A man spends $48 to buy 6 hamburgers and 8 colas for his office workers. The next day, he buys 5 hamburgers and 4

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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 constant, what is the price of one hamburger and one cola?

A. $6
B. $7
C. $8
D. $9
E. $10


OA B

Source: EMPOWERgmat

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BTGmoderatorDC wrote:
Sun Jul 25, 2021 8:14 pm
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 constant, what is the price of one hamburger and one cola?

A. $6
B. $7
C. $8
D. $9
E. $10


OA B

Source: EMPOWERgmat
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

Answer: B

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Brent
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BTGmoderatorDC wrote:
Sun Jul 25, 2021 8:14 pm
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 constant, what is the price of one hamburger and one cola?

A. $6
B. $7
C. $8
D. $9
E. $10


OA B

Source: EMPOWERgmat
Let price of one hamburger be \(x\) and that of cola be \(y\)

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

\(5x+4y=32 \qquad (2)\)

Subtracting equation \((1)\) from \((2)\) to get \(2x=8\) or \(x=4\)

Substituting the value of \(x\) in any equation \(((1)\) or \((2))\), we get \(y=3\)

Hence, \(x+y=4+3=7\)

Therefore, B