Mixture Problem: Equal Removal from Two Containers

Mixture & Concentration 7th-8th Grade

Mixture Problem: Equal Removal from Two Containers

Mixture & Concentration7th-8th Grade
PROBLEM
Container A has 18.6L of water. Container B has 10.3L of water. After an equal amount of water was removed from each container, container A now has 5 times as much water as container B. How many liters of water was removed from each container?

What This Problem Teaches

  • Setting up equations when equal quantities are removed from different starting amounts
  • Understanding how ratios change when you subtract the same value from different quantities
  • Recognizing that the difference between containers stays constant during equal removal
  • Translating "times as much" language into algebraic relationships
  • Verifying solutions by checking both the arithmetic and the logical constraints

Solution: Method 1 — The Direct Ratio Approach

This is a classic "equal removal" problem where the key insight is that removing the same amount from different quantities changes their ratio. Let's work through this systematically.

Step 1 — Define the variable

Let x = the number of liters removed from each container.

Step 2 — Express the final amounts

After removing x liters from each container:

  • Container A will have: 18.6 - x liters
  • Container B will have: 10.3 - x liters

Step 3 — Set up the ratio equation

We're told that after removal, container A has 5 times as much water as container B. This gives us:

18.6 - x = 5(10.3 - x)

Step 4 — Solve the equation

Expand the right side:

18.6 - x = 51.5 - 5x

Add 5x to both sides:

18.6 - x + 5x = 51.5 - 5x + 5x
18.6 + 4x = 51.5

Subtract 18.6 from both sides:

4x = 51.5 - 18.6
4x = 32.9
x = 8.225
8.225 liters was removed from each container

Solution: Method 2 — The Constant Difference Strategy

Here's a more elegant approach that uses the fact that the difference between the containers stays the same when we remove equal amounts.

Step 1 — Find the initial difference

The difference between the containers initially is:

18.6 - 10.3 = 8.3 liters

Step 2 — Use the final ratio to find the individual amounts

After removal, if container B has y liters, then container A has 5y liters. The difference must still be 8.3 liters:

5y - y = 8.3
4y = 8.3
y = 2.075

Step 3 — Find how much was removed

Container B ends with 2.075 liters, so the amount removed was:

10.3 - 2.075 = 8.225 liters

We can verify: Container A ends with 5 × 2.075 = 10.375 liters, so 18.6 - 10.375 = 8.225 liters was also removed from A.

Verification

Let's check our answer by substituting back into the original conditions:

After removing 8.225 liters from each container:

  • Container A: 18.6 - 8.225 = 10.375 liters
  • Container B: 10.3 - 8.225 = 2.075 liters

Check the ratio:

10.375 ÷ 2.075 = 5 ✓

Perfect! Container A indeed has exactly 5 times as much water as container B.

Common Pitfalls

✗ Assuming the ratio stays the same

Students often think: "A starts with about 1.8 times as much as B, so after equal removal it should still be 1.8 times as much." This misses the key point that equal subtraction changes ratios dramatically.

✗ Setting up the wrong equation

Writing 18.6/10.3 = 5 or (18.6-x)/(10.3-x) = 1.8 instead of (18.6-x) = 5(10.3-x). The phrase "5 times as much" means multiplication, not division.

✗ Forgetting that the same amount is removed from both

Using different variables like 18.6 - a = 5(10.3 - b). The problem specifically states "equal amount," so both containers lose the same quantity.

The Pattern Behind This

This problem follows the general pattern for "equal removal with a final ratio" problems. If you start with quantities A and B, remove x from each, and end with a ratio of r:1, then:

A - x = r(B - x)

Solving this general equation:

A - x = rB - rx
A - rB = x - rx
A - rB = x(1 - r)
x = (A - rB)/(1 - r)
Quick check: For our problem, A = 18.6, B = 10.3, r = 5:
x = (18.6 - 5×10.3)/(1 - 5) = (18.6 - 51.5)/(-4) = -32.9/(-4) = 8.225 ✓

Does This Seem Reasonable?

Let's do a sanity check on our answer of 8.225 liters removed:

Boundary thinking: Container B starts with only 10.3L, so we can't remove more than that. Our answer of 8.225L leaves B with about 2L, which seems reasonable for the smaller container in a 5:1 ratio.

Also consider: we're removing more than 80% of the original water from both containers. This makes sense because we're going from a ratio of about 1.8:1 down to 5:1 — a dramatic change requires dramatic removal.

If we'd only removed a small amount, the ratio would have stayed close to the original 1.8:1. To get to 5:1, we need to remove most of the water, leaving container B nearly empty.

What If?

1
Different Final Ratio
Container A has 18.6L of water. Container B has 10.3L of water. After removing an equal amount from each container, A now has 3 times as much water as B. How much was removed?
Step 1 — Set up the equation

Let x be the amount removed. After removal: 18.6 - x = 3(10.3 - x)

Step 2 — Expand and solve

18.6 - x = 30.9 - 3x
18.6 - x + 3x = 30.9
18.6 + 2x = 30.9
2x = 12.3
x = 6.15

Step 3 — Verify

Container A: 18.6 - 6.15 = 12.45L
Container B: 10.3 - 6.15 = 4.15L
Ratio: 12.45 ÷ 4.15 = 3 ✓

Answer

6.15 liters was removed from each container.

2
Find the Ratio
Container A has 20L, Container B has 8L. After removing 3 liters from each container, how many times as much water does A have compared to B?
Step 1 — Calculate final amounts

Container A: 20 - 3 = 17 liters
Container B: 8 - 3 = 5 liters

Step 2 — Find the ratio

17 ÷ 5 = 3.4

Step 3 — Interpret

Container A has 3.4 times as much water as container B.

Note

Initially A had 20 ÷ 8 = 2.5 times as much as B. Equal removal increased the ratio from 2.5:1 to 3.4:1.

3
Unequal Removals
Container A has 18.6L, B has 10.3L. We remove 2 liters more from A than from B. After removal, A has 4 times as much as B. How much was removed from B?
Step 1 — Define variables

Let x = liters removed from B
Then x + 2 = liters removed from A

Step 2 — Set up the equation

After removal: 18.6 - (x + 2) = 4(10.3 - x)
Simplify: 16.6 - x = 4(10.3 - x)

Step 3 — Solve

16.6 - x = 41.2 - 4x
16.6 - x + 4x = 41.2
16.6 + 3x = 41.2
3x = 24.6
x = 8.2

Step 4 — Verify

Removed from B: 8.2L → B has 10.3 - 8.2 = 2.1L
Removed from A: 10.2L → A has 18.6 - 10.2 = 8.4L
Ratio: 8.4 ÷ 2.1 = 4 ✓

Answer

8.2 liters was removed from container B.

4
Adding Instead
Container A has 18.6L, B has 10.3L. After adding an equal amount to each container, A has 2 times as much as B. How much was added to each?
Step 1 — Set up the equation

Let x = amount added to each
After addition: 18.6 + x = 2(10.3 + x)

Step 2 — Expand and solve

18.6 + x = 20.6 + 2x
18.6 + x - 2x = 20.6
18.6 - x = 20.6
-x = 2
x = -2

Step 3 — Interpret the negative result

The negative answer means this scenario is impossible! Container A starts with more than twice what B has (18.6 > 2 × 10.3 = 20.6 is false, but 18.6 > 2 × 10.3 = 20.6 let me recalculate: 2 × 10.3 = 20.6, and 18.6 < 20.6).

Correct calculation

Actually: 18.6 + x = 20.6 + 2x
18.6 - 20.6 = 2x - x
-2 = x
This means we'd need to remove 2L to achieve a 2:1 ratio, not add.

Answer

Impossible as stated. A already has less than twice B's amount, so adding equal amounts would move further from a 2:1 ratio.

Frequently Asked Questions

Set up an equation using the final relationship. If x is removed from each container, express both final amounts in terms of x and the given ratio. In this problem: (18.6 - x) = 5(10.3 - x), meaning container A's remainder equals 5 times container B's remainder.
Removing equal amounts changes ratios because subtraction isn't proportional. If you start with 20 and 10 (2:1 ratio) and remove 5 from each, you get 15 and 5 (3:1 ratio). The difference between the quantities stays constant, but their ratio changes. This is why these problems work algebraically.
The difference between the containers remains constant throughout the process. Here, containers A and B start with a difference of 8.3L. After removal, A has 5 times as much as B, so if B has y liters, then A has 5y liters, and 5y - y = 4y must equal 8.3L.
NJ
Neven Jurkovic, PhD

Professor of Computer Science, Palo Alto College, Alamo Colleges District, San Antonio, TX

Developer of Algebrator

Contact

This solution was prepared with AI assistance and reviewed by Dr. Jurkovic for mathematical accuracy and pedagogical clarity.

2026-09-14