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23 May 2020, 04:28
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If a and b are negative numbers, is b > a ?
(1) a/b > 1
(2) b/a < 1
DS21187
(1) a/b > 1
(2) b/a < 1
DS21187
23 May 2020, 05:49
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We can multiply the inequality in Statement 1 by b on both sides, reversing the inequality because b is negative, to find that
a < b
so Statement 1 is sufficient. We can multiply the inequality in Statement 2 by a on both sides, reversing the inequality because a is negative, to find that
b > a
so Statement 2 is also sufficient, and the answer is D.
a < b
so Statement 1 is sufficient. We can multiply the inequality in Statement 2 by a on both sides, reversing the inequality because a is negative, to find that
b > a
so Statement 2 is also sufficient, and the answer is D.
Siak0730
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23 May 2020, 07:03
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Bunuel
If a and b are negative numbers, is b > a ?
(1) a/b > 1
(2) b/a < 1
PS21187
(1) a/b > 1
(2) b/a < 1
PS21187
is b>a??
statement 1: a/b > 1
given : a, b are negative numbers, meaning they can be integers or fractions.
case 1: when a and b both are negative integers
lets take a=-3, b=-2 ; a/b >1
we get => b>a : sufficient
case 2: when a, b both are negative fractions
lets take a= -1/2, b= -1/3 ; a/b>1
we get => b>a : sufficient
for every a/b>1 where a, b are negative; absolute value of a > absolute value of b (lal > lbl)
which means : actual a, which is negative < actual b, which is negative ( a<b; where b and a are negative)
therefore, sufficient
statement 2: b/a < 1
given : a, b are negative numbers, meaning they can be integers or fractions.
case 1: when a and b both are negative integers
lets take a=-3, b=-2 ; b/a <1
we get => b>a : sufficient
case 2: when a, b both are negative fractions
lets take a= -1/2, b= -1/3 ; b/a<1
we get => b>a : sufficient
for every b/a<1 where a, b are negative; absolute value of b < absolute value of a (lbl < lal)
which means: actual b, which is negative > actual a, which is negative ( b>a; where b and a are negative)
therefore, sufficient
hence D
