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# A soda machine sells both bottles and cans, and no other items.

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Re: A soda machine sells both bottles and cans, and no other items. [#permalink]
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SW4 wrote:
A soda machine sells both bottles and cans, and no other items. Bottles cost $1.50 each, while cans cost$0.75 each. If on one day, the soda machine sold 250 total beverages and yielded $315, how many more bottles than cans were sold? A) 60 B) 80 C) 90 D) 115 E) 170 Let B = # of bottles sold Let C = # of cans sold If on one day, the soda machine sold 250 total beverages... We can write: B + C = 250 ...and yielded$315.
So: 1.5B + 0.75C = 315
We can eliminate the decimals by multiplying both sides by 4 to get: 6B + 3C = 1260

How many more bottles than cans were sold?
So, we must determine the value of B - C.

We have the system:
B + C = 250
6B + 3C = 1260

When we solve this system for B and C , we get: B = 170 and C = 80
So, B - C = 170 - 80 = 90

Cheers,
Brent
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Re: A soda machine sells both bottles and cans, and no other items. [#permalink]
SW4 wrote:
A soda machine sells both bottles and cans, and no other items. Bottles cost $1.50 each, while cans cost$0.75 each. If on one day, the soda machine sold 250 total beverages and yielded \$315, how many more bottles than cans were sold?

A) 60
B) 80
C) 90
D) 115
E) 170

We can let b = the number of bottles of soda sold and c = the number of cans of soda sold, and create the equations:

b + c = 250

and

1.5b + 0.75c = 315

Multiplying the first equation by 3, we have 3b + 3c = 750…….[Eq. 1]

Multiplying the second equation by 4, we have 6b + 3c = 1260…... [Eq. 2]

Subtracting Eq. 1 from Eq. 2, we have 3b = 510. So b = 510/3 = 170.

Since b + c = 250, so c = 250 - 170 = 80.

Therefore, the number of bottles sold is 170 - 80 = 90 more than the number of cans sold.

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Re: A soda machine sells both bottles and cans, and no other items. [#permalink]
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Re: A soda machine sells both bottles and cans, and no other items. [#permalink]
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