A precast plant has $79,000 fixed annual cost, $29.00 variable cost per unit, and sells units at $40.00. What annual output breaks even?

Start with the thinking

• Break-even sets total revenue equal to total cost.
• The contribution margin is p − v.

Step-by-step solution

1. Balance

   $$pQ = F + vQ$$

2. Rearrange

   $$Q = F/(p - v)$$

3. Contribution margin — p − v = $40.00 − $29.00 = $11.00/unit

4. Substituting — Q = $79,000/$11.00 = 7,182 units/yr

Why the other options are there

• 1,975 units (variable cost ignored)
• 2,724 units (price ignored)

Reference: FE Reference Handbook — Engineering Economics → MARR ����Minimum acceptable/attractive rate of return

A contractor invests $182,000 in equipment that returns $37,000 per year for 10 years with no salvage. Determine the rate of return, and estimate the after-tax return at a 34% tax rate on the net income.

Start with the thinking

• The rate of return is the interest rate that makes present worth zero — solved by trial or by the calculator's IRR.
• A quick after-tax screen scales the annual return by (1 − tax rate) and re-solves.

Step-by-step solution

1. Formula

   $$0 = -P + A(P/A, i, n)$$

2. Set up

   $$(P/A, i, 10) = P/A = 182000/37000 = 4.9189$$

3. Solve for i

   $$the rate satisfying [1 - (1+i)^-10]/i = 4.9189 is i = 15.53$$

4. After-tax cash flow — A_at = A(1 − t) = $37,000(1 − 0.34) = $24,420

5. After-tax (P/A) required — 7.4529

6. Solve again — i_at ≈ 5.74%

Why the other options are there

• 103.3% (simple total return)
• 20.3% (ignored the time value of money)

Reference: FE Reference Handbook — Engineering Economics → MARR ����Minimum acceptable/attractive rate of return

A precast plant has $116,000 fixed annual cost, $27.00 variable cost per unit, and sells units at $44.00. What annual output breaks even?

Start with the thinking

• Break-even sets total revenue equal to total cost.
• The contribution margin is p − v.

Step-by-step solution

1. Balance

   $$pQ = F + vQ$$

2. Rearrange

   $$Q = F/(p - v)$$

3. Contribution margin — p − v = $44.00 − $27.00 = $17.00/unit

4. Substituting — Q = $116,000/$17.00 = 6,824 units/yr

Why the other options are there

• 2,636 units (variable cost ignored)
• 4,296 units (price ignored)

Reference: FE Reference Handbook — Engineering Economics → MARR ����Minimum acceptable/attractive rate of return

A contractor invests $50,000 in equipment that returns $11,000 per year for 10 years with no salvage. Determine the rate of return, and estimate the after-tax return at a 24% tax rate on the net income.

Start with the thinking

• The rate of return is the interest rate that makes present worth zero — solved by trial or by the calculator's IRR.
• A quick after-tax screen scales the annual return by (1 − tax rate) and re-solves.

Step-by-step solution

1. Formula

   $$0 = -P + A(P/A, i, n)$$

2. Set up

   $$(P/A, i, 10) = P/A = 50000/11000 = 4.5455$$

3. Solve for i

   $$the rate satisfying [1 - (1+i)^-10]/i = 4.5455 is i = 17.68$$

4. After-tax cash flow — A_at = A(1 − t) = $11,000(1 − 0.24) = $8,360

5. After-tax (P/A) required — 5.9809

6. Solve again — i_at ≈ 10.63%

Why the other options are there

• 120.0% (simple total return)
• 22.0% (ignored the time value of money)

Reference: FE Reference Handbook — Engineering Economics → MARR ����Minimum acceptable/attractive rate of return

A precast plant has $94,000 fixed annual cost, $15.00 variable cost per unit, and sells units at $30.00. What annual output breaks even?

Start with the thinking

• Break-even sets total revenue equal to total cost.
• The contribution margin is p − v.

Step-by-step solution

1. Balance

   $$pQ = F + vQ$$

2. Rearrange

   $$Q = F/(p - v)$$

3. Contribution margin — p − v = $30.00 − $15.00 = $15.00/unit

4. Substituting — Q = $94,000/$15.00 = 6,267 units/yr

Why the other options are there

• 3,133 units (variable cost ignored)
• 6,267 units (price ignored)

Reference: FE Reference Handbook — Engineering Economics → MARR ����Minimum acceptable/attractive rate of return

A contractor invests $108,000 in equipment that returns $23,000 per year for 10 years with no salvage. Determine the rate of return, and estimate the after-tax return at a 32% tax rate on the net income.

Start with the thinking

• The rate of return is the interest rate that makes present worth zero — solved by trial or by the calculator's IRR.
• A quick after-tax screen scales the annual return by (1 − tax rate) and re-solves.

Step-by-step solution

1. Formula

   $$0 = -P + A(P/A, i, n)$$

2. Set up

   $$(P/A, i, 10) = P/A = 108000/23000 = 4.6957$$

3. Solve for i

   $$the rate satisfying [1 - (1+i)^-10]/i = 4.6957 is i = 16.78$$

4. After-tax cash flow — A_at = A(1 − t) = $23,000(1 − 0.32) = $15,640

5. After-tax (P/A) required — 6.9054

6. Solve again — i_at ≈ 7.37%

Why the other options are there

• 113.0% (simple total return)
• 21.3% (ignored the time value of money)

Reference: FE Reference Handbook — Engineering Economics → MARR ����Minimum acceptable/attractive rate of return

A precast plant has $95,000 fixed annual cost, $15.50 variable cost per unit, and sells units at $23.00. What annual output breaks even?

Start with the thinking

• Break-even sets total revenue equal to total cost.
• The contribution margin is p − v.

Step-by-step solution

1. Balance

   $$pQ = F + vQ$$

2. Rearrange

   $$Q = F/(p - v)$$

3. Contribution margin — p − v = $23.00 − $15.50 = $7.50/unit

4. Substituting — Q = $95,000/$7.50 = 12,667 units/yr

Why the other options are there

• 4,130 units (variable cost ignored)
• 6,129 units (price ignored)

Reference: FE Reference Handbook — Engineering Economics → MARR ����Minimum acceptable/attractive rate of return

A contractor invests $185,000 in equipment that returns $37,000 per year for 5 years with no salvage. Determine the rate of return, and estimate the after-tax return at a 25% tax rate on the net income.

Start with the thinking

• The rate of return is the interest rate that makes present worth zero — solved by trial or by the calculator's IRR.
• A quick after-tax screen scales the annual return by (1 − tax rate) and re-solves.

Step-by-step solution

1. Formula

   $$0 = -P + A(P/A, i, n)$$

2. Set up

   $$(P/A, i, 5) = P/A = 185000/37000 = 5.0000$$

3. Solve for i

   $$the rate satisfying [1 - (1+i)^-5]/i = 5.0000 is i = 0.01$$

4. After-tax cash flow — A_at = A(1 − t) = $37,000(1 − 0.25) = $27,750

5. After-tax (P/A) required — 6.6667

6. Solve again — i_at ≈ 0.01%

Why the other options are there

• 0.0% (simple total return)
• 20.0% (ignored the time value of money)

Reference: FE Reference Handbook — Engineering Economics → MARR ����Minimum acceptable/attractive rate of return

A precast plant has $98,000 fixed annual cost, $15.50 variable cost per unit, and sells units at $32.50. What annual output breaks even?

Start with the thinking

• Break-even sets total revenue equal to total cost.
• The contribution margin is p − v.

Step-by-step solution

1. Balance

   $$pQ = F + vQ$$

2. Rearrange

   $$Q = F/(p - v)$$

3. Contribution margin — p − v = $32.50 − $15.50 = $17.00/unit

4. Substituting — Q = $98,000/$17.00 = 5,765 units/yr

Why the other options are there

• 3,015 units (variable cost ignored)
• 6,323 units (price ignored)

Reference: FE Reference Handbook — Engineering Economics → MARR ����Minimum acceptable/attractive rate of return

A contractor invests $114,000 in equipment that returns $31,000 per year for 10 years with no salvage. Determine the rate of return, and estimate the after-tax return at a 23% tax rate on the net income.

Start with the thinking

• The rate of return is the interest rate that makes present worth zero — solved by trial or by the calculator's IRR.
• A quick after-tax screen scales the annual return by (1 − tax rate) and re-solves.

Step-by-step solution

1. Formula

   $$0 = -P + A(P/A, i, n)$$

2. Set up

   $$(P/A, i, 10) = P/A = 114000/31000 = 3.6774$$

3. Solve for i

   $$the rate satisfying [1 - (1+i)^-10]/i = 3.6774 is i = 24.04$$

4. After-tax cash flow — A_at = A(1 − t) = $31,000(1 − 0.23) = $23,870

5. After-tax (P/A) required — 4.7759

6. Solve again — i_at ≈ 16.32%

Why the other options are there

• 171.9% (simple total return)
• 27.2% (ignored the time value of money)

Reference: FE Reference Handbook — Engineering Economics → MARR ����Minimum acceptable/attractive rate of return