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CAIIB - ABM- CASE STUDIES / NUMERICAL QUESTIONS

The amount of term loan installment is Rs 10000/- per month, monthly average interest on TL is Rs 5000/-. If the amount of depreciation is Rs 30000/- p.a and PAT is Rs 270000/-. What would be the DSCR?

a. 1.75
b. 2
c. 1.65
d. 1.33

Ans - b

Let me Explain

Since DSCR = (interest + PAT + Depriciation) / (interest + instalment of TL)
Hence (5000×12 + 270000 + 30000)/(5000×12 + 10000×12)
i.e 360000/18000
i.e 2
.............................................

The amount of term loan instalment is Rs 15000/- per month, Monthly average interest on TL is Rs 10000/-. If the amount of depreciation is Rs 30000/- p.a and PAT is Rs 300000/-. What would be the DSCR?

a. 1
b. 1.5
c. 2
d. 2.5

Ans - c

Let me Explain

Since DSCR = (interest + PAT + Depriciation) / ( interest + instalment of TL )
= (10000×12 + 300000 + 30000)/(10000×12 + 15000×12)
= (120000 + 330000) / (120000 + 180000)
= 450000/300000
= 1.5
.............................................

The amount of term loan installment is Rs 15000/- per month, monthly average interest on TL is Rs 7500/-. If the amount of depreciation is Rs 100000/- p.a and PAT is Rs 350000/-. What would be the DSCR?

a. 1.75
b. 2
c. 1.65
d. 1.33

Ans - b

Let me Explain

Since DSCR = (interest + PAT + Depriciation) / (interest + instalment of TL)
DSCR = (7500×12 + 350000 + 100000)/(7500×12 + 15000×12)
= (90000 + 350000 + 100000) / (90000 + 180000)
= 540000 / 270000
= 2
.............................................

The amount of term loan instalment is Rs 30000/- per month, Monthly average interest on TL is Rs 15000/-. If the amount of depreciation is Rs 100000/- p.a and PAT is Rs 800000/-. What would be the DSCR?

a. 1
b. 1.5
c. 2
d. 2.5

Ans - c

Let me Explain

Since DSCR = (interest + PAT + Depriciation) / (interest + instalment of TL)
= (15000×12 + 800000 + 100000)/(15000×12 + 30000×12)
= (180000 + 800000 + 100000) / (180000 + 360000)
= 1080000 / 540000
= 2

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