---
title: "Salt tank mixing problem"
subject: "Mathematics I"
chapter: 4
key: "semester_1/mathematics_i:app-salt-tank-5000-gm"
source: https://www.bctengineer.com/q/semester_1/mathematics_i/app-salt-tank-5000-gm
site: BCT Engineering
---

# Salt tank mixing problem

Mathematics I · Chapter 4: Ordinary Differential Equations and their Applications

- Asked in: 2069 Asar
- Video: [Mixing Salt and Water - First Order Differential Equations](https://www.youtube.com/watch?v=aFfAz9wnoyY) — MasterWuMathematics
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/mathematics_i/app-salt-tank-5000-gm
- Whole chapter: https://www.bctengineer.com/md/semester_1/mathematics_i/chapter-4

## Question

A tank contains 1000 liters of fresh water. Salt water which contains 150 gms of salt per liter runs into it at the rate of 5 liter per minute and well-stirred mixture runs out of it at the same rate. When will the tank contain 5000 gms of salt?

## Short answer

Let x(t) be grams of salt. The rate equation dx/dt = 750 - x/200 with x(0)=0 solves to x=150000(1-e^(-t/200)). Setting x=5000 gives t = 200 ln(30/29) ≈ 6.78 minutes.

## Answer

Let x(t) be the amount of salt (in gm) in the tank after t minutes.

#### Rate equation
The volume stays at 1000 litres because inflow = outflow = 5 L/min.

- Salt in: 150 gm/L × 5 L/min = 750 gm/min.
- Salt out: concentration x/1000 gm/L × 5 L/min = x/200 gm/min.

dx/dt = 750 − x/200

#### Solve
dx/dt + x/200 = 750, IF = e^(t/200)

x e^(t/200) = 750 · 200 e^(t/200) + C = 150000 e^(t/200) + C

x = 150000 + C e^(−t/200)

At t = 0, x = 0 (fresh water): C = −150000.

**x(t) = 150000(1 − e^(−t/200)) gm**

#### Time to reach 5000 gm
5000 = 150000(1 − e^(−t/200))

e^(−t/200) = 1 − 1/30 = 29/30

t = 200 ln(30/29) = 200 × 0.033902

**t ≈ 6.78 minutes** (about 6 min 47 s).

(The salt content tends to 150000 gm = 1000 L × 150 gm/L, the concentration of the incoming brine.)

## Check yourself

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

1. The salt rate equation is:
   A. dx/dt = 150 - x/200
   B. dx/dt = 750 + x/200
   C. dx/dt = 750 - x/200
   D. dx/dt = 750 - x/1000
   Answer: C. In: 150 × 5 = 750; out: (x/1000) × 5. (Key point: Rate equation)

2. Why does volume stay at 1000 L?
   A. No water leaves
   B. Inflow equals outflow
   C. Salt does not take volume
   D. Tank overflows
   Answer: B. Both are 5 L/min. (Key point: Constant volume)

3. With x(0) = 0 the solution is:
   A. 150000(1 - e^(-t/200))
   B. 150000 e^(-t/200)
   C. 750(1 - e^(-t/200))
   D. 750 t
   Answer: A. x = 150000 + C e^(-t/200), C = -150000. (Key point: Solution)

4. Time for 5000 gm is about:
   A. 200 min
   B. 33.3 min
   C. 6.78 min
   D. 1.0 min
   Answer: C. t = 200 ln(30/29). (Key point: Time)

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