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Intern  Joined: 24 May 2010
Posts: 12
At a certain restaurant, the total price for one hamburger a  [#permalink]

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1
1 00:00

Difficulty:   15% (low)

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.
Math Expert V
Joined: 02 Sep 2009
Posts: 58320
Re: Kaplan Premier Data Sufficiency Test, Problem #36  [#permalink]

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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.

Hope it helps.
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Non-Human User Joined: 09 Sep 2013
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Re: At a certain restaurant, the total price for one hamburger a  [#permalink]

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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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