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Albert sells chocolate ice-cream for $0.15 per cup and

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Albert sells chocolate ice-cream for $0.15 per cup and [#permalink] New post 23 May 2008, 02:51
Albert sells chocolate ice-cream for $0.15 per cup and vanilla ice-cream for $0.14 per cup. If Albert earned $5 during his day's work, what is the least number of vanilla cups he could have sold?


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I need your reasoning! Thanks!
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Re: M16-33 [#permalink] New post 23 May 2008, 03:12
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C

we can write out: 0.15x+0.14y=5 or 15x+14y=500
1. 15x and 500 is divisible by 5, so 14y has to be also divisible by 5
2. It is possible for y=0,5,10,15.... or y=5k where k in an integer.
3. 15x+14*5k=500 --> 3x=100-14k
4. find the least number k that 100-14k is divisible by 3: 100 is not, 86 is not, 72 - is divisible (k=2) and therefore, x=10.
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Re: M16-33 [#permalink] New post 23 May 2008, 05:26
sondenso wrote:
Albert sells chocolate ice-cream for $0.15 per cup and vanilla ice-cream for $0.14 per cup. If Albert earned $5 during his day's work, what is the least number of vanilla cups he could have sold?


0
5
10
15
20

I need your reasoning! Thanks!



Let,
No. of chocolate ice-cream for $0.15 per cup = C
No. of vanilla ice-cream for $0.14 per cup = V

15*C + 14* V = 500 ($5 total)
So,

Plug options where you get C as an integral value, start from ..0
15*C = 500 X
15*C = 500 - 14*5 = 430 X
15*C = 500 - 14*10 = 360 Done!

Answer C
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Re: M16-33 [#permalink] New post 23 May 2008, 19:06
walker wrote:
C

we can write out: 0.15x+0.14y=5 or 15x+14y=500
1. 15x and 500 is divisible by 5, so 14y has to be also divisible by 5
2. It is possible for y=0,5,10,15.... or y=5k where k in an integer.
3. 15x+14*5k=500 --> 3x=100-14k
4. find the least number k that 100-14k is divisible by 3: 100 is not, 86 is not, 72 - is divisible (k=2) and therefore, x=10.


Yeah, so cool this reasoning! Many thanks, Walker!
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Re: M16-33   [#permalink] 23 May 2008, 19:06
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