Math, asked by keahav4708, 9 months ago

The price of land grows by 10% every year.
If the price in 1995 was ₹10,000 find its
price in 1998.

Answers

Answered by bhagyashreechowdhury
4

Given:

The price of land grows by 10% every year.

If the price in 1995 was ₹ 10,000

To find:

Its  price in 1998

Solution:

The price of the land in 1995 = ₹ 10000

The rate of increase in the price of the land every year = 10%

No. of years = 1998 - 1995 = 3 years

We know the formula for the increase in the price of an asset is as follows:

\boxed{\bold{A = P [1+ \frac{R}{100} ]^n}}

where

"A" → the price after n years

"P" → the current price

"R" → the rate of increase

"n" → the no. of years

Now, by substituting the values in the above formula, we get

Price\:after \:3 \:years = 10000 [1 + \frac{10}{100} ]^3

\implies Price\:after \:3 \:years = 10000 [\frac{110}{100}  } ]^3

\implies Price \:after \:3\:years = 10000\times \frac{110}{100}\times \frac{110}{100}\times \frac{110}{100}

\implies Price\:after\:3\:years = 110 \times 11\times 11

\implies Price \:after\: 3 \:years = 13310

Thus, the price of the land in 1998 is → ₹ 13310.

---------------------------------------------------------------------------------------

Also View:

the piece of land increase by 9% every year if the present price is Rupees 1188 1 what was its price 2 years ago?

https://brainly.in/question/2835858

Subba Rao started work in 1995 at an annual salary of Rs 5000 and received an increment of Rs 200 each year. In which year did his income reach Rs 7000?

https://brainly.in/question/1345107

Answered by sheikhshifani074
0

Answer:

Given:

The price of land grows by 10% every year.

If the price in 1995 was ₹ 10,000

To find:

Its price in 1998

Solution:

The price of the land in 1995 = ₹ 10000

The rate of increase in the price of the land every year = 10%

No. of years = 1998 - 1995 = 3 years

We know the formula for the increase in the price of an asset is as follows:

\boxed{\bold{A = P [1+ \frac{R}{100} ]^n}}

A=P[1+

100

R

]

n

where

"A" → the price after n years

"P" → the current price

"R" → the rate of increase

"n" → the no. of years

Now, by substituting the values in the above formula, we get

Price\:after \:3 \:years = 10000 [1 + \frac{10}{100} ]^3Priceafter3years=10000[1+

100

10

]

3

\implies Price\:after \:3 \:years = 10000 [\frac{110}{100} } ]^3

\implies Price \:after \:3\:years = 10000\times \frac{110}{100}\times \frac{110}{100}\times \frac{110}{100}⟹Priceafter3years=10000×

100

110

×

100

110

×

100

110

\implies Price\:after\:3\:years = 110 \times 11\times 11⟹Priceafter3years=110×11×11

\implies Price \:after\: 3 \:years = 13310⟹Priceafter3years=13310

Thus, the price of the land in 1998 is → ₹ 13310.

Step-by-step explanation:

Given:

The price of land grows by 10% every year.

If the price in 1995 was ₹ 10,000

To find:

Its price in 1998

Solution:

The price of the land in 1995 = ₹ 10000

The rate of increase in the price of the land every year = 10%

No. of years = 1998 - 1995 = 3 years

We know the formula for the increase in the price of an asset is as follows:

\boxed{\bold{A = P [1+ \frac{R}{100} ]^n}}

A=P[1+

100

R

]

n

where

"A" → the price after n years

"P" → the current price

"R" → the rate of increase

"n" → the no. of years

Now, by substituting the values in the above formula, we get

Price\:after \:3 \:years = 10000 [1 + \frac{10}{100} ]^3Priceafter3years=10000[1+

100

10

]

3

\implies Price\:after \:3 \:years = 10000 [\frac{110}{100} } ]^3

\implies Price \:after \:3\:years = 10000\times \frac{110}{100}\times \frac{110}{100}\times \frac{110}{100}⟹Priceafter3years=10000×

100

110

×

100

110

×

100

110

\implies Price\:after\:3\:years = 110 \times 11\times 11⟹Priceafter3years=110×11×11

\implies Price \:after\: 3 \:years = 13310⟹Priceafter3years=13310

Thus, the price of the land in 1998 is → ₹ 13310.

Similar questions