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12 Days of Christmas GMAT Competition - Day 11: If a*b 0, is (a + 5)

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12 Days of Christmas GMAT Competition - Day 11: If a*b 0, is (a + 5) [#permalink] 12 Days of Christmas GMAT Competition - Day 11: If a*b 0, is (a + 5)   24 Dec 2022, 07:00
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12 Days of Christmas GMAT Competition with Lots of Fun

If \(a*b ≠ 0\), is \(\frac{a + 5}{b} > \frac{a}{b}\) ?

(1) \(a < 0\)
(2) \(b < 0\)



 


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Re: 12 Days of Christmas GMAT Competition - Day 11: If a*b 0, is (a + 5) [#permalink] 12 Days of Christmas GMAT Competition - Day 11: If a*b 0, is (a + 5)   24 Dec 2022, 07:07
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a/b + 5/b > a/b ?

5/b > 0?

If b < 0 No
If b > 0 Yes

(1) Not sufficient
(2) Sufficient. (a+5)/b < a/b

B is the answer

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Re: 12 Days of Christmas GMAT Competition - Day 11: If a*b 0, is (a + 5) [#permalink] 12 Days of Christmas GMAT Competition - Day 11: If a*b 0, is (a + 5)   24 Dec 2022, 07:09
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Asked: If \(a*b ≠ 0\), is \(\frac{a + 5}{b} > \frac{a}{b}\) ?

Is \(\frac{a + 5}{b} > \frac{a}{b}\) ?
Is \(\frac{a + 5}{b} - \frac{a}{b} > 0?\)
Is \(\frac{5}{b} > 0?\)

(1) \(a < 0\)
Since \(\frac{5}{b}\) is independent of a.
NOT SUFFICIENT

(2) \(b < 0\)
\(\frac{5}{b} < 0\)
SUFFICIENT

IMO B
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Re: 12 Days of Christmas GMAT Competition - Day 11: If a*b 0, is (a + 5) [#permalink] 12 Days of Christmas GMAT Competition - Day 11: If a*b 0, is (a + 5)   24 Dec 2022, 07:42
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Using backtracking approach:
S1)
a<0:
we know
=> 5>0
Adding a :
=>5+a > a

if b >0 => (5+a)/b > a/b
if b<0 => (5+a)/b < a/b



Not Suff

S2) b<0:
Multiplying by 5
=> 5b<0;
Adding ab both sides:
=> ab+5b<ab
=> b(a+5)<ab
Dividing by b^2
=> (a+5)/b < a/b

S2 is suff


Hence B)
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Re: 12 Days of Christmas GMAT Competition - Day 11: If a*b 0, is (a + 5) [#permalink] 12 Days of Christmas GMAT Competition - Day 11: If a*b 0, is (a + 5)   24 Dec 2022, 08:14
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The easiest way to deal with this task is to selectively look through some numbers for A and B, depending on the conditions.
Quote:
(1) \(a<0\)

This gives us four potential scenarios, based on whether \(a+5\) is still below zero or becomes positive:
  • \(a<-5\) and \(b<0\), for instance \(a=-10\) and \(b=-1\)
    \(\frac{-5}{-1}\)<\(\frac{-10}{-1}\)
  • \(-5<a<0\) and \(b<0\), for instance \(a=-3\) and \(b=-1\)
    \(\frac{2}{-1}\)<\(\frac{-3}{-1}\)
  • \(a<-5\) and \(b>0\), for instance \(a=-10\) and \( b=1\)
    \(\frac{-5}{1}\)>\(\frac{-10}{1}\)
  • \(-5<a<0\) and \(b>0\), for instance \(a=-3\) and \(b=1\)
    \(\frac{2}{1}\)>\(\frac{-3}{1}\)

So, as we can see, we have both options here - both 'more' and 'less', as highlighted in yellow. Therefore, Condition 1 by itself is not sufficient.

Quote:
(2) \(b<0\)

In this case, because we have sort of 'fixed' B, we need to look into three options for A - below -5, above -5 and above 0:
  • \(a<-5\) and \(b<0\), for instance \(a=-10\) and \(b=-1\)
    \(\frac{-5}{-1}\)<\(\frac{-10}{-1}\)
  • \(-5<a<0\) and \(b<0\), for instance \(a=-3\) and \(b=-1\)
    \(\frac{2}{-1}\)<\(\frac{-3}{-1}\)
  • \(0<a\) and \(b<0\), for instance \(a=3\) and \(b=-1\)
    \(\frac{8}{-1}\)<\(\frac{3}{-1}\)
So, as can bee seen, ALL the signs here are 'less' which basically means that Condition 2 by itself is sufficient.

Therefore, the correct answer is B.
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Re: 12 Days of Christmas GMAT Competition - Day 11: If a*b 0, is (a + 5) [#permalink] 12 Days of Christmas GMAT Competition - Day 11: If a*b 0, is (a + 5)   24 Dec 2022, 19:35
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Simplifying the condition to verify, (a+5)/b > (a/b):

(a+5)b > ab

ab + 5b > ab

5b > 0

Hence the result only depend on, is b > 0?

Condition (1):

a < 0


We can't determine b here. So option A and D is out.
Keep BCE.

Condition (2):

b < 0


This option says b is negative. Hence it solves the question whether b is greater than zero.
Hence the answer is NO to the original question. Keep B.

So the best answer choice is B.
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Re: 12 Days of Christmas GMAT Competition - Day 11: If a*b 0, is (a + 5) [#permalink]
24 Dec 2022, 19:35
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