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

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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

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hamburgers and colas

by GMATGuruNY » Wed Jul 31, 2019 2:43 am
BTGmoderatorDC wrote: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
If 6h+8c = 48 and 5h+4c = 32, what is the value of h+c?
In

In the blue equation, the coefficient for h is LESS than the coefficient for c.
In the red equation, the coefficient for h is GREATER than the coefficient for c.

Given this situation, we can use the following approach to determine the value of h+c:
1. Alter the equation(s) so that the difference between the coefficients in each equation is THE SAME
2. Add the two equations
3. Divide as necessary to determine the value of h+c

In the blue equation, the difference between the coefficients = 8-6 = 2.
In the red equation, the difference between the coefficients = 5-4 = 1.
If we divide the blue equation by 2, the difference between the coefficients will decrease to 1:
6h+8c = 48 --> 3h+4c = 24 --> 4-3 = 1

Adding together 3h+4c = 24 and 5h+4c = 32, we get:
8h+8c = 56
Dividing both sides by 8, we get:
h+c = 7

The correct answer is B.
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by swerve » Thu Aug 01, 2019 4:25 pm
Let the price of one hamburger be \(x\) and that of cola is \(y\)

\(6x+8y=48 \Rightarrow 3x+4y=24\qquad (1)\)
\(5x+4y=32\qquad (2)\)

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, \(x+y=4+3=7\Rightarrow \) __B__.

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by Scott@TargetTestPrep » Mon Aug 05, 2019 4:11 pm
BTGmoderatorDC wrote: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
Letting h = the cost of a hamburger and c = the cost of a cola, we can create the equations:

48 = 6h + 8c

24 = 3h + 4c

and

32 = 5h + 4c

Subtracting equation one from equation two, we have:

8 = 2h

4 = h

So c is:

32 = 20 + 4c

12 = 4c

3 = c

So the cost of one cola and one hamburger is c + h = $7.

Answer: B

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