EACB

Plz put OA

Thanks!

vishalranka wrote:Can please someone tell me what will be the answers with explanation of the foll DS quests

Thank u

Is |v - x| < 8?
(1) v and x are integers.
(2) |v| = 4 and |x| = 6


If x and y are positive, is x^3 > y?
(1) Sq rt x > y
(2) x > y


Is |x| < 1?
(1) |x + 1| = 2|x - 1|
(2) |x - 3| not equal to 0


If a, b, c, and d are positive integers, is (a/b) (c/d) > c/b?
(1) c > b
(2) a > d

First of all Do not post multiple questions in one post - Defeats the purpose in terms of searchability
Second of all please search before you post - there is a lot of wisdom in these forums and the chances that someone else posted this question before you is high

2. https://www.beatthegmat.com/if-x-and-y-a ... 38860.html

Vishalranka , in the second question , is it 3X or x^3 ?


St 1) root x > y , squaring both sides : x>y^2

if x=2 , and y =1 Answer is Yes

if x = 1/2 and y = 1/3 the answer is NO . not sufficient

2nd statement could be proven insufficient by the same numbers.

we have two assumptions that satisfies both statemnts and gives an answer of a yes and a no. so E

Question 3 :

we have to test a +/- case here

|x+1| = 2 |x-1|

when |x+1| is positive : x+1 = 2x-2 so x = 3 , check for validity in the original equation , 3 is valid and the answer to the question is NO

when | x+1 | is -ve : x+1 = - [ 2 ( x-1 ) ] yields x = 1/3 , after checking for validity of the value , 1/3 is a valid value of x and the answer to the question stem is Yes

so St 1 : not sufficient

St ;2 obviously insufiecient on its own , eleminates one of the 2 values, it says that X = 3 +/- some value so X is not equal 3 ,

Both statements are needed to reach a conclusion that x = 1/3

C



Question 4:

Notice that a/b * c/d is the same as a/d * c/b

since every thing is positive we can rephrase as follows "


is a/d * c/b > c/b

divide both sides by the positive value of c/b : is a/d > 1 ?

multiply both sides by D:

finally : is A > D ?

St 1 : irrelevant

st 2 ; sufficient.

B

My answers are EBAB what are the OA .