If ab > 9, is a < 1? (1) b < 9 (2) 1 < b < 8

Hi,

(1)b < 9

Consider b=-8 and a=-2 & ab>9 is a < 1--Yes

Now Consider b=4 and a=4 & ab>9 is a < 1--No

So Statement 1 is not suff

(2) 1 < b < 8

In this case consider extreme b=7.99 then to prove ab>9 a has to be greater than 1, hence gives ans no hence suff

Hence B

Hi AbhishekGopal,

Here is how D is incorrect:

ab > 9

ab/9 > 1

a(b/9) > 1

stmt-1: b < 9

if b < 9 then b/9 < 1

a*something-less-than-1 > 1 shows that a has to be greater than 1 i.e. a > 1. so you say this sentence is fine. is this sufficient?

stmt-2:

1 < b < 8

again it means b/9 < 1 and reasoning is same a*something-less-than-1 > 1 shows that a has to be greater than 1.

but wait we are getting the same answer from both the statements but are the ranges provided for b by both the statements SAME? No!

in stmt-1 b can be -ve also.

all right so if b is negative then a also has to be -ve for a(b/9) > 1 to be true. which means a < 1. so this proves that stmt-1 is not sufficient.

I wanted to prove two things here:

#1. If i followed the process of evaluating the statement very carefully then i would not have gotten into the trap initially in statement-1. but now that i was trapped:
#2. Take help from the other statement. Great, this told me that there was sometthing fishy about statement-1.

Trap - We generally forget to test the -ve values. Learning!

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