# Depreciation or Decay

📌 Example 1. The value of an auto mobile depreciate at the rate of 15% per year. What will be the value of an automobile 3 years hence which is now purchased for \$45,000?
✍ Solution:
a= 45,000 = purchased value of automobile
The amount depreciate at the end of 1st year =a(15/100)=0.15a
The value of automobile at the end of 1st year =a-0.15a=a(1-0.15)=a(0.85)
The value of automobile at the end of 2nd year

=a(0.85)(1-0.15)=a(0.85)(0.85)=a(0.85)2

The value of automobile at the end of 3rd year
=a(0.85)3=45,000(0.85)3=45,000 (0.614125)
=\$27635.63

📌 Ex2. (Depreciation) Tammy’s car is expected to depreciate at a rate of 15% per year. If her car is currently valued at \$24, 000, to the nearest dollar, how much will it be worth in 6 years?
✍ Solution:
Substitute 0.15, 6 and 24000 for r, t and P in P(1-r)t then evaluate.

24000(1-0.15)6=24000(0.85)6
≈9052.

The worth of the car will be about \$9052 after 6 years.

📌 Ex3. What is the value of a car after ten years, if it is initially worth \$30 000 and is depreciated at 6% p.a.?
✍ Solution:
Here P=30000, r=0.06 and n=10. Thus, after ten years, the car is worth 30000(1-0.06)10≈\$16 158.

📌 Ex4. (Accounting) Julian Rockman is an accountant for a small company. On January 1, 2014, the company purchased \$50,000 worth of computers, printers, scanners, and hardware. Because this equipment is a company asset, Mr. Rockman needs to determine how much the computer equipment is presently worth. He estimates that the computer equipment depreciates at a rate of 45% per year. What value should Mr. Rockman assign the equipment in his 2019 year-end accounting report?
✍ Solution:
The equipment is originally worth \$50,000, so a1=50,000. Because the equipment depreciates at a rate of 45% per year, the value of the equipment on a given year will be 100%-45% or 55% of the value the previous year. So, r=0.55. The first term a1 corresponds to the year 2014, so the year 2019 corresponds to a6.

Use the formula for the nth term of a geometric sequence to find the a6.

an=a1r(n-1)
a6=50,000⋅(0.55)(6-1)
a6=50,000⋅(0.55)5
a6≈2516.42.

Therefore, the value of the equipment in 2019 is about \$2516.42.

📌 Ex5. A manufacturer reckons that the value of a machine, which costs him \$15625, will depreciate each year by 20%. Find the estimated value at the end of 5 years.
✍ Solution:
Cost of machine =\$15625. Machine depreciates by 20% every year.
Therefore, its value after every year is 80% of the original cost i.e., ⅘ of the original cost.
∴ Value at the end of 5 years =15625×(⅘×⅘×⅘×⅘×⅘)=5×1024=5120
Thus, the value of the machine at the end of 5 years is \$5120.

📌 Ex6. At the end of each year the value of a certain machine has depreciated by 20% of its value at the beginning of that year. If its initial value was USD1,250, find the value at the end of 5 years.
✍ Solution:
After each year the value of the machine is 80% of its value the previous year so at the end of 5 years the machine will depreciate as many times as 5.
Hence, we have to find the 6th term of the GP. whose first term a1 is 1250 and common ratio r is .8.
Hence, value at the end 5 years =a6=a1r5=1250⋅(.8)5=409.6

📌 Ex7. (Chemistry) Radon has a half-life of about 4 days. This means that about every 4 days, half of the mass of radon decays into another element. How many grams of radon remain from an initial 60 grams after 4 weeks?
✍ Solution:
Given a0=60 g, r=50% or 0.5, n=7.

an=a0rn
a7=60(0.5)7
=0.46875.

After 4 weeks the amount of radon remain is about 0.46875 g.

📌 Ex8. (Vacuums) A vacuum claims to pick up 80% of the dirt every time it is run over the carpet. Assuming this is true, what percent of the original amount of dirt is picked up after the seventh time the vacuum is run over the carpet?
✍ Solution:
Let 100% of dirt be in the carpet.
That is, a0=1.
A vacuum claims to pick up 80% of the dirt every time it is run over the carpet. That is 20% of dirt still in the carpet.
Therefore, r=100%-80%=20% or 0.2.
The number of times (n) is 7.

an=a0rn
a7=1(0.2)7
=0.0000128.

Therefore, after seven times of cleaning, 1-0.0000128=0.9999872 or 99.99872% of the original dirt is picked up.