Swagatalaxmi
In a photo contest, prize money of $30, $20, and $10 was awarded to the first-, second-, and third-place photos, respectively, in several categories. No other prize money was awarded. Photographers could submit an unlimited number of photos to the contest but could enter each photo in only 1 category. Julia received a total of $110 in prize money from the contest. How many of her photos were awarded second place in their categories?
(1) Twice as many of Julia's photos were awarded third place as were awarded first place in their categories.
(2) A total of 6 of Julia's photos received awards.
30f + 20s + 10t = 110
f, s and t cannot be negative but some of them can be 0. We need the value of s.
(1) Twice as many of Julia's photos were awarded third place as were awarded first place in their categories.30f + 20s + 20f = 110
50f + 20s = 110
My first instinct is that this equation still have 2 variables but second instinct is that it could have a unique solution since there are constraints on the values of f and s.
f = 1, s = 3 is one possible solution. Using our concept of integer solutions, we know that no other solution is possible.
Here is a post that discusses this concept:
https://anaprep.com/algebra-integer-sol ... variables/Sufficient alone.
(2) A total of 6 of Julia's photos received awards.This is harder but we can take a cue from the above solution we got.
30f + 20s + 20f = 110
We know that one possible solution is f = 1, s = 3, t = 2 where the total number of photos is 6. If we reduce f by 1, I will need to compensate for reduction of $30. I cannot do it by simply adding 1 to s. But I can reduce f and t both by 1 and increase s by 2 to compensate for the loss of 30 + 10 by adding gain of 20 + 20. This will be a valid solution so f = 0, s = 5, t = 1 works too.
Not sufficient alone.
Answer (A)
A couple of interesting Data Sufficiency questions:
https://youtu.be/vFNiowuBVEY
https://youtu.be/yzeh388tkLQ