Shin visits an upscale fast-food restaurant where he can purchase a combination of burgers, shakes, and colas. He notices that a meal consisting of 3 burgers, 7 shakes, and one cola costs $120, while a meal with 4 burgers, 10 shakes, and one cola costs $164.50. What is the cost of a single meal that includes one burger, one shake, and one cola at this restaurant? A. $21 B. $27 C. $31 D. $41 E. It cannot be determined

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Gemmie

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01 Jul 2024, 11:03

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3B + 7S + C = 120 (A)
4B + 10S + C = 164.5 (B)

(A) * 3: 9B + 21S + 3C = 360
(B) * 2: 8B + 20S + 2C = 329

B + S + C = 31

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16 Dec 2024, 23:06

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This needs more traction. Great solution!

Anshkarha

buan15

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.