Any Math Wizards?

Question

I need a detailed explanation for this problem because I am studying for my math test tomorrow by practicing with some problems, but I need a little help =]

1. A restaurant sells large and small submarine sandwiches. Rolls for the sandwiches are ordered from a baker. The roll for a large sandwich costs $0.25 and the roll for a small sandwich costs. $0.15. Melissa, the manager of the restaurant, ordered 130 more large rolls than small rolls. What was the greatest number of large rolls Melissa received if she spent less than $63?

Answer 1

Let n be the number of large rolls
Let n - 130 be the number of small rolls

0.25n + 0.15(n - 130) < 63

Now expand out:
0.25n + 0.15n - 19.5 < 63

I like multiplying by 100 to get rid of fractions:
25n + 15n - 1950 < 6300
40n - 1950 < 6300

Add 1950 to both sides:
40n < 8250

Divide both sides by 40
n < 206.25

So the biggest integer is 206.

206 large rolls (76 small rolls)

As a double check:
206 x 0.25 = $51.50
76 x 0.15 = $11.40
= $62.90

So the answer is correct, Melissa bought 206 large rolls.

Answer 2

Let L = No. of large rolls
then L-130 = No. of small rolls

0.25 L + 0.15(L-130) ≤63.00

Solve for the equality first

0.40L -19.5 = 63

0.40L = 82.5

L = 206.25

The nearest integer smaller than 206.25 is 206 large rolls.

Answer 3

let larger rolls ordered = x

so smaller rolls ordered = x - 130

The amount spent on smaller rolls = $ 0.15(x-130)

=>0.15x - 19.5

The amount spent on larger rolls = $0.25x

The inequality equation is

0.25x + 0.15x - 19.5 < 63

0.4x - 19.5 <63

0.4x < 82.5

divide by 0.4

x < 206.25

so the gratest number of larger rolls that can be ordered are 206

Answer 4

L = large
s = small

Write equations for what you know.
L = 130 + S
63 = .25L + .15S
sub in L
63 = .25(130 + s) + .15s
63 = 32.5 + .25s + .15s
63 = 32.5 + .4s
.4s = 30.5
s = 76.25

round down because you can't go over $63
sub in s now that you know s
L = 130 + s
L = 130 + 76
L = 206 rolls