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(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, \(Y<Z\). Sufficient
The correct answer is C. _________________
I'm happy, if I make math for you slightly clearer And yes, I like kudos:)
Statement 1 : y+z = 1 , This can be possible in different cases : Case 1 : 0<y<1 and 0<z<1 Case 2 : y>=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 y<z, statement 1 is not sufficient
Statement 2 : y^2 < z^2 , again which is possible in many cases case 1 : z>y>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 y<z, statement 2 is also not sufficient
By combining two statements and observing all the cases from both statements we can eliminate case 4 and case 3 from statement 2 as they violate the condition y + z = 1 from statement 1. So from the remaining cases which satisfy both statements we can clearly observe z>y . Hence we can say yes 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.
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.
Check out this awesome article about Anderson on Poets Quants, http://poetsandquants.com/2015/01/02/uclas-anderson-school-morphs-into-a-friendly-tech-hub/ . Anderson is a great place! Sorry for the lack of updates recently. I...