BTGmoderatorDC wrote: ↑Wed Mar 11, 2020 5:36 pm
Leon, Billy, and Jim ate an astounding total of 78 hot dogs in a hot dog eating contest. If Jim ate four times as many hot dogs as Leon but half as many hot dogs as Billy, how many more hot dogs did Billy eat than Leon?
A. 13
B. 20
C. 39
D. 42
E. 60
OA
D
Source: Veritas Prep
Solution:
We can let the number of hot dogs eaten by Leon, Billy, and Jim = L, B, and J, respectively, and create the equations:
L + B + J = 78
and
J = 4L
J/4 = L
and
J = (1/2)B
2J = B
Substituting, we have:
J/4 + 2J + J = 78
J/4 + 3J = 78
Multiplying by 4 we have:
J + 12J = 312
13J = 312
J = 24
Thus, B = 2(24) = 48 and L = 24/4 = 6.
So Billy ate 48 - 6 = 42 more hot dogs than Leon.
Alternate Solution:
Let n represent the number of hot dogs eaten by Leon. Then, Jim ate 4n hot dogs and Billy ate 8n hot dogs. In terms of n, the three of them ate n + 4n + 8n = 13n hot dogs. Setting this equal to 78, we obtain:
13n = 78
n = 6
In terms of n, Billy ate 8n - n = 7n more hot dogs than Leon. Thus, Billy ate 7n = 7(6) = 42 more hot dogs.
Answer: D
