Math, asked by Rudranil420, 3 months ago

The price of a machine in a leather factory depreciates at the rate of 10% every year. If the present price of the machine be Rs 100000, let us calculate what will be the price of that machine after 3 years. ​

Answers

Answered by Saby123
42

Solution :

Here , the present price of the machine is Rs. 100, 000 .

The price decreases regularly at a rate of 10% per annum .

We need to find what will be it's price after 3 years .

So , in the present , the price is Rs. 100,000

The next year , it will be decreased by 10%

Or in other terms , it will be 90% of what it was previously ; 90% of the price in the previous year.

So ,

In Year 1 :

New Price -

=> 90% of orginal price

=> 90% of 100,000

=> 90 × 1000

=> 90,000

In the second year the same thing will repeat

New price = 90% of what it was in the first year

=> 90% of 90,000

=> 90 × 900

=> 81, 000

In the third year

Price -

=> 90% of the price in the second year

=> 90% of 81,000

=> 90 × 810

=> 72900

Thus , the required price after 3 years is 72,900.

This is the required answer.

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Anonymous: Nice
sainiinswag: perfectly explained
jaydip1118: good
AdorableMe: Wow :P
Saby123: Thanks
Answered by BrainlyHero420
97

Answer:

Given :-

  • The price of a machine in a leather factory depreciates at the rate of 10% every year and the present price of the machine be Rs 100000.

To Find :-

  • What is the price of that machine after 3 years.

Formula Used :-

\sf\boxed{\bold{\large{A\: =\: P\bigg(1 - \dfrac{r}{100}\bigg)^{n}}}}

where,

  • A = Amount
  • P = Principal
  • r = Rate of Interest
  • t = Time

Solution :-

Given :

  • Principal = Rs 100000
  • Rate of Interest = 10%
  • Time = 3 years

According to the question by using the formula we get,

\sf A\: =\: 100000\bigg(1 - \dfrac{10}{100}\bigg)^{3}

\sf A\: =\: 100000\bigg(1 - \dfrac{\cancel{10}}{\cancel{100}}\bigg)^{3}

\sf A\: =\: 100000\bigg(1 - \dfrac{1}{10}\bigg)^{3}

\sf A\: =\: 100000\bigg(\dfrac{9}{10}\bigg)^{3}

\sf A\: =\: 100000 \times \dfrac{9}{10} \times \dfrac{9}{10} \times \dfrac{9}{10}

\sf A\: =\: \dfrac{72900000}{1000}

\sf A\: =\: \dfrac{\cancel{72900000}}{\cancel{1000}}

\sf\red{A\: =\: Rs\: 72900}

\therefore The price of that machine after 3 years is Rs 72900 .


jaydip1118: great
jaydip1118: nice
AdorableMe: :kaala_dil: :)
Saby123: :unamused:
Saby123: No more unnecessary comments.
razaali31: no comments
razaali31: thanks
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