CAT Quantitative Aptitude Questions | CAT Algebra - Inequalities questions

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Explanatory Answer

Method of solving this CAT Question from Algebra - Inequalities : How many negative and positive terms would be there for every value of x?

(x + 2) (x + 4) (x + 6) ........(x + 100) < 0
Now, the above expression will be zero for x = –2, –4, –6, – 8…..–100.
For x > – 2 all the terms will be positive and so, the product will be positive.
For x < – 100, all the terms will be negative and since there are 50 terms (even number), the product will be positive.

Now, if x = – 99, the term x + 100 would be positive, everything else would be negative, so the expression would have 49 negative terms and one positive term. So the product would be negative.

Overall the expression will be negative if there are exactly 49 negative terms, or exactly 47 negative terms, or exactly 45 terms…. Or so on, up to exactly one negative term.

Exactly 49 negative terms= x = – 99
Exactly 47 negative terms= x= – 95
Exactly 45 negative terms= x = – 91
...............
Exactly 1 negative term=. x = – 3

So, x can take values {–3, –7, –11, –15, –19…. –99}. We need to compute how many terms are there in this list.
In other words, how many terms are there in the list {3, 7, 11, ....99}. Now, these terms are separated by 4, so we can write each term as as multiple of 4 + ‘some constant’.

Or 3=0 * 4 + 3
7=1 * 4 + 3
11=2 * 4 + 3
......................
99=24 * 4 + 3
We go from 0 * 4 + 3 to 24 * 4 + 3, a total of 25 terms.

The question is "For how many integer values does the given inequality hold good? (x + 2) (x + 4) (x + 6)........(x + 100) < 0?"

Hence the answer is "25"

Choice A is the correct answer.