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Math Expert V
Joined: 02 Sep 2009
Posts: 56300

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13 00:00

Difficulty:   95% (hard)

Question Stats: 45% (02:24) correct 55% (02:52) wrong based on 65 sessions

### HideShow timer Statistics 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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Math Expert V
Joined: 02 Sep 2009
Posts: 56300

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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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Intern  B
Joined: 09 Feb 2015
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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: 185
GMAT Date: 08-04-2015

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I think this is a high-quality question and I agree with explanation. Great Q.
Thanks,

Binit.
Board of Directors P
Joined: 17 Jul 2014
Posts: 2539
Location: United States (IL)
Concentration: Finance, Economics
GMAT 1: 650 Q49 V30 GPA: 3.92
WE: General Management (Transportation)

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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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Intern  B
Joined: 14 May 2016
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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!
Intern  Joined: 13 Jul 2016
Posts: 1
GMAT 1: 740 Q49 V41 ### Show Tags

Is there any way for solving 3 variable equations as fast as possible.
Intern  B
Joined: 17 Dec 2016
Posts: 13
Location: United States (NY)
Concentration: Sustainability
GPA: 3.76
WE: Other (Military & Defense)

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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  B
Joined: 14 Jun 2016
Posts: 65
Location: India
GMAT 1: 610 Q49 V21 WE: Engineering (Manufacturing)

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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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If you appreciate my post then please click +1Kudos Math Expert V
Joined: 02 Sep 2009
Posts: 56300

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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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Intern  B
Joined: 27 Sep 2016
Posts: 1
Location: India
Concentration: Strategy, General Management
WE: Information Technology (Computer Software)

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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

##############################
Intern  S
Joined: 20 Sep 2012
Posts: 6
Location: Russian Federation
Concentration: Finance, General Management
GMAT 1: 710 Q49 V38 GPA: 3.5

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I think this is a high-quality question and I agree with explanation.
Intern  Joined: 10 Sep 2017
Posts: 1

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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  B
Joined: 22 Jul 2018
Posts: 5

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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.
Intern  Joined: 06 May 2018
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How does one identify such questions?

Are there places where we can practice more such questions?

Regards,
Math Expert V
Joined: 02 Sep 2009
Posts: 56300

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Siddhant26930 wrote:
How does one identify such questions?

Are there places where we can practice more such questions?

Regards,

7. Algebra

15. Word Problems

For more check Ultimate GMAT Quantitative Megathread

Hope it helps.
_________________ Re: M00-01   [#permalink] 25 Feb 2019, 01:43
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