Let's reevaluate the problem given that the price of one pen is known to be ₹ 2.25.

Let:
- \( e \) be the price of one eraser.
- \( p \) be the price of one pen.

Given:
1. The cost of 6 erasers and 2 pens is ₹ 15:
   \[
   6e + 2p = 15
   \]
2. The cost of more than 4 erasers (assuming this means 5 erasers) is ₹ 14, and pens for ₹ 9 (which means the remaining balance).

We already have:
\[
p = 2.25
\]

Substituting \( p = 2.25 \) into the first equation:
\[
6e + 2(2.25) = 15
\]
\[
6e + 4.5 = 15
\]
\[
6e = 15 - 4.5
\]
\[
6e = 10.5
\]
\[
e = \frac{10.5}{6}
\]
\[
e = 1.75
\]

So, the price of one eraser is ₹ 1.75, and the price of one pen is ₹ 2.25.

### Summary:
- Price of one eraser (\( e \)): ₹ 1.75
- Price of one pen (\( p \)): ₹ 2.25

Would you like further details or have any questions?

### Relative Questions:
1. How do we verify the correctness of these prices with the given conditions?
2. If the price of pens were different, how would it affect the price of erasers?
3. Can we interpret "more than 4 erasers" in another way for different solutions?
4. How can we apply these prices to different quantities for total cost?
5. What if the given total cost changed; how would it affect our equations?
6. How can this problem-solving method be applied to other similar pricing problems?
7. What assumptions are crucial in solving this type of algebraic problem?
8. How would rounding the prices to the nearest integer affect our results?

**Tip:** When solving for multiple variables, double-check each step for consistency and accuracy with the given conditions.

Learn how to solve a linear equation problem involving the prices of erasers and pens. Discover the step-by-step solution to find the prices of erasers and pens given specific conditions. Suitable for Grades 7-9 students studying algebraic equations.

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