+R100)nGiven:-P=4000R=7.5n=2⇒A=4000(1+7.5100)2⇒A=4000(107.5100)2⇒A=0.4×11556.25=4622.5

Solve for A:
$A = P {(1 + \frac{R}{100})}^{n}$
$P = 4000$
$R = 7.5$
$n = 2$

$A = P {(1 + \frac{R}{100})}^{n}$
Given:-
$P = 4000$
$R = 7.5$
$n = 2$
$\Rightarrow A = 4000 {(1 + \frac{7.5}{100})}^{2}$
$\Rightarrow A = 4000 {(\frac{107.5}{100})}^{2}$
$\Rightarrow A = 0.4 \times 11556.25 = 4622.5$