What’s the CATch- 4?

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.
If xyz is a number which ends in a digit that is neither prime nor composite, where x, y and z are distinct decimal digits and x:y:z=1:2:3, then how many values of the ordered triplet (x,y,z) are possible?
2. 2
3. 3
4. Can’t be determined
Topic : Numbers
Approach :xyzends in 1 (neither prime nor composite)
The three possibilities on the basis of the second condition are: 123, 246, 369
However, 246does not end in 1, hence ruled out.
Out of 123and 369 , 123 definitely ends in 1.
We just need to check for 369.
Now, an even digit (2,4,6,8) raised to the power of 2 or more will always be a multiple of 4. So, 369=34k
Apply cyclicity concept. Cycle for 3 is 3,9,7,1
Thus 34k ends in 1. Thus 369 also ends in 1.
Answer : Option 2
Challenge : Understanding the application of the concept of cyclicity and indices
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In 1937 a man who was then less than 100 years stated that he was ‘x’ years old in the year x2. He further said “If the number of years in the year x2 be added to the serial number of my birth month (counting January as the first month), the result equals the square of the day of the month in which I was born”. When was he born?
1. 7th May, 1936
2. 7th May, 1892
3. 5th July, 1936
4. Can’t be determined
Topic : Numbers
Approach: He was x years old in the year x2; so check for squares close to 1937.
Now 442= 1936, 432= 1849
If he was 43 years in the year 1849, then in 1957 he would be more than 100 years, hence it is ruled out. Thus he was 44 years in the year 1936. Thus his year of birth is 1936-44=1892
Now 44+m=d2, where m is the serial number of the birth month and d is the date.
Now m has to be less than equal to 12. Thus m=5, d=7
The birth date is 7th May, 1892
Answer : Option 2
Challenge : Comfort level with squares of numbers and ability to decode logical propositions into numerical correlations.
A and B enter into a partnership. A puts in the whole capital of Rs. 45,000 on the condition that the profits will be equally divided after which B will pay A interest on half the capital at 10% p.a. and receive Rs 60 per month from A for carrying on the concern. What is the yearly profit if B’s income is half of A’s income?
1. 9180
2. 4590
3. 8460
4. None of these
Topic : Partnerships
Approach: Let x be the annual total profit. B’s income will be:
x/2 - 2250 +720
And A’s income will be:
x/2 + 2250 -720
Thus x/2 - 2250 +720 = ½ (x/2 + 2250 -720)
Solve for x.
X = 9180
Answer: Option 3
Challenge: Ability to interpret algebraic statements into discrete equations
The average age of a group of n persons is 75. Two persons with ages between 56 and 52 leave the group. A third person with age between 80 and 89 joins the group. The new average age of the group is a prime number. What is the possible age of the person who joined the group?
1. 81
2. 87
2. 86
4. 88
Topic: Averages
Approach: Let y be the new average age of the group, and x be the age of the person who joins the group.
Then y= (75n-(56+52) +x)/(n-2+1) = (75n-108+x)/(n-1)
Rearranging the terms, the equation can be written as:
y= (75 (n-1) + (x-33) )/(n-1) = 75 + (x-33)/(n-1)
Now check for options
If x=81, then y = 75 + 48(n-1). For y to be prime, it has to be a natural number. Thus n-1 has to be a factor of 48 and can take values 2,3,4,6,8,12,16,24,48. When n-1 is 6, y is 83, which is a prime number. Hence x=81 is a possible value.
Note that other options will not satisfy this condition. For example,
If x=87, then y=75+ 54/(n-1). For n-I to be a factor of 54 it can take values 2,3,6,9,18,27,54. For none of these, y is a prime number.
Answer : Option 1
Challenge : Application of averages to questions with higher levels of difficulty, more in the context of multiple possibilities and elimination of ineligible options
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The speed of a railway engine is 42 km/hr when no compartment is attached, and the reduction in speed is directly proportional to the square root of the number of compartments attached. If the speed of the train carried by this engine is 24 km/hr when 9 compartments are attached, what is the maximum number of compartments that can be carried by the engine?
1. 49
2. 48
2. 0
4. None of these
Topic : Variation & Proportion
Approach: The reduction in speed r is directly proportional to the square root of the number of compartments attached. Thus r = k n1/2
Now 42- kn1/2is the net speed when n compartments are attached
Also, 42- k91/2= 24
Therefor k = 6.
So net speed is 42 – 6n1/2
Now 42 -6 nmax1/2is greater than zero
Thus nmax is less than 49
Or nmax = 48
Answer: Option 2
Challenge: Ability to apply the concept of variation and understanding the fact that nmax cannot be 49, because at 49, the engine ceases to move. Here it is more about managing the emotion of instant gratification as reaching the value of 49 is not much of a difficulty, but marking 48 as the right answer is the deciding point.
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