ifra94
Is there a faster way to do this? I simplified the equation to simpler numbers :
25x+8y+ 3z = 347
Then I wrote down all possible combinations of y and z = selected the one that would give me a multiple of 25.
A certain contest awarded $125 for each first prize, $40 for each second prize, and $15 for each third prize. What is the fewest number of prizes that could be awarded for a total of $1,735 ?(A) 12
(B) 14
(C) 15
(D) 17
(E) 18
Assume x $125 prizes, y $40 prizes, and z $15 prizes were given. Then, we'd have:
125x + 40y + 15z = 1,735
Reduce by 5:
25x + 8y + 3z = 347
Our goal is to minimize x + y + z. To achieve this, we need to maximize the number of the largest prize, x. The value of x cannot be 14 or higher, so we try x = 13. In this case, we have:
25*13 + 8y + 3z = 347
8y + 3z = 22
Now, let's maximize the number of the next largest prize, y. The maximum value of y is 2, making z = 2.
Therefore, x + y + z = 13 + 2 + 2 = 17.
Answer: D.
Similar question to practice:
https://gmatclub.com/forum/the-first-pr ... 21161.html