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.

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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.

Thus, total players who play for only one team = (k-2)n

One player is common in T_{1}& T_{2}, one in T_{2}& T_{3} and so on.

Number of such players = number of pairs =n

So, total players = (k-2)n+n = n(k-1)

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.

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)

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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.