Is y

Question

Is y < z ?

(1) y + z = 1

(2) y^2 < z^2

smyarga · Answer

(1) For both Y=0.3, Z=0.7 and Y=0.7, Z=0.3 we have Y+Z=1 . Insufficient (1) For both Y=0.3, Z=0.7 or Y=0.3, Z=-0.7 we have Y^2 < Z^2 . Insufficient (1)+(2) From Statement (1) Z=1-Y , and put in Statement (2): Y^2 < (1-Y)^2 Y^2 < 1-2Y+Y^2 2Y< 1 Y<1/2 Therefore, -Y>-1/2 and 1-Y>1-1/2 and Z=1-Y>1/2 . Hence, Y

meghaswaroop · Answer

Statement 1 : y+z = 1 , This can be possible in different cases : Case 1 : 0=1 and as Z = 1 - y so z<=0 Case 3 : Z>=1 and as y = 1-z so y<=0 As we can't conclude whether y>z or yy>0 case 2 : z>0>y and z>|y| case 3 : y>0>z and y<|z| case 4 : 0>y>z As we can't conclude whether y>z or yy . Hence we can say yes y

sunaimshadmani · Answer

Bunuel can you explain it more clearly plz

niyantg · Answer

Ans: C

Statement1 : Y & Z can take any values , so Definitely it cannot give the answer

Statement 2 : If Y = 5 & Z = 10 Then Y < Z
 But if Y= 0.5 & Z = 0.2 Then y >Z

1+2 :
 Y^2 < Z^2

Y^2 - Z^2 < 0

(Y-Z)(Y+Z)<0

From statement 1 : Y+Z =1

Putting the value  Y-Z< 0

Y < Z

For any positive value of y+z this will be true.

Answer C

Bunuel · Answer

Is y < z ?

(1) y + z = 1. The sum of two numbers is 1. Obviously from this info we cannot say which one is bigger. Not sufficient.

(2) y^2 < z^2 --> take the square root from both sides (we can safely do that since we know that both sides are non-negative): |y| < |z| --> z is farther from 0 than y is. Not sufficient.

(1)+(2) From (2): (y - z)(y + z) < 0. This implies that y - z and y + z have opposite signs. Since we know from (1) that y + z is positive, then y - z must be negative: y - z < 0 --> y < z. Sufficient.

Answer: C.

Hope it's clear.

mvictor · Answer

1. y=-3, z=4 => y+z=1 or y=4, z=-3 =y+z=1. so 1 alone is insufficient. 2. y^2 < z^2 y=2, z=-3 -> y>z or y=-2, z=3 -> y

AbdurRakib · Answer

Is y < z ?

(1) y + z = 1

(2)  y^2  <  z^2

Bunuel · Answer

Merging topics. Please refer to the discussion above.

unverifiedvoracity · Answer

1) y + z = 1: Scenario 1 - y= 0.3, z = 0.7, y + z = 1 and y < z; Scenario 2 - when y = 4, z = -3, y > z. Hence, not sufficient. 2) y^2 < z^2 , From this statement, we can only that absolute value of y is greater than absolute value of z. y =-4 and z = 5, and y^2 = 16 and z^2 = 25, y^2 < z^2 , but z > y. Not sufficient. Consider both statements together. when both y and z are positive, y< z as y^2 < z^2 . when only 1 of y and z are positive, y + z = 1 -> y = 1 -z . If y is negative, z has to be positive, so y < z when only 1 of y and z are positive, y + z = 1 -> y = 1 -z , if y is positive, z has to negative. If y = 4, z = -3. If y = 5, z = -4. If y=6, z=-5. As you can see |y| > |z|, but from statement 2, y^2 < z^2 or |y| < |z|. So, we can never have a scenario where y is positive and z is negative. Note that y and z both cannot be negative as y + z = 1.

Divyadisha · Answer

(1) y + z = 1 If y = 5 z=-4 then y>z if y = 0 Z= 1 z>y Not sufficient y^2 < z^2 if y = 4 z = 5 then yes if y= -4 z= -5 then no if y= -4 z=5 then yes if y= 4 z= -5 then no if y=0 z= 1 then yes Combining both statements y+z= 1 and y^2 < z^2 only if y= -4 z=5 or if y=0 z= 1 is possible hence z>y C is the answer

bumpbot · Answer

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