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M16-33

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M16-33 [#permalink]

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Question Stats:

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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
[Reveal] Spoiler: OA

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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
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: M16-33 [#permalink]

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New post 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.

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Re: M16-33 [#permalink]

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New post 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.

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.


Testing values is pretty much it. BTW it's not vlear what you mean by the red part above.
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

Kudos [?]: 139329 [0], given: 12783

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Re: M16-33 [#permalink]

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New post 17 Dec 2017, 01:12
Hello amargad0391,

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.


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.

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.

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