A man spends $48 to buy 6 hamburgers and 8 colas for his

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AAPL: Moderator; Posts: 2749; Joined: Sun Oct 29, 2017 2:08 pmFollowed by: 2 members

by AAPL » Mon Apr 15, 2019 4:53 am

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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

Source: Beat The GMAT — Problem Solving | Mon Apr 15, 2019 4:53 am

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Brent@GMATPrepNow: GMAT Instructor; Posts: 16206; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

by Brent@GMATPrepNow » Mon Apr 15, 2019 6:23 am

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

Answer: B

Cheers,
Brent

Brent Hanneson - Creator of GMATPrepNow.com

swerve: Legendary Member; Posts: 2604; Joined: Sun Oct 29, 2017 2:04 pmFollowed by: 6 members

by swerve » Tue Apr 16, 2019 10:50 am

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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