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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BTGmoderatorDC: Moderator; Posts: 7187; Joined: Thu Sep 07, 2017 4:43 pmFollowed by: 23 members

by BTGmoderatorDC » Sun Jul 25, 2021 8:14 pm

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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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Source: Beat The GMAT — Problem Solving | Sun Jul 25, 2021 8:14 pm

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

by Brent@GMATPrepNow » Mon Jul 26, 2021 5:28 am

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

Cheers,
Brent

Brent Hanneson - Creator of GMATPrepNow.com

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

by swerve » Mon Jul 26, 2021 11:54 am

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

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