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KanishkM
Joined: 09 Mar 2018
Last visit: 18 Dec 2021
Posts: 755
Given Kudos: 123
Location: India
Schools: DeGroote'21 Desautels '21 NUS '21
01 Jan 2019, 21:13
Kudos
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
51% (01:54) correct
49%
(02:09)
wrong
based on 188
sessions
History
Date
Time
Result
Not Attempted Yet
02 Jan 2019, 00:55
Kudos
Bookmarks
If a < b , is a > 0 ?
(1) \(a^2<b^2\)
case(i)
Given a<b,
So, if a < 0 then \(a^2 > b^2\)
example a = -2 and b = -1, where a<b
\(a^2\) =4
\(b^2\) = 1
\(a^2 > b^2\)
But this is contradiction to (1) a^2<b^2 so a is non-negative
case (ii) if a = 0 or a lies between 0 to 1
then b>0 as b>a, which in turn implies \(b^2 > a^2\)
case (ii) a>0 (1) satisfies
for a = 0 it satisfies so I can not strictly say a>0
Insufficient
(2) \(a^2<ab<b^2\)
\(a^2 < ab\)
\(a^2-ab\) < 0
a*(a-b)<0 ---- equation 1
given a < b => (a-b) < 0
since a-b < 0 => a >0 in equation 1 [as (-ve)*(+ve) < 0]
Sufficient
Option B is correct
(1) \(a^2<b^2\)
case(i)
Given a<b,
So, if a < 0 then \(a^2 > b^2\)
example a = -2 and b = -1, where a<b
\(a^2\) =4
\(b^2\) = 1
\(a^2 > b^2\)
But this is contradiction to (1) a^2<b^2 so a is non-negative
case (ii) if a = 0 or a lies between 0 to 1
then b>0 as b>a, which in turn implies \(b^2 > a^2\)
case (ii) a>0 (1) satisfies
for a = 0 it satisfies so I can not strictly say a>0
Insufficient
(2) \(a^2<ab<b^2\)
\(a^2 < ab\)
\(a^2-ab\) < 0
a*(a-b)<0 ---- equation 1
given a < b => (a-b) < 0
since a-b < 0 => a >0 in equation 1 [as (-ve)*(+ve) < 0]
Sufficient
Option B is correct
02 Jan 2019, 04:51
Kudos
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KanishkM
St 1 :
a < b
\(a^2 < b^2\)
Case 1 : a = -2 , b = 3 No
Case 2 : a = 3 , b = 4 Yes
Not Sufficient
St 2 :
a < b .... Ineq1
\(a^2 <ab\)
i.e a * a < a * b
We have multiplied both sides of Ineq1 by a and the sign has not changed. So, a > 0
Sufficient
Choice B
