alactraz

Guys, How long dose it take to get your rescoring scores?

Agree with B

Agree with E

IMO B that the Australian egg-laying mammals of today are a branch of... rather than a type that....

Another vote for E

OA is C. Thx 4 help me

Nice approach dwarrior

I am sorry guys, but didn't statement 2 just says that: All of the prime factors of z are also factors of y Is it possible, if y=4 , z will be 8, since all the only prime factor of z and y is 2 and if x=16, than x/yz will not be an integer

I will tell the OA later John is chosing a number n number randomly from all integers from 56 to 150 inclusive. What is the probability that the number he chooses will be one where n(n+1) is divisible by 5? (A) 1/5 (B) 19/95 © 2/5 (D) 19/24 (E) 3/5

If x,y, and z are integers, is x/yz an integer? (1) y is a factor of x more than once (2) All of the prime factors of z are also factors of y

If k=20x+50y and x+y=1, is k (1) y>1/2 (2) y>x

I am agree with above explanation Statement 2 alone is not sufficient, maybe he bought 2, 3, 4,..., etc stamps x(0.15) and y (0.15) Thanks guy

statement 1 alone insufficient x>y+4 for x=25 and y=1, satisfied the inequality x>y+4, 25>1+4 put the value x and y to 3x>7y? then 3(25)>7(1)? Yes for x=12 and y=7 --> 12>7+4 then the qestion is 3(12)>7(7)? since 36 hence, statement 1 insufficient once again for B

In my humble opinion, there is only one way to make 0.15x+0.29y=4,40, x and y must be 10, unless x and y is not and integer, and you can't buy a half stamp.

I use Permutation, not combination for example: for 5, 0, 0 : 3P!/2! for 4, 1, 0 : 3P! for 3, 1, 1 : 3P!/2!

there are 21 ways, (5,4,3,2,1 and 0 is represent of how many donuts each men have) LMD: 500, 050, 005 LMD: 410, 401, 041, 140, 014, 104 LMD: 320, 302, 032, 230, 023, 203 LMD: 311, 131, 113 LMD: 221, 212, 122

There are 21 ways (5,4,3,2,1,0 are the number how many donuts each men have)500, 050, 005410, 401, 041, 140, 014, 104,320, 302, 032, 230, 023, 230,311, 131, 113,221, 212, 122

Agree with B

A is sufficient, to make 0.15x + 0.29y=4.40 y must be multiple of 5 or 10 y must be 10 and x=10, it can't be any number statement 2 is also sufficient since x=y=10 D is the answer

If w and c are integers , is w>0 ? 1. w+c > 50 w could be -2 and c=53 or w could be 2 and 4 50 insufficient 2. c>48, didn't say anything about w, so statement 2 is insufficient if we combined both statement c is an integer, for example c=49 at least w=2 to make w+c>50 if c=59, w could be -2 or 2 to make w+c>50 Insufficient IMO E

Imo B

Agree with A, "likely to be" is the correct idiom

Imo E

Agree with vbhup2 second statement is sufficient. Another vote for B

OA is D