interesting DS, thanks for posting

st(1) offers conditional losses while we have no restriction per win/loose Only, what if draw -Not Sufficient;
st(2) Sufficient, as 0.8*40 + 0.5x +18 = 0.8*40 + 0.8x and we solve for x original wins ...
0.3x=18, x=60

hja379 wrote:After winning 80 percent of the first 40 matches he played, Igby won 50 percent of his remaining matches. How many total matches did he win?
(1) If Igby had won 50 percent of the total number of matches he played, he would have lost 12 more total matches.
(2) If Igby had won 80 percent of the total number of matches he played, he would have won 18 more total matches.

OA after discussion.

Night reader wrote:interesting DS, thanks for posting

st(1) offers conditional losses while we have no restriction per win/loose Only, what if draw -Not Sufficient;
st(2) is Not Sufficient, as 0.8*40 + 0.5x = 0.2*40 + 0.5x + 18 and we solve for x original wins ...
0.5x are canceled LHS-RHS - we are left with no variables

Combined st(1&2): conditionally 12 more losses and 18 more wins; since we don't know how many games Igby has lost Not Sufficient

IOM E

hja379 wrote:After winning 80 percent of the first 40 matches he played, Igby won 50 percent of his remaining matches. How many total matches did he win?
(1) If Igby had won 50 percent of the total number of matches he played, he would have lost 12 more total matches.
(2) If Igby had won 80 percent of the total number of matches he played, he would have won 18 more total matches.

OA after discussion.

I've updated the OA. It is not E.

very tricky problem to me, initially i got D, but then found my mistake
let it be x the number of remaining games, so he win in total
0,8*40+0,5*x=32+0,5x
he lost in total 0,2*40+0,5x=8+0,5x
and the total number of all games he played=40+x, clearly we need to find x, after that we can find everything
(1) 0,5*(40+x)-If Igby had won 50 percent of the total number of matches he played
(8+0,5x+12)=20+0,5x-he would have lost 12 more total matches.
and in total he played 40+x games
win+lost=total, (20+0,5x)+(20+0,5x)=40+x, we already have this info from this we got that 40+x=40+x, no way to find the value of x, insuff
(2)
0,8(40+x)=32+0,8x-If Igby had won 80 percent of the total number of matches he played
(32+0,5x)+18- he would have won 18 more total matches.
32+0,8x=50+0,5x, 0,3x=18, x=60. total number if winning games =32+0,5*60=32+30=62 suff

@clock, is not that we make assumption here about no draws in matches/games in D st(1)? what's the source?

initially, i put 0.5x on both sides, i don't why - but now i see the variables are not canceled and it's solvable for x=60

answer would be 60/2+32=62 games?

Hi Reader
your note about draws is quite reasonable, i also thought of it with draws existing the answer would be E