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Chemical Engineering Plant Economics

A series of equal payments (e.g., deposit or cost) made at equal intervals of time is known as

Question: A series of equal payments (e.g., deposit or cost) made at equal intervals of time is known as
[A].

perpetuity

[B].

capital charge factor

[C].

annuity

[D].

future worth

Answer: Option C

Explanation:

No answer description available for this question.

A series of equal payments (e.g., deposit or cost) made at equal intervals of time is known as Read More »

CHEMICAL ENGINEERING, Chemical Engineering Plant Economics

If an amount R is paid at the end of every year for ‘n’ years, then the net present value of the annuity at an interest rate of i is

Question: If an amount R is paid at the end of every year for ‘n’ years, then the net present value of the annuity at an interest rate of i is
[A].

[B].

[C].

R(1 + i)n

[D].

R/(1 + i)n

Answer: Option B

Explanation:

No answer description available for this question.

If an amount R is paid at the end of every year for ‘n’ years, then the net present value of the annuity at an interest rate of i is Read More »

CHEMICAL ENGINEERING, Chemical Engineering Plant Economics

A present sum of Rs. 100 at the end of one year, with half yearly rate of interest at 10%, will be Rs.

Question: A present sum of Rs. 100 at the end of one year, with half yearly rate of interest at 10%, will be Rs.
[A].

121

[B].

110

[C].

97

[D].

91

Answer: Option A

Explanation:

No answer description available for this question.

A present sum of Rs. 100 at the end of one year, with half yearly rate of interest at 10%, will be Rs. Read More »

CHEMICAL ENGINEERING, Chemical Engineering Plant Economics

‘P’ is the investment made on an equipment, ‘S’ is its salvage value and ‘n is the life of the equipment in years. The depreciation for rath year by the sum-of years digit method will be

Question: ‘P’ is the investment made on an equipment, ‘S’ is its salvage value and ‘n is the life of the equipment in years. The depreciation for rath year by the sum-of years digit method will be
[A].

[B].

[C].

[D].

Answer: Option D

Explanation:

No answer description available for this question.

‘P’ is the investment made on an equipment, ‘S’ is its salvage value and ‘n is the life of the equipment in years. The depreciation for rath year by the sum-of years digit method will be Read More »

CHEMICAL ENGINEERING, Chemical Engineering Plant Economics

An investment of Rs. 1000 is carrying an interest of 10% compounded quarterly. The value of the investment at the end of five years will be

Question: An investment of Rs. 1000 is carrying an interest of 10% compounded quarterly. The value of the investment at the end of five years will be
[A].

[B].

1000 (1 + 0.1)20

[C].

[D].

Answer: Option A

Explanation:

No answer description available for this question.

An investment of Rs. 1000 is carrying an interest of 10% compounded quarterly. The value of the investment at the end of five years will be Read More »

CHEMICAL ENGINEERING, Chemical Engineering Plant Economics

A reactor having a salvage value of Rs. 10000 is estimated to have a service life of 10 years. The annual interest rate is 10%. The original cost of the reactor was Rs. 80000. The book value of the reactor after 5 years using sinking fund depreciation method will be Rs.

Question: A reactor having a salvage value of Rs. 10000 is estimated to have a service life of 10 years. The annual interest rate is 10%. The original cost of the reactor was Rs. 80000. The book value of the reactor after 5 years using sinking fund depreciation method will be Rs.
[A].

40096

[B].

43196

[C].

53196

[D].

60196

Answer: Option D

Explanation:

No answer description available for this question.

A reactor having a salvage value of Rs. 10000 is estimated to have a service life of 10 years. The annual interest rate is 10%. The original cost of the reactor was Rs. 80000. The book value of the reactor after 5 years using sinking fund depreciation method will be Rs. Read More »

CHEMICAL ENGINEERING, Chemical Engineering Plant Economics

An investment of Rs. 100 lakhs is to be made for construction of a plant, which will take two years to start production. The annual profit from the operation of the plant is Rs. 20 lakhs. What will be the pay back time ?

Question: An investment of Rs. 100 lakhs is to be made for construction of a plant, which will take two years to start production. The annual profit from the operation of the plant is Rs. 20 lakhs. What will be the pay back time ?
[A].

5 years

[B].

7 years

[C].

12 years

[D].

10 years

Answer: Option B

Explanation:

No answer description available for this question.

An investment of Rs. 100 lakhs is to be made for construction of a plant, which will take two years to start production. The annual profit from the operation of the plant is Rs. 20 lakhs. What will be the pay back time ? Read More »

CHEMICAL ENGINEERING, Chemical Engineering Plant Economics

For a typical project, the cumulative cash flow is zero at the

Question: For a typical project, the cumulative cash flow is zero at the
[A].

end of the project life.

[B].

break even point.

[C].

start up.

[D].

end of the design stage.

Answer: Option B

Explanation:

No answer description available for this question.

For a typical project, the cumulative cash flow is zero at the Read More »

CHEMICAL ENGINEERING, Chemical Engineering Plant Economics

In a manufacturing industry, break even point occurs, when the

Question: In a manufacturing industry, break even point occurs, when the
[A].

total annual rate of production equals the assigned value.

[B].

total annual product cost equals the total annual sales.

[C].

annual profit equals the expected value.

[D].

annual sales equals the fixed cost.

Answer: Option B

Explanation:

No answer description available for this question.

In a manufacturing industry, break even point occurs, when the Read More »

CHEMICAL ENGINEERING, Chemical Engineering Plant Economics