What’s the CATch- 5?

In this article certain questions have been randomly taken from different topics of the quantitative section of CAT. Each question has been selected keeping in mind certain challenges you need to manage as a test taker. These challenges may range from concept basics to application orientation and test taking strategies.
What are the last two digits of  72008?
1. 21
2. 61
3. 01
4. 41
Topic : Numbers
Approach: Last two digits are determined by the remainder obtained when the number is divided by 100
71---- last 2 digits are 01
72---- last 2 digits are 49
73---- last 2 digits are 43
74---- last 2 digits are 01
Now this process has a cyclicity of 4 with the pattern 01, 49, 43 and 01.
Divide the index (2008) by 4 and the remainder decides the last two digits. Since 2008 is perfectly divisible, we are the 4th set which corresponds to the digits 01. Hence the answer is 01.
Answer : Option 3
Challenge : Application of the concept of cyclicity and understanding symmetry in numbers
The integers 1,2,3….40 are written on a blackboard. The following operation is then repeated 39 times. In each repetition, any two numbers, say a and b, currently on the blackboard are erased and a new number a+b-1 is written. What will be the number left on the board at the end?
1. 820
2. 821
3. 781
4. 819
Topic : Numbers
Approach: In such questions we start by taking a smaller set of numbers on the same pattern as the given set. Let the numbers be 1,2,3,4,5. We have 5 numbers, so we perform the operation 4 times.
Now remove any two numbers, say 3 and 5 and replace these with 3+5-1=7. Thus the set is 1,2,4,7
Now remove any two numbers, say 1 and 7. The new set is 2,4,7.
Now remove any two numbers, say 4 and 7. The new set is 2,10.
Now remove both and we are left with 11 as the last number.
If you note vigilantly, you can appreciate that the number left (11) is the sum of the first 4 numbers of the set plus 1. Validate the pattern through another set of numbers.
Now if we repeat this with the given set, the number left will be:
(1+2+3+….39)+1 = (39*40)/2 + 1= 781
Answer: Option 3
Challenge:  Ability to analyse trends in smaller sets of numbers  and then scaling up the concept
The number of common terms in the two sequences 17,21,25…..417 and 16,21,26….466 is
1. 78
2. 19
3. 20
4. 77
Topic : Sequences & Series
Approach :The first sequence is an A.P with 4 as the common difference and the second sequence is an A.P with 5 as the common difference. L.C.M of 4 and 5 is 20. So the common term will be an A.P with 21 as the first term and 20 as the common difference.
Thus the general term of the common sequence will be
21+ (n-1) 20 = 20n+1 and this will be less than equal to the lesser of the last term of the two sequences i.e. 417.
Thus, n is less than equal to 416/20.
So, n= 20
Answer : Option 3
Challenge :Ability to correlate two number series
Suppose the seed of any positive integer is defined as seed(n)=n if n is less than 10 or seed(s(n)) otherwise, where s(n) indicates the sum of digits of n. How many such positive integers n are there such that n is less than 500 and will have seed(n)=9?
1. 39
2. 72
3. 81
4. 55
Topic : Numbers
Approach :The first such number will be 9 as seed(9)=9.
The second such number will be 18 as seed(18)= seed(1+8)=seed(9)=9
Thereafter all subsequent numbers will be multiples of 9 only as divisibility of a number by 9 is determined by the sum of the digits of the number.
Hence the required numbers will be:
9,18,27……..495
This is an A.P whose first term is 9 and last is 495 with common difference as 9.
Thus 495= 9+ (n-1)9
Or n= 55
Answer : Option 4
Challenge :  Ability to apply logical propositions in the context of numbers
How many integers greater than 999 but not greater than 4000 can be formed with the digits 0,1,2,3 and 4 if repetition of digits is allowed?
1. 499
2. 500
3. 375
4. 376
Topic : Permutations & Combinations
Approach :Integers greater than 999 and not greater than 4000 will imply integers from 1000 to 4000, both inclusive.
Such numbers will have a format ABCD
A can be occupied by 1,2 and 3 (leaving 4000 aside). Thus there are 3 ways of occupying A
B, C and D can be occupied by all 0,1,2,3,4. Thus there are 5 ways each of occupying B,C and D
So the total number of integers will be 3*5*5*5= 375
However 4000 is also a possibility. So the total number of integers will be 375+1 =376
Answer : Option 4
Challenge :  Ability to think laterally and explore multiple possibilities
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