 # What’s the CATch? – Part 3

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.
1. Find the total number of integers k between 20 and 40 such that (k-1)! Is not divisible by k.
1. 6
2. 2
3. 5
4. None of these
Topic : Numbers
Approach: (k-1)! will NOT be divisible by k only when k is prime.
If k is composite than its prime factors are contained in (k-1)!
Hence the question simply translates into finding the number of prime numbers between 20 and 40, which are : 23, 29, 31 and 37.
Answer : Option 4 (there are 4 such integers)
Challenge : Application of the concept of factorials and prime factors
2. A die bearing the numbers 0,1,2,3,4 and 5 on its faces is repeatedly thrown until the total of the throws exceeds 12. What is the most likely total that will be thus obtained?
1. 14
2. 13
3. 15
4. None of these
Topic : Probability
Approach : Consider the throw before the last throw. The total after that throw must be 8, 9, 10, 11 or 12.
If the total is 8 then the final result will be 13. If it is 9, then the final result will be 13 or 14 with an equal chance of each. If it is 10, then the final result will be 13,14 or 15 with an equal chance of each. If it is 11, then the final result will be 13,14, 15 or 16 with an equal chance of each. If it is 12, then the final result will be 13,14, 15 or 16 with an equal chance of each.
The most likely final result is 13.
Challenge : Conceptualizing the penultimate throw situation and correlating it with the given information
3. If x is an even number divisible by 3, then 111111..... upto x digits is always divisible by .....
1. Only 7
2. only 11
3. Both 7 and 11
4. None of these
Topic : Numbers
Approach :Since x is an even number and a multiple of 6, it is also a multiple of 6. Thus x=6k, where k is a positive integer.
So we have to check the divisibility of 111…..upto 6/12/18 etc digits by 7 and 11.
It is known that a number of the form xyzxyz is divisible by 1001.
1001= 7*11*13
Thus the given number will be divisible by both 7 and 11
Challenge : Understanding symmetry in numbers and application of divisibility rules
4. ABCDEF is a six digit number with distinct decimal digits. ABCDEF is divisible by 11, and the sum of its digits is 16. It is also known that A>C>E and B>D>F. Then A+B is always ....
1. 5
2. 10
3. 1
4. None of these
Topic : Numbers
Approach :(A+C+E)- (B+D+F) = 11k, where k=0,1,2 etc
Let A+C+E = x and B+D+F = y, then
x – y = 11k and x + y= 16
K can’t be 1 or more; if k=1 then x=8.5 which is not possible, and if it is 2 then the value of y comes out to be negative which is not possible. Higher values of k are visibly ruled out. Hence k = 0
Thus x-y=0 or x= y and from x+y=16, we get x=y=8
A, C, E will be 6,2,0 or 4,3,1 and B, D, F will be 4,3,1 or 6,2,0 respectively. Thus A+B= 4+6 or 6+4 =10.
Challenge : Applying divisibility rule for 11 and doing a feasibility check on permissible values
5. A and B start running along a circular track from the same point at the same time in the same direction. To cover the whole track A needs 5 seconds less than B. They meet for the first time after 30 seconds. How many times will they meet in a minute if they run in opposite directions.
1. 18
2. 15
3. 5
4. None of these
Topic : Time-Speed-Distance
Approach :Let the speeds of A and B be m meters/s and n meters/s respectively and l meters be the circumfrence. Then time taken by A and B respectively to complete one round will be l/m and l/n seconds. Thus (l/n – l/m)= 5.
If they meet after 30 s, then l/(m-n)=30 or (m-n)/l=1/30
Let l/m=x and l/n=y. Thus 1/y – 1/x =5 and x-y=1/30
or xy = 1/150. Now (x-y)2= x2+y2-2xy or x2+y2= 13/900
(x+y)2= x2+y2+2xy or x+y= 6 or l/(m+n)= 6 seconds =1/10 minutes.
Thus they will meet 10 times in minute.