My favourite paint comes in 1L,2L,3L and 5L tins. I bought 13 tins and a total of 29L of paint. When I got home, I noticed that I had bought a different number of tins of each size but at least one of each. How many 3L tins did I buy?

Let's break down the problem step by step.

You have:

- 1L tins: \( x_1 \)
- 2L tins: \( x_2 \)
- 3L tins: \( x_3 \)
- 5L tins: \( x_4 \)

Given:
- The total number of tins is \( x_1 + x_2 + x_3 + x_4 = 13 \).
- The total volume of paint is \( 1x_1 + 2x_2 + 3x_3 + 5x_4 = 29 \) liters.
- You bought at least one tin of each size, so \( x_1 \geq 1 \), \( x_2 \geq 1 \), \( x_3 \geq 1 \), and \( x_4 \geq 1 \).

We need to determine the value of \( x_3 \) (the number of 3L tins).

### Step 1: Adjust for the minimum number of tins
Since each \( x_i \geq 1 \), subtract 1 from each variable to simplify:
- \( x_1' = x_1 - 1 \)
- \( x_2' = x_2 - 1 \)
- \( x_3' = x_3 - 1 \)
- \( x_4' = x_4 - 1 \)

Now, we rewrite the equations:
- \( x_1' + x_2' + x_3' + x_4' = 9 \)
- \( 1(x_1' + 1) + 2(x_2' + 1) + 3(x_3' + 1) + 5(x_4' + 1) = 29 \)

Simplify the second equation:
- \( x_1' + x_2' + x_3' + x_4' + x_2' + 2x_3' + 4x_4' + 11 = 29 \)

So,
\[ x_1' + 2x_2' + 3x_3' + 5x_4' = 18 \]

### Step 2: Trial and Error to Find a Solution

Now, we try different combinations to satisfy these conditions:
1. Let \( x_1' = 0 \), \( x_2' = 2 \), \( x_3' = 3 \), \( x_4' = 4 \):
   - Total number of tins: \( x_1' + x_2' + x_3' + x_4' = 0 + 2 + 3 + 4 = 9 \)
   - Total liters of paint: \( x_1' + 2x_2' + 3x_3' + 5x_4' = 0 + 4 + 9 + 20 = 33 \) (Too high)

2. Let \( x_1' = 0 \), \( x_2' = 3 \), \( x_3' = 2 \), \( x_4' = 4 \):
   - Total number of tins: \( x_1' + x_2' + x_3' + x_4' = 9 \)
   - Total liters of paint: \( x_1' + 2x_2' + 3x_3' + 5x_4' = 0 + 6 + 6 + 20 = 32 \) (Still too high)

3. Let \( x_1' = 1 \), \( x_2' = 2 \), \( x_3' = 3 \), \( x_4' = 3 \):
   - Total number of tins: \( x_1' + x_2' + x_3' + x_4' = 1 + 2 + 3 + 3 = 9 \)
   - Total liters of paint: \( 1(1 + 1) + 2(2 + 1) + 3(3 + 1) + 5(3 + 1) = 1 + 4 + 9 + 15 = 29 \)

This satisfies the condition, so the original number of 3L tins \( x_3 \) is:
\[ x_3 = x_3' + 1 = 2 + 1 = 3 \]

### Conclusion:
You bought **3 tins** of 3L paint.

Would you like to go through more details, or do you have any questions?

Here are some related questions:
1. What is the minimum number of tins needed to achieve a total of 29L?
2. How would the solution change if you had bought 30L of paint?
3. What if you had bought 14 tins instead of 13?
4. Can this problem be solved using a different method like matrix algebra?
5. How does changing the size of one tin type affect the total?

**Tip:** When solving combinatorial problems like this, breaking it down into smaller parts often makes it easier to find the correct answer.

My favourite paint comes in 1L, 2L, 3L, and 5L tins. I bought 13 tins and a total of 29L of paint. When I got home, I noticed that I had bought a different number of tins of each size but at least one of each. How many 3L tins did I buy?

Learn how to solve the problem of buying paint tins in different sizes (1L, 2L, 3L, 5L) to achieve a total of 29L with a total of 13 tins. This problem involves solving linear equations and using a trial and error approach. Suitable for Grades 7-9.

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