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

Say Leon, Billy, and Jim ate L, B, and J no. of hot dogs.

Thus, L + B + J = 78.

We have to get the value of B – L.

From the information, we get J = 4L = B/2.

=> J = 4L and B = 8L

From L + B + J = 78, we have L + 8L + 4L = 78 => L = 6 and B = 8*6 = 48

Thus, B – L = 48 – 6 = 42

The correct answer: D

Hope this helps!

-Jay
_________________
Manhattan Review

Locations: Manhattan Review Chennai | GMAT Prep Himayatnagar | GRE Prep Hyderabad | Bangalore GRE Coaching | and many more...

Schedule your free consultation with an experienced GMAT Prep Advisor! Click here.

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