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Joined: 02 Sep 2009
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Question Stats:
53% (01:44) correct 47% (01:59) wrong based on 72 sessions
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At an upscale fastfood restaurant, Shin can buy 3 burgers, 7 shakes, and one cola for $120. At the same place it would cost $164.50 for 4 burgers, 10 shakes, and one cola. How much would it cost for a meal of one burger, one shake, and one cola? A. $21 B. $27 C. $31 D. $41 E. It cannot be determined
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Re M0001 [#permalink]
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16 Sep 2014, 00:14
Official Solution:At an upscale fastfood restaurant, Shin can buy 3 burgers, 7 shakes, and one cola for $120. At the same place it would cost $164.50 for 4 burgers, 10 shakes, and one cola. How much would it cost for a meal of one burger, one shake, and one cola? A. $21 B. $27 C. $31 D. $41 E. It cannot be determined Let's suppose that the price of a burger is \(B\), of a shake  \(S\) and that of a cola is \(C\). We can then construct these equations: \(3B+7S+C = 120\) \(4B+10S+C = 164.5\) Subtracting the first equation from the second gives us \(B+3S=44.5\). Now if we subtract the new equation two times from first or 3 times from second we will get \(B+S+C=31\). In any case, there is no necessity to know each item's price, just the sum. Answer: C
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Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
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Re: M0001 [#permalink]
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18 May 2015, 06:50
Another solution: Given: 3B+ 7S + C = 120 Multiply by 3: 9B+21S+3C=360 (1) 4B + 10S + C = 164.5 Multiply by 2: 8B + 20S + 2C = 329 (2) => (1)  (2) = 31.



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Joined: 02 Nov 2014
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Re M0001 [#permalink]
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13 Oct 2015, 04:50
I think this is a highquality question and I agree with explanation. Great Q. Thanks,
Binit.



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Re: M0001 [#permalink]
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14 Oct 2015, 12:27
Multiply first equation by 2: 6B+14S+2C=240
subtract second equation from the new one 6B+14S+2C=240 _ 4B+10S+1C=164.50
get: 2B+4S+1C=75.50
now, if we subtract from the second equation the first one before we multiply, we get: 4B+10S+1C=164.50 _ 3B+7S+1C=120 and get B+3S = 44.50
ok, now we're getting to something. 2B+4S+1C=75.50 and B+3S = 44.50 substract the second one from the first one and get B+S+C=31. answer choice C.



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Re: M0001 [#permalink]
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28 Aug 2016, 12:46
I understood it when I solved it again without any time constraints... Here's what I learned from this:
When writing more than one linear equation (with the same variables), write them one under the other for the facilitation of a quick insight!



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Re: M0001 [#permalink]
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27 Dec 2016, 01:16
Is there any way for solving 3 variable equations as fast as possible.



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Re M0001 [#permalink]
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25 Jul 2017, 20:12
I think this is a highquality question and I agree with explanation. I was looking for some type of hook. Very good question!



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Re: M0001 [#permalink]
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22 Aug 2017, 11:04
Very difficult to answer under time constraints; any simple way to figure out or hit and trial is the only option?
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Re: M0001 [#permalink]
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22 Aug 2017, 11:21



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Re: M0001 [#permalink]
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02 Dec 2017, 02:06
I did it like this
3B + 7S+ Z= 120 1
4B + 10S +Z = 164.50 2
Multiply eq1 by 3 & eq2 by 2
9B + 21S+ 3Z= 360
8B + 20S+ 2Z = 329
Now subtract above eqns B + S+ Z= 31
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Re M0001 [#permalink]
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28 Dec 2017, 07:07
I think this is a highquality question and I agree with explanation.



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Re M0001 [#permalink]
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15 Jan 2018, 18:54
I think this is a highquality question and I don't agree with the explanation. I have now understood the equations underlying.. Perhaps, it should have been explained. Nevertheless, a quality question !










