Hello to all,
i have a query abtfinding out the rate of depreciation..
rate of depreciation=1-(2.5/10)^1/10
Ans. is coming to 12.95%BUt its getting really difficult to solve above equation with d use of logarithm...If anyone can solve it using log technique..as scientific calc is not allowed in CA exam..
THIs case is taken form CA FINal study material-new course(chap-valuation of tangible fixed assets...pn.9.15)
waiting eagerly for reply..
REgards,
Manasvi
| Steps to find Depreciation rates under WDV method, with the given |
| depreciation rate under SLM and life of the asset. |
| (i) Take Depreciation Rate under SLM |
| (ii) Multiply with life of asset |
| (iii) Divide by 100 |
| (iv) Deduct from 1 |
| (v) Find the root of step (iv) to the power of step (ii). (Here the power is the given life of asset) |
| (vi) Deduct the above found result from 1 |
| (vii) Multiply the above with 100 |
| (viii) The result is the Depreciation Rate under WDV Method. |
| Example - Cost = 10 L |
| - RV = 2.5 L |
| - Life = 10 Y |
| Steps = |
| SLM Rate is (10-2.5)/10*100 = 7.5% |
| 7.5*10 = 75 |
| 75/100 = .75 |
| 1-.75 = .25 |
| root of .25 to the power of 10 (life) = .87055 |
| 1-.87055 = .12945 |
| .12945*100 = 12.945% (WDV Rate) |
| Every subsequent yr.'s depreciation will be (100-12.945)% of the preceeding year's depreciation. In this ex. 2nd yr. depn. is 112693 and 1st yr. depn. is 129450. (i.e) 112693/129450*100 = 87.055 which is nothing but (100-WDV rate). Reducing portion is WDV rate between intermediate depreciations. |
THanks Sathish,
But what i wanna know that how to find out root of 0.25 to d power of 10(either using normal calc or using log)
Waiting for ur reply.
No, it's not possible to find root of 0.25 to the power of 10 by using normal calculator. i have tried it on my own before posting this reply. Better you obtain help of how to find the solution using log from any science student. I hope the reply would have given you some idea about the relationship between SLM & WDV method.
ya...but i tried to get it solved by my sis who is a science students...many of my friends tried but noone managed to get that answer even using log...
my main concern is if in exam sum like thi is asked, then i wud b in trouble..
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