Question

The cost of the scooter depreciates every year by 10% of its value at the
beginning of the year . If the present cost of scooter is ₹ 8000.Find its cost after 2
years.

Answer 1

Answer:

Rate of depreciation (R) = 10%

Present cost (P) =  Rs. 8000

Time (T) = 2 years

Cost after 2 years (P t) =  P * ( 1- R /100 ) ^T

  = 8000* ( 1-10/100) ^2

  = 8000 *( 9/10 )^2

  = 8000 *81 /100

    = Rs, 6480

Answer 2

Given:-

• The cost of the scooter depreciates every year by 10% of its original value
• The present cost of scooter is ₹ 8000.

To Find:-

• The cost of the scooter after 2 years

Assumptions:-

• Let it's cost after 2 years be considered as the amount

Formula used:-

• ${\small{\underline{\boxed{\pmb{\sf{ A = P\bigg(1 - \frac{r}{100} \bigg)^n}}}}}}$  

Where:-

• A stands for Amount
• P stands for Principal
• R stands for Rate of Interest
• N stands for No of years

• Provided Values :

• Principal Amount =  ₹ 8000
• Rate of interest = 10% p.a
• Time Period = 2 years

Substituting we get:-

${:\implies} \bf A = P\bigg[1 - \dfrac{r}{100} \bigg]^n$

${:\implies} \bf A =8000\bigg[1 - \dfrac{10}{100} \bigg]^2$

${:\implies} \bf A =8000\bigg[ \dfrac{100}{100} - \dfrac{10}{100} \bigg]^2$

${:\implies} \bf A =8000\bigg[ \dfrac{90}{100} \bigg]^2$

${:\implies} \bf A =8000\bigg[ \dfrac{9}{10} \bigg]^2$

${:\implies} \bf A =8000\bigg[ \dfrac{81}{100} \bigg]$

${:\implies} \bf A =80\bigg[ \dfrac{81}{1} \bigg]$

${:\implies} \bf A = 80 \times 81$

${:\implies} {\purple{\underline{\boxed{\frak{ Amount = 6480 }}}\bigstar}}$

Therefore :-

• The cost of the scooter after 2 years is ₹ 6480

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