Nine books are to be bought by a student. Some cost $6.00 each and the remainder cost $6.50 each. If the total amount spent was $56, how many of each book are bought?
Buy n books at $6. The total cost is
6n + 6.5(9-n) = 56
6n + 58.5 - 6.5n
.5n = 2.5
n = 5
5 at $6 + 4 at $6.50 = 30 + 26 = $56
You can also do that in your head as follows:
Books cost $6 and $6.50, the difference is $0.50.
9 books at $6 each cost $54.
You then have $2 left, good for exchanging for 4 books at $6.50. So a 4 at $6.50 and (9-4)=5 at $6.00.
Let's assume that the number of books costing $6.00 each is x.
Therefore, the number of books costing $6.50 each would be 9 - x (since there are a total of 9 books).
Now let's calculate the total cost of books costing $6.00 each: 6 * x = 6x
And the total cost of books costing $6.50 each: 6.5 * (9 - x) = 58.5 - 6.5x
Given that the total amount spent was $56, we can set up the equation:
6x + (58.5 - 6.5x) = 56
Now let's solve this equation:
6x + 58.5 - 6.5x = 56
Combine like terms:
-0.5x + 58.5 = 56
Subtract 58.5 from both sides:
-0.5x = 56 - 58.5
-0.5x = -2.5
Divide by -0.5:
x = -2.5 / -0.5
x = 5
So, the student bought 5 books costing $6.00 each, and therefore bought 9 - 5 = 4 books costing $6.50 each.
To solve this problem, let's assume that the student bought π₯ books that cost $6.00 each and π¦ books that cost $6.50 each.
We are given that the total amount spent was $56. So we have the equation:
6π₯ + 6.50π¦ = 56 (Equation 1)
We also know that the student bought a total of nine books. So we have the equation:
π₯ + π¦ = 9 (Equation 2)
To find the values of π₯ and π¦, we need to solve this system of equations.
Let's solve Equation 2 for π₯:
π₯ = 9 β π¦
Now substitute this value of π₯ in Equation 1:
6(9 β π¦) + 6.50π¦ = 56
Expand and simplify:
54 β 6π¦ + 6.50π¦ = 56
0.50π¦ = 2
π¦ = 4
Substitute the value of π¦ back into Equation 2 to find π₯:
π₯ + 4 = 9
π₯ = 5
Therefore, the student bought 5 books that cost $6.00 each and 4 books that cost $6.50 each.