---
title: "Modified B/C ratio for 100,000 project"
subject: "Engineering Economics"
chapter: 3
key: "semester_6/engineering_economics:modified-bcr-100000-25000"
source: https://www.bctengineer.com/q/semester_6/engineering_economics/modified-bcr-100000-25000
site: BCT Engineering
---

# Modified B/C ratio for 100,000 project

Engineering Economics · Chapter 3: Basic Methodologies of Engineering Economic Analysis

- Asked in: 2068 Jestha, 6 marks
- Video: [Benefit cost ratio | Present Worth calculation |Engineering Economics |Board Exam Solution in NEPALI](https://www.youtube.com/watch?v=Nl1CN7e--7U) — OPengineer
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/engineering_economics/modified-bcr-100000-25000
- Whole chapter: https://www.bctengineer.com/md/semester_6/engineering_economics/chapter-3

## Question

Evaluate the modified B/C ratio for the following project: initial investment = Rs 100,000, salvage value = Rs 10,000, annual benefits = Rs 25,000, annual expenses = Rs 3,000, useful life = 10 years, MARR = 10%. (The paper refers to the data of Q4(b).)

## Short answer

Modified B/C = (PW of benefits – PW of O&M) / (I – PW of salvage) = 135,180 / 96,145 = 1.41. Since >1, the project is acceptable.

## Answer

The modified B/C ratio treats operating and maintenance cost as a deduction from benefits (net benefit) and keeps only the investment (less salvage) in the denominator:

Modified B/C = [PW(B) − PW(O&M)] / [I − PW(S)]

```text
 yr 0: -1,00,000     yr 1..10: +25,000 benefit, -3,000 O&M     yr 10: +10,000
```
Factors at 10%, 10 years: (P/A) = 6.1446, (P/F) = 0.3855

- PW(B) = 25,000 × 6.1446 = Rs 153,614
- PW(O&M) = 3,000 × 6.1446 = Rs 18,434
- PW(S) = 10,000 × 0.3855 = Rs 3,855

Modified B/C = (153,614 − 18,434) / (1,00,000 − 3,855) = 135,180 / 96,145 = **1.406**

For comparison, the conventional ratio is 153,614 / (1,00,000 + 18,434 − 3,855) = 1.341. Both are above 1.

**Answer: modified B/C = 1.41 > 1, so the project is acceptable.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What is the formula?
   A. [PW(B) - PW(O&M)] / [I - PW(S)]
   B. PW(B)/[I + PW(O&M) - PW(S)]
   C. [PW(B)+PW(O&M)]/I
   D. PW(B)/PW(O&M)
   Answer: A. O&M is netted from benefits. (Key point: Modified B/C)

2. What is PW(O&M)?
   A. 3,000 x 10
   B. 3,000 x 0.3855
   C. 3,000 x 6.1446 = 18,434
   D. 25,000 x 6.1446
   Answer: C. Uniform annual cost uses P/A at 10%, 10 years. (Key point: PW of O&M)

3. What is modified B/C?
   A. 1.34
   B. 1.41
   C. 0.71
   D. 2.00
   Answer: B. 135,180 / 96,145 = 1.406. (Key point: Result)

4. How does conventional ratio compare?
   A. Higher, 1.50
   B. Equal
   C. Below 1
   D. Lower, 1.34; both above 1
   Answer: D. Conventional = 153,614/114,579 = 1.341. (Key point: Comparison)

## Related in Engineering Economics

- [Sensitivity on investment, revenue and life; interpret](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Asensitivity-30-initial-100000-useful-10) — Engineering Economics, chapter 6
- [PW sensitivity to investment, income, life by ±25%](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Asensitivity-pw-400000-60000-25) — Engineering Economics, chapter 6
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- [Sensitivity on initial cost and life; interpret](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Asensitivity-30-initial-cost-life-25) — Engineering Economics, chapter 6

## Related in other subjects

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