When the values of two groups are averaged, the result is a weighted average; it's called weighted average because instead of being exactly in the middle of the two groups, it is closer to the value of the larger group; there is a proportional relationship to the sizes of the groups: if group 1 is twice as big as group 2, the weighted average will be twice as close to group 1's value.
All this means that there are just 4 variables in a weighted avg question between 2 groups; the values of the individual groups, the relative weights of the groups (ratio of group sizes), and the weighted average. Knowing 3 of these unknowns is enough to discover the 4th.
Let's solve:
Statement 1:
The prompt gives us 2 values up front; the values of the individual groups (price of choc chip and price of brownies). Statement 1 gives us the weighted average, so we have 3 out of 4 unknowns; that's enough to find the 4th unknown, the ratio of group sizes.
SUFFICIENT
Statement 2:
If we call b and c the number of brownies and cookies, we can write the equation 1.3c + 1.5b = 14.2. Simplify to:
13c + 15b = 142
In general, when you have two variables and one equation you will not have enough data to solve. There is at least one important exception:
✔ If your equation is of the form Ax + By = C where A, B and C are known, and
✔ if your unknowns are positive integers, and
✔ if C < the least common multiple of A & B, then
the equation has only 1 solution! Since our equation (13c + 15b = 142) meets all three criteria I can tell at a glance that it is SUFFICIENT.
The answer is D.
