English
Maharashtra State BoardSSC (English Medium) 8th Standard

The cost price of a machine is 2,50,000. If the rate of depreciation is 10% per year, find the depreciation in price of the machine after two years. - Mathematics

Advertisements
Advertisements

Question

The cost price of a machine is 2,50,000. If the rate of depreciation is 10% per year, find the depreciation in price of the machine after two years.

Sum
Advertisements

Solution

Here, P = Cost price of the machine = 2,50,000

A = Cost price after 2 years

I = Depreciation in price after 2 years

R = Rate of depreciation = 10 %

N = 2 years

A = P `(1 + "R"/100)^"N"`

= 2,50,000 `(1 + (-10)/100)^2`

= 2,50,000 `(1 - 1/100)^2`

= 2,50,000 `((100 - 10)/100)^2`

= 2,50,000 `(90/100)^2`

= 2,50,000 `(9/10)^2`

= 2,50,000 `(81/100)`

= 2,500 × 81

= 2,02,500

Depreciation in price = Cost price (P) – Depreciated price (A)

= 2,50,000 − 2,02,500

= 47,500

∴ The depreciation in the price of the machine after 2 years would be 47,500.

shaalaa.com
  Is there an error in this question or solution?
Chapter 14: Compound interest - Practice Set 14.2 [Page 93]

APPEARS IN

Balbharati Mathematics [English] Standard 8 Maharashtra State Board
Chapter 14 Compound interest
Practice Set 14.2 | Q 4 | Page 93
Balbharati Mathematics Integrated [English] Standard 8 Maharashtra State Board
Chapter 14 Compound Interest
Practice Set 14.2 | Q 4. | Page 48

RELATED QUESTIONS

Compute the amount and the compound interest in  the following by using the formulae when:
Principal = Rs 3000, Rate = 5%, Time = 2 years


Compute the amount and the compound interest in  the following by using the formulae when:
Principal = Rs 2000, Rate = 4 paise per rupee per annum, Time = 3 years


The compound interest on Rs 1800 at 10% per annum for a certain period of time is Rs 378. Find the time in years.


The present population of a town is 25000. It grows at 4%, 5% and 8% during first year, second year and third year respectively. Find its population after 3 years.


The population of a town increases at the rate of 50 per thousand. Its population after 2 years will be 22050. Find its present population.


The count of bacteria in a culture grows by 10% in the first hour, decreases by 8% in the second hour and again increases by 12% in the third hour. If the count of bacteria in the sample is 13125000, what will be the count of bacteria after 3 hours?


The population of a town increases at the rate of 40 per thousand annually. If the present population be 175760, what was the population three years ago.


The value of a machine depreciates at the rate of 10% per annum. It was purchased 3 years ago. If its present value is Rs 43740, find its purchase price.


Find the amount and the compound interest on ₹ 10,000 in 3 years, if the rates of interest for the successive years are 10%, 15%, and 20% respectively.


The population of a town was decreasing every year due to migration, poverty and unemployment. The present population of the town is 6,31,680. Last year the migration was 4% and the year before last, it was 6%. What was the population two years ago?


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×