A pencil, eraser, and notebook cost $1 in total. A notebook costs more than 2 pencils, 3 pencils cost more than 4 erasers, and 3 erasers cost more than a notebook. How much does each item cost?
Pencil costs $0.26.
Eraser costs $0.19.
Notebook costs $0.55.
Solution of Diophantic equation
p + e + n = 100
with constraints
n > 2 p
3 p > 4 e
3 e > n
Sophia spent $7.30 for school supplies. She spent $3.00 for a notebook and $1.75 for a pen. She also bought 3 large erasers. If each eraser had the same cost, how much did she spend for each eraser?
Let's assume the cost of a pencil is "x" dollars, the cost of an eraser is "y" dollars, and the cost of a notebook is "z" dollars.
Given that the total cost of a pencil, eraser, and notebook is $1, we can write the equation:
x + y + z = 1 ---(equation 1)
We are also given the following information:
1. A notebook costs more than 2 pencils
z > 2x ---(equation 2)
2. 3 pencils cost more than 4 erasers
3x > 4y ---(equation 3)
3. 3 erasers cost more than a notebook
3y > z ---(equation 4)
Now we can solve equations 1, 2, 3, and 4 to find the values of x, y, and z.
From equation 4, we can express z in terms of y:
z = 3y/1 --(equation 5)
Substitute equation 5 into equation 2:
3y/1 > 2x
Simplifying:
3y > 2x --(equation 6)
Substitute equation 5 into equation 3:
3x > 4y/1
Simplifying:
3x > 4y --(equation 7)
We need to find values for x, y, and z that satisfy equations 1, 6, and 7.
Now let's test different values to find a suitable solution. We will start with small integers.
Let's assume:
x = 1
y = 2
z = 3
By substituting these values into equation 1, we can check if they satisfy the given information:
1 + 2 + 3 = 6
Based on this assumption, the total cost is $6, which is not equal to $1 as given. So, these values do not satisfy the given information.
Let's try again.
Assume:
x = 2
y = 1
z = 3
By substituting these values into equation 1:
2 + 1 + 3 = 6
Again, the total cost is $6, which is not equal to $1 as given. So, these values do not satisfy the given information either.
One more attempt.
Assume:
x = 0.50 (50 cents)
y = 0.20 (20 cents)
z = 0.30 (30 cents)
By substituting these values into equation 1:
0.50 + 0.20 + 0.30 = 1
Now the total cost is $1, which matches the given information.
So, the cost of a pencil is $0.50, the cost of an eraser is $0.20, and the cost of a notebook is $0.30.
To solve this problem, let's assign variables to each of the items:
Let's say the cost of a pencil is "p" dollars.
The cost of an eraser is "e" dollars.
The cost of a notebook is "n" dollars.
According to the given information:
1. A pencil, eraser, and notebook cost $1 in total:
p + e + n = 1
2. A notebook costs more than 2 pencils:
n > 2p
3. 3 pencils cost more than 4 erasers:
3p > 4e
4. 3 erasers cost more than a notebook:
3e > n
We can use these equations to solve for each variable.
Let's start by solving equation 3:
3p > 4e
Rearrange this equation to isolate p:
p > (4/3)e
Now, let's solve equation 4:
3e > n
Rearrange this equation to isolate e:
e < (1/3)n
We can now substitute (4/3)e from equation 3 into equation 2:
n > 2(4/3)e
n > (8/3)e
From equation 1, we know the total cost is $1:
p + e + n = 1
We can substitute the expressions for n and e we found into this equation:
p + (1/3)n + (8/3)e = 1
Now, let's solve this equation for p:
p = 1 - (1/3)n - (8/3)e
We can now use this equation to find the values of p, e, and n that satisfy all the given conditions.
With this solution method, we have set up a system of equations that we can solve to find the values of the variables p, e, and n.