CAT 2021 Question Paper | Quant Slot 3

CAT 2021 Quant was dominated by Arithmetic followed by Algebra. In Arithmetic, the questions were dominated by topics like Speed-time-distance, Mixture and Alligations. This year, there was a surprise. The questions from Geometry were relatively on the lower side as compared to the previous years. There were 8 TITA Qs this year. Overall this section was at a medium level of difficulty.

Question 18 : The number of distinct pairs of integers (m, n) satisfying |1+mn| < |m+n| < 5 is

Video Explanation

Explanatory Answer

|1 + mn| < |m + n| < 5
For two numbers ‘a’ and ‘b’,
|a| < |b| is equivalent to a² < b²


So, we can say that:
(1 + mn)² < (m + n)²
1 + 2mn +m²n² < m² + n² + 2mn
1 - n² - m² + m²n² < 0
(1 - n²) - m²(1 - n²) < 0
(1 - m²)(1 - n²) < 0


For the product to be negative, either one of the two terms has to be negative.
But they cannot simultaneously be 0.
The only possibility for either of the two terms to be positive is when
n = 0 and |m| > 1, or |n| > 1 and m = 0


Now for the case when m = 0 and |n| > 1
|m + n| < 5
|0 + n| < 5
So n can be \( \pm \)2, \( \pm \)3, \( \pm \)4
Which are 6 cases

Similarly for the case when n = 1 and |m| > 1
|m + n| < 5
|0 + m| < 5
So m can be \( \pm \)2, \( \pm \)3, \( \pm \)4
Again we have 6 cases.
Hence the answer is 12.

The question is " The number of distinct pairs of integers (m, n) satisfying |1+mn| < |m+n| < 5 is "

Hence, the answer is '12'