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At an upscale fast-food restaurant, Shin can buy 3 burgers, 7 shakes,

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Director
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At an upscale fast-food restaurant, Shin can buy 3 burgers, 7 shakes,  [#permalink]

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New post 22 Jul 2007, 13:59
2
12
00:00
A
B
C
D
E

Difficulty:

  85% (hard)

Question Stats:

44% (01:12) correct 56% (01:05) wrong based on 176 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

This is a challenge problem, but from the FREE PRACTICE BIN, so don't freak out on my for posting a challenge problem on the forum. I've never seen this type of explanation before. I always assumed that if you had 3 variables, you needed 3 equations to solve. I've also never seen a solution where you subtract multiple times from the same equation. Is this even kosher? Is there a way to solve this without subtracting multiple times by the same #?

Explanation

Let's suppose that the price of burger is $B, of shake $S and cola's price is $C. We can then construct equations:
{3B+7S+C=$120
{4B+10S+C=$164.5

Subtracting first equation from the second, gives us B+3S=$44.5.
Now if we subtract 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 items price, just the sum.
:evil:


M00-01
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Re: At an upscale fast-food restaurant, Shin can buy 3 burgers, 7 shakes,  [#permalink]

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New post 22 Jul 2007, 14:50
That's really helpful! I've never seen that type of solution before either, but there's no reason why it shouldn't work. If B+32=$44.50 is a true statement, there's no reason why it couldn't be applied as many times as needed to get to the answer.
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Re: At an upscale fast-food restaurant, Shin can buy 3 burgers, 7 shakes,  [#permalink]

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New post 08 Sep 2014, 03:20
7
6
3B + 7S + 1C = 120
4B + 10S + 1C = 164.50

Subtracting 1 from 2

B + 3S = 44.50

Now MULTIPLY ABOVE by 2

2B + 6S = 89

First equation can be written as:

B + 2B+6S + S + C = 120

OR

B + S + C + 89 = 120

B + S + C = 31
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Re: At an upscale fast-food restaurant, Shin can buy 3 burgers, 7 shakes,  [#permalink]

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New post 13 Nov 2014, 01:03
1
anceer wrote:
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

price of one burger=x, price of one shake= y, and price of one cola= z

3x+7y+z = 120------------1)
4x+10y+z=164.5-------------2)

subtracting 1 from 2 we have
x+3y=44.5

now multiply both sides by 3,
3x+9y=133.5------------------3)

from 2 we have
x+3x +9y+y+z=164.5
substitute the value of 3x+9y in 3 we have
x+y+z=164.5-133.5
=31
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Re: At an upscale fast-food restaurant, Shin can buy 3 burgers, 7 shakes,  [#permalink]

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New post 13 Nov 2014, 02:58
3b + 7s + 1c = 120 ......... (1)

4b + 10s + 1c = 164.5 ........ (2)

(2) - (1)

1b + 3s = 44.5

3b + 9s = 44.5*3 .............. (3)

Rearranging equation (2)

(3b + 9s) + (1a + 1s + 1c) = 164.5

1a + 1s + 1c = 164.5 - 133.5 = 31

Answer = C
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At an upscale fast-food restaurant, Shin can buy 3 burgers, 7 shakes,  [#permalink]

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New post 13 Nov 2014, 06:21
2
1
jimmyjamesdonkey wrote:
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

This is a challenge problem, but from the FREE PRACTICE BIN, so don't freak out on my for posting a challenge problem on the forum. I've never seen this type of explanation before. I always assumed that if you had 3 variables, you needed 3 equations to solve. I've also never seen a solution where you subtract multiple times from the same equation. Is this even kosher? Is there a way to solve this without subtracting multiple times by the same #?

Explanation

Let's suppose that the price of burger is $B, of shake $S and cola's price is $C. We can then construct equations:
{3B+7S+C=$120
{4B+10S+C=$164.5

Subtracting first equation from the second, gives us B+3S=$44.5.
Now if we subtract 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 items price, just the sum.
:evil:


M00-01


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.


Answer: C.
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Re: At an upscale fast-food restaurant, Shin can buy 3 burgers, 7 shakes,  [#permalink]

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Re: At an upscale fast-food restaurant, Shin can buy 3 burgers, 7 shakes, &nbs [#permalink] 27 Aug 2018, 01:48
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