hi all. it seems that it is kind of problems in which plugging and obtaining yes and no answer works better than any other ways, i honestly tried to simplify but looks like it does not work here at first we need to determine does ((2a-b)/(2b-a))>((2b-a)/2a-b)) after simplify i got does (a^2-b^2)/((2...

hi aspirant i understand your doubts about D, (D) that have a comparable size, and also they emit far less carbon dioxide gas and other gases that have as i think, far less here refers not only to carbon dioxide, but far less is implied before other gases, imagine that gasoline engines emit some gas...

hi aspirant, i saw this problem earlier. solved it wrong picking B,but think that i can explain something C)far fewer carbon dioxide. carbon dioxide is uncountable so we need far less. also they is dubious here, but i believe that it is a small problem D) i don`t like: that have comparable size.agai...

hi aspirant, great unpleasant sentence it can`t be B they denounce.....but supporting, not || it can`t be C they denounce....., say... between denounce and say we need conjunction, perhaps denounce and say will be better version. it can`t be E while they are denouncing-....supporting...the main prob...

for sure i am wrong but my pick is A we have set of integers 1.a2.a3.a4.a5, and need to find a5 -greatest (1)1+a2+a3+a4+a5=20, with median a3=4, to make a5 greatest, first least value for a2 is 2, and least for a4=5, the remainder for a5=20-(1+2+4+5)=8,so given set looks like 1,2,4,5,8 as for st 2 r...

vote for C

Ozlemg wrote:IMO D.

Statement 1 helps us to figure out that AC // DE. So we can solve the prolem.

Statement 2 is also sufficient. (Property of similarity in triangles)

Thanks

hi all, i want to believe that the answer is D
(1)r+s>2t, and t>s, if we sum both we get
r+s+t>2t+s. canceling s, left us with r+t>2t. and finally r>t suff
(2)the same way
r+s>2t, and r>s, again sum, 2r+s>2t+s, cancel s and left with 2r>2t, r>t also suff