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605-655 (Medium)|   Algebra|      
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SW4
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A soda machine sells both bottles and cans, and no other items. 13 Nov 2016, 11:16
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

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45% (medium)

Question Stats:

74% (02:35) correct 26% (02:55) wrong based on 544 sessions

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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
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C
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A soda machine sells both bottles and cans, and no other items. 13 Nov 2016, 13:08
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SW4
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
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b+c=250_______(1)

1.5b+0.75c=315

2b+c=420_______(2)


(2)-(1)

b=170

c=80

b-c=170-80=90

C



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A soda machine sells both bottles and cans, and no other items. 13 Nov 2016, 23:45
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SW4
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
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\(\frac{3b}{2} + \frac{3c}{4}\) = \(315\)

\(b + c = 250\)

Now, \(\frac{3b}{2} + \frac{3c}{4}\) = \(315\)

Or, \(6b + 3c = 1260\)

NOw, \(b + c = 250\) \(* 6\) => \(6b + 6c = 1500\)

Or, 1500 - 6c + 3c = 1260

Or, 3c = 240

So, c = 80 and b = 170

Thus 90 more bottles were sold than cans , answer will be (C) 90
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A soda machine sells both bottles and cans, and no other items. 28 May 2018, 06:11
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SW4
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
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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

Answer: C

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Brent
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A soda machine sells both bottles and cans, and no other items. 30 May 2018, 17:16
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SW4
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
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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.

Answer: C
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A soda machine sells both bottles and cans, and no other items. 10 Sep 2025, 12:32
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315/250 = $1.26 per beverage

Use alligation:

1.5............1.26........0.75

B/C = 1.26-0.75/1.5-1.26

B/C = 0.51/0.24 = 51/24 = 17/8

Total bottle sold =17k
Total cans sold = 8k

25k= 250 => k= 250/25 = 10

Q: 17k-8k = 9k = 90

Option C
SW4
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
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