3. In how many different ways can R20 be obtained from R5; and 20c coins?

We want to obtain R20 using only R5 coins and 20¢ coins. To set up the problem, let: - \( x \) be the number of R5 coins. - \( y \) be the number of 20¢ coins. Since R20 means 20 rupees, and knowing that 1 rupee = 100 cents, we have: - R5 = \( 5 \text{ rupees} = 500 \text{ cents} \). - 20¢ = \( 20 \text{ cents} \). - R20 = \( 2000 \text{ cents} \). The total value, in cents, can be written as: \[ 500x + 20y = 2000 \] ### Step 1. Simplify the Equation Divide the entire equation by 10: \[ 50x + 2y = 200 \] Then divide by 2: \[ 25x + y = 100 \] ### Step 2. Express \( y \) in Terms of \( x \) Solve for \( y \): \[ y = 100 - 25x \] ### Step 3. Find the Possible Values for \( x \) Since \( y \) must be a nonnegative integer: \[ 100 - 25x \geq 0 \quad \Rightarrow \quad 25x \leq 100 \quad \Rightarrow \quad x \leq 4 \] Also, \( x \) must be nonnegative, so \( x \geq 0 \). Thus, \( x \) can take the values: \[ x = 0, 1, 2, 3, 4 \] ### Step 4. Count the Number of Solutions For each valid value of \( x \), there is a corresponding value of \( y \): - If \( x = 0 \), then \( y = 100 - 25(0) = 100 \). - If \( x = 1 \), then \( y = 100 - 25(1) = 75 \). - If \( x = 2 \), then \( y = 100 - 25(2) = 50 \). - If \( x = 3 \), then \( y = 100 - 25(3) = 25 \). - If \( x = 4 \), then \( y = 100 - 25(4) = 0 \). There are 5 possible pairs \((x, y)\). ### Final Answer There are \( \boxed{5} \) different ways to obtain R20 from R5 coins and 20¢ coins.