Solving the Blue and Yellow Colour Ratio Problem

This problem involves understanding ratios and how they change when a whole is divided into parts. We are given the ratio of blue and yellow colour in a complete picture and the ratio in its upper half. We need to find the ratio in the lower half.

Understanding the Given Ratios

Let's break down the information provided:

• Total Picture Ratio: Blue : Yellow = 4 : 3
• Upper Half Ratio: Blue : Yellow = 2 : 3
• Goal: Find the ratio Blue : Yellow in the lower half.

We can assume that the total quantity of colour in the picture is divided equally between the upper and lower halves. Let the total quantity of blue colour be \(B_{total}\) and the total quantity of yellow colour be \(Y_{total}\).

Setting up the Problem with Variables

From the total picture ratio 4 : 3, we can write:

\(B_{total} = 4k\)

\(Y_{total} = 3k\)

for some constant \(k\). The total amount of colour in the picture is \(T = B_{total} + Y_{total} = 4k + 3k = 7k\).

Assuming the total quantity of colour is equally divided, the upper half contains \(T/2 = 7k/2\) and the lower half contains \(T/2 = 7k/2\).

In the upper half, the ratio of blue to yellow is 2 : 3. Let \(B_{upper}\) be the amount of blue and \(Y_{upper}\) be the amount of yellow in the upper half. So:

\(B_{upper} : Y_{upper} = 2 : 3\)

This means \(B_{upper} = 2m\) and \(Y_{upper} = 3m\) for some constant \(m\).

The total colour in the upper half is \(B_{upper} + Y_{upper} = 2m + 3m = 5m\).

We know the total colour in the upper half is \(7k/2\). Therefore:

\(5m = \frac{7k}{2}\)

\(m = \frac{7k}{10}\)

Now we can find the actual amounts of blue and yellow in the upper half in terms of \(k\):

\(B_{upper} = 2m = 2 \times \frac{7k}{10} = \frac{14k}{10} = \frac{7k}{5}\)

\(Y_{upper} = 3m = 3 \times \frac{7k}{10} = \frac{21k}{10}\)

Finding the Colour Amounts in the Lower Half

Let \(B_{lower}\) and \(Y_{lower}\) be the amounts of blue and yellow colour in the lower half.

The total amount of blue colour in the picture is the sum of blue colour in the upper and lower halves:

\(B_{total} = B_{upper} + B_{lower}\)

\(4k = \frac{7k}{5} + B_{lower}\)

To find \(B_{lower}\), subtract \(7k/5\) from \(4k\):

\(B_{lower} = 4k - \frac{7k}{5} = \frac{20k - 7k}{5} = \frac{13k}{5}\)

Similarly, the total amount of yellow colour is the sum of yellow colour in the upper and lower halves:

\(Y_{total} = Y_{upper} + Y_{lower}\)

\(3k = \frac{21k}{10} + Y_{lower}\)

To find \(Y_{lower}\), subtract \(21k/10\) from \(3k\):

\(Y_{lower} = 3k - \frac{21k}{10} = \frac{30k - 21k}{10} = \frac{9k}{10}\)

Calculating the Ratio in the Lower Half

The ratio of blue to yellow colour in the lower half is \(B_{lower} : Y_{lower}\).

Ratio = \(\frac{13k}{5} : \frac{9k}{10}\)

To simplify this ratio, we can divide both sides by \(k\) (assuming \(k \neq 0\)) and then multiply both sides by the least common multiple (LCM) of the denominators, which is 10.

Ratio = \(\frac{13}{5} \times 10 : \frac{9}{10} \times 10\)

Ratio = \(13 \times 2 : 9 \times 1\)

Ratio = \(26 : 9\)

Thus, the ratio of blue to yellow colour in the lower half is 26 : 9.

Conclusion

By calculating the total amounts of blue and yellow colour based on the initial ratio and then subtracting the amounts found in the upper half, we determined the quantities in the lower half. The ratio of these quantities gives us the final answer.

Revision Table: Key Ratios and Amounts

Additional Information on Ratio and Proportion

Ratios are used to compare two or more quantities. A ratio \(a:b\) can be written as a fraction \(\frac{a}{b}\). Proportions are equations stating that two ratios are equal.

Key concepts used in solving this ratio problem:

• Representing quantities using a common multiple (\(k\)).
• Understanding that the total quantity is the sum of quantities in parts.
• Simplifying ratios by finding a common factor or multiplying by a common multiple to remove fractions.

Ratio and proportion problems often involve finding unknown quantities based on given relationships between known quantities. This problem highlights how ratios can be applied to combined or divided quantities.