Belling the CAT
Appropriate preparation and application of the right thought process can help you clear the CAT exam without much difficulty. Arun Sharma presents the QA section of the CAT 2007 exam with tips on how to approach each question and increase your chances of passing
One of the things that has amazed experts and academicians about the CAT for a long time now, is that the exam continues to retain its aura of invincibility. This, in spite of the fact that most questions, which appear in the exam, are elementary class ten questions.
The quantitative aptitude (QA) section of the CAT 2007 paper presented below further reiterates the fact, that answering each question in a logical and methodical manner is the only way to clear the exam.
Reminders (about the paper) before proceeding further
- Test takers who were able to attempt seven to eight questions of the total 25 questions in the paper received an excess of 95 percentile (for this section). Students were given 50 minutes to complete each section in the exam. Hence, even if a student answered one question correct every 6 minutes, he/ she would easily be one of the top 95 percentile holders in the section.
- At zero marks in this paper, students reported a 20 percentile score. This in effect means that one in five test takers scored negative marks!
- The QA section in the CAT 2007 paper has widely been considered as one of the toughest QA papers of all time. Presented below are simplistic solutions that can be taught to children as young as those in class five. It is advisable to look at the provided thought process and solutions only after trying to solve the questions before.
Q) How many pairs of positive integers m, n satisfy 1/ m + 4/ n = 1/ 12 where n is an odd integer less than 60?
(a) 7 (b) 5 (c) 3 (d) 6 (e) 4
Deduction 1: Since two positive fractions on the LHS equals 1/ 12 on the right hand side, the value of both these fractions must be less than 1/ 12. Hence, n can only take the values 49,51,53,55,57 and 59.
Deduction 2: After this, students need to check which of the possible values of n would give an integral value of m.
The equation can be transformed to 1/ 12 – 4/ n = 1/ m –> (n – 48)/ 12n = 1/ m.
On reading this equation one should realize that for ‘m’ to be an integer, the LHS must be able to yield a ratio in the form of 1/ x. It can be easily seen that this occurs for n = 49, n =51 and n =57. Hence, there are only three pairs.
Maximum solution time: 90 seconds
Q) 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?
(a) 18 (b) 15 (c) 19 (d) 17 (e) 16
Deduction 1: If you were to use two 50 misos notes, you would only pay the remaining 7 misos through one miso notes.
Deduction 2: If you were to use only one 50 miso note, you could use 10 miso notes in 6 different ways (from zero to five).
Deduction 3: If you were not to use any 50 miso notes, you could use 10 miso notes in 11 different ways (from zero to 10).Hence, the required answer is 1+ 6+ 11=18.Maximum solution time: 45 seconds
Q) In a tournament, there are n teams T1, T2, …, Tn, with n > 5. Each team consists of k players, k > 3. The following pairs of teams have one player in common
T1 and T2, T2 and T3, …, Tn (1 and Tn, and Tn and T1)
No other pair of teams has any player in common. How many players are participating in the tournament, considering all the h teams are together?
(a) n (k ( 2) (b) k (n ( 2) (c) (n ( 1) (k ( 1)
(d) n (k ( 1) (e) k (n (1)
If one considers six teams and four players per team, one would get four players in T1 (each one of them unique), three more players in T2 (since one player of T2 would be shared with T1), three more players in T3 (since one player of T3 would be shared with T2), three more players in T4 (since one player of T4 would be shared with T3), three more players in T5 (since one player of T5 would be shared with T4) and two more players in T6 (since one player of T6 would be shared with T5 and one with T1). Hence, there would be a total of 18 (4+ 3+ 3+ 3+ 3+ 2) players with n = 6 and k = 4. Only option four would provide 18 as the solution.
Maximum solution time: 60 seconds.
Q) 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?
(a) 4 (b) 0 (c) 1 (d) 3 (e) 2
A number of students got stuck on this question for over five to seven minutes in the exam, since they tried to find out the squares of all two-digit numbers starting from 32. However, if one is well aware of the logic of finding squares of two digit numbers, one would realise that only three two-digit numbers after 32 have the last two digits in their squares equal (38, 62 and 88). Hence, there is no need to check any other numbers apart from these three. After checking these one would get the square of 88 as 7744. And hence, there is only one such number.
Maximum solution time: 60 seconds
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