{"id":4623,"date":"2020-07-17T16:20:55","date_gmt":"2020-07-17T10:50:55","guid":{"rendered":"https:\/\/online.2iim.com\/cat-exam\/blogs\/?p=4623"},"modified":"2020-07-17T16:51:21","modified_gmt":"2020-07-17T11:21:21","slug":"intriguing-inequalities-questions-cat-quant-solutions","status":"publish","type":"post","link":"https:\/\/online.2iim.com\/cat-exam\/blogs\/cat-quantitative-aptitude\/intriguing-inequalities-questions-cat-quant-solutions\/","title":{"rendered":"Intriguing Inequalities Questions | CAT Quant Solutions"},"content":{"rendered":"\n
\"Intriguing<\/figure>\n\n\n\n

Here are a set of 3 interesting CAT level Inequalities questions with detailed solutions for the same. You might also want to check out this wonderful article on preparing for Quant from July<\/a>. Happy solving!<\/p>\n\n\n\n\n\n

Question 1:<\/h3>\n\n\n\n

How many positive integer values can x take that satisfy the inequality: (x \u2013 8) (x \u2013 10) (x \u2013 12)\u2026\u2026.(x \u2013 100) < 0<\/em>?<\/p>\n\n\n\n

Solution: <\/h3>\n\n\n\n

Let us try out a few values to see if that gives us anything.

When x = 8, 10, 12, \u2026.100<\/em> this goes to zero. So, these cannot be counted.
When x = 101, 102 or beyond<\/em>, all the terms are positive, so the product will be positive. 

Straight-away we are down to numbers 1, 2, 3, \u20267 and then odd numbers from there to 99.

Let us substitute x =1<\/em>,

All the individual terms are negative. There are totally 47 terms in this list (How? Figure that out). Product of 47 negative terms will be negative. So, x = 1 <\/em>works. So will x =2, 3, 4, 5, 6, and 7<\/em>.
<\/p>\n\n\n\n

\"Inequalities<\/figure><\/div>\n\n\n\n

What else can we latch on to?<\/h3>\n\n\n\n

Remember, product of an odd number of negative terms is negative; product of even number of negative terms is positive. Now, this idea sets up the rest of the question and is useful when it comes to inequalities questions.

When x = 9<\/em>, there is one positive terms and 46 negative terms. So, the product will be positive. 
When x = 11<\/em>, there are two positive terms and 45 negative terms. So, the product will be negative. 
When x = 13<\/em>, there are three positive terms and 44 negative terms. So, the product will be positive and so on.

Essentially, alternate odd numbers need to be counted, starting from 11.

So, the numbers that will work for this inequality are 1, 2, 3, 4, 5, 6, 7\u2026and then 11, 15, 19, 23, 27, 31,\u2026.. and so.

What will be the last term on this list? 99, because when x = 99<\/em>, there are 46 positive terms and 1 negative term. 

So, we need to figure out how many terms are there in the list 11, 15, 19,\u2026.99. These can be written as: 
4 * 2 + 3, 
4 * 3 + 3, 
4 * 4 + 3
\u2026
4 * 24 + 3

A set of 23 terms. So, total number of values = 23 + 7 = 30. <\/p>\n\n\n\n

30 positive integer values of x exist satisfying the condition. <\/em><\/p>\n\n\n\n

Question 2: <\/h3>\n\n\n\n

Find the range of x for which (x + 2) (x + 5) > 40.<\/em><\/p>\n\n\n\n

\"\"<\/figure><\/div>\n\n\n\n

Solution:<\/h3>\n\n\n\n

There are two ways of trying this one:<\/p>\n\n\n\n

First Method:<\/h4>\n\n\n\n

We can expand and simplify this algebraically as x^2 + 7x + 10 > 40<\/em> or x^2 + 7x \u2013 30 > 0(x + 10) (x -3) > 0<\/em><\/p>\n\n\n\n

The roots are -10 and +3.
=> x should lie outside the roots.<\/p>\n\n\n\n

Now, what is this based on? There is a simple thumb rule for solving quadratic inequality, which is that for any quadratic inequality ax2<\/sup>+ bx + c < 0,<\/em> factorize it as a(x \u2013 p) ( x \u2013 q) < 0<\/em><\/p>\n\n\n\n

Whenever a is greater than 0, the above inequality will hold good if x lies between p and q. 

a(x \u2013 p) (x \u2013 q)<\/em> will be greater than 0, whenever x does not lie between p and q. In other words x should lie in the range ( -infinity, p) or (q, infinity)<\/p>\n\n\n\n

Now, coming back to the question’s solution: <\/p>\n\n\n\n

(x+10) (x -3) > 0<\/em> or, x < -10 or x > 3<\/em> <\/p>\n\n\n\n

Second method: <\/h4>\n\n\n\n

5 * 8 = 40, -8 * -5 = 40; this essentially means if x + 2 > 5<\/em> this will hold good => x > 3<\/em><\/p>\n\n\n\n

If x+ 2 < -8<\/em> also, this will hold good => x < -10<\/em>. <\/p>\n\n\n\n

The first method is far more robust.<\/p>\n\n\n\n

\"\"<\/figure><\/div>\n\n\n\n

Question 3:<\/h3>\n\n\n\n

Find the range of x where ||x \u2013 3| \u2013 4| > 3<\/em>.<\/p>\n\n\n\n

Solution:<\/h3>\n\n\n\n

If we have an inequality |y| > 3<\/em>, this will be satisfied if => y > 3<\/em> or y < -3<\/em>.
Hence, the above inequality simplifies to two inequalities:<\/p>\n\n\n\n

1. | x \u2013 3| \u2013 4 > 3<\/em> <\/h4>\n\n\n\n

| x \u2013 3| \u2013 4 > 3 => | x \u2013 3 | > 7<\/em>
 x \u2013 3 > 7<\/em> or x \u2013 3 < -7<\/em>
Or, x > 10<\/em> or x < -4<\/em>

x lies outside of  -4 and 10. <\/p>\n\n\n\n

Or, x can lie in the range ( \u2013 infinity, -4) or ( 10, infinity) <\/strong><\/em><\/p>\n\n\n\n

2. |x -3| \u2013 4 < -3<\/em><\/h4>\n\n\n\n

|x \u2013 3 | \u2013 4 < \u2013 3<\/em>
=> | x \u2013 3 | < 1<\/em>
=> -1 < x \u2013 3 < 1<\/em> or x lies between 2 and 4<\/p>\n\n\n\n

The final solution is given by the range( \u2013 infinity, -4) or (2,4) or ( 10, infinity)<\/em><\/p>\n\n\n\n

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