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# M16-33

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Math Expert
Joined: 02 Sep 2009
Posts: 52971

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16 Sep 2014, 00:00
00:00

Difficulty:

45% (medium)

Question Stats:

69% (01:43) correct 31% (01:19) wrong based on 87 sessions

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? A. 0 B. 5 C. 10 D. 15 E. 20 _________________ Math Expert Joined: 02 Sep 2009 Posts: 52971 Re M16-33 [#permalink] ### Show Tags 16 Sep 2014, 00:00 Official Solution: 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?

A. 0
B. 5
C. 10
D. 15
E. 20

Let $$C$$ and $$V$$ denote the number of chocolate ice-creams and vanilla ice-creams sold during the day. The question is what is the least integer $$V$$ such that $$0.15C + 0.14V = 5$$ or $$15C + 14V = 500$$. $$V$$ has to end with either 0 or 5 (otherwise $$500 - 14V$$ will not end with 0 or 5 and so will not be divisible by 15). $$V$$ cannot be 0 because 500 is not divisible by 15. $$V$$ cannot be 5 because $$500 - 14*5 = 430$$ is not divisible by 15. $$V$$ can be 10 as $$500 - 14*10 = 360$$ is divisible by 15.

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Joined: 09 Sep 2015
Posts: 21

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05 Apr 2017, 20:07
Bunuel wrote:
Official Solution:

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? A. 0 B. 5 C. 10 D. 15 E. 20 Let $$C$$ and $$V$$ denote the number of chocolate ice-creams and vanilla ice-creams sold during the day. The question is what is the least integer $$V$$ such that $$0.15C + 0.14V = 5$$ or $$15C + 14V = 500$$. $$V$$ has to end with either 0 or 5 (otherwise $$500 - 14V$$ will not end with 0 or 5 and so will not be divisible by 15). $$V$$ cannot be 0 because 500 is not divisible by 15. $$V$$ cannot be 5 because $$500 - 14*5 = 430$$ is not divisible by 15. $$V$$ can be 10 as $$500 - 14*10 = 360$$ is divisible by 15. Answer: C Hi Bunuel, Is there any other way to solve these kind of problems ? May be by considering linear equations and solving for required answer. Math Expert Joined: 02 Sep 2009 Posts: 52971 Re: M16-33 [#permalink] ### Show Tags 05 Apr 2017, 20:44 amargad0391 wrote: Bunuel wrote: Official Solution: 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?

A. 0
B. 5
C. 10
D. 15
E. 20

Let $$C$$ and $$V$$ denote the number of chocolate ice-creams and vanilla ice-creams sold during the day. The question is what is the least integer $$V$$ such that $$0.15C + 0.14V = 5$$ or $$15C + 14V = 500$$. $$V$$ has to end with either 0 or 5 (otherwise $$500 - 14V$$ will not end with 0 or 5 and so will not be divisible by 15). $$V$$ cannot be 0 because 500 is not divisible by 15. $$V$$ cannot be 5 because $$500 - 14*5 = 430$$ is not divisible by 15. $$V$$ can be 10 as $$500 - 14*10 = 360$$ is divisible by 15.

Hi Bunuel,

Is there any other way to solve these kind of problems ? May be by considering linear equations and solving for required answer.

Testing values is pretty much it. BTW it's not vlear what you mean by the red part above.
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Manager
Joined: 12 Jun 2016
Posts: 215
Location: India
WE: Sales (Telecommunications)

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17 Dec 2017, 01:12
1

One way to solve problems such as these is by using 'diophantine equations' approach.

This approach is explained very nicely by mau5 in a similar question. The link to mau5's explanation is - https://gmatclub.com/forum/joanna-bough ... l#p1201409

Applying this approach , we have

Let $$C$$ = Number of Chocolate ice creams sold and $$V$$ = Number of Vanilla ice creams sold.

$$15C+ 14V = 500$$ <---Converting dollars into cents for ease of calculation.

$$C = \frac{{500-14V}}{15}$$

Substituting the answer choices, we can see that the smallest value for $$V$$ for which $$C$$ is an integer is $$V = 10$$ (In this case, $$C = 24$$).

Hence the required answer is C.

I hope you find this useful.

Bunuel wrote:
Official Solution:

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?

A. 0
B. 5
C. 10
D. 15
E. 20

Let $$C$$ and $$V$$ denote the number of chocolate ice-creams and vanilla ice-creams sold during the day. The question is what is the least integer $$V$$ such that $$0.15C + 0.14V = 5$$ or $$15C + 14V = 500$$. $$V$$ has to end with either 0 or 5 (otherwise $$500 - 14V$$ will not end with 0 or 5 and so will not be divisible by 15). $$V$$ cannot be 0 because 500 is not divisible by 15. $$V$$ cannot be 5 because $$500 - 14*5 = 430$$ is not divisible by 15. $$V$$ can be 10 as $$500 - 14*10 = 360$$ is divisible by 15.

Hi Bunuel,

Is there any other way to solve these kind of problems ? May be by considering linear equations and solving for required answer.

_________________

My Best is yet to come!

Re: M16-33   [#permalink] 17 Dec 2017, 01:12
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# M16-33

Moderators: chetan2u, Bunuel

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