A engineering economics problem uses Uniform-series present worth. Given annual amount (A) = 48,300 $/yr; interest rate (i) = 0.0250 1/yr; years (n) = 25.0000 yr, determine the present worth (P) in $.

Start with the thinking

• The governing relation printed in this handbook section is Uniform-series present worth.
• Everything except P is given, so isolate P symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Engineering Economics items reward recognising the unknown before touching a calculator.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$P = A [ (1+i)^n - 1 ] / [ i (1+i)^n ]$$

2. Step 2 — Rearrange the relation so that P stands alone on the left-hand side.

3. Step 3 — List the givens: annual amount (A) = 48,300 $/yr, interest rate (i) = 0.0250 1/yr, years (n) = 25.0000 yr.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$P = 889897\ \text{\$}$$

6. Step 6 — Check: returning P = 889,897 $ to

   $$P = A [ (1+i)^n - 1 ] / [ i (1+i)^n ]$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 1,779,795 — kept a factor of two that cancels in the correct rearrangement.
• 444,949 — dropped that same factor in the other direction.
• 978,887 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Engineering Economics → Non-Annual Compounding

A engineering economics problem uses Uniform-series present worth. Given interest rate (i) = 0.0550 1/yr; years (n) = 7.0000 yr; present worth (P) = 316,023 $, determine the annual amount (A) in $/yr.

Start with the thinking

• The governing relation printed in this handbook section is Uniform-series present worth.
• Everything except A is given, so isolate A symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Engineering Economics items reward recognising the unknown before touching a calculator.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$P = A [ (1+i)^n - 1 ] / [ i (1+i)^n ]$$

2. Step 2 — Rearrange the relation so that A stands alone on the left-hand side.

3. Step 3 — List the givens: interest rate (i) = 0.0550 1/yr, years (n) = 7.0000 yr, present worth (P) = 316,023 $.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$A = 55609\ \text{\$/yr}$$

6. Step 6 — Check: returning A = 55,609 $/yr to

   $$P = A [ (1+i)^n - 1 ] / [ i (1+i)^n ]$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 111,218 — kept a factor of two that cancels in the correct rearrangement.
• 27,804 — dropped that same factor in the other direction.
• 61,170 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Engineering Economics → Non-Annual Compounding

A engineering economics problem uses Uniform-series present worth. Given annual amount (A) = 47,700 $/yr; interest rate (i) = 0.0750 1/yr; years (n) = 25.0000 yr, determine the present worth (P) in $.

Start with the thinking

• The governing relation printed in this handbook section is Uniform-series present worth.
• Everything except P is given, so isolate P symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Engineering Economics items reward recognising the unknown before touching a calculator.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$P = A [ (1+i)^n - 1 ] / [ i (1+i)^n ]$$

2. Step 2 — Rearrange the relation so that P stands alone on the left-hand side.

3. Step 3 — List the givens: annual amount (A) = 47,700 $/yr, interest rate (i) = 0.0750 1/yr, years (n) = 25.0000 yr.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$P = 531709\ \text{\$}$$

6. Step 6 — Check: returning P = 531,709 $ to

   $$P = A [ (1+i)^n - 1 ] / [ i (1+i)^n ]$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 1,063,419 — kept a factor of two that cancels in the correct rearrangement.
• 265,855 — dropped that same factor in the other direction.
• 584,880 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Engineering Economics → Non-Annual Compounding

A engineering economics problem uses Uniform-series present worth. Given interest rate (i) = 0.1000 1/yr; years (n) = 29.0000 yr; present worth (P) = 461,833 $, determine the annual amount (A) in $/yr.

Start with the thinking

• The governing relation printed in this handbook section is Uniform-series present worth.
• Everything except A is given, so isolate A symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Engineering Economics items reward recognising the unknown before touching a calculator.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$P = A [ (1+i)^n - 1 ] / [ i (1+i)^n ]$$

2. Step 2 — Rearrange the relation so that A stands alone on the left-hand side.

3. Step 3 — List the givens: interest rate (i) = 0.1000 1/yr, years (n) = 29.0000 yr, present worth (P) = 461,833 $.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$A = 49291\ \text{\$/yr}$$

6. Step 6 — Check: returning A = 49,291 $/yr to

   $$P = A [ (1+i)^n - 1 ] / [ i (1+i)^n ]$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 98,581 — kept a factor of two that cancels in the correct rearrangement.
• 24,645 — dropped that same factor in the other direction.
• 54,220 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Engineering Economics → Non-Annual Compounding

A engineering economics problem uses Uniform-series present worth. Given annual amount (A) = 41,800 $/yr; interest rate (i) = 0.0300 1/yr; years (n) = 17.0000 yr, determine the present worth (P) in $.

Start with the thinking

• The governing relation printed in this handbook section is Uniform-series present worth.
• Everything except P is given, so isolate P symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Engineering Economics items reward recognising the unknown before touching a calculator.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$P = A [ (1+i)^n - 1 ] / [ i (1+i)^n ]$$

2. Step 2 — Rearrange the relation so that P stands alone on the left-hand side.

3. Step 3 — List the givens: annual amount (A) = 41,800 $/yr, interest rate (i) = 0.0300 1/yr, years (n) = 17.0000 yr.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$P = 550344\ \text{\$}$$

6. Step 6 — Check: returning P = 550,344 $ to

   $$P = A [ (1+i)^n - 1 ] / [ i (1+i)^n ]$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 1,100,688 — kept a factor of two that cancels in the correct rearrangement.
• 275,172 — dropped that same factor in the other direction.
• 605,378 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Engineering Economics → Non-Annual Compounding

A engineering economics problem uses Uniform-series present worth. Given interest rate (i) = 0.0750 1/yr; years (n) = 23.0000 yr; present worth (P) = 102,243 $, determine the annual amount (A) in $/yr.

Start with the thinking

• The governing relation printed in this handbook section is Uniform-series present worth.
• Everything except A is given, so isolate A symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Engineering Economics items reward recognising the unknown before touching a calculator.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$P = A [ (1+i)^n - 1 ] / [ i (1+i)^n ]$$

2. Step 2 — Rearrange the relation so that A stands alone on the left-hand side.

3. Step 3 — List the givens: interest rate (i) = 0.0750 1/yr, years (n) = 23.0000 yr, present worth (P) = 102,243 $.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$A = 9461\ \text{\$/yr}$$

6. Step 6 — Check: returning A = 9,461 $/yr to

   $$P = A [ (1+i)^n - 1 ] / [ i (1+i)^n ]$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 18,922 — kept a factor of two that cancels in the correct rearrangement.
• 4,731 — dropped that same factor in the other direction.
• 10,407 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Engineering Economics → Non-Annual Compounding

A engineering economics problem uses Uniform-series present worth. Given annual amount (A) = 3,100 $/yr; interest rate (i) = 0.1000 1/yr; years (n) = 23.0000 yr, determine the present worth (P) in $.

Start with the thinking

• The governing relation printed in this handbook section is Uniform-series present worth.
• Everything except P is given, so isolate P symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Engineering Economics items reward recognising the unknown before touching a calculator.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$P = A [ (1+i)^n - 1 ] / [ i (1+i)^n ]$$

2. Step 2 — Rearrange the relation so that P stands alone on the left-hand side.

3. Step 3 — List the givens: annual amount (A) = 3,100 $/yr, interest rate (i) = 0.1000 1/yr, years (n) = 23.0000 yr.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$P = 27538\ \text{\$}$$

6. Step 6 — Check: returning P = 27,538 $ to

   $$P = A [ (1+i)^n - 1 ] / [ i (1+i)^n ]$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 55,076 — kept a factor of two that cancels in the correct rearrangement.
• 13,769 — dropped that same factor in the other direction.
• 30,292 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Engineering Economics → Non-Annual Compounding

A engineering economics problem uses Uniform-series present worth. Given interest rate (i) = 0.0450 1/yr; years (n) = 10.0000 yr; present worth (P) = 478,105 $, determine the annual amount (A) in $/yr.

Start with the thinking

• The governing relation printed in this handbook section is Uniform-series present worth.
• Everything except A is given, so isolate A symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Engineering Economics items reward recognising the unknown before touching a calculator.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$P = A [ (1+i)^n - 1 ] / [ i (1+i)^n ]$$

2. Step 2 — Rearrange the relation so that A stands alone on the left-hand side.

3. Step 3 — List the givens: interest rate (i) = 0.0450 1/yr, years (n) = 10.0000 yr, present worth (P) = 478,105 $.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$A = 60422\ \text{\$/yr}$$

6. Step 6 — Check: returning A = 60,422 $/yr to

   $$P = A [ (1+i)^n - 1 ] / [ i (1+i)^n ]$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 120,845 — kept a factor of two that cancels in the correct rearrangement.
• 30,211 — dropped that same factor in the other direction.
• 66,465 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Engineering Economics → Non-Annual Compounding

A engineering economics problem uses Uniform-series present worth. Given annual amount (A) = 44,500 $/yr; interest rate (i) = 0.1150 1/yr; years (n) = 11.0000 yr, determine the present worth (P) in $.

Start with the thinking

• The governing relation printed in this handbook section is Uniform-series present worth.
• Everything except P is given, so isolate P symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Engineering Economics items reward recognising the unknown before touching a calculator.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$P = A [ (1+i)^n - 1 ] / [ i (1+i)^n ]$$

2. Step 2 — Rearrange the relation so that P stands alone on the left-hand side.

3. Step 3 — List the givens: annual amount (A) = 44,500 $/yr, interest rate (i) = 0.1150 1/yr, years (n) = 11.0000 yr.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$P = 270104\ \text{\$}$$

6. Step 6 — Check: returning P = 270,104 $ to

   $$P = A [ (1+i)^n - 1 ] / [ i (1+i)^n ]$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 540,208 — kept a factor of two that cancels in the correct rearrangement.
• 135,052 — dropped that same factor in the other direction.
• 297,114 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Engineering Economics → Non-Annual Compounding

A engineering economics problem uses Uniform-series present worth. Given interest rate (i) = 0.0550 1/yr; years (n) = 16.0000 yr; present worth (P) = 259,390 $, determine the annual amount (A) in $/yr.

Start with the thinking

• The governing relation printed in this handbook section is Uniform-series present worth.
• Everything except A is given, so isolate A symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Engineering Economics items reward recognising the unknown before touching a calculator.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$P = A [ (1+i)^n - 1 ] / [ i (1+i)^n ]$$

2. Step 2 — Rearrange the relation so that A stands alone on the left-hand side.

3. Step 3 — List the givens: interest rate (i) = 0.0550 1/yr, years (n) = 16.0000 yr, present worth (P) = 259,390 $.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$A = 24793\ \text{\$/yr}$$

6. Step 6 — Check: returning A = 24,793 $/yr to

   $$P = A [ (1+i)^n - 1 ] / [ i (1+i)^n ]$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 49,586 — kept a factor of two that cancels in the correct rearrangement.
• 12,397 — dropped that same factor in the other direction.
• 27,272 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Engineering Economics → Non-Annual Compounding