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D & E Test #1
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Question 1 of 6
1. Question
1 pointsCategory: EASYA person wishes to have a future sum of Rs. 1,00,000 for his son’s education after 10 years from now. What is the singlepayment that he should deposit now so that he gets the desired amount after 10 years? The bank gives 15% interest rate compounded annually.
Correct
F = Rs. 1,00,000 , i = 15%, compounded annually , n = 10 years
P = F/(1 + i)^n = F(P/F, i, n)
= 1,00,000 (P/F, 15%, 10)
= 1,00,000 × 0.2472
= Rs. 24,720
The person has to invest Rs. 24,720 now so that he will get a sum of Rs. 1,00,000 after 10 years at 15% interest rate compounded annually.Incorrect
F = Rs. 1,00,000 , i = 15%, compounded annually , n = 10 years
P = F/(1 + i)^n = F(P/F, i, n)
= 1,00,000 (P/F, 15%, 10)
= 1,00,000 × 0.2472
= Rs. 24,720
The person has to invest Rs. 24,720 now so that he will get a sum of Rs. 1,00,000 after 10 years at 15% interest rate compounded annually. 
Question 2 of 6
2. Question
1 pointsCategory: EASYA person deposits a sum of Rs. 20,000 at the interest rate of 18%compounded annually for 10 years. Find the maturity value after 10 years.
Correct
P = Rs. 20,000 , i = 18% compounded annually , n = 10 years
F =P(1 + i)^n= P(F/P, i, n) = 20,000 (F/P, 18%, 10)
= 20,000 × 5.234 = Rs. 1,04,680
The maturity value of Rs. 20,000 invested now at 18% compounded yearly is equal to Rs. 1,04,680 after 10 years.
Incorrect
P = Rs. 20,000 , i = 18% compounded annually , n = 10 years
F =P(1 + i)^n= P(F/P, i, n) = 20,000 (F/P, 18%, 10)
= 20,000 × 5.234 = Rs. 1,04,680
The maturity value of Rs. 20,000 invested now at 18% compounded yearly is equal to Rs. 1,04,680 after 10 years.

Question 3 of 6
3. Question
1 pointsCategory: EASYA person who is now 35 years old is planning for his retired life. He plans to invest an equal sum of Rs. 10,000 at the end of every year for the next 25 years starting from the end of the next year. The bank gives 20% interest rate, compounded annually. Find the maturity value of his account when he is 60 years old.
Correct
A = Rs. 10,000 , n = 25 years , i = 20% , F = ?
The corresponding cash flow diagram is shownCash flow diagram of equalpayment series compound amount
The future sum of the annual equal payments after 25 years is equal to Rs. 47,19,810.
Incorrect
A = Rs. 10,000 , n = 25 years , i = 20% , F = ?
The corresponding cash flow diagram is shownCash flow diagram of equalpayment series compound amount
The future sum of the annual equal payments after 25 years is equal to Rs. 47,19,810.

Question 4 of 6
4. Question
1 pointsCategory: EASYA company has to replace a present facility after 15 years at an outlay of Rs.5,00,000. It plans to deposit an equal amount at the end of every year for the next 15 years at an interest rate of 18% compounded annually. Find the equivalent amount that must be deposited at the end of every year for the next 15 years.
Correct
F = Rs. 5,00,000 , n = 15 years , i = 18% , A = ? , The corresponding cash flow diagram is shown
Cash flow diagram of equalpayment series sinking fund.
The annual equal amount which must be deposited for 15 years is Rs. 8,200.
Incorrect
F = Rs. 5,00,000 , n = 15 years , i = 18% , A = ? , The corresponding cash flow diagram is shown
Cash flow diagram of equalpayment series sinking fund.
The annual equal amount which must be deposited for 15 years is Rs. 8,200.

Question 5 of 6
5. Question
1 pointsCategory: EASYA company wants to set up a reserve which will help the company to have an annual equivalent amount of Rs. 10,00,000 for the next 20 years towards its employees welfare measures. The reserve is assumed to grow at the rate of 15% annually. Find the singlepayment that must be made now as the reserve amount.
Correct
A= Rs. 10,00,000 , i = 15% , n = 20 years , P = ? The corresponding cash flow diagram is illustrated
Cash flow diagram of equalpayment series present worth amount
= 10,00,000 × (P/A, 15%, 20)
= 10,00,000 × 6.2593
= Rs. 62,59,300Incorrect
A= Rs. 10,00,000 , i = 15% , n = 20 years , P = ? The corresponding cash flow diagram is illustrated
Cash flow diagram of equalpayment series present worth amount
= 10,00,000 × (P/A, 15%, 20)
= 10,00,000 × 6.2593
= Rs. 62,59,300 
Question 6 of 6
6. Question
1 pointsCategory: EASYAlpha Industry is planning to expand its production operation. It has identified three different technologies for meeting the goal. The initial outlay and annual revenues with respect to each of the technologies are summarized in Table Suggest the best technology which is to be implemented based on the present worth method of comparison assuming 20% interest rate, compounded annually.
Correct
In all the technologies, the initial outlay is assigned a negative sign and the annual revenues are assigned a positive sign.
TECHNOLOGY 1 Initial outlay, P = Rs. 12,00,000 Annual revenue, A = Rs. 4,00,000 , Interest rate, i = 20%, compounded annually Life of this technology, n = 10 years The cash flow diagram of this technology is as shown
The present worth expression for this technology is PW(20%)1=–12,00,000 + 4,00,000 × (P/A, 20%, 10)
= –12,00,000 + 4,00,000 × (4.1925)
= –12,00,000 + 16,77,000
= Rs. 4,77,000TECHNOLOGY 2 Initial outlay, P = Rs. 20,00,000 Annual revenue, A = Rs. 6,00,000 Interest rate, i = 20%, compounded annually Life of this technology, n = 10 years The cash flow diagram of this technology is shown
The present worth expression for this technology i PW(20%)2 = – 20,00,000 + 6,00,000 × (P/A, 20%, 10) = – 20,00,000 + 6,00,000 × (4.1925) = – 20,00,000 + 25,15,500 =Rs. 5,15,500
TECHNOLOGY 3 Initial outlay, P = Rs. 18,00,000 Annual revenue, A = Rs. 5,00,000 Interest rate, i = 20%, compounded annually Life of this technology, n = 10 years The cash flow diagram of this technology is shown
The present worth expression for this technology is PW (20%)3 = –18,00,000 + 5,00,000 × (P/A, 20%, 10)
= –18,00,000 + 5,00,000 × (4.1925)
= –18,00,000 + 20,96,250
= Rs. 2,96,250From the above calculations, it is clear that the present worth of technology 2 is the highest among all the technologies. Therefore, technology 2 is suggested for implementation to expand the production
Incorrect
In all the technologies, the initial outlay is assigned a negative sign and the annual revenues are assigned a positive sign.
TECHNOLOGY 1 Initial outlay, P = Rs. 12,00,000 Annual revenue, A = Rs. 4,00,000 , Interest rate, i = 20%, compounded annually Life of this technology, n = 10 years The cash flow diagram of this technology is as shown
The present worth expression for this technology is PW(20%)1=–12,00,000 + 4,00,000 × (P/A, 20%, 10)
= –12,00,000 + 4,00,000 × (4.1925)
= –12,00,000 + 16,77,000
= Rs. 4,77,000TECHNOLOGY 2 Initial outlay, P = Rs. 20,00,000 Annual revenue, A = Rs. 6,00,000 Interest rate, i = 20%, compounded annually Life of this technology, n = 10 years The cash flow diagram of this technology is shown
The present worth expression for this technology i PW(20%)2 = – 20,00,000 + 6,00,000 × (P/A, 20%, 10) = – 20,00,000 + 6,00,000 × (4.1925) = – 20,00,000 + 25,15,500 =Rs. 5,15,500
TECHNOLOGY 3 Initial outlay, P = Rs. 18,00,000 Annual revenue, A = Rs. 5,00,000 Interest rate, i = 20%, compounded annually Life of this technology, n = 10 years The cash flow diagram of this technology is shown
The present worth expression for this technology is PW (20%)3 = –18,00,000 + 5,00,000 × (P/A, 20%, 10)
= –18,00,000 + 5,00,000 × (4.1925)
= –18,00,000 + 20,96,250
= Rs. 2,96,250From the above calculations, it is clear that the present worth of technology 2 is the highest among all the technologies. Therefore, technology 2 is suggested for implementation to expand the production