हिंदी

A Piece of Equipment Cost a Certain Factory Rs 60,000. If It Depreciates in Value, 15% the First, 13.5% the Next Year, 12% the Third Year, and So On. What Will Be Its Value at the End of 10 Years, - Mathematics

Advertisements
Advertisements

प्रश्न

A piece of equipment cost a certain factory Rs 60,000. If it depreciates in value, 15% the first, 13.5% the next year, 12% the third year, and so on. What will be its value at the end of 10 years, all percentages applying to the original cost?

योग
Advertisements

उत्तर

In the given problem,

Cost of the equipment = Rs 600,000

It depreciates by 15% in the first year. So,

Depreciation in 1 year

= 600000 − 495000

= 105000

= 90000

It depreciates by 13.5% of the original cost in the 2 year. So,

Depreciation in 2 year  `= (13.5)/100 (600000) = 81000`

Further, it depreciates by 12% of the original cost in the 3 year. So,

Depreciation in 3 year  `= 12/100 (600000)=72000` 

 

So, the depreciation in value of the equipment forms an A.P. with first term as 90000 and common difference as −9000.

So, the total depreciation in value in 10 years can be calculated by using the formula for the sum of n terms of an A.P.

`S_n = n/2 [2a + (n-1) d]`

We get,

`S_n = 10/2 [2(90000) +(10-1)(-9000)]`

      `=10/2 [180000 + (9)(-9000)]`

      `=5(180000 - 81000)`  

     ` = 5(99000)`

       = 495000

So, the total depreciation in the value after 10 years is Rs 495000.

Therefore, the value of equipment = 600000 − 495000 = 105000 

So, the value of the equipment after 10 years is Rs 105,000.

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 5: Arithmetic Progression - Exercise 5.6 [पृष्ठ ५४]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 10
अध्याय 5 Arithmetic Progression
Exercise 5.6 | Q 68 | पृष्ठ ५४

संबंधित प्रश्न

Find the number of natural numbers between 101 and 999 which are divisible by both 2 and 5.


If the mth term of an A.P. is 1/n and the nth term is 1/m, show that the sum of mn terms is (mn + 1)


The sum of n, 2n, 3n terms of an A.P. are S1 , S2 , S3 respectively. Prove that S3 = 3(S2 – S1 )


The sum of the first p, q, r terms of an A.P. are a, b, c respectively. Show that `\frac { a }{ p } (q – r) + \frac { b }{ q } (r – p) + \frac { c }{ r } (p – q) = 0`


In an AP, given a = 7, a13 = 35, find d and S13.


In an AP given an = 4, d = 2, Sn = −14, find n and a.


The first term of an A.P. is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.


Which term of the AP 3,8, 13,18,…. Will be 55 more than its 20th term?


Find the 6th  term form the end of the AP 17, 14, 11, ……, (-40).


Find the value of x for which (x + 2), 2x, ()2x + 3) are three consecutive terms of an AP.


Which term of the AP 21, 18, 15, … is zero?


Choose the correct alternative answer for  the following question .

 If for any A.P. d = 5 then t18 – t13 = .... 


Find the sum:  1 + 3 + 5 + 7 + ... + 199 .


​The first and the last terms of an A.P. are 7 and 49 respectively. If sum of all its terms is 420, find its common difference. 


For what value of p are 2p + 1, 13, 5p − 3 are three consecutive terms of an A.P.?

 

If four numbers in A.P. are such that their sum is 50 and the greatest number is 4 times, the least, then the numbers are


If Sn denote the sum of the first terms of an A.P. If S2n = 3Sn, then S3n : Sn is equal to


Find the sum of first 20 terms of an A.P. whose first term is 3 and the last term is 57.


In an AP, if Sn = n(4n + 1), find the AP.


The ratio of the 11th term to the 18th term of an AP is 2 : 3. Find the ratio of the 5th term to the 21st term, and also the ratio of the sum of the first five terms to the sum of the first 21 terms.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×