A fashion buyer for a large retail store purchased 315 items directly from the manufacturer for a total of $6000. Some of the items were dresses purchased for $25 each, and the rest were shirts purchased for $10 each. How many more dresses than shirts did the buyer purchase?

The correct answer is: 65

HINT: Frame 2 equations with the provided information and then solve.
Complete step by step solution:
Total number of items purchased = 315
Total price for which 315 items were purchased = $6000
Let the number of dresses purchased = x
Let the number of shirts purchased = y
From this, we get the equation x + y = 315…(i)
Price of one dress = $ 25
Price of x dresses = $ 25x
Price of one shirt = $ 10
Price of y shirts = $ 10y
From this, we get the equation 25x + 10y = 6000…(ii)
Multiply (i) with 10 to make the coefficients of y to be the same.
On multiplying (i) with 10, we get 10x + 10y = 3150..(iii)
Subtract (iii) from (ii),
On subtracting, we get LHS to be 25x - 10x =15x and RHS to be 6000 - 3150 = 2850
On equating LHS and RHS, we get 15x = 2850
Thus, we get the value of x = 190.
On substituting the value of x = 190 in (i), we get y = 315 - 190 = 125
So total number of dresses = 190 and total number of shirts = 125
Number of dresses, the buyer purchased more than number of shirts = 190 - 125 = 65