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bmwhype2
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Danny sells A for 15 cents each. Danny sells B for 14 cents Updated on: 06 Nov 2007, 11:52
Originally posted by bmwhype2 on 06 Nov 2007, 10:40.
Last edited by bmwhype2 on 06 Nov 2007, 11:52, edited 1 time in total.
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Danny sells A for 15 cents each.
Danny sells B for 14 cents each.
If he earned 5 dollars total, what is the least number of b he could have sold?


0
5
10
15
20



Can we solve this without looking at the answer choices?



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GMAT TIGER
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Danny sells A for 15 cents each. Danny sells B for 14 cents 06 Nov 2007, 11:17
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bmwhype2
Danny sells A for 15 cents each.
Danny sells B for 14 cents each.
If he earned 5 dollars total, what is the least number of vanilla cups he could have sold?


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Can we solve this without looking at the answer choices?
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imo, we can but if you make clear what is Vanilla cup? A? B?
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sportyrizwan
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Danny sells A for 15 cents each. Danny sells B for 14 cents 06 Nov 2007, 11:50
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If the Vanila cup are either of A or B
least number of vanila cups he could have sold would be 10...
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sevenplus
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Danny sells A for 15 cents each. Danny sells B for 14 cents 06 Nov 2007, 12:11
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Least number of b vanilla cups sold = 10
15a+14b=500
Cups sold can only be integer
Plug in values, least value of b for which above equation holds good is 10
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Danny sells A for 15 cents each. Danny sells B for 14 cents 06 Nov 2007, 12:26
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bmwhype2
Danny sells A for 15 cents each.
Danny sells B for 14 cents each.
If he earned 5 dollars total, what is the least number of b he could have sold?


0
5
10
15
20



Can we solve this without looking at the answer choices?
Show more


15A+14B=500

i think we will need to have two equations with the same variables in order not to pick answer choices
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Danny sells A for 15 cents each. Danny sells B for 14 cents 06 Nov 2007, 13:13
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Ravshonbek
bmwhype2
Danny sells A for 15 cents each.
Danny sells B for 14 cents each.
If he earned 5 dollars total, what is the least number of b he could have sold?


0
5
10
15
20



Can we solve this without looking at the answer choices?

15A+14B=500

i think we will need to have two equations with the same variables in order not to pick answer choices
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It can be solved.

First step- 500/15 = 33x15( reminder 5)
So there are 33As possible in 500 cents but it leaves only 5. So we will have to reduce number of As till the reminder become a multiple of 14.
Complicated? lol okay...
33(15)-------5
32(15)-------5+ 15=20 ( still not a multiple of 14)
Common sense tells us that since we are dealing with numbers 0f 5 (15 and 5) first number that is the multiple of both 5 and 14 will be 70 but it wont work. Next number is 140
Lets try 140 = 9(15)+5. Thats it!
So, we have to reduce 9 As from 33 As and the reminder ( 33-9=24 As) will be the maximum number of A's sold and 10 will be the minimum number of B sold.
Simple :P lol it is always easy to pick numbers in such questions. This method is easy too but need a bit practice. Hope, I have been able to explain.



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Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
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