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47%
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At a certain restaurant, burgers, fries, and soft drinks are sold individually but can also be combined in a value meal consisting of a burger, an order of fries, and a soft drink. The cost of the value meal is less than the costs of the three items combined. If Hubert bought 3 value meals, 2 individual burgers, and an individual soft drink for a total of $24, and if all individual items and value meals cost whole dollar amounts, what is the price of one value meal?
(1) The total cost for an individual burger and an individual soft drink is $5.
(2) An order of fries is more expensive than a soft drink and less expensive than a burger.
(1) The total cost for an individual burger and an individual soft drink is $5.
(2) An order of fries is more expensive than a soft drink and less expensive than a burger.
28 May 2013, 06:06
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At a certain restaurant, burgers, fries, and soft drinks are sold individually but can also be combined in a value meal consisting of a burger, an order of fries, and a soft drink. The cost of the value meal is less than the costs of the three items combined. If Hubert bought 3 value meals, 2 individual burgers, and an individual soft drink for a total of $24, and if all individual items and value meals cost whole dollar amounts, what is the price of one value meal?
Cost of a burger = B;
Cost of fries = F;
Cost of a drink = D;
Cost of a value meal = V.
Given: 3V+2B+D=24 (V=8-(2B+D/3)) and B+F+D<V.
Question: V=8-(2B+D/3)=?
(1) The total cost for an individual burger and an individual soft drink is $5 --> B+D=5 --> (B, D) could be (1, 4), (2, 3), (3, 2), or (4, 1).
Now, if (B, D) is (2, 3) or (3, 2), then V=8-(2B+D/3) won't be an integer, thus these values are not possible.
Therefore, valid solutions for (B, D) are (1, 4) or (4, 1).
Therefore V=8-(2B+D/3)=6 or V=8-(2B+D/3)=5.
Not sufficient.
(2) An order of fries is more expensive than a soft drink and less expensive than a burger --> D<F<B. Not sufficient.
(1)+(2) Since from (2) D<B, then D=1 and B=4 --> V=8-(2B+D/3)=5. Sufficient.
Answer: C.
Hope it's clear.
Cost of a burger = B;
Cost of fries = F;
Cost of a drink = D;
Cost of a value meal = V.
Given: 3V+2B+D=24 (V=8-(2B+D/3)) and B+F+D<V.
Question: V=8-(2B+D/3)=?
(1) The total cost for an individual burger and an individual soft drink is $5 --> B+D=5 --> (B, D) could be (1, 4), (2, 3), (3, 2), or (4, 1).
Now, if (B, D) is (2, 3) or (3, 2), then V=8-(2B+D/3) won't be an integer, thus these values are not possible.
Therefore, valid solutions for (B, D) are (1, 4) or (4, 1).
Therefore V=8-(2B+D/3)=6 or V=8-(2B+D/3)=5.
Not sufficient.
(2) An order of fries is more expensive than a soft drink and less expensive than a burger --> D<F<B. Not sufficient.
(1)+(2) Since from (2) D<B, then D=1 and B=4 --> V=8-(2B+D/3)=5. Sufficient.
Answer: C.
Hope it's clear.
11 Feb 2014, 20:48
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Bunuel
Bunuel,
Infact when you consider the case (B,D) = (2,3) then V=8-(2B+D/3)=5 will be V=8-(2*2+3/3)=3-->Integer. On the contrary, how did you arrive at integer values for (1,4) and (4,1). Both 4 and 1 are not divisible by 3.
Could you help me?
