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Joined: 02 Sep 2009
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A cafeteria served hamburgers and hot dogs on a certain day. If the co [#permalink]
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01 Aug 2017, 11:24
Question Stats:
88% (00:59) correct 13% (00:26) wrong based on 30 sessions
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Re: A cafeteria served hamburgers and hot dogs on a certain day. If the co [#permalink]
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01 Aug 2017, 11:47
Let, hamburger=x hotdogs=y We have to find 5x+3y=?
Statement 1: x+y=130 Individually x and y can have infinite values to add upto 130. Insufficient.
Statement 2: x/y=3/2. > x=(3y)/2
Combining 1 and 2: (3y)/2+ y=130 We get value of y and hence also value of x...finially giving the ultimate result for 5x+3y.
Ans: C



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Re: A cafeteria served hamburgers and hot dogs on a certain day. If the co [#permalink]
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01 Aug 2017, 11:49
Answer is C
Simplifying the question stem gives us the following equation 5x +3y = ?. Where x is the number of hamburgers and y is the number of hot dogs. We need to find what "?" Is.
Statement 1 gives us the equation x+y=130 but this could mean x=100 and y=30 or x=30 and y=100, each of those gives us a different answer for "?" Therefore this statement is NOT SUFFICIENT
Statement 2 tells us that the Ham:Hot is 3:2, this again doesn't give us enough information since x could be 3 and y could be 2 or x=30 and y=20. Both of those would give different answers for "?" Therefore this statement is NOT SUFFICIENT
1+2, now that we know the total number of hamburgers and hamburgers and the ratio for each, we can find what x and y are since we can find the common multiplier for the ratio z. 3z + 2z = 5z => 5z = 130 => z=26. 3(26) = 78 hamburgers and 2(26) = 52 Hot dogs. Now we can find the total cost. Therefore these two statements together are sufficient




Re: A cafeteria served hamburgers and hot dogs on a certain day. If the co
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01 Aug 2017, 11:49






