 # 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 a2 + b2 + c2 + d2 is 4m2 – 2m + 1

(2) The minimum possible value of a2 + b2 + c2 + d2 is 4m2 + 2m + 1

(3) The maximum possible value of a2 + b2 + c2 + d2 is 4m2 – 2m +1

(4) The maximum possible value of a2 + b2 + c2 + d2 is 4m2 + 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+.

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### 4 Responses to Solved Paper of CAT 2003- Part I

1. nazma June 1, 2013 at 12:48 pm #

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

2. vishal August 21, 2013 at 10:50 am #

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….

3. Ravi Handa August 25, 2013 at 8:04 am #

@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.

4. shiksha February 17, 2020 at 9:42 am #

answer number 2 is wrong may be correct answer would be 31

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