Question

Sunil bought a 3 year old machine from Anil at a price of Rs. 7,29,000. Its value
decreases by 10 p.c.p.a. What was the price of the machine when Anil bought it ?

Answer 1

The price of the machine when Anil bought is was Rs.1000000

-Lets start by considering the price Anil bought as x

-After 1st year , the value decreased by 10% of x

i.e., price after 1st year = x - 10%x = 0.9x

-After 2nd year , the value decreased by 10% of 0.9x

i.e., price after 1st year = 0.9x - 10%(0.9x) = 0.81x

-After 3rd year , the value decreased by 10% of 0.81x

i.e., price after 1st year = 0.81x - 10%(0.81x) = 0.729x

-Sunil bought after 3 years for Rs.729000

-Therefore we have 0.729x=729000

=> x = 729000/0.729

we get x=1000000 , which was Anil's price

Answer 2

$\small{\textsf{\textbf{\underline{\underline{The price of the machine when Anil bought it is (2) Rs 10,00,000. \::}}}}}$

Step-by-step explanation:

Given that:

• Sunil bought a 3-year-old machine from Anil for Rs 7,29,000.
• Its value decreases by 10% compounded per annum.

To Find:

• What was the price of the machine when Anil bought it?

Formula used:

• In compound interest.
• A = P(1 ± R/100)ᵀ

Where,

A = Amount

P = Principal

R = Rate of interest

T = Time

+ = For increased rate

– = For decreased rate

Let us assume:

• The price of machine for Anil be x.

We have:

• Time = 3 years
• Amount = Rs 7,29,000
• Rate = 10% c.p.a
• Principal = x

Finding the price of the machine:

⟶ 729000 = x(1 - 10/100)³

⟶ 729000 = x(1 - 0.1)³

⟶ 729000 = x(0.9)³

⟶ 729000 = x × 0.9 × 0.9 × 0.9

⟶ 729000 = 0.729x

⟶ x = 729000/0.729

⟶ x = 1000000

• ∴ The price of the machine = Rs 10,00,000.