The price of a new car is Rs. $2500$ . The price depreciates by $12 %$ each year (p.a). Find its value at the end of $3$ years.

1703.68

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2703.68

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3703.68

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4703.68

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Solution

The correct option is C $1703.68$
Given, $P = 2500, r = 12 %, n = 3$ years
We know, $A = P [{(1 - \frac{r}{100})}^{n}]$
$\Rightarrow A = 2500 [{(1 - \frac{12}{100})}^{3}]$
$\Rightarrow A =$ Rs. $1703.68$
Thus, the value of a car at the end of $3$ years is Rs. $1703.68$ .

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The price of a new car is Rs. 2500. The price depreciates by 12% each year (p.a). Find its value at the end of 3 years.

The correct option is C 1703.68Given, P=2500,r=12%,n=3 yearsWe know, A=P[(1−r100)n]⇒A=2500[(1−12100)3]⇒A= Rs. 1703.68Thus, the value of a car at the end of 3 years is Rs. 1703.68.

The price of a new car is Rs. $2500$ . The price depreciates by $12 %$ each year (p.a). Find its value at the end of $3$ years.

The correct option is C $1703.68$
Given, $P = 2500, r = 12 %, n = 3$ years
We know, $A = P [{(1 - \frac{r}{100})}^{n}]$
$\Rightarrow A = 2500 [{(1 - \frac{12}{100})}^{3}]$
$\Rightarrow A =$ Rs. $1703.68$
Thus, the value of a car at the end of $3$ years is Rs. $1703.68$ .