Depreciation Solved Examples for CA Foundation - Part 1

Posted on - 01-03-2025

ca-foundation-accounts

Straight Line, No Salvage

Example 1:

A machine is purchased for ₹1,00,000. Its useful life is 5 years, and there is no salvage value. Compute the annual depreciation using the straight-line method.

Solution

Straight Line Method (SLM) formula: Annual Depreciation=Cost−Salvage Value\Useful Life
Cost = 1,00,000, Salvage = 0, Life = 5.

=1,00,000/05=₹20,000 per year

Depreciation Schedule Table:

Year	Opening Value (₹)	Depreciation (₹)	Closing Value (₹)
1	1,00,000	20,000	80,000
2	80,000	20,000	60,000
3	60,000	20,000	40,000
4	40,000	20,000	20,000
5	20,000	20,000	0

Straight Line with Salvage

Example 2:

A vehicle costs ₹2,50,000, salvage value is ₹50,000, and useful life is 4 years. Find the annual depreciation (SLM).

Solution

Depreciable amount = (Cost – Salvage) = (2,50,000 – 50,000) = 2,00,000.
Useful life = 4 years.
Annual depreciation = 2,00,000 / 4 = ₹50,000.

Table:

Year	Opening (₹)	Dep. (₹)	Closing (₹)
1	2,50,000	50,000	2,00,000
2	2,00,000	50,000	1,50,000
3	1,50,000	50,000	1,00,000
4	1,00,000	50,000	50,000 (salvage)

Hinglish Explanation:

Paas ek vehicle hai jiska end mein 50k salvage rahega.
Har saal hum 50k depreciate karte hain, 4 saal ke baad closing value ₹50k ban jaayega.

Reducing Balance (WDV), No Salvage

Example 3:

Machinery purchased for ₹1,00,000. Depreciation rate is 20% p.a. on Written Down Value. Compute depreciation for 3 years.

Solution

Reducing Balance ya WDV method: Har saal opening value pe 20% nikalte hain.

Calculation:

Year 1 depreciation = 1,00,000 × 20% = 20,000 → Closing = 80,000.
Year 2 opening = 80,000 → 80,000 × 20% = 16,000 → closing = 64,000.
Year 3 opening = 64,000 → 64,000 × 20% = 12,800 → closing = 51,200.

Table:

Year	Opening (₹)	Dep. @20% (₹)	Closing (₹)
1	1,00,000	20,000	80,000
2	80,000	16,000	64,000
3	64,000	12,800	51,200

Hinglish Explanation:

Pehle saal 1 lakh par 20k, agle saal 80k par 16k, phir 64k par 12.8k.
Har saal depreciation kum hota jaata hai kyunki base value reduce hoti jaati hai.

Reducing Balance with Salvage (Target)

Example 4:

An asset costs ₹2,00,000. The firm wants a salvage of ₹50,000 after 3 years. Rate is 15% on WDV. Show the depreciation for 3 years.

Solution

1. Year 1: opening 2,00,000 → depreciation 15% = 30,000 → closing = 1,70,000.

2. Year 2: opening 1,70,000 → dep 25,500 → closing 1,44,500.

3. Year 3: opening 1,44,500 → dep 21,675 (approx) → closing ~1,22,825.

But yahan salvage 50,000 mention hai. If the method is pure WDV, we do not forcibly end up with 50k. The final value might differ.

Exact:

Let’s see the final figure: 2,00,000 → 1,70,000 → 1,44,500 → (1,44,500 – 21,675) = 1,22,825.
If the question specifically needs exactly 50k salvage, we might have to recalculate rate or do a balancing figure.

Table (basic WDV at 15%):

Year	Opening	Dep @15%	Closing
1	2,00,000	30,000	1,70,000
2	1,70,000	25,500	1,44,500
3	1,44,500	21,675	1,22,825 (not 50k!)

Hinglish Explanation:

Pure WDV par exact 50k salvage achieve karna tough ho sakta hai unless hum rate adjust karein.
Basic calculation se 3rd year end me ~1,22,825 bachta hai.
(If the question was purely "Rate 15%, what's salvage after 3 years?" then yeh final answer hoga. If we want 50k guaranteed, we must solve for a different rate.)

Sum-of-the-Years’-Digits (SYD)

Example 5:

A machine cost ₹2,40,000, life = 3 years, no salvage. Compute the annual depreciation using Sum-of-the-Years’-Digits method.

Solution

SYD for 3 yrs = 1+2+3 = 6.
Depreciation factor each year:

Year 1: 3/6
Year 2: 2/6
Year 3: 1/6

1. Year 1: 2,40,000 × (3/6) = 1,20,000

2. Year 2: 2,40,000 × (2/6) = 80,000

3. Year 3: 2,40,000 × (1/6) = 40,000
Total = 2,40,000.

Table:

Year	SYD Fraction	Depreciation (₹)	Closing Value (₹)
1	3/6 = 0.5	1,20,000	1,20,000
2	2/6 = 0.3333	80,000	40,000
3	1/6 = 0.1667	40,000	0

Hinglish Explanation:

SYD method me sabse pehle “(1+2+3+…+n)” se fraction banta hai.
Yahan 3 years ke liye total 6.
Sabse upar fraction year 1 me apply hota hai, phir kam hota hai.

Units of Production

Example 6:

A machine costs ₹3,00,000 with no salvage, expected to produce 1,00,000 units in total. In Year 1, it produces 25,000 units; in Year 2, 35,000 units; in Year 3, 40,000 units. Compute depreciation for each year by units-of-production method.

Solution

Rate per unit = 3,00,000 / 1,00,000 = ₹3 per unit.
Year 1: 25,000 units × 3 = 75,000
Year 2: 35,000 units × 3 = 1,05,000
Year 3: 40,000 units × 3 = 1,20,000

Check total = 3,00,000.

Table:

Year	Units Produced	Rate/Unit	Depreciation	Closing Value (From 3,00,000)
1	25,000	₹3	₹75,000	2,25,000
2	35,000	₹3	₹1,05,000	1,20,000
3	40,000	₹3	₹1,20,000	0

Hinglish Explanation:

Is method me jitna production, utna depreciation. Total cost across entire units = 3 lakh, per unit cost 3.
Har saal ke actual units ke base par multiply karte hain.

Partial Year (Straight Line)

Example 7:

A machinery is purchased on 1st July for ₹1,20,000, with a 5-year life and no salvage. The accounting year ends on 31st March. Find the depreciation for the first year.

Solution

If full year depreciation = (1,20,000 / 5) = ₹24,000.
But asset used for 9 months (July to March).
So pro-rata: 24,000 × (9/12) = ₹18,000 for the first year.

Table:

Particulars	Calculation	Amount (₹)
Annual Depreciation (full year)	1,20,000 / 5 = 24,000	24,000
First year usage (9 months)	24,000 × 9/12	18,000
Depreciation for Year 1	--	18,000

Hinglish Explanation:

Straight line me hum normal annual figure nikalte hain, phir jitne mahine use hua, utna proportionate kar dete hain.

Half-Yearly Depreciation Convention

Example 8:

Company policy: any asset purchased in the first half (April–Sept) gets full year depreciation; purchased in the second half (Oct–Mar) gets half-year depreciation. On 1st Nov, a machine is bought for ₹1,00,000, 10% SLM. Find depreciation for that year.

Solution

1. Policy says second half purchase → half-year depreciation only.

2. Normal annual depreciation @10% of 1,00,000 = ₹10,000.

3. But half-year convention → 50% of 10,000 = ₹5,000.

Table:

Particulars	Calculation	Amount (₹)
Full-year Depreciation @10%	1,00,000 × 10%	10,000
Half-year Depreciation (policy)	10,000 × 50%	5,000

Hinglish Explanation:

Kai firms aisa rule rakhte hain ki second half me liye asset pe sirf aadha depreciation charge hoga. Is example me Nov is second half, so ₹5k.

Revision of Useful Life (Straight Line)

Example 9:

A machine cost ₹1,50,000 with original life = 5 years. After 2 years, it’s found the machine can last for 6 more years (total 8). No salvage. Show new depreciation from year 3 onward.

Solution

1. Original annual depreciation: 1,50,000 ÷ 5 = 30,000.

2. 2 years done => total depreciation 60,000. Carrying value = 90,000.

3. Now new life = 6 more years, so total = 8, but 2 used up.

4. Revised depreciation from year 3 = 90,000 ÷ 6 = ₹15,000 per year.

Table:

Period	Calculation	Depreciation or Value
Years 1–2	30,000 each year = 60,000 total	CV now 90,000
Year 3 onwards	90,000 ÷ 6 = 15,000 / year	–

Hinglish Explanation:

2 saal ho gaye, ab pata chala life lambi hai. Aage ke liye hum new life pe carry forward value ko spread kar dete hain.

Change in Salvage Value

Example 10:

An asset cost ₹2,00,000, salvage was originally ₹20,000, life 4 years (SLM). After 1 year, salvage revised to ₹10,000. Recompute depreciation from year 2 onward.

Solution

1. Original depreciation = (2,00,000 – 20,000)/4 = ₹45,000/yr.

2. 1 year done, depreciation 45,000. CV = 1,55,000.

3. New salvage = 10,000, so new depreciable = 1,55,000 – 10,000 = 1,45,000.

4. Remaining life = 3 years.

5. New annual depreciation = 1,45,000 ÷ 3 = ₹48,333 approx.

Table:

Period	Calculation	Value
Year 1 Dep	(2,00,000 – 20,000)/4 = 45,000	CV = 1,55,000
Revised salvage	10,00 (instead of 20,000)	–
Dep. from Year 2	(1,55,000 – 10,000)/3 = 48,333 approx / year	–

Hinglish Explanation:

Pehle saal me hum old salvage se depreciation kar chuke the. Baaki 3 saal ke liye salvage kam ho gaya, toh depreciation zyada ho jaayega.