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Amy's sells kale at x dollar per pound for the first 20 poun [#permalink]

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22 Jun 2011, 07:02

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69% (04:12) correct
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Amy's sells kale at x dollar per pound for the first 20 pounds and .8x for every subsequent pound. Lucia's price is x per pound for the first 14 pounds and .9x for subsequent pounds. What is the minimum number of pounds over 15 for which Amy's becomes an equal or better deal?

A 24 B 25 C 26 D 27 E 28

I know we can solve this question by solving 20x + (y - 20)(.8)x = 14x + (y - 14)(.9)x

Amy's sells kale at x dollar per pound for the first 20 pounds and .8x for every subsequent pound. Lucia's price is x per pound for the first 14 pounds and .9x for subsequent pounds. What is the minimum number of pounds over 15 for which Amy's becomes an equal or better deal?

A 24 B 25 C 26 D 27 E 28

I know we can solve this question by solving 20x + (y - 20)(.8)x = 14x + (y - 14)(.9)x

Is there a better way to do this quickly?

i think plugging in can be an option but I guess algebra is best for this question.

Amy's sells kale at x dollar per pound for the first 20 pounds and .8x for every subsequent pound. Lucia's price is x per pound for the first 14 pounds and .9x for subsequent pounds. What is the minimum number of pounds over 15 for which Amy's becomes an equal or better deal?

A 24 B 25 C 26 D 27 E 28

I know we can solve this question by solving 20x + (y - 20)(.8)x = 14x + (y - 14)(.9)x

Is there a better way to do this quickly?

If I follow the wordings of the question, the answer should be 11.

If the question were: "What is the minimum number of pounds for which Amy's becomes an equal or better deal" 26 lbs would be the answer.

Equation: 20+(k-5)0.8=14+(k+1)0.9

Alternative approach can be substitution. Is it better, don't know!!! _________________

Re: Amy's sells kale at x dollar per pound for the first 20 poun [#permalink]

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28 Jun 2013, 08:05

fozzzy wrote:

Would plugin be a better strategy for this question or the algebraic approach

My personal opinion? No

A 24 B 25 C 26 D 27 E 28

The answer choices are too close, number picking works best when the choices are more different. If you cannot set up an equation, plug numbers and see what you find, but I would not consider it a good strategy here.

My opinion _________________

It is beyond a doubt that all our knowledge that begins with experience.

Re: Amy's sells kale at x dollar per pound for the first 20 poun [#permalink]

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28 Jun 2013, 08:55

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udaymathapati wrote:

Amy's sells kale at x dollar per pound for the first 20 pounds and .8x for every subsequent pound. Lucia's price is x per pound for the first 14 pounds and .9x for subsequent pounds. What is the minimum number of pounds over 15 for which Amy's becomes an equal or better deal?

A 24 B 25 C 26 D 27 E 28 I know we can solve this question by solving 20x + (y - 20)(.8)x = 14x + (y - 14)(.9)x

Is there a better way to do this quickly?

fozzzy wrote:

Would plugin be a better strategy for this question or the algebraic approach

The given values in the problem can help one do this problem logically, without the aid of any equations/pluggin-in:

Till 14 pounds, both Amy and Lucia would sell at 1 dollar each. After this, for each pound, Lucia offers a discount of 0.1 dollars as compared to the price offered by Amy. Thus, a person buying from both would have saved 0.6 dollars for 20 pounds with respect to Lucia.

After 20 pounds, for each pound, Amy offers a discount of 0.1 dollars as compared to the price offered by Lucia. Thus, to offset the initial discount of 0.6 dollars from Lucia, the person should buy atleast 6 more pounds, to exactly equalize the offered price between Amy and Lucia.

Thus the minimum pounds required to exactly match the price is 20+6 = 26 pounds. _________________

Re: Amy's sells kale at x dollar per pound for the first 20 poun [#permalink]

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28 Jun 2013, 09:17

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udaymathapati wrote:

Amy's sells kale at x dollar per pound for the first 20 pounds and .8x for every subsequent pound. Lucia's price is x per pound for the first 14 pounds and .9x for subsequent pounds. What is the minimum number of pounds over 15 for which Amy's becomes an equal or better deal?

A 24 B 25 C 26 D 27 E 28

I know we can solve this question by solving 20x + (y - 20)(.8)x = 14x + (y - 14)(.9)x

Is there a better way to do this quickly?

For Amy's deal to be better, the cost has to be less or equal to Lucia's

Assuming 'n' is the number of pounds of kale, the equation is

Re: Amy's sells kale at x dollar per pound for the first 20 poun [#permalink]

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18 Nov 2013, 17:52

I think simple logic will do the trick here,

From 15th to 20th pound Lucia loses 0.1S to Amy, therefore she is basically 0.6$ behind Amy when Amy starts her sale. When Amy starts her Sale she loses 0.1$ to Lucia, therefore, Lucia will Cover the 0.6$ when Amy has sold 6 pounds of the stuff. i.e. 20+6 =26

Re: Amy's sells kale at x dollar per pound for the first 20 poun [#permalink]

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19 Nov 2013, 20:14

udaymathapati wrote:

Amy's sells kale at x dollar per pound for the first 20 pounds and .8x for every subsequent pound. Lucia's price is x per pound for the first 14 pounds and .9x for subsequent pounds. What is the minimum number of pounds over 15 for which Amy's becomes an equal or better deal?

A 24 B 25 C 26 D 27 E 28

I know we can solve this question by solving 20x + (y - 20)(.8)x = 14x + (y - 14)(.9)x

Is there a better way to do this quickly?

Since the Amy's deal need to be equal/ better it not possible for the first 20 pounds . Hence calculate the Lucia deal for first 20 pounds which comes to be 19.4 x. Now the difference of margin between the two deals is 20x-19.4x = 0.6x. Now the difference between above 20 pounds for both is 0.9 x -0.8 x = 0.1 x. So to cover the 0.6 x we will have to purchase 0.6x/0.1x= 6 more pounds. Hence the answer is 6. _________________

Re: Amy's sells kale at x dollar per pound for the first 20 poun [#permalink]

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