In this article, we will appreciate the application of creativity to solve questions in the CAT and validate the fact that a lateral way of thinking can help establish an immediate connect with certain questions. For this purpose we have taken three levels of questions - easy, moderate and difficult, in terms of the difficulty index. We understand that individual perception and personal skills may result in differences in the way these questions are deemed to be in a certain level of difficulty, but our segregation is based on averaging the feedback of CAT aspirants over the years.
In a hemispherical igloo, an Eskimo's head just touches the roof when he stands erect at the centre of the floor, but his son can play over an area of 9856 sq. units of the floor without stooping. If the Eskimo's height is 65 units, what is the son's height?
4. None of these
Let r be the radius of the circular path over which the son plays. Then (22/7)*r2 = 9856, which gives the value of r as 56. The Eskimo's head just touches the roof when he stands erect at the centre of the floor; so the Eskimos height corresponds to the radius of the semicircular section of the Igloo. Let the Eskimos height be represented by OA where O is the centre of the semicircular section. If CB is the son's height, where C is a point on the semicircle and B is a point on the diameter of the semicircle (beyond C, the son has to stoop), then OB = 56 (as calculated above). By a simple application of Pythagoras, OC2 = OB2 + CB2. From here we get the son's height CB as 33. Hence the answer is option 1. The creativity here is to form a visual imagery of the set up and place the Eskimo and the son in that set up, which leads to us to connect them through Pythagoras theorem.
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Ram leaves home with "x" flowers, and then goes to the bank of a nearby river. On the bank of the river, there are four places of worship in a row. He dips all the "x" flowers into the river. The number of flowers doubles. Then he enters the first place of worship, offers "y" flowers to the deity. He dips the remaining flowers into the river and again the number of flowers doubles. He goes to the second place of worship, offers "y" flowers to the deity. He dips the remaining flowers into the river, and again the number of flowers doubles. He goes to the third place of worship, offers "y" flowers to the deity. He dips the remaining flowers into the river, and again the number of flowers doubles. He then goes to the fourth place of worship, offers "y" flowers to the deity. Now he is left with no flowers in hand. What is the minimum number of flowers that could be offered to each deity?
The starting point for this question is the fact that the number of flowers doubles upon dipping the flowers in the river, thus becoming 2x. After y flowers are offered to the deity, he is left with 2x-y flowers. Then again the number doubles to 4x-2y; he is left with 4x-3y as y are offered to the deity. Again the number doubles to 8x-6y; he is left with 8x-7y as y are offered to the deity. Again the number doubles to 16x-14y; and he is finally left with 16x-15y after y flowers are offered at the last deity. But this should be equated to zero as he is left with no more flowers after making the last offering. Thus, 16x-15y= 0. This gives (x/y) = (15/16). Now in this form, (x/y) is in its least reduced form and cannot be reduced any further, which means that the minimum value of x is 15 and that of y is 16. Hence the answer is 16 (option 3). The length of the question is what places it in the moderate level of difficulty. Further, the application of creativity is manifested in the way you ensure a closure upon reaching the equation 16x-15y = 0 and concluding that this equation itself yields the minimum values of x and y.
The average age of a group of "n" persons is 75. Two persons with ages 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. Which of the following is a possible age of the person who joined the group?
The sum of the ages of all the n persons is 75n; from this we subtract 108 (the total of the ages of the two people who leave the group), add k (the age of the person who joins the group) and divide by n-1 (two leave and one joins the group) to get the new average age of the group, which becomes (75n-108+k)/(n-1). On simplifying this becomes 75 + ((k-33)/n-1)). Now we just need to go to the options (which are different values of k) and check which one gives the overall value of the expression a prime number. For example if k= 87, then the expression becomes 75 + (54/n-1); now n-1 has to be a factor of 54 for it to perfectly divide 54. Putting different values of n-1 as different factors of 54 (1, 2, 3, 6, 9, 18, 27, 54), we find that none of them yields a value which when added to 75 gives a prime number. Through trial and error, we discover that for k=81, the expression becomes 75 + (48/n-1). Putting n-1 =6 (one of the factors of 48), we get the overall value of the expression as 83, which is a prime number. No other value of k will give the new average age of the group as a value which is a prime number. Hence the answer is option 3. The creativity part of the challenge was to split the expression for the new average age of the group as 75 + ((k-33)/n-1)) and proceed there onwards.
In conclusion, it can be said that creativity enables the candidate to look at the question from a divergent angle, and facilitates a faster treatment of steps which are required to solve the question.
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