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Updated on: 24 Jun 2017, 14:35
Originally posted by MBAhereIcome on 30 Aug 2011, 03:32.
Last edited by Bunuel on 24 Jun 2017, 14:35, edited 3 times in total.
Last edited by Bunuel on 24 Jun 2017, 14:35, edited 3 times in total.
Renamed the topic and edited the question.
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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:
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77% (01:29) correct
23%
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If \(|a| > |b|\), which of the following must be true?
A. \(ab > 0\)
B. \(|a| + b > 0\)
C. \(a + |b| > 0\)
D. \(\frac{|b|}{a} > 0\)
E. \(|a|*b > 0\)
A. \(ab > 0\)
B. \(|a| + b > 0\)
C. \(a + |b| > 0\)
D. \(\frac{|b|}{a} > 0\)
E. \(|a|*b > 0\)
Show SpoilerSolution
|a|>|b| so a is a bigger number than b, regardless of their signs. e.g. a=5, b=-2 OR a=-8, b=7.
a) not true if b<0 & a>0
b) TRUE. e.g.|5|-4>0 OR |-8|+7>0
c) incorrect if a<0. the sum will always be negative in this case
d) not true if a<0
e) not true if b<0
a) not true if b<0 & a>0
b) TRUE. e.g.|5|-4>0 OR |-8|+7>0
c) incorrect if a<0. the sum will always be negative in this case
d) not true if a<0
e) not true if b<0
24 Jun 2017, 14:29
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Mansoor50
\(|a| = a\), when \(a \geq 0\) and \(|a| = -a\), when \(a\leq 0\). You cannot simply plug a or -a if you don't know which case you have.
If \(|a| > |b|\), which of the following must be true?
A. \(ab > 0\)
B. \(|a| + b > 0\)
C. \(a + |b| > 0\)
D. \(\frac{|b|}{a} > 0\)
E. \(|a|*b > 0\)
\(|a| > |b|\) means that a is further from 0 than b is. So, we can have one of the following 4 cases:
1. ----a----b----0---------------
2. ----a---------0----b----------
3. ---------b----0---------a-----
4. --------------0----b----a-----
A. \(ab > 0\) --> Not true for cases in which a and b have different signs (cases 2 and 3). Discard.
B. \(|a| + b > 0\) --> true for all cases.
C. \(a + |b| > 0\) --> not true for cases in which a is negative (cases 1 and 2). Discard.
D. \(\frac{|b|}{a} > 0\) --> not true for cases in which a is negative (cases 1 and 2). Discard.
E. \(|a|*b > 0\) --> not true for cases in which b is negative (cases 1 and 3). Discard.
Answer: B.
Hope it helps.
30 Aug 2011, 04:07
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Hello,
lets take each one given |a| > |b|
a) ab > 0
Not allways . If a,b are both negative then true, either positive then fails.
b) |a|+b > 0
Since we know that absolute value of a is greater than absolute value of b , we can for sure say that |a| > b
The above would fail only if |a| = |b| or |b| > |a| so B is the answer because -- given |a| > |b|
c) a+|b| > 0
Not always consider a = -7 and b = -5 (given |a| > |b|)
-7 + 5 = -2 <0 hence not possible ;
d) |b|/a > 0
not always if a is positive then true, if a is negative then false because for all values of {b} except zero |B| is always positive
e) |a|*b > 0
not always if b is positive then true, if b is negative then false because for all values of {a} except zero |a| is always positive
Hope my explanation was helpful. OA is B
Regards
Raghav.V
Consider Kudos if you think my explanation was helpful
lets take each one given |a| > |b|
a) ab > 0
Not allways . If a,b are both negative then true, either positive then fails.
b) |a|+b > 0
Since we know that absolute value of a is greater than absolute value of b , we can for sure say that |a| > b
The above would fail only if |a| = |b| or |b| > |a| so B is the answer because -- given |a| > |b|
c) a+|b| > 0
Not always consider a = -7 and b = -5 (given |a| > |b|)
-7 + 5 = -2 <0 hence not possible ;
d) |b|/a > 0
not always if a is positive then true, if a is negative then false because for all values of {b} except zero |B| is always positive
e) |a|*b > 0
not always if b is positive then true, if b is negative then false because for all values of {a} except zero |a| is always positive
Hope my explanation was helpful. OA is B
Regards
Raghav.V
Consider Kudos if you think my explanation was helpful
