 # What’s the CATch- 6?

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.
Consider the set s=(2,3,4……..2n+1),where ‘n’ is a positive integer larger than 2007. Define X as the average of the odd integers in S and Y as the average of the even integers in S. What is the value of X-Y?
1. 0
2. 1
3.n/2
4. (n+1)/2n
Topic : Numbers
Approach : Since the last integer in S is 2n+1, it implies that the last integer is odd. Also we can see that the first integer is even (2). This implies that the total number of integers in S is 2n which is even.
Further the integers are all consecutive numbers. Hence the difference of the average of the odd and even integers will be one.
Alternatively,
Sum of odd integers in S = X=n/2(2*3 + (n-1)*2)= n+2
Sum of even integers in S= Y=n/2(2*2+(n-1)*2)= n+1
Thus X-Y= (n+2)-(n+1) = 1
Challenge : Ability to apply the concept of averages in numbers
How many pairs of positive integers m,n satisfy 1/m + 4/n = 1/12, where n is an odd integer less than 60?
1. 6
2. 4
3. 7
4. 3
Topic : Numbers
Approach : Given 1/m + 4/n = 1/12, where n is odd integer less than 60
Rearranging, 1/m = 1/12 – 4/n
Solving, m= 12n/(n-48)
For m to be an integer, (n-48) has to be a factor of 12n
Thus n take only three values– 49, 51 and 57.
Hence there are three pairs of (m,n) which satisfy the given conditions.
Challenge : Ability to rearrange data; understanding the application of factors
Suppose you have a currency named Miso, in three denominations: 1 Miso, 10 Misos and 50 Misos. In how many ways can you pay a bill of 107 Misos?
1. 17
2. 16
3. 18
4. 15
Topic : Numbers
Approach : Let the number of currency of 1 Miso, 10 Misos and 50 Misos be x, y and z respectively
Thus x + 10y + 50z = 107
Now the possible values of z could be 0, 1 and 2
For z=0, the equation will be x+10y= 107
Number of integer pairs of values of x and y that satisfy the equation will be (7,10), (17,9), (27,8)…..(107,0)- a total of 11 values
For z=1, the equation will be x+10y+50=107 or x+10y=57
Number of integer pairs of values of x and y that satisfy the equation will be (7,5), (17,4), (27,3)…..(57,0)- a total of 6 values
For z=2, the equation will be x+10y+100=107 or x+10y=7
There is only one integer value of x and y that satisfies the equation and that is (7,0)
The total number of values is 11+6+1= 18.
Challenge : Ability to form/solve discrete equations from word problems
Directions for the next two questions: Each 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
Consider integers x,y,z. What is the minimum possible value of x2 + y2 +z2 ?
A. x+y+z = 89
B. Among x, y, z, two are equal
Topic : Numbers
Approach : Using A, x=30, y=30, z=29, will give the minimum value
If the sum is fixed, the numbers have to be as close to each other as possible, for minimum value of the sum of their squares
Using B, nothing specific can be derived about the relation between x, y and z.