Question

1 मशीन प्राइस ₹48000 एवरी ईयर प्राइस डाउन बाय 5% रेट them आफ्टर 2 ईयर्स हाउ मेनी प्राइस ऑफ मशीन​

Answer 1

Question:

One machine price Rs. 48000. Every year price down by 5% rate then after 2 years how many price of machine?

Answer:

• The price of the machine is Rs 43320.

Step-by-step explanation:

Given that:

• Principal = Rs. 48000
• Rate = 5% per annum
• Time = 2 years

To Find:

• The price of machine.

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

Finding the price of the machine:

⟶ A = 48000(1 - 5/100)²

⟶ A = 48000(1 - 0.05)²

⟶ A = 48000(0.95)²

⟶ A = 48000 × 0.95 × 0.95

⟶ A = 43320

∴ The price of the machine = Rs 43320

Answer 2

${\pmb{\underline{\sf{ Required \ Solution ... }}}}$

• Machine's Price = ₹ 48000
• Decreasing Rate at 5%
• Time Period = 2 years

$\\ {\pmb{\underline{\sf{Conditions ... }}}}$

As We know that If Price of any thing or any OBJECT (x) is decreasing at some rate (y) Year by year. So, Rate of Decreasing will always present in (—ve) Negative.

$\\ {\pmb{\underline{\sf{Final \ Solution ... }}}}$

If Price is decreasing at Definite rate so we can apply this to get required value for desired number of Years.

As we know that:

$\colon\Rightarrow \ {\underline{\boxed{\sf\large{ x \left( 1 - \dfrac{y}{100} \right)^n }}}} \\ \\ \\ \colon\implies{\sf{ 48000 \left( 1 - \dfrac{5}{100} \right)^2 }} \\ \\ \\ \colon\implies{\sf{ 48000 \left( \cancel{ \dfrac{95}{100} } \right)^2 }} \\ \\ \\ \colon\implies{\sf{ 48000 \left( \dfrac{19}{20} \right)^2 }} \\ \\ \\ \colon\implies{\sf{ \cancel{48000} \times \dfrac{19}{ \cancel{20} } \times \dfrac{19}{ \cancel{20} } }} \\ \\ \\ \colon\implies{\sf{ 120 \times 19 \times 19 }} \\ \\ \\ \colon\implies{\underline{\boxed{\sf\large\pink{ 43320 _{ ( Rs. )} }}}}$

Hence,

${\pmb{\underline{\sf{ The \ Price _{(Machine)} \ after \ two \ years \ will \ be \ Rs. \ 43320. }}}} \bigstar$