Question

Rahul bought a car Rs. 4 lakhs in 2018. The rate of
depreciation is 10% per year. Find the depreciation
value of the car in the year 2020?

Solution P=4,00,000 R= 10%, N=_years.
A=P(_______) (Formula)
On substituting the values of P, N, and R solving we get
A=4,00,000 X _______ square
= 4000 X (_______)
= Rs.________
Depreciation in price = Cost price - A
= Rs.________​


koi nahi aa raha ha

Answer 1

Answer:

• The depreciation value of the car in the year 2020 is Rs 76,000.

Step-by-step explanation:

Given that:

• Rahul bought a car Rs. 4 lakhs in 2018.
• The rate of depreciation is 10% per year.

To Find:

• The depreciation value of the car in the year 2020?

Formula used:

In compound interest.

• A = P(1 ± R/100)ᵀ

Where,

• A = Amount
• P = Principal = Rs 4,00,000
• R = Rate = 10% per year
• T = Time = 2020 - 2018 = 2 years
• + = Increased rate
• - = Decreased rate

Finding the amount:

⟶ A = P(1 - R/100)ᵀ

⟶ A = 400000(1 - 10/100)²

⟶ A = 400000(1 - 0.1)²

⟶ A = 400000(0.9)²

⟶ A = 400000 × 0.9 × 0.9

⟶ A = 324000

∴ Amount = Rs 3,24,000

Finding the depreciation value of the car:

Depreciation value = Actual cost of car - Amount

Depreciation value = 400000 - 324000

Depreciation value = 76000

Hence,

• The depreciation value of the car in the year 2020 is Rs 76,000.

Answer 2

Answer:

• Rs.76,000

Step-by-step explanation:

Given

• Principal (P) = Rs.4,00,000
• Rate of interest (R) = 10 %
• Time period (T) = 2020 - 2018 = 2 years

To find

• Amount (A)
• Simple interest (SI)

Solution

In the question it is given that Rahul bought a car for 4 lakhs, that is here 4 lakhs is the principal and the rate of depreciation is 10 % per year. That means the rate of interest is 10 %, and the rate of interest shall be taken negative that means we shall minus rate of interest (multiply) principal, he bought the car in 2018 and we are asked to find the value of car in the year 2020.

↪ P = 4,00,000

↪ R = 10 %

↪ T = 2 years

↪ A = P(1 - R/100)ᵀ

• 400000(1 - 10/100)²
• 400000(0.9)²
• 324000

∴ Amount is Rs.3,24,000.

↪ A = 3,24,000

↪ P = 4,00,000

↪ S.I = P - A

↪ Depreciation in price = Cost prize - Amount

• 400000 - 324000
• 76000

∴ Simple interest or the depreciation in price is Rs.76,000

∴ Depreciation value of the car is Rs.76,000.