Question

Rahul purchased a two-wheeler for ₹62,000. Its value is depreciating at the rate of 10% p.a. find its value after 2 years. ​

Answer 1

Given :

• ➻ Rahul purchased a two wheeler for Rs. 62000 .Its value is deprecating at the rate of 10 % pa.

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To Find :

• ➻ Value after 2 years.

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Solution :

${\pink{❒}}$Formula Used :

$\large{\color{cyan}{\bigstar}} \: \: {\underline{\boxed{\color{orange}{\sf{ C.I = P \bigg\lgroup 1 + \dfrac{R}{100} \bigg\rgroup^T - P }}}}}$

Where :

• ➵ C.I = Depreciation amount
• ➵ P = Current cost
• ➵ R = Depreciation Rate
• ➵ T = Time

$\qquad{━━━━━━━━━━━━━━━━━━━━}$

${\pink{❒}}$Calculating the Cost after 2 years :

Depreciation :

${:\implies{\qquad{\sf{ C.I= P \bigg\lgroup 1 + \dfrac{R}{100} \bigg\rgroup ^T - P}}}} \\ \\ \ {:\implies{\qquad{\sf{ C.I = 62000 \bigg\lgroup 1 + \dfrac{10}{100} \bigg\rgroup ^2 - 62000}}}} \\ \\ \ {:\implies{\qquad{\sf{ C.I = 62000 \bigg\lgroup \dfrac{110}{100} \bigg\rgroup ^2 - 62000 }}}} \\ \\ \ {:\implies{\qquad{\sf{ C.I = 62000 \bigg\lgroup \cancel\dfrac{110}{100} \bigg\rgroup ^2 - 62000 }}}} \\ \\ \ {:\implies{\qquad{\sf{ C.I= 62000 \bigg\lgroup 1.10 \bigg\rgroup ^2 - 62000 }}}} \\ \\ \ {:\implies{\qquad{\sf{ C.I = 62000 \times 1.10 \times 1.10 - 62000}}}} \\ \\ \ {:\implies{\qquad{\sf{ C.I = 62000 \times 1.21 - 62000 }}}} \\ \\ \ {:\implies{\qquad{\sf{ C.I = 75020 - 62000 }}}} \\ \\ \ {\qquad{\sf{Depreciation \: Amount \: = {\blue{\sf{ ₹ \: 13020 }}}}}}$

Value after 2 years :

${:\implies{\qquad{\rm{ Cost{\small_{(2 \: years) }} = Current \: cost - Depreciation \: amount }}}} \\ \\ \ {:\implies{\qquad{\rm{ Cost{\small_{(2 \: years) }} = 62000 - 13020}}}} \\ \\ \ {\qquad{\sf{Cost \: after \: 2 \: years \: = {\red{\sf{ ₹ \: 48980 }}}}}}$

$\qquad{━━━━━━━━━━━━━━━━━━━━}$

${\pink{❒}}$Therefore :

❝ Cost of two - wheeler after 2 years will be ₹ 48980 .❞

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Answer 2

✰ Given :-

• Current cost of the two - wheeler = Rs. 62,000
• Rate at which its value is depreciating = 10 % per annum
• Time = 2 years

✰ To Find :-

• Its value after 2 years

✰ Formula Used :-

$\bigstar \: \boxed{ \sf \: A = P - ( 1 + \frac{ R}{100} ) ^{T} - P}$

Where,

• A = its value after 2 years
• T = Time
• R = Rate of depreciation
• P = Current Cost

✰ Solution :-

$: \implies \sf \: A = 62000(1 + \frac{10}{100}) ^{2} - 62000\: \\ : \implies \sf \: A = 62000( \frac{110}{100} ) ^{2} - 62000 \: \: \: \: \: \: \: \: \: \\ : \implies \sf \: A = 62000 \times 1.1 \times 1.1 - 62000\\ : \implies \sf \: A = 75020 - 62000\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:\\ : \implies \sf \: A = 13020 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

We know that,

Cost of two wheeler after two years will be current cost - Depreciation amount

${ \sf \orange{\therefore \: cost \: of \: two - wheeler_{(2 years)} }= \pink{Rs. 48980}}$