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07 Jan 2020, 07:29
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E
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Difficulty:
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Question Stats:
51% (01:23) correct
49%
(01:07)
wrong
based on 70
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Updated on: 17 Jul 2021, 09:34
Originally posted by BrushMyQuant on 07 Jan 2020, 08:09.
Last edited by BrushMyQuant on 17 Jul 2021, 09:34, edited 3 times in total.
Last edited by BrushMyQuant on 17 Jul 2021, 09:34, edited 3 times in total.
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I have written a detailed post on How to Solve: Inequality Problems - Algebra and Sine Wave/Wavy Method. You can go through it to get more details about how to approach these kind of problems in general.
abx > aby? Can be rearranged as abx - aby >0 => ab(x-y) > 0
so, if we are able to prove that either two of the three terms a,b and (x-y) are negative or all three are positive then we can prove that ab(x-y) > 0
STAT 1:
ax > ay => ax - ay>0 => a(x-y) > 0
If product of two things are positive then both have same sign. Either both are positive or both are negative.
=> Either a and (x-y) are negative ....(1)
a and (x-y) are positive. ...(2).
In any case, we do not know sign of b so not sufficient
STAT2:
bx > by => bx - by>0 => b(x-y) > 0
If product of two things are positive then both have same sign. Either both are positive or both are negative.
=> Either b and (x-y) are negative or ....(3)
b and (x-y) are positive. ...(4)
In any case, we do not know sign of a so not sufficient
Combining both of them:
Now, when we combine both of them then (x-y) becomes the common link.
So, there will be two cases
Case 1: (x-y) is positive
if (x-y) is positive then a is +ve from (2) and b is also +ve from (4)
Making a*b*(x-y) +ve
Case 2: (x-y) is negative
if (x-y) is negative then a is -ve from (1) and b is also -ve from (3)
Making a*b*(x-y) Negative
Since we are getting +ve and -ve answer so together also they are not sufficient.
So, Answer will be E
Hope it helps!
Watch the following video to learn Basics of Inequalties
abx > aby? Can be rearranged as abx - aby >0 => ab(x-y) > 0
so, if we are able to prove that either two of the three terms a,b and (x-y) are negative or all three are positive then we can prove that ab(x-y) > 0
STAT 1:
ax > ay => ax - ay>0 => a(x-y) > 0
If product of two things are positive then both have same sign. Either both are positive or both are negative.
=> Either a and (x-y) are negative ....(1)
a and (x-y) are positive. ...(2).
In any case, we do not know sign of b so not sufficient
STAT2:
bx > by => bx - by>0 => b(x-y) > 0
If product of two things are positive then both have same sign. Either both are positive or both are negative.
=> Either b and (x-y) are negative or ....(3)
b and (x-y) are positive. ...(4)
In any case, we do not know sign of a so not sufficient
Combining both of them:
Now, when we combine both of them then (x-y) becomes the common link.
So, there will be two cases
Case 1: (x-y) is positive
if (x-y) is positive then a is +ve from (2) and b is also +ve from (4)
Making a*b*(x-y) +ve
Case 2: (x-y) is negative
if (x-y) is negative then a is -ve from (1) and b is also -ve from (3)
Making a*b*(x-y) Negative
Since we are getting +ve and -ve answer so together also they are not sufficient.
So, Answer will be E
Hope it helps!
Watch the following video to learn Basics of Inequalties
08 Jan 2020, 05:58
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From statement (1): x>y & a>0 OR x<y & a<0 , no info about b --> insufficient
From statement (2): x>y & b>0 OR x<y & b<0 , no info about a --> insufficient
By combining,
if x>y & a>0 & b>0 , the answer is yes
if x<y & a<0 & b<0 , the answer is no
E
From statement (2): x>y & b>0 OR x<y & b<0 , no info about a --> insufficient
By combining,
if x>y & a>0 & b>0 , the answer is yes
if x<y & a<0 & b<0 , the answer is no
E
