# Solved Paper of CAT 2003- Part I

*Continuing with our efforts to provide articles for the student community, we’ll provide solved papers of CAT of previous years. Mr. Ravi Handa, an alumnus of IIT Kharagpur who has taught Quantitative Aptitude at various coaching institutes for seven years, has agreed to kickstart this series.
*

In this particular series of posts, I will provide previous years’ solved paper of CAT on video. I am starting of with CAT 2003 solutions for **Number System** questions.

## Solved Paper of CAT 2003- Number System

**Q1. How many even integers n, where 100 <= n <= 200, arc divisible neither by seven nor by nine?**

(1) 40

(2) 37

(3) 39

(4) 38

Video Solution:

**Q2. A positive whole number M less than 100 is represented in base 2 notation, base 3 notation, and base 5 notation. It is found that in all three cases the last digit is 1, while in exactly two out of the three cases the leading digit is 1. Then M equals**

(1) 31

(2) 63

(3) 75

(4) 91

**Solution**:

**Q3. Let a. b, c, d be four integers such that a + b + c + d = 4m + l where m is a positive integer. Given m, which one of the following is necessarily true?**

(1) The minimum possible value of a^{2} + b^{2} + c^{2} + d^{2} is 4m^{2} – 2m + 1

(2) The minimum possible value of a^{2} + b^{2} + c^{2} + d^{2} is 4m^{2} + 2m + 1

(3) The maximum possible value of a^{2} + b^{2} + c^{2} + d^{2} is 4m^{2} – 2m +1

(4) The maximum possible value of a^{2} + b^{2} + c^{2} + d^{2} is 4m^{2} + 2m + 1

**Solution**:

**Q4. If the product of n positive real numbers is unity, then their sum is necessarily**

(1) A multiple of n

(2) Equal to n + 1/n

(3) Never less than n

(4) A positive integer

**Solution**:

I hope you enjoyed this post and liked it. If you found it useful, let me know through your comments so that I continue with this series. Don’t forget to share it with your friends on Facebook, Twitter & Google+.

the way you approach the problem is very simple and it is really useful

answer number 2 is worng…… the correct answer would be 31

the method you are using is not correct in last lcm section. if you apply same lcm method 91 will not be the answer.. try it….

@Vishal – I don’t understand what you are trying to say. 31 could not be the answer because

31 in base 2 is 11111

31 in base 3 is 1011

31 in base 5 is 111

The leading digit is one in all these case. We want it to be 1 in two of these cases.