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A juice company is creating a new fruit punch that is 60% orange juice

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A juice company is creating a new fruit punch that is 60% orange juice  [#permalink]

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New post 29 Dec 2017, 09:48
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  85% (hard)

Question Stats:

50% (02:51) correct 50% (03:21) wrong based on 109 sessions

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A juice company is creating a new fruit punch that is 60% orange juice, 25% grape juice, and 15% apple juice by weight. If the cost for the company is x cents per kg for orange juice, y cents per kg for grape juice, and z cents per kg for apple juice, is the cost of k kg of the fruit punch more than d dollars?

1. \(12x+5y+3z >\) \(\frac{2000d}{k}\)

2. \(10d\) \(=\)\(2^\)\(\sqrt{3x}\) \(=\) \(\sqrt{5y}\)\(=\)\(\sqrt{3z}\)\(=\) \(\frac{100}{k}\)

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Re: A juice company is creating a new fruit punch that is 60% orange juice  [#permalink]

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New post 29 Dec 2017, 10:24
QZ wrote:
A juice company is creating a new fruit punch that is 60% orange juice, 25% grape juice, and 15% apple juice by weight. If the cost for the company is x cents per kg for orange juice, y cents per kg for grape juice, and z cents per kg for apple juice, is the cost of k kg of the fruit punch more than d dollars?

1. \(12x+5y+3z >\) \(\frac{2000d}{k}\)

2. \(10d\) \(=\)\(2^\)\(\sqrt{3x}\) \(=\) \(\sqrt{5y}\)\(=\)\(\sqrt{3z}\)\(=\) \(\frac{100}{k}\)


60% orange juice, 25% grape juice, and 15% apple juice by weight.
Means per kg cost = 0.6x+0.25y+0.15z.... is it >100d/k..
So 0.6x+0.25y+0.15z>100d/k.....
\(12x+5y+3z>\frac{20*100d}{k}\)

Let's see what each statement tells us..

Statement I tells us exactly that..
Sufficient

Second statement tells us 12x=5y=3z=10d*10d..
Therefore 12x+5y+3z=3*10d*10d=3*10d*100/k=3000d/k which is >2000d/k
Ans YES
Sufficient

D
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Re: A juice company is creating a new fruit punch that is 60% orange juice  [#permalink]

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New post 31 Dec 2017, 08:29
I thought of a slightly different approach. Please give me kudos if it you felt this method was useful.
Basically the stem asks if K > D? (kg or dollar)

1) 12x + 5y + 3z > 2000D/ k
subsituting values from above --> 12[0.6] + 5[0.25] + 3[0.15]> 2000D/K

[8.9] > 2000DK

so K [8.9]> 2000K --> if ~9 of k> 2000 * D --> of course K>D . suff

2) we can directly ignore the variables in the middle and see 100 d^2 = 100^2/ K --> d = 10/ K. D = K = 1 cannot happen but DK = 10 can. In each case [1,10,5,2..fitting back into D = 10/ K] D < K.

Answer D.
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Re: A juice company is creating a new fruit punch that is 60% orange juice  [#permalink]

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New post 04 May 2019, 22:45
chetan2u wrote:
QZ wrote:
A juice company is creating a new fruit punch that is 60% orange juice, 25% grape juice, and 15% apple juice by weight. If the cost for the company is x cents per kg for orange juice, y cents per kg for grape juice, and z cents per kg for apple juice, is the cost of k kg of the fruit punch more than d dollars?

1. \(12x+5y+3z >\) \(\frac{2000d}{k}\)

2. \(10d\) \(=\)\(2^\)\(\sqrt{3x}\) \(=\) \(\sqrt{5y}\)\(=\)\(\sqrt{3z}\)\(=\) \(\frac{100}{k}\)


60% orange juice, 25% grape juice, and 15% apple juice by weight.
Means per kg cost = 0.6x+0.25y+0.15z.... is it >100d/k..
So 0.6x+0.25y+0.15z>100d/k.....
\(12x+5y+3z>\frac{20*100d}{k}\)

Let's see what each statement tells us..

Statement I tells us exactly that..
Sufficient

Second statement tells us 12x=5y=3z=10d*10d..
Therefore 12x+5y+3z=3*10d*10d=3*10d*100/k=3000d/k which is >2000d/k
Ans YES
Sufficient

D


Hi,

Could you explain the detailed working out of the below:

"Second statement tells us 12x=5y=3z=10d*10d..
Therefore 12x+5y+3z=3*10d*10d=3*10d*100/k=3000d/k which is >2000d/k"
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Re: A juice company is creating a new fruit punch that is 60% orange juice   [#permalink] 04 May 2019, 22:45
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