in the TV store, the price-quality ratio of a computer is 3/5 - the ratio of a television and a refrigerator is 5/7. how much does a refrigerator cost if the television is more expensive than a computer by 460000?

To solve this problem, we will first set up the relationships using algebraic expressions.

Let:
- \( C \) represent the price of the computer.
- \( T \) represent the price of the television.
- \( R \) represent the price of the refrigerator.

We are given:
1. The price-quality ratio of a computer is \( \frac{3}{5} \).
2. The price-quality ratio of a television and a refrigerator is \( \frac{5}{7} \).
3. The television is more expensive than a computer by 460,000 units of currency.

From the given data, we can write the equations:

\[ T = C + 460,000 \]

We need to find the price of the refrigerator \( R \). Using the given ratio relationships, we have:

\[ \frac{T}{R} = \frac{5}{7} \]

This can be rearranged to:

\[ T = \frac{5}{7} R \]

Now, substitute \( T = C + 460,000 \) into \( T = \frac{5}{7} R \):

\[ C + 460,000 = \frac{5}{7} R \]

Next, solve for \( R \):

\[ R = \frac{7}{5} (C + 460,000) \]

We don't have the actual value of \( C \) (the price of the computer), but we can express \( R \) in terms of \( C \):

\[ R = \frac{7}{5} C + \frac{7}{5} \times 460,000 \]

\[ R = \frac{7}{5} C

Learn how to solve a ratio and proportion problem involving prices of a TV, computer, and refrigerator. This problem includes algebraic expressions and ratio relationships. Find out how much a refrigerator costs given specific price-quality ratios and price differences. Suitable for Grades 9-12.

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