Jess saved $500 working during the summer. He plans to buy school clothes with his money. He found jeans he liked for $30 per pair, including tax, and shirts for $17, including twice. If he buys exactly twice as many shirts as jeans, how many pairs of jeans can jess buy without exceeding $500
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To determine how many pairs of jeans Jess can buy without exceeding $500, we'll set up an equation based on the given information.
Let's say the number of pairs of jeans is represented by 'x'. Since he plans to buy exactly twice as many shirts as jeans, the number of shirts bought can be represented as '2x'.
The cost of each pair of jeans is $30, including tax. So, the total cost of jeans can be calculated as 30x.
The cost of each shirt is $17, including tax. So, the total cost of shirts can be calculated as 17(2x) = 34x.
According to the question, the total amount Jess can spend without exceeding $500 should be equal to or less than $500.
Therefore, the equation can be set up as follows:
30x + 34x ≤ 500
Combining like terms, we get:
64x ≤ 500
Dividing both sides of the inequality by 64:
x ≤ 500/64
x ≤ 7.81 (approximately)
Since we can't have a fraction or a decimal value for the number of pairs of jeans, Jess can buy a maximum of 7 pairs of jeans without exceeding $500.
So, the answer is that Jess can buy a maximum of 7 pairs of jeans.
To find out how many pairs of jeans Jess can buy without exceeding $500, we need to set up an equation based on the given information.
Let's assume the number of jeans Jess can buy is 'x'.
The cost of each pair of jeans, including tax, is $30.
So, the total cost of buying x pairs of jeans would be 30x dollars.
The cost of each shirt, including tax, is $17.
Since Jess plans to buy exactly twice as many shirts as jeans, he would buy 2x shirts.
So, the total cost of buying 2x shirts would be 17 * 2x dollars, which simplifies to 34x dollars.
According to the problem, Jess saved $500 and wants to buy clothes without exceeding that amount.
The total cost of jeans and shirts should be less than or equal to $500.
Therefore, we can set up the equation: 30x + 34x ≤ 500.
Combine like terms: 64x ≤ 500.
Divide both sides by 64: x ≤ 500/64.
Calculating: x ≤ 7.8125.
Since Jess cannot buy a fraction of a pair of jeans, he can buy a whole number of jeans that is less than or equal to 7.
Therefore, Jess can buy a maximum of 7 pairs of jeans without exceeding $500.
2S = J
30J + 17S ≤ 500
Substitute 2S for J in the second equation and solve for S. Insert that value into the first equation to solve for J. Check by putting both values into the second equation.