Strategy is a critical differentiator in the CAT; a well laid out strategy can give the test taker a potent competitive advantage. Here you need to appreciate that the CAT is not merely a test of Maths and English, it is a test of managerial skills assessed through questions on Quantitative and Verbal ability. These skills facilitate your role as a manager and strategy is core to this skill set.
At the onset it is imperative to understand strategy as a sequence of steps required to solve a problem. However, you also need to realize that an increased number of steps will lead to more time to accomplish the task and reduce the efficiency. Hence an optimal number of steps are advocated in the right application of strategy. Let's apply this to two levels of questions - an easy level and a difficult level (categorization based on student feedback over the years).
Consider the following question, which has been rated by the student community as a relatively easy question:
Q 1. 'ab' is a two digit number and 'ccb' is a three digit number such that (ab)2 = ccb, where 'ccb' is greater than 300. What is the value of 'b'?
Let's try to look at the various steps required to solve the question. The starting point is the fact that b2 ends in b ; so you should be able to conclude that the possible values which 'b' can take are 0,1,5 and 6, as these are the only digits in the decimal system which will end in the same respective digit when squared. At this juncture you need to inspect the options. If only one of these four values is given in the options, then that particular value qualifies as the answer and the question is solved. However, if more than one of these four values are profiled in the options (which is true in the given case), then you need to graduate to the next step. The next step is base on the number 'ccb' which is a three digit perfect square more than 300 - the first such number is 324 and the last such number is 961, which are the squares of 18 and 31 respectively. Hence you may infer that 'ab' is a number between 18 and 31, such that 'b' is one of 0, 1, 5, and 6. The only numbers in the consideration set that satisfy this condition are 20, 21, 25, 26 and 30, out of which only 21 satisfies the condition that the first two digits of the square are the same. Hence the equation is 212 = 441, and the values of 'a', 'b' and 'c' are 2, 1 and 4 respectively. So the answer is option 2.
Now let us take a question which has been ranked by the student community as a tough one.
Q2. If a!+b!+c!=abc, where a, b & c are distinct decimal digits, find b.
This question is popularly attempted by students because of its favourable length (just two lines to be read) .Further the values in the options are single digit numbers, making the overall package an attractive one for potential students. However, as you start processing the information in the question, there seems to be an apparent insufficiency of data- the foremost challenge is identifying the starting point. If you look at the equation more meticulously, you will realize that none of 'a', 'b' or 'c' can take a value of 7 or more, because the right hand side is a three digit number and 7 onwards will yield four digit numbers. Hence the variables 'a', 'b' and 'c ' can take only values less than 7. Here you need to look at the options; if only option is less than 7, then that qualifies as the answer and the next step is not required. However in the given case, all options are numbers less than 7, hence the need to move on to the next step. Now check the feasibility of the next number in the descending order, which is '6'. Factorial of 6 is 720, to which if you add two more factorials, you get a number starting with a 7, 8 or 9, all of which are not possible (as discussed above). Hence you should be able to conclude that 6 is also not a possible value on the left hand side. Now check the possibility for the next number, which is 5, and you should be quick to infer that 5 is mandatory in the system, because if 5 is not there then the maximum value which 'abc' can take will be 4! + 3! + 2! = 32, which is a two digit number. Thus one of 'a', 'b' and 'c' will be equal to 5. With 5 as one of the values the maximum value of 'abc' will be 5! + 4! + 3! = 150, and 150 is also not possible because it does not satisfy the given equation. So you should be able to conclude that 'abc' is a three digit number less than 150, which implies that the value of 'a' is 1. Now the next step is to determine which out of 'b' and 'c' will be 5. Since 'abc' is less than 150, 'b' can't be 5, hence 'c' will be equal to 5. Based on the above you get the equation: 1! + b! + 5! = 1b5, where 'b' can take the values 0, 2, 3 and 4. The last step requires you to check for these values by trial and error and conclude that only 4 fit in the given equation, which finally becomes 1! +4! +5! = 145. Hence the values of a, b and c are 1, 4 and 5 respectively. So the answer is option 3.
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