---
title: "MPN per 100 ml numerical"
subject: "Water Supply Engineering"
chapter: 4
key: "semester_5/water_supply_engineering:mpn-numerical-k78"
source: https://www.bctengineer.com/q/semester_5/water_supply_engineering/mpn-numerical-k78
site: BCT Engineering
---

# MPN per 100 ml numerical

Water Supply Engineering · Chapter 4: Quality of Water

- Asked in: 2078 Kartik, 4 marks
- Video: [Most Probable Number (MPN) Method for Coliform Detection in Water and Food samples](https://www.youtube.com/watch?v=t0mU6eaFObI) — Ghadir Ali
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/water_supply_engineering/mpn-numerical-k78
- Whole chapter: https://www.bctengineer.com/md/semester_5/water_supply_engineering/chapter-4

## Question

What will be the MPN/100 ml, if 6 tubes for each diluted samples of 1.0 ml, 0.1 ml, 0.01 ml and 0.001 ml are taken and the no. of positive tubes are found to be 3, 4, 2 and 0 respectively in a multiple tube fermentation technique of coliform analysis?

## Short answer

Using Thomas' formula: MPN/100 mL = (number of positive tubes × 100) / √[(mL in negative tubes) × (mL in all tubes)]. Total positive = 9, negative volume = 3.246, all volume = 6.666, so MPN = 900 / √(3.246×6.666) = 193.5 ≈ 193 per 100 mL.

## Answer

Use Thomas' formula for the Most Probable Number:

$$
\text{MPN}/100\ \text{ml} = \frac{\text{No. of positive tubes}\times100}{\sqrt{(\text{ml of sample in negative tubes})\times(\text{ml of sample in all tubes})}}
$$

| Dilution (ml) | Tubes | Positive | Negative | ml in negative tubes | ml in all tubes |
|---|---|---|---|---|---|
| 1.0 | 6 | 3 | 3 | 3.000 | 6.000 |
| 0.1 | 6 | 4 | 2 | 0.200 | 0.600 |
| 0.01 | 6 | 2 | 4 | 0.040 | 0.060 |
| 0.001 | 6 | 0 | 6 | 0.006 | 0.006 |
| **Total** | | **9** | | **3.246** | **6.666** |

$$
\text{MPN}/100\ \text{ml} = \frac{9\times100}{\sqrt{3.246\times6.666}} = \frac{900}{4.652} = 193.5
$$

**Answer: MPN $\approx 193$ per 100 ml.**

## Check yourself

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

1. Thomas' formula uses:
   A. positive tubes x 100 / sqrt(ml in negative tubes x ml in all tubes)
   B. positive tubes / negative tubes
   C. negative tubes x 100 / sqrt(ml in positive)
   D. positive tubes x 100 x ml in all tubes
   Answer: A. This is the formula used. (Key point: Formula)

2. The total positive tubes for 3, 4, 2, 0 are:
   A. 6
   B. 9
   C. 24
   D. 12
   Answer: B. 3+4+2+0 = 9. (Key point: Total positives)

3. ml of sample in negative tubes is:
   A. 6.666
   B. 3.0
   C. 3.246
   D. 0.246
   Answer: C. 3.000+0.200+0.040+0.006 = 3.246. (Key point: Negative ml)

4. MPN per 100 ml is about:
   A. 9
   B. 650
   C. 19
   D. 193
   Answer: D. 900 / sqrt(3.246 x 6.666) = 900/4.652 = 193.5. (Key point: Result)

## Related in Water Supply Engineering

- [Multiple tube fermentation (MPN) technique](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Amtf-mpn-procedure) — Water Supply Engineering, chapter 4, asked 5×
- [Indicator organisms and membrane tube fermentation](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Aindicator-organisms-mtf) — Water Supply Engineering, chapter 4, asked 3×

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