Coin Denominations and Ratio Problem

This problem involves finding the minimum amount of money a person can have, given the ratio of the products of the number of coins of different denominations. The denominations are Rs. 10, Rs. 5, and Rs. 2.

Setting up the Ratios

Let the number of coins of Rs. 10, Rs. 5, and Rs. 2 denominations be \(n_{10}\), \(n_5\), and \(n_2\), respectively. These must be non-negative integers. Since we are looking for a minimum amount, we assume the number of coins is positive.

The ratio of the products is given as:

\(n_{10} n_5 : n_5 n_2 : n_2 n_{10} = 3 : 4 : 2 \)

Finding the Relationship between Coin Counts

From the given ratio, we can establish relationships between the individual coin counts. Consider the ratios of consecutive terms:

• The ratio of the first two products:

  \(\frac{n_{10} n_5}{n_5 n_2} = \frac{3}{4}\)

  Assuming \(n_5 \neq 0\), this simplifies to:

  \(\frac{n_{10}}{n_2} = \frac{3}{4}\)

• The ratio of the second and third products:

  \(\frac{n_5 n_2}{n_2 n_{10}} = \frac{4}{2}\)

  Assuming \(n_2 \neq 0\), this simplifies to:

  \(\frac{n_5}{n_{10}} = 2\)

• The ratio of the third and first products:

  \(\frac{n_2 n_{10}}{n_{10} n_5} = \frac{2}{3}\)

  Assuming \(n_{10} \neq 0\), this simplifies to:

  \(\frac{n_2}{n_5} = \frac{2}{3}\)

We now have relationships:

\(\frac{n_{10}}{n_2} = \frac{3}{4}\)

\(\frac{n_5}{n_{10}} = 2\)

\(\frac{n_2}{n_5} = \frac{2}{3}\)

Determining the Minimum Number of Coins

Let's express the number of coins in terms of a common variable. From \(\frac{n_5}{n_{10}} = 2\), we get \(n_5 = 2 n_{10}\).

From \(\frac{n_{10}}{n_2} = \frac{3}{4}\), we get \(n_2 = \frac{4}{3} n_{10}\).

Let's check consistency with the third ratio: \(\frac{n_2}{n_5} = \frac{\frac{4}{3} n_{10}}{2 n_{10}} = \frac{4/3}{2} = \frac{4}{6} = \frac{2}{3}\). This is consistent.

So, the number of coins must be in the proportions \(n_{10} : n_5 : n_2 = n_{10} : 2 n_{10} : \frac{4}{3} n_{10}\).

To get integer values for the number of coins, \(n_{10}\) must be a multiple of 3. Let \(n_{10} = 3m\) for some positive integer \(m\) (since we need coins).

Then, the number of coins are:

• \(n_{10} = 3m\)
• \(n_5 = 2 \times (3m) = 6m\)
• \(n_2 = \frac{4}{3} \times (3m) = 4m\)

The coin counts are \(3m\), \(6m\), and \(4m\). For these to be the minimum number of coins (greater than zero), the smallest positive integer value for \(m\) is 1.

When \(m=1\), the number of coins are:

• Number of Rs. 10 coins \(n_{10} = 3 \times 1 = 3\)
• Number of Rs. 5 coins \(n_5 = 6 \times 1 = 6\)
• Number of Rs. 2 coins \(n_2 = 4 \times 1 = 4\)

Let's verify the product ratios with these minimum counts:

• \(n_{10} n_5 = 3 \times 6 = 18\)
• \(n_5 n_2 = 6 \times 4 = 24\)
• \(n_2 n_{10} = 4 \times 3 = 12\)

The ratio \(18 : 24 : 12\) simplifies to \(3 : 4 : 2\) by dividing by 6. This matches the given ratio.

Calculating the Minimum Total Amount

The total amount of money is calculated by summing the value of coins of each denomination:

Total Amount = \((\text{Number of Rs. 10 coins} \times 10) + (\text{Number of Rs. 5 coins} \times 5) + (\text{Number of Rs. 2 coins} \times 2)\)

Using the minimum number of coins (\(n_{10}=3, n_5=6, n_2=4\)):

Total Amount = \((3 \times 10) + (6 \times 5) + (4 \times 2)\)

Total Amount = \(30 + 30 + 8\)

Total Amount = \(68\)

The minimum amount of money this person could have is Rs. 68.

Revision Table: Coin Problem Concepts

Additional Information: Ratio and Proportion Basics

A ratio is a comparison of two quantities. For example, the ratio \(a:b\) means the quantity \(a\) compared to the quantity \(b\). Ratios can be written as fractions (\(\frac{a}{b}\)). A proportion is an equality between two ratios.

In this problem, we were given ratios of products of quantities. By setting up proportions using these product ratios, we were able to determine the ratios between the individual quantities (the number of coins). This allowed us to express the number of coins as multiples of a common factor (\(m\)) and find the smallest integer values.

When dealing with problems involving ratios and minimum values for integer quantities, it is often helpful to represent the quantities as multiples of a variable and find the smallest integer value of that variable that satisfies the conditions.