Question

unil bought a 3-year-old machine from Anil for 7.29.000. Its value decreases
by 10 p.c.p.a. What was the price of the machine when Anil bought it? (2019)
1) 9.00,000 (2) 10,00.000 (3) 8.00.000 (4) 12,00.000​

Answer 1

Answer:

• 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

Answer 2

Question:-

• Unil bought a 3-year-old machine from Anil for 7.29.000. Its value decreases by 10 p.c.p.a. What was the price of the machine when Anil bought it?

To Find:-

• Find the price of the machine when Anil bought it.

Solution:-

Formula to be Used:-

• ${\boxed{ \sf \red{ A } = \red{{P(1 ± \dfrac{ R }{ 100 } )}^{ T } }}}$

Now ,

We have to find the price of the machine:-

• Let the price of machine be ' x '

$\longrightarrow\sf \: 729000 = x{( 1 - \dfrac { 10 } { 100 } ) }^{ 3 }$

$\longrightarrow\sf \: 729000 = x{ ( 0.9 ) }^{ 3 }$

$\longrightarrow\sf \: 729000 = 0.729x$

$\longrightarrow\sf \: x = \cancel\dfrac{ 729000 } { 0.729 }$

$\longrightarrow\sf \: x = 10,00,000$

Hence ,

• Price of machine is Rs 10,00,000