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Gejo’s Way Of Approaching The Quants Section

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This is a useful article, by one of IMS Faculty. I thought of sharing with you guys. Have a look

Gejo’s way of approaching the Quants Section!!!

The quant section of the CAT is feared by many – just the look of a question can cause worry in the mind of the test taker. The best way to tackle a CAT Quant question is to solve them logically than mathematically. Of course, you need to have your basic fundas in place to solve any quant question using logic.

The quant section of the CAT is feared by many – just the look of a question can cause worry in the mind of the test taker. The best way to tackle a CAT Quant question is to solve them logically than mathematically. Of course, you need to have your basic fundas in place to solve any quant question using logic.

Let us look at some examples – these are a few questions which are based on actual CAT questions.

1. Let n be the total number of different 5-digit numbers with all distinct digits, formed using 2, 3, 4, 5 and 6 and divisible by 4. What is the value of n?

1] 44 2] 32 3] 36 4] 38 5] 40

Permutations & Combinations is probably the most ‘hated’ topic. However, if you understand the basics and use logic, then it is the most fun-filled topic of all. Let us get to this question. As mentioned, we need to find out the 5 digit numbers divisible by 4 formed by the digits given in the question. To be divisible by 4, the last 2 digits should be divisible by 4. So to arrive at the answer, the first step is to find out combinations of the 2 digits from 2,3,4,5 & 6 that are divisible by 4 – eg: 24. Then, for each such combination, the last 2 digits are fixed. The remaining 3 digits can be arranged in 3! Ways = 6 ways. So the answer would be – 6 multiplied by the total no. of combinations of 2 digits divisible by 4. The answer necessarily should be a multiple of 6 and therefore the answer is 36 – option [3]. We just got lucky here, by the way, since there is just one option that is divisible by 6!

Of course, you have to be strong in the concepts so that you can solve this question in less than 10 seconds. I must remind you that short cuts happen when your concepts are strong.
2. The set of all integer values of x such that 3 × 5x –5|x|+1> 1 is:

1] x  >  –1  and  x <  5 / 3
2] x  >  1
3] ε I
4] x  < 5 / 3
4] No solution

Here is a question that looks scary! To solve this question, I am going to do a little manipulation. The equation given in the question is 3 × 5x –5|x|+1> 1. Now, to make it little more friendly, let me change it to   5 × 5x –5|x|+1> 1 (If it has to work for 3 × 5x it must work for 5 × 5x. Please note that this cannot be used everywhere). Now, the question changes to

5 × 5x –5|x|+1> 1
= 5x+1 –5|x|+1> 1

The above cannot have a solution because at best 5x+1 will be equal to 5|x|+1 when x is positive. When x is negative 5|x|+1 will be greater than 5x+1. Since 3 × 5x is less than 5 × 5x, 3 × 5x – 5|x|+1 > 1 will have no solution. Hence option [4]

Sometimes, we could change the question a bit without changing the answer outcome.
Most of the time test takers avoid dangerous looking question, which is a bad idea. You must read every question and give it a fair shot. You leave a question only after this. Be prudent not to waste a question just to save time! But also remember, not to get stuck on a question for long.

3. The capacity of tank B is 1.5 times the capacity of tank A. One tap fills tank A in 9 hrs and other tap fills tank B in 11 hrs. Both the taps are started at the same time initially. After 7 hrs, both of them are closed. Then remaining part of tank B is filled with the water taken from tank A. After this, how much time will it take to fill tank A with its tap?

1]  9.4 hrs       2]  2.1 hrs      3]  5.8 hrs      4]  6.9 hrs      5]  1.7 hrs
A – In less than 30 secs, if you apply logic, you can eliminate all options but 4. To explain this, it will take lot of words. But let me try.

Here is the story. There are two tanks, tank A & tank B and two taps, I will call them tap A & tap B [you will know why]. Given the capacity of tank B is 1.5 times the capacity of tank A. Also given tap A takes 9 hours to fill tank A and tap B takes 11 hrs to fill tank B. Both the tanks have been filled for 7 hrs. At this point, tank A needs 2 more hrs of water from tap A and tank B needs 4 more hrs of water from tap B. Now, the tank B is filled using water in tank A. We need to find out how much time will it take for tap A to fill tank A.

The answer will be 2 hrs + tap A time equivalent to 4 hr of tap B.

If the above statement seems confusing, let me explain it a bit. If the water was not transferred to tank B, then the tap A would have taken 2 hours (9hrs – 7hrs). That is the 2 hrs. Now the second term – tap B needed 4 hours of water from tap B (11hrs – 7 hrs). This is filled by tank A. 4 hrs of tap B water is filled by the tank A, therefore, tap A would need to fill water equivalent to 4 hr of tap B. [Confusing? Read it again, slowly!]

Let us now eliminate some options – Option 1 is out since tap A will take only 9 hours to fill tank A. Option 5 is also definitely out.  Option 2 seems to be out, at this moment, let us not eliminate it.

The fight is between 6.9 hrs, 5.8 hrs & 2.1 hrs. Look at this – tank B is 1.5 times bigger than tank A. While tap B takes 11 hrs to fill tank B & tap A takes 9 hrs to fill tank A. If the flow of tap A & B were same then tap A : tap B should have been  1: 1.5. However, it is 9: 11. You can see that tap B is faster than tap A. So 4 hrs of tap B > 4 hrs of tap A. Therefore, the tap A time equivalent of 4 hr of tap B > 4 hrs. So the answer has to be greater than 6. Hence Option [4]

The shorter way to solve this question seems so long, that is only because I am explaining the logic to a 3rd person. While reading this question, it is quite natural that you would straight away want to apply work, pipe cistern formula. In this case, the question can be solved using ratios.
Capacity Ratio : 1: 1.5
Tap ratio 9: 11
So for every 1 hr of tap B = (1.5 X 9)11 hr for tap A = 1.23 hr for tap A
4 hrs of tap for tap B = 1.23 X 4 = 4.9 hrs.
Therefore, the answer is 2 + 4.9 = 6.9 hrs.
After reading a question, take a moment to think and understand the question thoroughly before solving it

4. Consider two different cloth-cutting processes. In the first one, n circular pieces are cut from a square piece of side a in the following steps. The original square of side a is divided into n smaller squares, not necessarily of the same size; then a circle of maximum possible area is cut from each of the smaller squares. In the second process, only one circle of maximum possible area is cut from the square of side a and the process ends there. The cloth pieces remaining after cutting the circles are scrapped in both the processes. The ratio of the total area of scrap cloth generated in the former to that in the latter is:

1] 1  : 1     2]  √2  :  1     3]  n( 4 – π )  :  4n – π     4] 4n – π  :  n(4 – π)

For ease of calculation, let  4x = a

For process 2,

Radius = a/2 = 2x

The area is   4πx2

Scrap clot area  =  a2 – 4πx2

For Process 1, assume n = 4

Radius of 1 circle = x

The area of each circle =  πx2

Total area of the 4 circles =  4πx2

Scrap clot area =  a2 –  4πx2

The answer has to be 1 : 1
I do not know how many would even try reading the question just because it is long question. Once you read the question and understand what needs to be done, then a little logic would help you reach the solution in no time.

5. 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 square?
1]  4      2]  0      3]  1       4]  4      5]  2
Here, we need to find a perfect square which looks like aabb [a & b are digits]. Now, we need to use a little logic to arrive at the fact that aabb = 11 X a0b [for eg. 11 X 102 = 1122, 11 X 304 = 3344]
For aabb to be a perfect square a0b should be of the form 11x2 so that aabb = 112 X  x2 . The first task is to list down all possibilities of a0b being divisible by 11. a+b should be equal to 11 or 0 (this is the divisibility rule for 11 : sum of odd digits – sum of even digits should be either 0 or 11). a+b cannot be equal to 0 (for this, a & b both have to be equal to 0). a+b = 11.

The possibilities for a0b that are divisible by 11 are

209 11 X 19

308 11 X 28

407 11 X 37

506 11 X 46

605 11 X 55

704 11 X 64 64 is a Perfect Square!

803 11 X 73

902 11 X 82

There is one solution – 704 X 11 = 7744 = 882
Ans: 3

Boom! One line question but not necessarily a one line answer. Many make this mistake of thinking that the level of difficulty of a question is directly proportional to the length of the question. There is NO such relation. You must solve each question on its merit. [not by the length or the look!]. This question needs you to first crack aabb = 11 X a0b. It may not come directly. If you crack this one step, you get the answer. This question becomes time consuming depending on the logic you use.


It is an application of a basic funda. Again, the question looks scary and many would miss it.
During the analysis of the SimCATs, I would suggest that before looking at the explanatory answers, solve each question yourself. Then you look at the explanatory answers and see if you can find alternative methods to solve every question. This will help you build on your ‘logical cells’. You still have good 3 months for the CAT and you can crack it – just ensure that you use logic & common sense more.


Written by Implex

August 29, 2008 at 10:26 am

Posted in General, Tips

Tagged with ,

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