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## A man spends $48 to buy 6 hamburgers and 8 colas for his ## Timer 00:00 ## Your Answer A B C D E ## Global Stats Difficult EMPOWERgmat A man spends$48 to buy 6 hamburgers and 8 colas for his office workers. The next day, he buys 5 hamburgers and 4 colas and spends $32. Assuming the prices of hamburgers and colas remain constants, what is the price of one hamburger and one cola? A.$6
B. $7 C.$8
D. $9 E.$10

OA B

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AAPL wrote:
EMPOWERgmat

A man spends $48 to buy 6 hamburgers and 8 colas for his office workers. The next day, he buys 5 hamburgers and 4 colas and spends$32. Assuming the prices of hamburgers and colas remain constants, what is the price of one hamburger and one cola?

A. $6 B.$7
C. $8 D.$9
E. $10 OA B Let H = price of one hamburger Let C = price of one cola A man spends$48 to buy 6 hamburgers and 8 colas for his office workers.
6H + 8C = 48

The next day, he buys 5 hamburgers and 4 colas and spends $32. 5H + 4C = 32 Assuming the prices of hamburgers and colas remain constant, what is the price of one hamburger and one cola? We have: 6H + 8C = 48 5H + 4C = 32 Take TOP equation and divide both sides by 2 to get: 3H + 4C = 24 5H + 4C = 32 Subtract the bottom equation from the top equation to get: -2H = -8 Solve: H = 4 Now plug H = 4 into 3H + 4C = 24 to get: 3(4) + 4C = 24 Simplify: 12 + 4C = 24 So: 4C = 12 Solve: C = 3 So the price of one hamburger and one cola = H + C = 4 + 3 =$7

Cheers,
Brent

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Let the price of one hamburger be $$x$$ and that of cola be $$y$$

$$6x+8y=48 \Rightarrow 3x+4y=24 \cdots (1)$$

$$5x+4y=32 \cdots (2)$$

Subtract equation $$(1)$$ from $$(2)$$ to get $$2x=8$$ or $$x=4$$

Substitute the value of $$x$$ in any equation to get $$y=3$$

Hence $$x+y=4+3=7\quad\Rightarrow\quad$$ Option B

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