a pencil, eraser and notebook together cost one dollar. a notebook costs more than two pencils, three pencils cost more than four erasers, and three erasers cost more than a notebook. how much does each item cost
To solve this problem, let's assign variables to the prices of the items.
Let's say the cost of a pencil is 'p', the cost of an eraser is 'e', and the cost of a notebook is 'n'.
Given:
1) a pencil, eraser, and notebook together cost one dollar, so we have the equation: p + e + n = 1.
Now let's analyze the given information:
2) a notebook costs more than two pencils, so we have the inequality: n > 2p.
3) three pencils cost more than four erasers, so we have the inequality: 3p > 4e.
4) three erasers cost more than a notebook, so we have the inequality: 3e > n.
Let's solve these equations step by step.
Firstly, let's address the inequality 3p > 4e. We can rearrange it to find the relationship between p and e: p > (4/3)e.
Next, let's consider the inequality 3e > n. Rearranging it, we have n < 3e.
Now, let's combine the inequalities n > 2p and n < 3e. This implies 2p < n < 3e, where p is the smallest value and e is the largest value.
Since we know that all three costs add up to $1, we can express n as (1 - p - e).
Combining the inequalities 2p < n < 3e and n = (1 - p - e), we get:
2p < (1 - p - e) < 3e.
Now, let's solve this inequality. Distribute -1 through the parentheses:
2p < 1 - p - e < 3e.
Combine like terms:
2p + p + e < 1 < 3e + p + e.
Simplify further:
3p + 2e < 1 < 4e + p.
To find the possible values for p and e, we need to find the value range in which these inequalities hold true.
Considering the lower limit, we have 3p + 2e < 1.
Taking into account that e is the largest value and p is the smallest value, we can assume the smallest possible values for p and e:
p = 0, and e = 0.
Substituting these values into the inequality:
3(0) + 2(0) < 1,
0 < 1.
The inequality holds true because 0 is indeed less than 1.
Now let's examine the upper limit, 1 < 4e + p.
For the largest possible values of e and p, we can assume e = 0.5 and p = 0.25 (since the cost of an item cannot exceed $1).
Substituting these values into the inequality:
1 < 4(0.5) + 0.25,
1 < 2 + 0.25,
1 < 2.25.
This inequality is false since 1 is not less than 2.25.
Therefore, we can conclude that the possible values for p and e lie between 0 (inclusive) and 0.5 (exclusive).
Considering all the information we have gathered:
The possible values:
- p: between 0 and 0.5 (inclusive-exclusive range)
- e: between 0 and 0.5 (inclusive-exclusive range)
- n: 1 - p - e since the sum of all three items is equal to 1.
By varying the values within this range, we can find different combinations satisfying the given conditions. For instance, one possible solution could be:
- p = 0.25 (25 cents)
- e = 0.4 (40 cents)
- n = 1 - p - e = 0.35 (35 cents).
Another solution could be:
- p = 0.3 (30 cents)
- e = 0.45 (45 cents)
- n = 1 - p - e = 0.25 (25 cents).
So, there can be multiple valid combinations of prices for the pencil, eraser, and notebook that satisfy the given conditions.