We have completed the concepts of Simple Interest and Compound Interest, learned more about them in our SI & CI Formula section, and practiced questions on Simple Interest, Compound Interest, Difference Between SI & CI, and Compound Growth.
In this section, we’re specifically covering Depreciation Questions, which are closely related to the concepts we have covered in those previous lessons. We will practice different types of depreciation questions covering various conditions, along with their answers and step-by-step solutions.
You can also explore our vast collection of maths learning resources:
- Maths Notes
- Maths Formula Section
- Practice Questions on All Chapters
- PYQs to get a real understanding of the types of questions asked in competitive exams.
Quick Note: Depreciation Formulas
| Term | Formula |
| Value after Depreciation | V = P × (1 – r/100)^n |
| Total Depreciation | D = P – V |
| Original Value | P = V / (1 – r/100)^n |
| Different Depreciation Rates | V = P × (1 – r1/100) × (1 – r2/100) × (1 – r3/100) |
| Overall Depreciation % | [(P – V) / P] × 100 |
| Value Remaining % | (V / P) × 100 |
| P = Original Value, V = Final/Depreciated Value, r = Rate of Depreciation, n = Number of Years. | |
You can also explore our Maths Syllabus for Competitive Exams, where we cover the Maths and Numerical Ability syllabus for various competitive exams.
And don’t forget to explore our Calculators, where you’ll find useful calculators related to different math chapters and concepts.
So, why wait? Let’s start solving Depreciation Questions with Answers and Solutions.
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Question 1: After joining the Railways as an Assistant Loco Pilot (ALP), Ravi bought a bike for ₹1,20,000. The value of the bike depreciates by 10% per annum. What will be the value of the bike after one year?
Solution:
Given:
Original value of the bike = ₹1,20,000
Rate of depreciation = 10% per annum
Time = 1 year
First, let us find the depreciation in the value of the bike after one year.
Depreciation = 10% of ₹1,20,000
= (10/100) × 1,20,000
= ₹12,000
Therefore, value of the bike after one year:
= Original value − Depreciation
= ₹1,20,000 − ₹12,000
= ₹1,08,000
Answer: The value of Ravi’s bike after one year will be ₹1,08,000.
Exam Tip: SSC, RRB, Defence, Banking, Police Exams
When an asset depreciates by 10%, its value becomes 90% of its previous value. Therefore, we can also calculate it directly:
Value after one year = ₹1,20,000 × (90/100)
= ₹1,08,000
Question 2: A piece of equipment at a Railway Coach Factory was purchased for ₹2,00,000. Its value depreciates at 10% per annum. Find its value after 2 years.
Solution:
Given:
Original value of the equipment = ₹2,00,000
Rate of depreciation = 10% per annum
Time = 2 years
Value after the first year:
Depreciation in the first year:
= 10% of ₹2,00,000
= (10/100) × ₹2,00,000
= ₹20,000
Value after the first year:
= ₹2,00,000 − ₹20,000
= ₹1,80,000
Value after the second year:
The depreciation in the second year will be calculated on the reduced value of ₹1,80,000.
Depreciation in the second year:
= 10% of ₹1,80,000
= (10/100) × ₹1,80,000
= ₹18,000
Value after the second year:
= ₹1,80,000 − ₹18,000
= ₹1,62,000
Answer: The value of the equipment after 2 years will be ₹1,62,000.
Connection with Compounding Decrease:
As we learned in Compounding Decrease, when a quantity decreases by the same percentage repeatedly, each decrease is calculated on the reduced value.
Therefore, we can also solve this directly using the compounding decrease formula:
Value after depreciation = Original Value × (1 − Rate/100)ⁿ
= ₹2,00,000 × (1 − 10/100)²
= ₹2,00,000 × (90/100)²
= ₹2,00,000 × 81/100
= ₹1,62,000
Thus, depreciation at a fixed percentage every year is an application of compounding decrease.
Exam Tip: SSC, RRB, Defence, Banking, Police Exams
For two successive depreciations of 10% each:
Net depreciation = 10 + 10 − (10 × 10)/100
= 20 − 1
= 19%
Therefore, 81% of the original value remains.
Value after 2 years:
= ₹2,00,000 × 81/100
= ₹1,62,000
Question 3: After one year, Rahul’s MacBook is valued at ₹80,000 after depreciating by 20%. What was the original price of the MacBook?
Solution:
Given:
Value of the MacBook after depreciation = ₹80,000
Rate of depreciation = 20%
Since the MacBook depreciated by 20%, its value after one year is:
= 100% − 20%
= 80% of its original value
Therefore,
80% of Original Value = ₹80,000
Original Value = ₹80,000 × (100/80)
= ₹1,00,000
Answer: The original price of Rahul’s MacBook was ₹1,00,000.
Exam Tip: SSC, RRB, Defence, Banking, Police Exams
A 20% depreciation means the MacBook retains 80% of its original value.
So, if 80% = ₹80,000,
1% = ₹80,000 ÷ 80 = ₹1,000
Therefore,
100% = ₹1,000 × 100
= ₹1,00,000
Question 4: A railway contractor purchased track-maintenance equipment for ₹5,00,000. Its value depreciates at 20% per annum. Find:
(a) its value after 3 years, and
(b) the total depreciation during these 3 years.
Solution:
Given:
Original value of the equipment = ₹5,00,000
Rate of depreciation = 20% per annum
Time = 3 years
Since the equipment depreciates by the same percentage every year, we can use the compounding decrease formula:
Value after depreciation = Original Value × (1 − Rate/100)ⁿ
Therefore,
Value after 3 years
= ₹5,00,000 × (1 − 20/100)³
= ₹5,00,000 × (80/100)³
= ₹5,00,000 × (4/5)³
= ₹5,00,000 × 64/125
= ₹2,56,000
Therefore,
(a) Value of the equipment after 3 years = ₹2,56,000
Now, total depreciation during the 3 years:
Total Depreciation = Original Value − Value after 3 years
= ₹5,00,000 − ₹2,56,000
= ₹2,44,000
Therefore,
(b) Total depreciation during the 3 years = ₹2,44,000
Answer:
(a) Value after 3 years = ₹2,56,000
(b) Total depreciation = ₹2,44,000
Shortcut for Competitive Exams:
A depreciation of 20% means 80% of the value remains each year.
So, the remaining value after 3 years is:
80% × 80% × 80%
= 0.8 × 0.8 × 0.8
= 0.512
= 51.2% of the original value
Therefore,
Value after 3 years
= 51.2% of ₹5,00,000
= ₹2,56,000
Hence, the total percentage depreciation over 3 years is:
= 100% − 51.2%
= 48.8%
Total depreciation
= 48.8% of ₹5,00,000
= ₹2,44,000
Question 5: An SSC coaching centre purchased computers worth ₹4,00,000. After 2 years, their total value decreased to ₹3,24,000. If the computers depreciate at the same rate every year, find the annual rate of depreciation.
Solution:
Given:
Original value of the computers = ₹4,00,000
Value after 2 years = ₹3,24,000
Time = 2 years
Let the annual rate of depreciation be r%.
Using the compounding decrease formula:
Value after depreciation = Original Value × (1 − r/100)ⁿ
Therefore,
₹3,24,000 = ₹4,00,000 × (1 − r/100)²
Dividing both sides by ₹4,00,000:
3,24,000/4,00,000 = (1 − r/100)²
81/100 = (1 − r/100)²
Taking the square root on both sides:
9/10 = 1 − r/100
Therefore,
r/100 = 1 − 9/10
= 1/10
r = 10%
Answer: The computers depreciate at 10% per annum.
Shortcut for Competitive Exams:
The value of the computers decreased from ₹4,00,000 to ₹3,24,000 in 2 years.
So, the fraction of value remaining after 2 years is:
= ₹3,24,000 / ₹4,00,000
= 81/100
Now, the same depreciation rate is applied in both years.
Therefore, if the fraction of value remaining after each year is x, then:
x × x = 81/100
x² = 81/100
x = √(81/100)
= 9/10
This means the computers retain 9/10 or 90% of their value each year.
Therefore:
Annual depreciation rate
= 100% − 90%
= 10%
Answer: The annual rate of depreciation is 10%.
Question 6: A Railway Section Controller bought a car for ₹8,00,000 after saving from his first year of salary. The car depreciated by 15% in the first year and 10% in the second year. What was the value of the car at the end of the second year?
Solution:
Given:
Original value of the car = ₹8,00,000
Depreciation in the first year = 15%
Depreciation in the second year = 10%
Since the rate of depreciation is different in each year, we calculate the reduced value successively.
Value after the first year:
After a depreciation of 15%, the car retains:
= 100% − 15%
= 85% of its value
Therefore,
Value after the first year
= ₹8,00,000 × 85/100
= ₹6,80,000
Value after the second year:
In the second year, the car depreciates by 10% on its reduced value of ₹6,80,000.
Therefore, it retains:
= 100% − 10%
= 90% of ₹6,80,000
Value after the second year
= ₹6,80,000 × 90/100
= ₹6,12,000
Answer: The value of the car at the end of the second year will be ₹6,12,000.
Shortcut for Competitive Exams:
For two successive decreases of 15% and 10%, the net percentage decrease is:
Net decrease = a + b − (ab/100)
= 15 + 10 − (15 × 10)/100
= 25 − 1.5
= 23.5%
Therefore, the car retains:
= 100% − 23.5%
= 76.5% of its original value
Value after 2 years:
= ₹8,00,000 × 76.5/100
= ₹6,12,000
Question 7: A railway signalling system worth ₹6,25,000 depreciates at 20% per annum. After how many years will its value become ₹3,20,000?
Solution:
Given:
Original value of the signalling system = ₹6,25,000
Value after depreciation = ₹3,20,000
Rate of depreciation = 20% per annum
Let the required time be n years.
Using the compounding decrease formula:
Value after depreciation = Original Value × (1 − Rate/100)ⁿ
Therefore,
₹3,20,000 = ₹6,25,000 × (1 − 20/100)ⁿ
₹3,20,000 = ₹6,25,000 × (80/100)ⁿ
₹3,20,000/₹6,25,000 = (4/5)ⁿ
Simplifying the fraction:
64/125 = (4/5)ⁿ
We know:
64/125 = 4³/5³
Therefore,
(4/5)³ = (4/5)ⁿ
Hence,
n = 3
Answer: The value of the railway signalling system will become ₹3,20,000 after 3 years.
Shortcut for Competitive Exams:
A depreciation of 20% means the asset retains 80%, or 4/5, of its value each year.
Starting value = ₹6,25,000
After the first year:
₹6,25,000 × 4/5 = ₹5,00,000
After the second year:
₹5,00,000 × 4/5 = ₹4,00,000
After the third year:
₹4,00,000 × 4/5 = ₹3,20,000
Therefore, the required time is 3 years.
Question 8: A railway track-maintenance vehicle was purchased for ₹10,00,000. Its value depreciated by 20% during the first year.
At the beginning of the second year, answer the following:
(a) What is the value of the vehicle after the first year’s depreciation?
(b) By what percentage must its depreciated value increase to become equal to its original purchase value of ₹10,00,000?
Suppose its value increases by 25% during the second year and then depreciates by 10% during the third year.
Find:
(c) its value at the end of the third year, and
(d) the overall percentage increase or decrease in its value compared with the original purchase value.
Solution:
Original value of the vehicle = ₹10,00,000
Depreciation during the first year = 20%
(a) Value after the first year
Depreciation:
= 20% of ₹10,00,000
= (20/100) × ₹10,00,000
= ₹2,00,000
Therefore,
Value after the first year:
= ₹10,00,000 − ₹2,00,000
= ₹8,00,000
Answer (a): ₹8,00,000
(b) Percentage increase required to return to the original value
Current value = ₹8,00,000
Original value = ₹10,00,000
Increase required:
= ₹10,00,000 − ₹8,00,000
= ₹2,00,000
The percentage increase must be calculated on the current value of ₹8,00,000, not on the original value.
Required percentage increase:
= (₹2,00,000 / ₹8,00,000) × 100
= 25%
Answer (b): The vehicle’s value must increase by 25%.
Remember: A 20% decrease requires a 25% increase to return to the original value. A 20% decrease followed by a 20% increase would not bring the value back to its starting point because the percentage bases are different.
(c) Value at the end of the third year
At the end of the first year, the vehicle is worth:
= ₹8,00,000
During the second year, its value increases by 25%.
Increase:
= 25% of ₹8,00,000
= ₹2,00,000
Value after the second year:
= ₹8,00,000 + ₹2,00,000
= ₹10,00,000
Notice that the 25% increase has restored the vehicle’s value to its original value.
During the third year, the value depreciates by 10%.
Third-year depreciation:
= 10% of ₹10,00,000
= ₹1,00,000
Therefore,
Value at the end of the third year:
= ₹10,00,000 − ₹1,00,000
= ₹9,00,000
Answer (c): ₹9,00,000
(d) Overall percentage change
Original value = ₹10,00,000
Final value = ₹9,00,000
Overall decrease:
= ₹10,00,000 − ₹9,00,000
= ₹1,00,000
Overall percentage decrease:
= (₹1,00,000 / ₹10,00,000) × 100
= 10%
Answer (d): The vehicle’s value decreased by 10% overall.
Shortcut for Competitive Exams
Follow the percentage changes directly:
Original value = 100%
After 20% depreciation:
= 100% × 80/100
= 80%
A 25% increase on this reduced value gives:
= 80% × 125/100
= 100%
So, the 20% depreciation followed by a 25% appreciation brings the value exactly back to its original value.
Now apply the third-year depreciation of 10%:
= 100% × 90/100
= 90%
Therefore, the final value is 90% of ₹10,00,000:
= ₹10,00,000 × 90/100
= ₹9,00,000
Hence, the overall depreciation is:
= 100% − 90%
= 10%
Final Answers:
(a) ₹8,00,000
(b) 25% increase
(c) ₹9,00,000
(d) 10% overall decrease
Question 9: Apsara bought a MacBook whose value depreciated by 20% in the first year, 15% in the second year and 10% in the third year. At the end of the third year, its value was ₹97,920.
What was the original price of the MacBook?
Solution:
Given:
Depreciation in the first year = 20%
Depreciation in the second year = 15%
Depreciation in the third year = 10%
Value after 3 years = ₹97,920
Let the original price of the MacBook be ₹P.
After a depreciation of 20% in the first year, 80% of the value remains.
Value after the first year:
= P × 80/100
In the second year, the value depreciates by 15%, so 85% of the first year’s value remains.
Value after the second year:
= P × 80/100 × 85/100
In the third year, the value depreciates by 10%, so 90% of the second year’s value remains.
Therefore, value after the third year:
= P × 80/100 × 85/100 × 90/100
Given that the final value is ₹97,920:
₹97,920 = P × 80/100 × 85/100 × 90/100
Simplifying:
₹97,920 = P × 0.8 × 0.85 × 0.9
₹97,920 = P × 0.612
Therefore:
P = ₹97,920 / 0.612
P = ₹1,60,000
Answer: The original price of Apsara’s MacBook was ₹1,60,000.
Shortcut for Competitive Exams
Instead of using decimals, convert the percentages into simple fractions of the value remaining.
After 20% depreciation → 80% remains = 4/5
After 15% depreciation → 85% remains = 17/20
After 10% depreciation → 90% remains = 9/10
Therefore,
Final Value = Original Value × 4/5 × 17/20 × 9/10
Working backwards,
Original Value
= ₹97,920 × 5/4 × 20/17 × 10/9
Cancel and simplify:
= ₹1,60,000
Answer: ₹1,60,000
Remember: When depreciation rates are different in different years, apply each percentage successively to the reduced value. Do not add the depreciation rates directly.
20% + 15% + 10% = 45%
The percentage of the original value remaining after 3 years is:
= 80% × 85% × 90%
= 61.2%
Therefore, the overall depreciation is:
= 100% − 61.2%
= 38.8%
Question 10: Rahul and Apsara each bought a MacBook for ₹1,50,000.
Rahul’s MacBook depreciates at a constant rate of 20% per annum for 3 years.
Apsara’s MacBook depreciates by 10% in the first year, 20% in the second year and 30% in the third year.
Find:
(a) the value of Rahul’s MacBook after 3 years,
(b) the value of Apsara’s MacBook after 3 years,
(c) whose MacBook depreciated more, and
(d) the difference between their values after 3 years.
Solution:
Original price of each MacBook = ₹1,50,000
(a) Value of Rahul’s MacBook after 3 years
Rahul’s MacBook depreciates by 20% every year.
Therefore, 80% of its value remains after each year.
Using the compounding decrease formula:
Value after 3 years
= ₹1,50,000 × (80/100)³
= ₹1,50,000 × 0.512
= ₹76,800
Answer (a): Rahul’s MacBook is worth ₹76,800 after 3 years.
(b) Value of Apsara’s MacBook after 3 years
Apsara’s MacBook depreciates at different rates in each of the three years.
After 10% depreciation, 90% remains.
After 20% depreciation, 80% remains.
After 30% depreciation, 70% remains.
Therefore:
Value after 3 years
= ₹1,50,000 × 90/100 × 80/100 × 70/100
= ₹1,50,000 × 0.9 × 0.8 × 0.7
= ₹1,50,000 × 0.504
= ₹75,600
Answer (b): Apsara’s MacBook is worth ₹75,600 after 3 years.
(c) Whose MacBook depreciated more?
Rahul’s total depreciation:
= ₹1,50,000 − ₹76,800
= ₹73,200
Apsara’s total depreciation:
= ₹1,50,000 − ₹75,600
= ₹74,400
Therefore, Apsara’s MacBook depreciated more.
(d) Difference between their values after 3 years
Difference:
= ₹76,800 − ₹75,600
= ₹1,200
Answer (d): The difference between their values after 3 years is ₹1,200.
Shortcut for Competitive Exams
Since both MacBooks originally cost the same amount, we can compare their percentage of value remaining without first calculating their actual prices.
Rahul’s remaining value:
= 80% × 80% × 80%
= 51.2%
Apsara’s remaining value:
= 90% × 80% × 70%
= 50.4%
Difference in remaining value:
= 51.2% − 50.4%
= 0.8%
Therefore, the difference in their values is simply:
= 0.8% of ₹1,50,000
= (0.8/100) × ₹1,50,000
= ₹1,200
Since Apsara’s MacBook retains a smaller percentage of its original value, Apsara’s MacBook has depreciated more.
Final Answers
(a) Rahul’s MacBook after 3 years = ₹76,800
(b) Apsara’s MacBook after 3 years = ₹75,600
(c) Apsara’s MacBook depreciated more
(d) Difference between their values = ₹1,200
What We Learned in This Section
- Calculating value after depreciation.
- Finding the original value and depreciation rate.
- Calculating depreciation over multiple years.
- Solving questions with different depreciation rates.
- Finding overall depreciation and remaining value.
- Calculating the increase required to recover from depreciation.
Related Chapters:
- Profit, Loss & Discount
- Ratio & Proportion
- Percentage
- Successive Percentage Questions with Answers and Solutions
- Percentage Decrease Questions with Answers and Solutions
Resources Related to Depreciation
- Simple Interest & Compound Interest
- Simple Interest Questions with Answers and Solutions
- Simple Interest & Compound Interest Formula with Examples and Explanations
- Difference Between Simple Interest and Compound Interest Questions with Answers and Solutions
- Compound Interest Questions with Answers and Solutions (15 Solved)
- Compound Growth Questions with Answers and Solutions
- Simple Interest & Compound Interest Practice Questions
- Simple Interest & Compound Interest PYQs SSC, RRB, Banking & Defence Exams With Solution.