If w and c are integers is w > 0 ? (1) w + c > 50 (2)

Bunuel, taking the qtn as you specified i.e. w+c<50 Please clarify why am i getting w < 2
1) \(w+c<50\)
2) \(c > 48\)

now 2) can be written as \(-c < -48\)
as both the inequalities have the same sign, 1) + 2)
==> \(w < 50-48\)
==> \(w< 2\)
why am i not getting \(w <= 0\) for the answer to be C

Regards,
Murali.

First of all:
You can only add inequalities when their signs are in the same direction:

If \(a>b\) and \(c>d\) (signs in same direction: \(>\) and \(>\)) --> \(a+c>b+d\).
Example: \(3<4\) and \(2<5\) --> \(3+2<4+5\).

You can only apply subtraction when their signs are in the opposite directions:

If \(a>b\) and \(c<d\) (signs in opposite direction: \(>\) and \(<\)) --> \(a-c>b-d\) (take the sign of the inequality you subtract from).
Example: \(3<4\) and \(5>1\) --> \(3-5<4-1\).

So, you can directly subtract \(c > 48\) from \(w+c<50\): \(w+c-c<50-48\) --> \(w<2\) (no need to rewrite and then add). But this approach won't work for this question: \(w<2\) would be correct if we were not told that both \(w\) and \(c\) are integers then for example c=48.5>48 and w=1>0 would be valid solutions but since both must be integers then \(c\) can be 49, 50, 51 ... and \(w\) can be 0, -1, -2, ... (so c=integer>48 and w=integer<1)

Hope it's clear.

oops. Those are integers. Yes this will not apply in case of integers.
I missed it out.

Thanks Bunuel. Kudos for u...

regards,
Murali.

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