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Amy's sells kale at x dollar per pound for the first 20 poun

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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, 06:02
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72% (02:58) correct 28% (03:08) wrong based on 309 sessions

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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

Is there a better way to do this quickly?
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Re: Amy's sells kale at x dollar per pound for the first 20 poun  [#permalink]

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28 Jun 2013, 07:55
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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.
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22 Jun 2011, 07:20
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?

i think plugging in can be an option but I guess algebra is best for this question.
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22 Jun 2011, 07:37
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?

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!!!
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22 Jun 2011, 16:09
20x+0.8x(p) = 14x+0.9x(p+6)

=> p =6 where is the pounds over 20 lbs

so total sold pounds = 26.

i agree with @fluke, that answer should be 11 , esp when the question is asking for how many pounds over 15..
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Re: Amy's sells kale at x dollar per pound for the first 20 poun  [#permalink]

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28 Jun 2013, 07:02
Would plugin be a better strategy for this question or the algebraic approach
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Re: Amy's sells kale at x dollar per pound for the first 20 poun  [#permalink]

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28 Jun 2013, 07:05
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
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Re: Amy's sells kale at x dollar per pound for the first 20 poun  [#permalink]

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28 Jun 2013, 08:17
3
3
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

20x + (n-20) (0.8x) <= 14x + (n-14)(0.9x)

Resolve it:

==> 20x + 0.8nx - 16x <= 14x + 0.9nx - 12.6x

==> 2.6x <=0.1nx

==> 26x < =nx

==> x (n-26) > = 0

as x cannot be 0,

==> n - 26 >=0

==> n > = 26

so the minimum value is 26

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

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16 Oct 2013, 00:19
Plugged in value of x = 100 & solved as per chart below:

Also note from chart that value 2000 is constant for Amy & 1400 is constant for Lucia.

So, once you have calculated for 24 (the value from where I started), just add 80 (for Amy) & 90 (for Lucia) for calculating for other options.
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Re: Amy's sells kale at x dollar per pound for the first 20 poun  [#permalink]

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18 Nov 2013, 16:52
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
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Re: Amy's sells kale at x dollar per pound for the first 20 poun  [#permalink]

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19 Nov 2013, 19:14
1
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.
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Re: Amy's sells kale at x dollar per pound for the first 20 poun  [#permalink]

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26 Jul 2015, 19:49
question should be about maximum not minimum pounds in which Amy wins

Inequality is:

20x+(y-20)*0.8x>=14x+(y-14)0.9x

2.6x>0.1xy => 26>=y

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

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18 Oct 2015, 01:14
starting form 14th pound one girl lose 1/10x, so until 20th she will lose 6*(1/10)x

starting from 20th pound another girl lose 1/10x. So to compensate 6 loses she needs 20+6=26 pounds

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

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05 Oct 2016, 06:28
did the straightforward way...
20x+0.8x(y-20)=14x+0.9x(y-14)
we can factor out x from both sides then divide by x.
6=0.9y - 63/5 - 0.8y +16
63/5 = 12.6
6=0.1y + 3.4
2.6 = 0.1y
y=26.
so when we have 26 pounds, Amy's deal is the same as Lucia's. If we buy more from Amy, we'll save more...
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Amy's sells kale at x dollar per pound for the first 20 poun  [#permalink]

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19 Dec 2016, 08:12
Can be solved using options. Let x = 10. and then verify the options. All the options are arranged in ascending order, so start with the middle value, i.e. option(c). In such a question always start with middle value, so we can eliminate at least two options in one goal.
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Re: Amy's sells kale at x dollar per pound for the first 20 poun  [#permalink]

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19 Dec 2016, 11:37
In my case I found simpler and faster just to plug in numbers. I use 1 for x and started from answer B)

Ans B) Amy: (20 * 1) + (5 * 0.8) = 24.0; Lucia: (14 * 1) + (11 * 0.9) = 23.9. Amy is slightly less convenient than Lucia
Ans C) Amy: (20 * 1) + (6 * 0.8) = 24.8; Lucia: (14 * 1) + (12 * 0.9) = 24.8. They have the same price - Correct answer
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Re: Amy's sells kale at x dollar per pound for the first 20 poun  [#permalink]

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19 Mar 2017, 17:58
Why pluging number for x doesnt work? for example i put 10 for x and solved the problem 20*10+8P>=14*10+9P and it gives p<= 60. Any number less than 60. And the minimum number among answer choices us 24. why doesnt it work?
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Re: Amy's sells kale at x dollar per pound for the first 20 poun  [#permalink]

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17 Dec 2017, 04:42
GMATSPARTAN wrote:
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

20x + (n-20) (0.8x) <= 14x + (n-14)(0.9x)

Resolve it:

==> 20x + 0.8nx - 16x <= 14x + 0.9nx - 12.6x

==> 2.6x <=0.1nx

==> 26x < =nx

==> x (n-26) > = 0

as x cannot be 0,

==> n - 26 >=0

==> n > = 26

so the minimum value is 26

'C' would be the correct answer

Dear Bunuel,
Can you help with this , in this post it has been mentioned that C.P. of Amy should be less than Lucy but the calculation has been done using S.P.
For a better deal Amy should be earning more than lucy
Actual equation should be 20x + (n-20) (0.8x) $$\geq$$14x + (n-14)(0.9x), after simplification we get n $$\leq$$ 26, so it seems as though 26 is the maximum value rather than the minimum value ! Please let me know if I have made any mistake anywhere. Thank you.
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Re: Amy's sells kale at x dollar per pound for the first 20 poun   [#permalink] 17 Dec 2017, 04:42
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