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A street vendor sells only hot dogs and hamburgers, and at the beginni
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10 Feb 2017, 19:24
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75% (02:15) correct 25% (02:13) wrong based on 118 sessions
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A street vendor sells only hot dogs and hamburgers, and at the beginning of the day has a ratio of two hot dogs for every one hamburger. At the end of the day in which he did not add any new items or sell any hamburgers, and only sold some of his hot dogs, his new ratio is one hot dog for every two hamburgers. Which of the following cannot represent the number of hot dogs he sold? (A) 2 (B) 3 (C) 6 (D) 9 (E) 24
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Re: A street vendor sells only hot dogs and hamburgers, and at the beginni
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10 Feb 2017, 23:58
ziyuenlau wrote: A street vendor sells only hot dogs and hamburgers, and at the beginning of the day has a ratio of two hot dogs for every one hamburger. At the end of the day in which he did not add any new items or sell any hamburgers, and only sold some of his hot dogs, his new ratio is one hot dog for every two hamburgers. Which of the following cannot represent the number of hot dogs he sold?
(A) 2 (B) 3 (C) 6 (D) 9 (E) 24 Begining of the day D : H = 2x : x End of the day \(\frac{2x  s}{x} = \frac{1}{2}\) Or, \(4x  2s = x\) Or, \(3x = 2s\) Or, \(s = \frac{3x}{2}\) Check carefully the number of hot dogs sols must be a multiple of 3 , among the given options none but (A) is correct...
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Re: A street vendor sells only hot dogs and hamburgers, and at the beginni
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11 Feb 2017, 05:32
ziyuenlau wrote: A street vendor sells only hot dogs and hamburgers, and at the beginning of the day has a ratio of two hot dogs for every one hamburger. At the end of the day in which he did not add any new items or sell any hamburgers, and only sold some of his hot dogs, his new ratio is one hot dog for every two hamburgers. Which of the following cannot represent the number of hot dogs he sold?
(A) 2 (B) 3 (C) 6 (D) 9 (E) 24 Here's an approach that uses three variables. Let D = # of hotdogs the vendor STARTED with Let H = # of hamburgers the vendor STARTED with Let x = # of hotdogs sold At the beginning of the day has a ratio of two hot dogs for every one hamburger.So, D/H = 2/1 Cross multiply to get: D = 2H At the end of the day in which he did not add any new items or sell any hamburgers, and only sold some of his hot dogs, his new ratio is one hot dog for every two hamburgers. So, (D  x)/H = 1/2 Cross multiply to get: 2(D  x) = HSo, we have the following system: D = 2H2(D  x) = HTake 2(D  x) = H and replace D with 2HWe get: 2(2H  x) = H Expand: 4H  2x = H Solve for H to get: H = 2x/3 Since H must be an INTEGER, we can see that x must be divisible by 3. Answer choice A is the only answer that is NOT divisible by 3. ALTERNATE APPROACH Once we get to the point where we have 4H  2x = H, we can also solve for x to get: x = 3H/2At that point, we can check each answer choice to see what happens when x = that certain amount. For example. let's check answer choice E first It tells us that x = 24 Plug x = 24 into our equation: 24 = 3H/2Multiply both sides by 2 to get: 48 = 3H Solve for H to get: H = 16 No problem. When x = 24, we get a nice integer value for the number of Hamburgers sold. Now let's check answer choice A It tells us that x = 2 Plug x = 2 into our equation: 2 = 3H/2Multiply both sides by 2 to get: 4 = 3H Solve for H to get: H = 4/3 Problem. When x = 2, we get a NONinteger value for the number of Hamburgers sold. So, it is impossible to sell 2 hotdogs. Cheers, Brent
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Re: A street vendor sells only hot dogs and hamburgers, and at the beginni
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11 Feb 2017, 05:55
Abhishek009 wrote: ziyuenlau wrote: A street vendor sells only hot dogs and hamburgers, and at the beginning of the day has a ratio of two hot dogs for every one hamburger. At the end of the day in which he did not add any new items or sell any hamburgers, and only sold some of his hot dogs, his new ratio is one hot dog for every two hamburgers. Which of the following cannot represent the number of hot dogs he sold?
(A) 2 (B) 3 (C) 6 (D) 9 (E) 24 Begining of the day D : H = 2x : x End of the day \(\frac{2x  s}{x} = \frac{1}{2}\) Or, \(4x  2s = x\) Or, \(3x = 2s\) Or, \(s = \frac{3x}{2}\) Check carefully the number of hot dogs sols must be a multiple of 3 , among the given options none but (A) is correct...I could not understand as to why s should be a multiple of 3? Should it not be a multiple of 2 so that 2 in the denominators cancels with the numerator leaving an integer as the answer. I know I am missing some point but not able to understand why multiple of 3 and not 2 since s = 3x/2?



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Re: A street vendor sells only hot dogs and hamburgers, and at the beginni
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12 Feb 2017, 07:22
grichagupta wrote: I could not understand as to why s should be a multiple of 3? Should it not be a multiple of 2 so that 2 in the denominators cancels with the numerator leaving an integer as the answer. I know I am missing some point but not able to understand why multiple of 3 and not 2 since s = 3x/2? Abhishek009 concluded that s = 3x/2, where s = # of hotdogs sold. In other words, s equals some multiple of 3 (which is also divided by 2). So, s is definitely a multiple of 3 ALSO, since we know that s must be an INTEGER, we can also conclude that x (not s) must be a multiple of 2 We can also convince ourselves of this fact by testing some values of x. For example, if x = 2, then s = 3 If x = 4, then s = 6 If x = 6, then s = 9 If x = 8, then s = 12 Also notice that if we let x = 5 (an odd number), then s = 7.5, which makes no sense. Cheers, Brent
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A street vendor sells only hot dogs and hamburgers, and at the beginni
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12 Feb 2017, 13:22
ziyuenlau wrote: A street vendor sells only hot dogs and hamburgers, and at the beginning of the day has a ratio of two hot dogs for every one hamburger. At the end of the day in which he did not add any new items or sell any hamburgers, and only sold some of his hot dogs, his new ratio is one hot dog for every two hamburgers. Which of the following cannot represent the number of hot dogs he sold?
(A) 2 (B) 3 (C) 6 (D) 9 (E) 24 let t=total beginning hot dogs and hamburgers let d=hot dogs sold (2/3*t)d=1/3*(td) d=t/2 because of 1:2 ratio, t must divide by 3 plugging in 2 as d, t=4, a nonmultiple of 3 A



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Re: A street vendor sells only hot dogs and hamburgers, and at the beginni
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13 Feb 2017, 21:33
Hi ziyuenlau, This question can be solved in a number of different ways  but you can avoid a lot of the complex math if you think in terms of the ratios that are involved and TEST VALUES. The question asks for what CANNOT be the number of hotdogs sold... We're told that the STARTING ratio of hotdogs to hamburgers is 2:1 and the ENDING ratio of hotdogs to hamburgers is 1:2. Since we only sold hotdogs, the number of hamburgers stays the same (and must be an EVEN number  since the ending ratio is 1:2). Thus, we could have 2, 4, 6, 8, 10, etc. hamburgers and the number of hotdogs that we START with is DOUBLE the number of hamburgers. We can TEST VALUES to define the pattern behind this question. IF... we start with 2 hamburgers, then we start with 4 hotdogs. To end with a 1:2 ratio, we had to sell 3 hotdogs. we start with 4 hamburgers, then we start with 8 hotdogs. To end with a 1:2 ratio, we had to sell 6 hotdogs. we start with 6 hamburgers, then we start with 12 hotdogs. To end with a 1:2 ratio, we had to sell 9 hotdogs. we start with 8 hamburgers, then we start with 16 hotdogs. To end with a 1:2 ratio, we had to sell 12 hotdogs. Etc. Notice how as we increase the number of hamburgers by 2, we increase the number of hotdogs sold by 3. Thus, we can "hit" every answer except for.... Final Answer: GMAT assassins aren't born, they're made, Rich
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Re: A street vendor sells only hot dogs and hamburgers, and at the beginni
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26 Mar 2018, 16:47
hazelnut wrote: A street vendor sells only hot dogs and hamburgers, and at the beginning of the day has a ratio of two hot dogs for every one hamburger. At the end of the day in which he did not add any new items or sell any hamburgers, and only sold some of his hot dogs, his new ratio is one hot dog for every two hamburgers. Which of the following cannot represent the number of hot dogs he sold?
(A) 2 (B) 3 (C) 6 (D) 9 (E) 24 We can let x = the number of hamburgers at the beginning of the day; thus, 2x = the number of hot dogs at the beginning of the day. We can create the equation in which n = the number of hot dogs sold. Thus we have: (2x  n)/x = 1/2 2(2x  n) = x 4x  2n = x 3x = 2n We see that n can’t be 2; otherwise 2n = 4. However, 3x can’t be equal to 4 since x is an integer. Answer: A
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