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14 Jun 2010, 16:39
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
5%
(low)
Question Stats:
85% (01:05) correct
15%
(01:31)
wrong
based on 143
sessions
History
Date
Time
Result
Not Attempted Yet
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.
(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.
14 Jun 2010, 17:18
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Bradrs24
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.
10 Aug 2021, 22:33
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Bradrs24
Let cost of one hamburger be x
Let cost of one drink be y.
Given : x + y = 3.5
Statement 1 : Sufficient
\(\frac{x}{5} = \frac{y}{2}\)
\(x = \frac{5y}{2}\)
Solving with first equation we get, y = $1
Statement 2 : Sufficient
\(x + 2y = 4.5\)
Solving with first equation, y = $1
Hence, OA is (D).
