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 four digit numbers for which the first two digits are equal and the last two digits are also equal. How many such numbers are perfect squares?
Topic : Numbers
Approach :Let the four digit number be denoted by aabb.
Then the number can be expressed as : 1000a+100a+10b+b
= 1100a+11b = 11(100a+b)
Since aabb is a perfect square, 100a+b should be 11k where k is a perfect square
Now 100a+b can be written as a0b, and for divisibility by 11, a+b=11. The various combinations of (a,b) are (2,9), (9,2), (3,8), (8,3), (4,7), (7,4), (5,6), (6,5). However out of these only (7,4) satisfies the second condition where k is a perfect square.
i.e. For a=7, b=4, 100a+b = 704= 11*64, and aabb= 7744, which is a perfect square. No other pair satisfies the given conditions.Hence only one number satisfies the given conditions.
Answer : Option 3
Challenge :Understanding of divisibility rules, perfect squares and expansion of numbers in the decimal system
In a tournament there are n teams T1, T2, T3....Tn with n greater than 5. Each team consists of k players where k is more than 3. The following pairs of teams have one player in common : T1& T2, T2& T3,....Tn-1& Tn and Tn& T1. No other pair of teams has any player in common. How many players are participating in the tournament, considering all the n teams together?
Topic : Logic
Approach :In each team Tj there are k players, out of which two are shared, one with Tj-I and the other with Tj+I. The remaining k-2 players are not shared with any team.
Thus, total players who play for only one team = (k-2)n
One player is common in T1& T2, one in T2& T3 and so on.
Number of such players = number of pairs =n
So, total players = (k-2)n+n = n(k-1)
Answer : Option 1
Challenge :Understanding logical propositions in word problems; cyclic relationships
The price of Darjeeling Tea (in rupees per kilogram) is 100+0.1n on the nth day of a non-leap year, where n=(1,2,3…….100) and then remains constant. On the other hand the price of Ooty Tea (in rupees per kilogram) is 89+0.15n on the nth day of the same year where n=(1,2,3……365). On which date of the year will the prices of the two varieties of tea be equal?
Topic : Equations
Approach :Price of Darjeeling Tea on the 100th day of the year will be 100+(0.1*100)= 110. This is the maximum price after which it becomes constant for the remaining part of the year.
Price of Ooty Tea on the 100th day will be 89+(0.15*100)=104
Now for the prices to be equal, Ooty Tea has to gain Rs. 6 because the price of the other variety is constant.
It gains Rs 0.15 every day after the 100th day. So it will gain Rs. 6 in 6/0.15= 40 days.
Thus the prices of the two varieties will be equal on the 140th day of the year.
Number of days in the months of January, February, March & April of a non-leap year will be 31+28+31+30=120
Therefore the prices will be equal on 20th May.
Answer : Option 1
Challenge :Understanding parity of arithmetic expressions
X and Y start running around a circular track simultaneously. They run in the same direction starting from the same point at the same time. X travels at 6 m/s and Y runs at s m/s. If they cross each other at exactly two points on the circular track and s is a natural number less than 30, how many values can s take?
Topic : Time-Speed- Distance
Approach : Let track length be equal to T.
Time taken to meet for the first time = T/relative speed=T/(6−s) or T/(s−6)
Time taken for a lap for X = T/6
Time taken for a lap for Y = T/s
So, time taken to meet for the first time at the starting point =LCM (T/6,T/s)=T/HCF(6,b)
Number of meeting points on the track = Time taken to meet at starting point/Time taken for first meeting = (T/Relative speed) / (T/HCF (6,s)) = Relative Speed/HCF(6,s)
So, in essence we have to find values for b such that (6−s)/HCF(6,s) = 2 or (s−6)/HCF(6,s) = 2
s = 2, 10, 18 satisfy this equation. So, there are three different values that s can take.
Answer : Option 1
Challenge :Application of time-speed-distance to circular motion