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The key to getting a good percentile in Quant or any other competitive exam for that matter is to solve the question in the quickest way possible. Using CAT Quant shortcuts is one such method of getting the question done and dusted easily. In this article, we will talk about using the answer options to your advantage. This may be a method familiar to many aspirants. However, for the benefit of everyone, let’s have a look at the technique.

Take this question from CAT 2006:

The sum of 4 consecutive 2-digit odd numbers, when divided by 10, becomes a perfect square. Which of the following can possibly be one of these 4 numbers?
(1) 21  (2) 25 (3) 41 (4) 67  (5) 73

Solution: Since, the sum of the 4 consecutive odd numbers should be divisible by 10, the last digit of the sum should be 0. Since, the numbers are all consecutive odd numbers, the last digit should be 1,3,5,7 or 3,5,7,9 – so on. Out of these combinations, the only one which gives the sum of the digits as 0 is 7,9,1,3. Hence the 4 consecutive numbers should end with these digits.

Look at the answer options now.
Option (2) can be eliminated upfront as the last digit cannot be 5.
Option (1): In this case, the numbers are 17, 19, 21 and 23. Adding up the numbers gives us 80. Division by 10 is 8 which is not a perfect square. Hence, this is eliminated
Option (3): In this case, the numbers are 37, 39, 41 and 43. Adding up the numbers gives us 160. Division by 10 is 16 which is a perfect square. Hence, this is the right answer.

The question can easily be solved within a minute by this method.

Lets take another example –  this time from CAT 2008:

Find the sum √(1 + 1/12 + 1/22) + √(1 + 1/22 + 1/32) + …. + √(1 + 1/20072 + 1/20082)
(1) 2008 – 1/2008  (2) 2007 – 1/2007 (3) 2007 – 1/2008 (4) 2008 – 1/2007 (5) 2008 – 1/2009

Solution: The answer options contain almost the same numbers as the numbers in the last term. The trick here is to frame another question out of the one given and then use the substitution method to solve it.

Let n=2008. In this case, the question becomes

find the sum √(1 + 1/12 + 1/22) + √(1 + 1/22 + 1/32) + …. + √(1 + 1/(n-1)2 + 1/n2)

and the answer options now are

(1) n- 1/n (2) (n-1) – 1/(n-1)  (3) (n-1) – 1/n (4) n- 1/(n-1) (5) n- 1/(n+1)

By doing this, we have reduced the complexity of the question enormously.  Now we can substitute a simple number instead of n and then check the option.

Let n=2

Therefore, the sum now is √(1 + 1/12 + 1/22)  which is 3/2 = 1.5.

Check the answer options now with the value n=2

Option (1): 2 – 1/2 = 3 /2 = 1.5 which is correct
Option (2): 1- 1/1 = 0 which is incorrect
Option (3): 1 – 1/2 = 1/2 which is incorrect
Option (4): 2 – 1/1 = 1 which is incorrect
Option (5): 2- 1/3 = 5/3 which is incorrect

Again, a question which looked difficult to solve has been cracked within a minute. Being clever by using the options in the question can help you save precious time during the exam. Hope this post has helped. Till next time!

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