Question

In the Ruchika's wallet there are only Rs. 16, consisting of 10 paise, 20 paise and Re. 1 coins. The ratio of no. of coins of 10 paise and 20 paise is \(6: 1\). The minimum no. of Re 1 coin is : (a) 5 (b) 12 (c) 4 (d) 8

Short answer

Answer: 4

Step-by-step solution

Define variables

Let's define the variables for the number of coins of each denomination: - x: number of 10 paise coins - y: number of 20 paise coins - z: number of Re. 1 coins

Write equations from given information

We have the following equations from the given information: 1. Total money in the wallet: \(10x + 20y + 100z = 1600\) paise (As, Rs. 16 = 1600 paise) 2. Ratio of the number of 10 paise coins and 20 paise coins: \(\frac{x}{y} = \frac{6}{1}\), which means \(x = 6y\)

Substitute and simplify

Substitute the value of x from equation (2) into equation (1): \(10(6y) + 20y + 100z = 1600\) Now, simplify the equation: \(60y + 20y + 100z = 1600\)

Find the minimum value of z

Combine like terms in the equation: \(80y + 100z = 1600\) Now, we want to find the minimum value of z that satisfies the equation. We can try dividing the equation by 20 to simplify it further: \(4y + 5z = 80\) For z to be minimum, let's try the smallest integer value greater than or equal to 0, which is z = 1 (since z cannot be 0, as there has to be AT LEAST one Re. 1 coin). Now, we will solve for y: \(4y + 5(1) = 80\) \(4y = 75\) \(y = \frac{75}{4}\) Since y must be an integer (coins cannot be in fractions), we will try the next smallest value of z: z = 2 \(4y + 5(2) = 80\) \(4y = 70\) \(y = \frac{70}{4}\) y is still not an integer. Let's try z = 3: \(4y + 5(3) = 80\) \(4y = 65\) \(y = \frac{65}{4}\) y is still not an integer. Let's try z = 4: \(4y + 5(4) = 80\) \(4y = 60\) \(y = \frac{60}{4}\) Now, y = 15, which is an integer.

Find the corresponding value of x

Now that we have found the value of y, we can find the value of x using the ratio equation (2): \(x = 6y\) \(x = 6(15)\) \(x = 90\) The minimum number of Re. 1 coins (z) in Ruchika's wallet is 4. Therefore, the correct answer is (c) 4.

Key concepts

Understanding Ratio and Proportion

In math problems, ratios and proportions help us relate quantities to each other. A ratio is a way to compare two quantities, expressing how many times one value contains or is contained within the other. For example, if you have a ratio of 6:1, it means for every 6 units of one thing, there is 1 unit of another.

In our problem, the ratio of 10 paise coins to 20 paise coins is given as 6:1. This translates into a simple fraction form, where \[ \frac{x}{y} = \frac{6}{1} \]This tells us that the number of 10 paise coins is 6 times that of the 20 paise coins. Hence, if we define the number of 20 paise coins as y, then the number of 10 paise coins will automatically be 6y.

Using these ratios helps us set up equations that model the real-world scenarios in a manageable way. This allows us to find unknown values and solve problems systematically.

Breaking Down Linear Equations

Linear equations are a type of algebraic equation where each term is either a constant or the product of a constant and a single variable. The general form of a linear equation in one variable looks like this:\[ ax + b = c \]where a, b, and c are constants.

In the exercise, we use a linear equation to represent the total value of coins in Ruchika's wallet: \[ 10x + 20y + 100z = 1600 \]This represents, in paise, the total of 10 paise coins, 20 paise coins, and Re. 1 coins which altogether add up to Rs. 16 (or 1600 paise).

To solve the problem, we manipulated the equation by substituting our ratio-derived relationship \(x = 6y\) into the total value equation. Then, we combined terms and simplified equations step by step, illustrating the power of linear equations in solving real-world problems.

Finding Integer Solutions

In many math problems, especially those involving coins or objects, we need whole-number solutions, or integer solutions. Coins cannot be cut into fractions; thus, our answers must be whole numbers.

When we dealt with the equation \[ 4y + 5z = 80 \]we aimed to find the smallest integer value for z that still allowed y to be an integer. This process involved checking various integer values for z until finding one that satisfied both conditions.

We found that only for \( z = 4 \), with \( y = 15 \), do we produce a valid integer solution for the number of coins. Always remember when solving for integer solutions, trial and error might be necessary, where you systematically test small values to ensure all conditions are met, particularly when dealing with practical scenarios like coin counting.