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At an upscale fastfood restaurant, Shin can buy 3 burgers, 7 shakes,
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22 Jul 2007, 13:59
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44% (01:12) correct 56% (01:05) wrong based on 176 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 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. M0001
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Re: At an upscale fastfood restaurant, Shin can buy 3 burgers, 7 shakes,
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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 fastfood restaurant, Shin can buy 3 burgers, 7 shakes,
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08 Sep 2014, 03:20
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 fastfood restaurant, Shin can buy 3 burgers, 7 shakes,
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13 Nov 2014, 01:03
anceer wrote: 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 price of one burger=x, price of one shake= y, and price of one cola= z 3x+7y+z = 1201) 4x+10y+z=164.52) subtracting 1 from 2 we have x+3y=44.5 now multiply both sides by 3, 3x+9y=133.53) from 2 we have x+3x +9y+y+z=164.5 substitute the value of 3x+9y in 3 we have x+y+z=164.5133.5 =31



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Re: At an upscale fastfood restaurant, Shin can buy 3 burgers, 7 shakes,
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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 fastfood restaurant, Shin can buy 3 burgers, 7 shakes,
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13 Nov 2014, 06:21
jimmyjamesdonkey wrote: 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 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: M0001 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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