## A man spends $48 to buy 6 hamburgers and 8 colas for his office workers. The next day, he buys 5 hamburgers and 4 ##### This topic has expert replies Moderator Posts: 5938 Joined: 07 Sep 2017 Followed by:19 members ### 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 BTGmoderatorDC » Sun Jul 25, 2021 8:14 pm

00:00

A

B

C

D

E

## Global Stats

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 ### GMAT/MBA Expert GMAT Instructor Posts: 15760 Joined: 08 Dec 2008 Location: Vancouver, BC Thanked: 5254 times Followed by:1266 members GMAT Score:770 ### 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 [email protected] » 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

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com

Legendary Member
Posts: 2031
Joined: 29 Oct 2017
Followed by:6 members

### 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

• Page 1 of 1