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At a certain restaurant, the total price for one hamburger a

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At a certain restaurant, the total price for one hamburger a  [#permalink]

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New post 14 Jun 2010, 16:39
1
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Question Stats:

85% (01:03) correct 15% (01:32) wrong based on 60 sessions

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At a certain restaurant, the total price for one hamburger and one drink is $3.50. How much does one drink cost?

(1) If the price of the drink were half of its present price, the price of a drink would be 1/5 of the price of a hamburger.

(2) The cost of one hamburger and two drinks is $4.50.
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Re: Kaplan Premier Data Sufficiency Test, Problem #36  [#permalink]

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New post 14 Jun 2010, 17:18
Bradrs24 wrote:
Hi, I was just wondering how you guys would solve this problem. I got it wrong because I didn't use Kaplan's n-variable, n-equations rule. Which basically states that if for every variable there is a linear equation , then it is sufficient.

At a certain restaurant, the total price for one hamburger and one drink is $3.50. How much does one drink cost?

Statement 1- If the price of the drink were half of its present price, the price of a drink would be 1/5 of the price of a hamburger.

Statement 2- The cost of one hamburger and two drinks is $4.50.

- I picked B, because I thought the problem couldn't be solved from statement one. But to some things up, I was just wondering if you guys apply the n-variable, n-equations rule to statement 1 and 2 and move on knowing you got the problem correctly.


This rule needs important restriction: equations for variables must be distinct, meaning that no equation can be derived with the help of others or by arithmetic operation (multiplication, addition).

For example:
\(x+y=2\) and \(3x+3y=6\) --> we do have two linear equations and two variables but we can not solve for \(x\) or \(y\) as the second equation is just the first one multiplied by 3 (basically we have only one distinct equation);
OR
\(x+y=1\), \(y+z=2\) and \(x+2y+z=3\) --> we have 3 linear equations and 3 variables but we can not solve for \(x\), \(y\) or \(z\) as the third equation can be derived with the help of first two if we sum them (basically we have only two distinct equation);

As for the original question:
Given: \(h+d=3.5\). Question: \(d=?\)

(1) \(\frac{d}{2}=\frac{h}{5}\). Two distinct linear equations for two variables, hence sufficient.

(2) \(h+2d=4.5\). Two distinct linear equations for two variables, hence sufficient.

Answer: D.

Hope it helps.
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Re: At a certain restaurant, the total price for one hamburger a  [#permalink]

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New post 20 Dec 2018, 00:30
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Re: At a certain restaurant, the total price for one hamburger a   [#permalink] 20 Dec 2018, 00:30
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