Bunuel wrote:

A pack of baseball cards consists of 12 outfielder cards and 8 infielder cards. What is the lowest number of outfielder cards that would have to be removed from the pack so that no more than 40 percent of the pack would be outfielder cards?

(A) 4

(B) 5

(C) 6

(D) 7

(E) 8

Kudos for a correct solution.

Let x be the number of outfirlder cards that must be removed.

Then,

\(\frac{12-x}{20-x}*100\leq{40}\)

\(\frac{12-x}{20-x}*5\leq{2}\)

\(60-5x\leq{40-2x}\)

\(20\leq{3x}\)

\(\frac{20}{3}\leq{x}\)

\(6.66\leq{x}\)

So, x should at least be 7

Answer:- D