What’s the CATch- 11?

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.
Find the remainder of 21040 divided by 131.
1. 1
2. 3
3. 5
4. 7
Topic : Numbers
Approach :This question can be best solved by the application of Euler’s Theorem.
Grab the Unbelievable Offer:
Kick start Your Preparations with FREE access to 25+ Mocks, 75+ Videos & 100+ Sectional/Area wise Tests
Sign Up Now
When yE (z) is divided by z, the remainder will always be 1, where E(z) is Euler number of z and y & z are co-prime to each other. Also, when yE (z)k is divided by z, where k is an integer, remainder will always be 1 i.e if the power is any multiple of the Euler number of the divisor, even in that case the remainder will be 1.

Note : The Euler number of a number x means the number of natural numbers which are less than x and are co-prime to x. Mathematically, the Euler number of a number z denoted by the symbol E(z) is calculated as E(z)= z(1- 1/p)(1- 1/q) (1- 1/r), where  p, q and r are the different prime factors of z. Euler number of a prime number is always 1 less than the number.
In this case 2 and 131 are co-prime, so we can apply Euler’s Formula.
Now, the E(131)= 130 (since 131 is a prime number).
21040 can be written as 2(130*8). So the question amounts to checking the remainder of 2(130*8) on divisibility by 13. Applying Euler’s Formula, the remainder is 1.
Answer : Option 1
Challenge :Application  of the concept of divisibility using a quicker approach
If mod(r-6)=11 and mod(2q-12)= 8,what is the minimum possible value of q/r?
1. 10/17
2. 2/17
3. -2/5
4.None of these
Topic : Numbers
Approach :mod (r-6) = 11
Thus, r-6= 11 or r-6 = -11
r = 17, -5
mod (2q-12) = 8
Thus, 2q-12 = 8 or 2q-12 = -8
q = 10, 2
There are 4 vales which q/r can take : 10/17, -2, 2/17 and -2/5
The least of these is -2.
Answer : Option 4
Challenge :Application of the concept of modulus of a number; exploring multiple possibilities in a consideration set
What is the remainder when 9+92+93+94+…..926345is divided by 6?
1. 3
2. 0
3. 1
4. 4
Topic : Numbers
Approach :We start checking for remainders as under:-
9/6 leaves a remainder of 3
(9+92)/6 leaves a remainder of 0
(9+92+93)/6 leaves a remainder of 3
(9+92+93+94)/6 leaves a remainder of 0
Thus, when the expression ends in an odd power of 9 remainder is 3, and when it ends in an even power of 9, the remainder is 0. In the given case the last term ends in an odd power of 9, hence the remainder is 3.
Answer : Option 1
Challenge :Understanding the concept of symmetry in numbers; application of trend analysis in divisibility questions
FORE
What is the remainder when 113246578-5 is divided by 8?
1. 4
2. 6
3. 3
4. 7
Topic : Numbers
Approach :We start checking for remainders as under:
(111-5)/8 leaves a remainder of 6
(112-5)/8 leaves a remainder of 4
(113-5)/8 leaves a remainder of 6
(114-5)/8 leaves a remainder of 4
Thus we observe that 11 raised to an odd power minus 5 leaves a remainder of 6, and 11 raised to an even power minus 5 leaves a remainder of 4. In the given case, the power of 11 is even, hence the answer is 4
Answer : Option 1
Challenge :Understanding the concept of symmetry in numbers; application of trend analysis in divisibility questions
The question is followed by two statements A and B.
Mark (1) if the questions can be answered using A alone but not using B alone
Mark (2) if the questions can be answered using B alone but not using A alone
Mark (3) if the questions can be answered using A and B together but not using A or B alone
Mark (4) if the questions cannot be answered even using A and B together
What are the respective values of X and Y given that they are both distinct whole numbers?
  1. X+Y is a common prime divisor of 56 and 112.
  2. X and Y are prime numbers
Free Master Classes:
Tips & Tricks to Crack CAT by our Star Faculty with 20+ years of experience.
Register Now
Topic : Numbers
Approach :Using  A, the common prime divisors of 56 and 112 are 2 and 7.
Thus X+Y=2 or X+Y=7
But this statement alone is not sufficient
From B, X and Y are prime numbers; which does not give any clue on their values.
Using A and B, X=Y=2 is ruled out as two prime numbers can’t add upto 2. If X+Y=7, then the only possibility is that the numbers are 2 and 5. However, which variable takes which value is not clear from here. Hence we can’t get the answer even if we use both the statements together.
Answer : Option 4
Challenge :Understanding data sufficiency challenges in numbers; ability to manage impulse 
Rate Us
Views:4261