depreciation - The reduction in value of an asset over timeA $654,000 property is depreciated for tax purposes by its owner with the straight-line depreciation
A $654,000 property is depreciated for tax purposes by its owner with the straight-line depreciation method. The value of the building, y, after x months of use is given by y = 654,000 ? 1800x dollars. After how many months will the value of the building be $409,200? We want to know x for the equation: 654000 - 1800x = 409200 To solve this equation for x, we [URL='https://www.mathcelebrity.com/1unk.php?num=654000-1800x%3D409200&pl=Solve']type it in our math engine[/URL] and we get: x = [B]136 months[/B]
A brand new car that is originally valued at $25,000 depreciates by 8% per year. What is the value of the car after 6 years? The Book Value depreciates 8% per year. We set up a depreciation equation: BV(t) = BV(0) * (1 - 0.08)^t The Book Value at time 0 BV(0) = 25,000. We want the book value at time 6. BV(6) = 25,000 * (1 - 0.08)^6 BV(6) = 25,000 * 0.92^6 BV(6) = 25,000 * 0.606355 BV(6) = [B]15,158.88[/B]
A car is purchased for $24,000 . Each year it loses 30% of its value. After how many years will the car be worth $7300 or less? (Use the calculator provided if necessary.) Write the smallest possible whole number answer. Set up the depreciation equation D(t) where t is the number of years in the life of the car: D(t) = 24,000 * (1 - 0.3)^t D(t) = 24000 * (0.7)^t The problem asks for D(t)<=7300 24000 * (0.7)^t = 7300 Divide each side by 24000 (0.7)^t = 7300/24000 (0.7)^t= 0.30416666666 Take the natural log of both sides: LN(0.7^t) = -1.190179482215518 Using the natural log identities, we have: t * LN(0.7) = -1.190179482215518 t * -0.35667494393873245= -1.190179482215518 Divide each side by -0.35667494393873245 t = 3.33687437943 [B]Rounding this up, we have t = 4[/B]
A car worth $43,000 brand new, depreciates at a rate of $2000 per year. What is the formula that describes the relationship between the value of the car (C) and the time after it has been purchased (t)? Let t be the number of years since purchase. Depreciation means the value decreases, so we have: [B]C = 43000 - 2000t[/B]
a machine has a first cost of 13000 an estimated life of 15 years and an estimated salvage value of 1000.what is the book value at the end of 9 years? Using [URL='https://www.mathcelebrity.com/depsl.php?d=&a=13000&s=1000&n=15&t=9&bv=&pl=Calculate']our straight line depreciation calculator[/URL], we get a book value at time 9, B9 of: [B]5,800[/B]
A new car worth $24,000 is depreciating in value by $3,000 per year , how many years till the cars value will be $9,000 We have a flat rate depreciation each year. Set up the function D(t) where t is the number of years of depreciation: D(t) = 24000 - 3000t The problem asks for the time (t) when D(t) = 9000. So we set D(t) = 9000 24000 - 3000 t = 9000 To solve for t, [URL='https://www.mathcelebrity.com/1unk.php?num=24000-3000t%3D9000&pl=Solve']we plug this function into our search engine[/URL] and we get: t = [B]5[/B]
A new car worth $30,000 is depreciating in value by $3,000 per year. After how many years will the cars value be $9,000 Step 1, the question asks for Book Value. Let y be the number of years since purchase. We setup an equation B(y) which is the Book Value at time y. B(y) = Sale Price - Depreciation Amount * y We're given Sale price = $30,000, depreciation amount = 3,000, and B(y) = 9000 30000 - 3000y = 9000 To solve for y, we [URL='https://www.mathcelebrity.com/1unk.php?num=30000-3000y%3D9000&pl=Solve']type this in our math engine[/URL] and we get: y = [B]7 [/B] To check our work, substitute y = 7 into B(y) B(7) = 30000 - 3000(7) B(7) = 30000 - 21000 B(7) = 9000 [MEDIA=youtube]oCpBBS7fRYs[/MEDIA]
A vehicle purchased for $25,000 depreciates at a constant rate of 5%. Determine the approximate value of the vehicle 11 years after purchase. Round to the nearest whole dollar. Depreciation at 5% means it retains 95% of the value. Set up the depreciation equation to get Book Value B(t) at time t. B(t) = $25,000 * (1 - 0.05)^t Simplifying, this is: B(t) = $25,000 * (0.95)^t The problem asks for B(11) B(11) = $25,000 * (0.95)^11 B(11) = $25,000 * 0.5688 B(11) = [B]$14,220[/B]
Free Activity Method Depreciation Calculator - Calculates the following: Depreciable Base, Depreciation per Unit, Depreciation for Period
Free Declining Balance Depreciation Calculator - Solves for Depreciation Charge, Asset Value, and Book Value using the Declining Balance Method
Free Double Declining Balance Depreciation Calculator - Calculates Depreciation and Book Value using the Double Declining Balance Depreciation Method.
Free Earnings Before Interest and Taxes (EBIT) and Net Income Calculator - Given inputs of sales, fixed costs, variable costs, depreciation, and taxes, this will determine EBIT and Net Income and Profit Margin
1. Spend 8000 on a new machine. You think it will provide after tax cash inflows of 3500 per year for the next three years. The cost of funds is 8%. Find the NPV, IRR, and MIRR. Should you buy it? 2. Let the machine in number one be Machine A. An alternative is Machine B. It costs 8000 and will provide after tax cash inflows of 5000 per year for 2 years. It has the same risk as A. Should you buy A or B? 3. Spend 100000 on Machine C. You will need 5000 more in net working capital. C is three year MACRS. The cost of funds is 8% and the tax rate is 40%. C is expected to increase revenues by 45000 and costs by 7000 for each of the next three years. You think you can sell C for 10000 at the end of the three year period. a. Find the year zero cash flow. b. Find the depreciation for each year on the machine. c. Find the depreciation tax shield for the three operating years. d. What is the projects contribution to operations each year, ignoring depreciation effects? e. What is the cash flow effect of selling the machine? f. Find the total CF for each year. g. Should you buy it?
In 2016, National Textile installed a new textile machine in one of its factories at a cost of $300,000. The machine is depreciated linearly over 10 years with a scrap value of $10,000. (a) Find an expression for the textile machines book value in the t?th year of use (0 ? t ? 10) We have a straight line depreciation. Book Value is shown on the [URL='http://www.mathcelebrity.com/depsl.php?d=&a=300000&s=10000&n=10&t=3&bv=&pl=Calculate']straight line depreciation calculator[/URL].
Free Percentage Depreciation Calculator - Solves for Book Value given a flat rate percentage depreciation per period
Free Sinking Fund Depreciation Method Calculator - Using the Sinking Fund method of Depreciation, this calculator determines the following:
* Depreciation at time t (Dt)
* Asset Value (A)
* Salvage Value (S)
* Book Value at time t (Bt)
Free Straight Line Depreciation Calculator - Solves for Depreciation Charge, Asset Value, Salvage Value, Time, N, and Book Value using the Straight Line Method.
Free Sum of the Years Digits (SOYD) Depreciation Calculator - Solves for Depreciation Charge, Asset Value, and Book Value using the Sum of the Years Digits Method
The club uses the function S(t) = -4,500t + 54,000 to determine the salvage S(t) of a fertilizer blender t years after its purchase. How long will it take the blender to depreciate completely? Complete depreciation means the salvage value is 0. So S(t) = 0. We need to find t to make S(t) = 0 -4,500t + 54,000 = 0 Subtract 54,000 from each side -4,500t = -54,000 Divide each side by -4,500 [B]t = 12[/B]
Today a car is valued at $42000. the value is expected to decrease at a rate of 8% each year. what is the value of the car expected to be 6 years from now. Depreciation at 8% per year means it retains (100% - 8%) = 92% of it's value. We set up our depreciation function D(t), where t is the number of years from right now. D(t) = $42,000(0.92)^t The problem asks for D(6): D(6) = $42,000(0.92)^6 D(6) = $42,000(0.606355) D(6) = [B]$25,466.91[/B]
Free Units of Output (Service Output) Depreciation Calculator - Given an asset value, salvage value, production units, and units per period, this calculates the depreciation per period using the units of output depreciation (service output depreciation)
You purchase a car for $23,000. The car depreciates at a rate of 15% per year. Determine the value of the car after 7 years. Round your answer to the nearest cent. Set up the Depreciation equation: D(t) = 23,000/(1.15)^t We want D(7) D(7) = 23,000/(1.15)^7 D(7) = 23,000/2.66002 D(7) = [B]8,646.55[/B]
You purchase a new car for $35,000. The value of the car depreciates at a rate of 8.5% per year. If the rate of decrease continues, what is the value of your car in 5 years? Set up the depreciation function D(t), where t is the time in years from purchase. We have: D(t) = 35,000(1 - 0.085)^t Simplified, a decrease of 8.5% means it retains 91.5% of it's value each year, so we have: D(t) = 35,000(0.915)^t The problem asks for D(5) D(5) = 35,000(0.915)^5 D(5) = 35,000(0.64136531607) D(5) = $[B]22,447.79[/B]
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