pratyoosh wrote:Hello,

Can anyone explain the resolution to the below problem:

In the month of June, a street vendor sold 10% more hot dogs than he sold in the month of May. How many total hot dogs did the vendor sell in May and June?
(1) The vendor sold 27 more hot dogs in June than in May.
(2) In July, the vendor sold 20% more hot dogs than he sold in May.

I often find it useful to use multiple variables to solve DS word problems.

Let M = number of hot dogs sold in May
Let J = number of hot dogs sold in June
Let L = number of hot dogs sold in July

Target question: How many total hot dogs did the vendor sell in May and June?

REPHRASED target question: What is the value of M + J?

Given: In June, the vendor sold 10% more hot dogs than he sold in the month of May.
We can write: J = (1.1)M
Rearrange to get: J - (1.1)M = 0

Statement 1: The vendor sold 27 more hot dogs in June than in May.
So, we can write J = M + 25
Rearrange to get: J - M = 25
We also know that: J - (1.1)M = 0
Now that we have 2 different linear equations with 2 variables (M and J) we COULD solve for M and J, which means we COULD definitely find the value of M + J [or course we're not going to solve the system for M and J, because this would be a waste of time. We need only determine whether the information is sufficient]
Since we could answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: In July, the vendor sold 20% more hot dogs than he sold in May.
In other words, X = 1.2M
We also know that: J - (1.1)M = 0
Here we have 2 equations with 3 variables.
This system cannot help us find the value of M + J
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Answer = A

Cheers,
Brent