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Joined: 02 Sep 2009
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15 Sep 2014, 23:14
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Difficulty:

95% (hard)

Question Stats:

48% (01:45) correct 52% (01:55) wrong based on 85 sessions

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At an upscale fast-food 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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15 Sep 2014, 23:14
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Official Solution:

At an upscale fast-food 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.

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18 May 2015, 05:50
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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.
Manager
Joined: 02 Nov 2014
Posts: 187
GMAT Date: 08-04-2015

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13 Oct 2015, 03:50
I think this is a high-quality question and I agree with explanation. Great Q.
Thanks,

Binit.
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Joined: 17 Jul 2014
Posts: 2599
Location: United States (IL)
Concentration: Finance, Economics
GMAT 1: 650 Q49 V30
GPA: 3.92
WE: General Management (Transportation)

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14 Oct 2015, 11:27
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.
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Joined: 14 May 2016
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28 Aug 2016, 11:46
1
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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Joined: 13 Jul 2016
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GMAT 1: 740 Q49 V41

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27 Dec 2016, 00:16
Is there any way for solving 3 variable equations as fast as possible.
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25 Jul 2017, 19:12
I think this is a high-quality question and I agree with explanation. I was looking for some type of hook. Very good question!
Manager
Joined: 14 Jun 2016
Posts: 67
Location: India
GMAT 1: 610 Q49 V21
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22 Aug 2017, 10:04
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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Joined: 02 Sep 2009
Posts: 52294

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22 Aug 2017, 10:21
buan15 wrote:
Very difficult to answer under time constraints; any simple way to figure out or hit and trial is the only option?

Check here: https://gmatclub.com/forum/at-an-upscal ... 49205.html
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Joined: 27 Sep 2016
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Concentration: Strategy, General Management
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02 Dec 2017, 01:06
I did it like this-

3B + 7S+ Z= 120 -------1

4B + 10S +Z = 164.50 -----------2

Multiply eq-1 by 3 & eq-2 by 2

9B + 21S+ 3Z= 360

8B + 20S+ 2Z = 329

Now subtract above eqns
B + S+ Z= 31

##############################
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Joined: 20 Sep 2012
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28 Dec 2017, 06:07
I think this is a high-quality question and I agree with explanation.
Intern
Joined: 10 Sep 2017
Posts: 1

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15 Jan 2018, 17:54
I think this is a high-quality 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 !
Intern
Joined: 22 Jul 2018
Posts: 6

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03 Sep 2018, 10:19
buan15 wrote:
Very difficult to answer under time constraints; any simple way to figure out or hit and trial is the only option?

There's a mechanical way of solving any question like this without relying on hit and trial based on linear algebra

The idea is that we are manipulating the equations by multiplying with a constant, and then adding the 2 equations to get our desired equation.
Our desired equation is one that has the same coefficient for B, S and C

eq 1 - 3B + 7S + C multiply this equation by x
eq 2 - 4B + 10S + C multiply this equation by y

Since we want the coefficients the same after adding, 7x+ 4y (coefficient of B) = 7x + 10y (coefficient of S) = x+y (coefficient of C)
this gives x = -3y/2
Since for simplicity we want x & y to be integers, we can use y = -2, so x = 3
So if we multiply eq 1 by 3 and eq 2 by -2, and add the two equations together we will get B + S + C.

Not sure if my explanation is complicated but in my opinion its a very easy and foolproof way to solve these type of questions.
Re: M00-01 &nbs [#permalink] 03 Sep 2018, 10:19
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