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A street vendor sells only hot dogs and hamburgers, and at..

Expert replies
by BTGmoderatorLU » Thu Mar 22, 2018 5:47 pm
A street vendor sells only hot dogs and hamburgers, and at the beginning of the day has a ratio of two hot dogs for every one hamburger. At the end of the day in which he did not add any new items or sell any hamburgers, and only sold some of his hot dogs, his new ratio is one hot dog for every two hamburgers. Which of the following cannot represent the number of hot dogs he sold?

A. 2
B. 3
C. 6
D. 9
E. 24

The OA is A.

I'm really confused by this PS question. Experts, any suggestion about how can I solve it? Thanks in advance.
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Source: — Problem Solving |

by GMATGuruNY » Thu Mar 22, 2018 6:08 pm
LUANDATO wrote:A street vendor sells only hot dogs and hamburgers, and at the beginning of the day has a ratio of two hot dogs for every one hamburger. At the end of the day in which he did not add any new items or sell any hamburgers, and only sold some of his hot dogs, his new ratio is one hot dog for every two hamburgers. Which of the following cannot represent the number of hot dogs he sold?

A. 2
B. 3
C. 6
D. 9
E. 24
Let x = the multiplier for the original ratio and s = the number of hot dogs sold.

Original ratio of hot dogs to hamburgers = 2x/x.
After s hot dogs are sold, the remaining number of hot dogs = 2x - s.
Since the resulting ratio of hot dogs to hamburgers is 1 to 2, we get:
(2x-s)/x = 1/2
4x - 2s = x
3x = 2s
s = (3/2)x.

If x = 2, then s = (3/2) * 2 = 3.
If x = 4, then s = (3/2) * 4 = 6.
If x = 6, then s = (3/2) * 6 = 9.
The resulting values of s indicate that the number of hot dogs sold can be any positive multiple of 3.
Eliminate B, C, D and E.

The correct answer is A.
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by Brent@GMATPrepNow » Thu Mar 22, 2018 6:43 pm
LUANDATO wrote:A street vendor sells only hot dogs and hamburgers, and at the beginning of the day has a ratio of two hot dogs for every one hamburger. At the end of the day in which he did not add any new items or sell any hamburgers, and only sold some of his hot dogs, his new ratio is one hot dog for every two hamburgers. Which of the following cannot represent the number of hot dogs he sold?

A. 2
B. 3
C. 6
D. 9
E. 24

The OA is A.

I'm really confused by this PS question. Experts, any suggestion about how can I solve it? Thanks in advance.
Here's an approach that uses three variables.

Let D = # of hotdogs the vendor STARTED with
Let H = # of hamburgers the vendor STARTED with
Let x = # of hotdogs sold

At the beginning of the day has a ratio of two hot dogs for every one hamburger.
So, D/H = 2/1
Cross multiply to get: D = 2H


At the end of the day in which he did not add any new items or sell any hamburgers, and only sold some of his hot dogs, his new ratio is one hot dog for every two hamburgers.
So, (D - x)/H = 1/2
Cross multiply to get: 2(D - x) = H

So, we have the following system:
D = 2H
2(D - x) = H

Take 2(D - x) = H and replace D with 2H
We get: 2(2H - x) = H
Expand: 4H - 2x = H
Solve for H to get: H = 2x/3

Since H must be an INTEGER, we can see that x must be divisible by 3.
Answer choice A is the only answer that is NOT divisible by 3.


-----ALTERNATE APPROACH-------------------------------------------
Once we get to the point where we have 4H - 2x = H, we can also solve for x to get: x = 3H/2
At that point, we can check each answer choice to see what happens when x = that certain amount.

For example. let's check answer choice E first
It tells us that x = 24
Plug x = 24 into our equation: 24 = 3H/2
Multiply both sides by 2 to get: 48 = 3H
Solve for H to get: H = 16
No problem. When x = 24, we get a nice integer value for the number of Hamburgers sold.


Now let's check answer choice A
It tells us that x = 2
Plug x = 2 into our equation: 2 = 3H/2
Multiply both sides by 2 to get: 4 = 3H
Solve for H to get: H = 4/3
Problem.
When x = 2, we get a NON-integer value for the number of Hamburgers sold.
So, it is impossible to sell 2 hotdogs.

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by swerve » Fri Mar 23, 2018 11:13 am
Hi LUANDATO,

Let t = total beginning hot dogs and hamburger.

Let d = hot dogs sold

(2/3*t)-d = 1/3*(t-d)

d = t/2 because of 1:2 ratio, t must divide by 3

Plugin in 2 in as d, t = 4, a non-multiple of 3. Therefore the correct answer is the Option A.

Regards!
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by Jeff@TargetTestPrep » Mon Mar 26, 2018 4:48 pm
LUANDATO wrote:A street vendor sells only hot dogs and hamburgers, and at the beginning of the day has a ratio of two hot dogs for every one hamburger. At the end of the day in which he did not add any new items or sell any hamburgers, and only sold some of his hot dogs, his new ratio is one hot dog for every two hamburgers. Which of the following cannot represent the number of hot dogs he sold?

A. 2
B. 3
C. 6
D. 9
E. 24
We can let x = the number of hamburgers at the beginning of the day; thus, 2x = the number of hot dogs at the beginning of the day. We can create the equation in which n = the number of hot dogs sold. Thus we have:

(2x - n)/x = 1/2

2(2x - n) = x

4x - 2n = x

3x = 2n

We see that n can't be 2; otherwise 2n = 4. However, 3x can't be equal to 4 since x is an integer.

Answer: A

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