Entries Tagged 'CAT' ↓

Exercise3 + Answers + Explanations

Question:
When 26854 and 27584 are divided by a certain two digit prime number, the remainder obtained is 47. Which of the following choices is a possible value of the divisor?

(1) 61
(2) 71
(3) 73
(4) 89

Correct Answer - (3)

Explanation:
When both 26854 and 27584 are divided by a divisor, the remainder obtained happens to be the same value 47.

Which means that the difference between 26854 and 27584 is a multiple of the divisor. Else the remainders would have been different.

=> 27584 - 26854 = 730

Now we need to find out a two digit prime number, whose multiple will be 730. 730 can be written as 73 * 10, where 73 is a prime number and greater than 47.
Hence, the two digit prime divisor is 73.

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Question:
How many times will the digit ‘0′ appear between 1 and 10,000?

(1) 4000
(2) 4003
(3) 2893
(4) 3892
Correct Answer - (3)

Explanation:

In two digit numbers, ‘0′ appears 9 times.
In 3-digit numbers, either 1 ‘0′ or 2 ‘0s’ can appear.
When one ‘0′ appears: The first digit of a 3-digit number can be any of the 9 digits other than ‘0′. The second digit can be any of the ‘9′ digits and the third digit be ‘0′.Hence there will 9*9 = 81 such combinations. The ‘0′ can either be the 2nd digit or the third digit, hence there are 2 arrangements possible.Therefore, there are a total of 81 * 2 = 162 three digit numbers with ‘0′ as one of the digits.

When two ‘0s’ appear: The first digit of the number can be any of the 9 digits. The second and third digit are 0s.
Hence there are 9 such numbers and 0 appears 9*2 = 18 times. Hence, ‘0′ appears 18 + 162 = 180 times in three digit numbers. In 4-digit numbers, 0 might be either 1 or 2 or 3 of the digits.

When one ‘0′ appears: The remaining three digits can be any of the 9 digits.i.e. 93 combinations = 729. The ‘0′ can either be the second or third or fourth digit. Hence 3*729 numbers = 2187.

When two 0s appear: The remaining two digits can be any of the 9 digits.Hence 92 = 81 possible combinations. The two 0s can be in two of the three positions = 3*81 = 243 such numbers and 2*243 = 486 number of times 0 will be appear.

When three 0s appear: There are 9 such numbers and 0 appears thrice in each. Hence 9*3 = 27 times.
Hence the number of 0s in 4-digit numbers = 2187 + 486 + 27 = 2700 times.
There is one 5-digit number - 10,000 and 0 appears four times in this number.
Hence the total number of times 0 appears between 1 and 10,000 = 9 + 180 + 2700 + 4 = 2893.

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Question:
What number should be subtracted from x^3 + 4x^2 – 7x + 12 if it is to be perfectly divisible by x + 3?

(1) 42
(2) 39
(3) 13
(4) None of these
Correct Answer - (1)

Explanation:
According to remainder theorem when, then the remainder is f(-a). In this case, as x + 3 divides x3 + 4×2 – 7x + 12 – k perfectly (k being the number to be subtracted), the remainder is 0 when the value of x is substituted by –3.

=> (-3)3 + 4(-3)2 – 7(-3) + 12 – k = 0
=> -27 + 36 + 21 + 12 = k
=> k = 42

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Question:
What is the value of M and N respectively? If M39048458N is divisible by 8 and 11; Where M and N are single digit integers?

(1) 7, 8
(2) 8, 6
(3) 6, 4
(4) 5, 4

Correct Answer - (3)

Explanation:
If the last three digits of a number is divisible by 8, then the number is divisible by 8 (test of divisibility by 8).
Here, last three digits 58N is divisible by 8 if N = 4. (Since 584 is divisible by 8.)

For divisibility by 11. If the digits at odd and even places of a given number are equal or differ by a number divisible by 11, then the given number is divisible by 11.

Therefore, (M+9+4+4+8)-(3+0+8+5+N)=(M+5) should be divisible by 11 => when M = 6.

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Question:
How many zeros contained in 100!?

(1) 100
(2) 24
(3) 97
(4) Cannot be determined

Correct Answer - (2)

Explanation:
If the number N = 2×1.3×2.5×3 ….. pxq. It is clear that each pair of prime factors 2 and 5 will generate one zero in the number N, because 10 = 2X5.

For finding out the number of zeros in the number it is sufficient to find out how many two’s and five’s appear in the expansion of each product N.

No. of five’s = No. of two’s = Hence it is evident that there are only 24 pairs of primes 2 and 5 and therefore 100! ends in 24 zeros.

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Exercise4 + Answers + Explanation

Question:
When 242 is divided by a certain divisor the remainder obtained is 8. When 698 is divided by the same divisor the remainder obtained is 9. However, when the sum of the two numbers 242 and 698 is divided by the divisor, the remainder obtained is 4. What is the value of the divisor?

(1) 11
(2) 17
(3) 13
(4) 23

Correct Answer - (3)

Explanation:
Let the divisor be d.
When 242 is divided by the divisor, let the quotient be ‘x’ and we know that the remainder is 8. Therefore, 242 = xd + 8

Similarly, let y be the quotient when 698 is divided by d.
Then, 698 = yd + 9.

242 + 698 = 940 = xd + yd + 8 + 9
940 = xd + yd + 17
As xd and yd are divisible by d, the remainder when 940 is divided by d should have been 17.
However, as the question states that the remainder is 4, it would be possible only when leaves a remainder of 4.
If the remainder obtained is 4 when 17 is divided by d, then d has to be 13.

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Question:
What is the total number of different divisors of the number 7200?

(1) 20
(2) 4
(3) 54
(4) 32

Correct Answer - (3)

Explanation:
Express 7200 as a product of prime numbers. We get 7200 = 25 * 32 * 52.
The number of different divisors of 7200 including 1 and 7200 are (5 + 1)(2 + 1)(2 + 1) = 6 * 3 * 3 = 54

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Question:
When a number is divided by 36, it leaves a remainder of 19. What will be the remainder when the number is divided by 12?

(1) 10
(2) 7
(3) 192
(4) None of these

Correct Answer - (2)

Explanation:
Let the number be ‘a’.
When ‘a’ is divided by 36, let the quotient be ‘q’ and we know the remainder is 19
=> and remainder is 19
=> a = 36q + 19

now, a is divided by 12

=> 36q is perfectly divided by 12
Therefore, remainder = 7

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Question:
A person starts multiplying consecutive positive integers from 20. How many numbers should he multiply before the will have result that will end with 3 zeroes?

(1) 11
(2) 10
(3) 6
(4) 5

Correct Answer - (3)

Explanation:
A number will end in 3 zeroes when it is multiplied by 3 10s. To get a 10, one needs a 5 and a 2.
Therefore, this person should multiply till he encounters three 5s and three 2s.
20 has one 5 (5 * 4) and 25 has two 5s (5 * 5).
20 has two 2s (5 * 2 * 2) and 22 has one 2 (11 * 2).

Therefore, he has to multiply till 25 to get three 5s and three 2s, that will make three 10s. So, he has to multiply from 20 to 25 i.e. 6 numbers.

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Question:
How many four digit numbers exist which can be formed by using the digits 2, 3, 5 and 7 once only such that they are divisible by 25?

(1) 4! - 3!
(2) 4
(3) 8
(4) 6

Correct Answer - (2)

Explanation:
The numbers that are divisible by 25 are the numbers whose last two digits are ‘00′, ‘25′, ‘50′ and ‘75′.
However as ‘0′ is not one of the four digits given in the problem, only numbers ending in ‘25′ and ‘75′ need to be considered.

Four digit number whose last two digits are ‘25′ using 3 and 7 as the other two digits are 3725 and 7325.
Similarly, four digit numbers whose last two digits are ‘75′ using 2 and 3 as the other two digits are 2375 and 3275.

Hence there are four 4-digit numbers that exist.

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Tips For Number System:

This rule applies to numbers whose units digits add up to ten (2 and 8 in this case) and the tenth digits are same.
If you have a pair of numbers like 22 and 28 and you need to multiply the two of them.

Step 1:
Multiply the unit digits
2 * 8 = 16

Step 2:
Increment the tenth digit by 1.
2 + 1 = 3

Step 3:
Multiply the incremented number with the original tenth digit number
3 * 2 = 6

Step 4:
Concatenate the results got from step 1 and step 3
6 and 16 => 616

Note: Remember to put the products of unit digits of the given numbers in the unit and tenth digit of your result.

Example 2: 36 * 34 = ?

Step 1: Multiply the unit digits
6 * 4 = 24

Step 2: Increment the tenth digit by 1.
3 + 1 = 4

Step 3: Multiply the incremented number with the original tenth digit number
4 * 3 = 12

Step 4: Concatenate the results got from step 1 and step 3
12 and 24 => 1224

Therefore, 36 * 34 = 1224

Example 3: 61 * 69 = ?

Step 1: Multiply the unit digits
1 * 9 = 9. In this case remember to write 9 as 09

Step 2: Increment the tenth digit by 1.
6 + 1 = 7

Step 3: Multiply the incremented number with the original tenth digit number
6 * 7 = 42

Step 4: Concatenate the results got from step 1 and step 3
42 and 09 => 4209 (Since, we have to put the product of the unit digits only in the unit and tenth digit of the result)

Therefore, 61 * 69 = 4209

Example 4: 127 * 123 = ?

Step 1: Multiply the unit digits
7 * 3 = 21

Step 2: Increment the tenth digit by 1.
12 + 1 = 13

Step 3: Multiply the incremented number with the original tenth digit number
12 * 13 = 156

Step 4: Concatenate the results got from step 1 and step 3
156 and 21 => 15621

Therefore, 127 * 123 = 15621

Try out the above technique with the following problems.

Questions:

1. 54 * 56
2. 79 * 71
3. 152 * 158
4. 33 * 37
5. 111 * 119

Correct Answers:

1. 3024
2. 5609
3. 24016
4. 1221
5. 13209

A collection of questions that typically appear in the Common Admission Test (CAT) from the topic Number Theory.

These questions will guide you through your CAT and other MBA entrance exam preparation.

1. When 26854 and 27584 are divided by a certain two digit prime number, the remainder obtained is 47. Which of the following choices is a possible value of the divisor?

2. If both 112 and 33 are factors of the number a * 43 * 62 * 1311, then what is the smallest possible value of ‘a’?

3. What is the remainder when 9^1 + 9^2 + 9^3 + …. + 9^8 is divided by 6?

2. A number when divided by a divisor leaves a remainder of 24. When twice the original number is divided by the same divisor, the remainder is 11. What is the value of the divisor?

4. ‘a’ and ‘b’ are the lengths of the base and height of a right angled triangle whose hypotenuse is ‘h’. If the values of ‘a’ and ‘b’ are positive integers, which of the following cannot be a value of the square of the hypotenuse?

5. Let n be the number of different 5 digit numbers, divisible by 4 with the digits 1, 2, 3, 4, 5 and 6, no digit being repeated in the numbers. What is the value of n?

6. What is the reminder when 91 + 92 + 93 + …… + 99 is divided by 6?

7. For what value of ‘n’ will the remainder of 351^n and 352^n be the same when divided by 7?

8. How many keystrokes are needed to type numbers from 1 to 1000?

9. When 242 is divided by a certain divisor the remainder obtained is 8. When 698 is divided by the same divisor the remainder obtained is 9. However, when the sum of the two numbers 242 and 698 is divided by the divisor, the remainder obtained is 4. What is the value of the divisor?

10. Find the greatest number of five digits, which is exactly divisible by 7, 10, 15, 21 and 28.

11.Anita had to do a multiplication. Instead of taking 35 as one of the multipliers, she took 53. As a result, the product went up by 540. What is the new product?

12. Find the value of 1.1! + 2.2! + 3.3! + ……+n.n!

13. Let x, y and z be distinct integers. x and y are odd and positive, and z is even and positive. Which one of the following statements cannot be true?

14. When a number is divided by 36, it leaves a remainder of 19. What will be the remainder when the number is divided by 12?

15. A person starts multiplying consecutive positive integers from 20. How many numbers should he multiply before the will have result that will end with 3 zeroes?

16. How many different factors are there for the number 48, excluding 1 and 48?

17. How many zeros contained in 100!?
18. Two numbers when divided by a certain divisor leave remainders of 431 and 379 respectively. When the sum of these two numbers is divided by the same divisor, the remainder is 211. What is the divisor?

19. What number should be subtracted from x^3 + 4x^2 – 7x + 12 if it is to be perfectly divisible by x + 3?

20. What is the least number that should be multiplied to 100! to make it perfectly divisible by 350?

Questions along with respective explanations.

Q.1. Train A traveling at 60 km/hr leaves Mumbai for Delhi at 6 P.M. Train B traveling at 90 km/hr also leaves Mumbai for Delhi at 9 P.M. Train C leaves Delhi for Mumbai at 9 P.M. If all three trains meet at the same time between Mumbai and Delhi, what is the speed of Train C if the distance between Delhi and Mumbai is 1260 kms?

(1) 60 km/hr (2) 90 km/hr (3) 120 km/hr (4) 135 km/hr

Correct Answer - (3)

Solution:
All three trains meet at the same time between Delhi and Mumbai. Which means Train A and Train B are at the same point at that time. This will happen when Train B is overtaking Train A.

Train A starts 3 hours before Train B. Therefore, by the time Train B leaves Mumbai, Train A has covered 3 * 60 = 180 kms.

The relative speed between Train A and Train B = 90 - 60 = 30 kmph. Therefore, Train B will overtake Train A in = 6 hours from the time Train B leaves Mumbai. That is at 3 A.M, Train B will overtake Train A. The point between Mumbai and Delhi at which Train B overtakes Train A will be 6*90=540 kms from Mumbai.

Train C will also be at that point at 3 A.M while Train B is overtaking Train A. And Train C would have travelled 1260-540 = 720 kms in these 6 hours. Therefore, the speed of Train C = 120 km/hr.

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Q.2 A train travels at an average speed of 90 km/hr without any stoppages. However, its average speed decrease to 60km/hr on account of stoppages. On an average, how many minutes per hour does the train stop?

(1) 12 minutes (2) 18 minutes (3) 24 minutes (4) 20 minutes

Correct Answer - (4)

Solution:
If it travelled at 90 km / hr, it would have crossed 90 kms in an hour. However, it covered only 60 kms due to stoppages.

The distance it covered decreased by 1/3 or it covered only 2/3rd of the distance that it can cover for which the traveling time would have been 2/3rd of an hour. The remaining 1/3rd of an hour was spent in stoppages. Therefore, the train stops on an average for 20 minutes every hour.
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Q.3 A man goes from city A to city B situated 60 kms apart by a boat. His onward journey was with the stream while the return journey was an upstream journey. It took him four and half hours to complete the round trip. If the speed of the stream is 10 km/hr, how long did it take him to complete the onward journey?
(1) 3 hours (2) 3.5 hours (3) 2.25 hours (4) 1.5 hours

Correct Answer - (4)

Solution:
The average speed for the round trip = km/hr
Let the speed during the onward journey be ‘D’ km/hr. Let the speed of the boat in still water be ‘B’ km/hr.
Therefore, D = B + S => D = B + 10 (As the speed of the stream is 10 km/hr).
Let the speed during the return journey be ‘U’ km/hr.
Therefore, U = B - S = B - 10

As the distance between A and B is the same as the distance between B and A, the average speed is given by the formula =
=> => 3B2 - 300 = 80B.
=> 3B2 - 80B - 300 = 0 => 3B2 - 90B + 10B - 300 = 0
=> 3B(B - 30)+10(B - 30) = 0
=> (B-30)(3B+10) = 0
=> B = 30 or B = -10/3

As speed is a positive quantity, B = 30.
Therefore, D = 30 + 10 = 40 km/hr and U = 30 - 10 = 20 km/hr.

His onward journey was done at a speed of 40 km/hr. The distance covered was 60 kms.
Therefore, the time taken for the onward journey = 1.5 hours

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Q. 4 The speed of a motor boat itself is 20 km/h and the rate of flow of the river is 4 km/h. Moving with the stream the boat went 120 km. What distance will the boat cover during the same time going against the stream?
(1) 80 km (2) 180 km (3) 60 km (4) 100 km

Correct Answer - (1)

Solution:
Let the distance to be covered by the boat when it is travelling against the stream be x.
The boat goes down the river at a speed of 20 + 4 = 24 km/h and up the river at a speed of 20 – 4 = 16 km/h.
Since the time taken is same 120/24 = x/16
Therefore, x = 80 km.

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Q. 5 A passenger train covers the distance between stations X and Y, 50 minutes faster than a goods train. Find this distance if the average speed of the passenger train is 60 kmph and that of goods train is 20 kmph.
(1) 20 kms (2) 25 kms (3) 45 kms (4) 40 kms
Correct Answer - (2)

Solution:
Let ‘d’ be the distance between the stations X and Y.
Time taken by the passenger train to cover the distance ‘d’ = d/60 hour
Time taken by the goods train to cover the distance ‘d’ = d/20 hour
Time difference between these two trains is given by 50 minutes or 50/60 hour

i.e., d/20 –d/60 = 50/60

d = 25kms.

How many number of times will the digit ‘7′ be written when listing the integers from 1 to 1000?

Sol:
7 does not occur in 1000. So we have to count the number of times it appears between 1 and 999. Any number between 1 and 999 can be expressed in the form of xyz where 0 < x, y, z < 9.

1. The numbers in which 7 occurs only once. e.g 7, 17, 78, 217, 743 etc. This means that 7 is one of the digits and the remaining two digits will be any of the other 9 digits (i.e 0 to 9 with the exception of 7). You have 1*9*9 = 81 such numbers. However, 7 could appear as the first or the second or the third digit. Therefore, there will be 3*81 = 243 numbers (1-digit, 2-digits and 3- digits) in which 7 will appear only once. In each of these numbers, 7 is written once. Therefore, 243 times.

2. The numbers in which 7 will appear twice. e.g 772 or 377 or 747 or 77 In these numbers, one of the digits is not 7 and it can be any of the 9 digits ( 0 to 9 with the exception of 7). There will be 9 such numbers. However, this digit which is not 7 can appear in the first or second or the third place. So there are 3 * 9 = 27 such numbers. In each of these 27 numbers, the digit 7 is written twice. Therefore, 7 is written 54 times.

3. The number in which 7 appears thrice - 777 - 1 number. 7 is written thrice in it. Therefore, the total number of times the digit 7 is written between 1 and 999 is 243 + 54 + 3 = 300
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A three digit number consists of 9,5 and one more number . When these digits are reversed and then subtracted from the original number the answer yielded will be consisting of the same digits arranged yet in a different order. What is the other digit?

Ans : 4

Sol. Let the digit unknown be n.

The given number is then 900+50+n=950+n.
When reversed the new number is 100n+50+9=59+100n.
Subtracting these two numbers we get 891-99n.
The digit can be arranged in 3 ways or 6 ways.
We have already investigated 2 of these ways.
We can now try one of the remaining 4 ways. One of these is n 95

100n+90+5=891-99n
or 199n =796
so, n=4
the unknown digit is 4.
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On sports day,if 30 children were made to stand in a column,16 columns could be formed. If 24 children were made to stand in a column, how many columns could be formed?

Ans. 20

Sol:

Total number of children=30*16=480
Number of columns of 24 children each =480/24=20.
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The accuracy is highest in such questions, and the time spent solving them can be quite less if one has done a good amount of practice. The questions that are asked in logical reasoning require one to find out whether the given conclusion is valid or not. For example:

1. All drunken drivers meet with an accident.
Salman is a drunken driver.
Conclusion – Salman will meet with an accident
This one is pretty straightforward. Since Salman belongs to a group, all members of which meet with an accident, he would meet with an accident.

2. Most males are intelligent.
Mr. X is a male
Conclusion – Mr. X may or may not be intelligent
Here there are two possibilities. Either Mr. X is a part of the group of intelligent males or he is not. Hence, it follows that he may or may not be intelligent.

In the exam the questions that are asked have two statements followed by two conclusions and you have to determine which conclusion is correct. Obviously the questions are not so easy as the above examples and may require one to draw venn diagrams to solve the questions. E.g.

1. Statements –
some students are smart
all students are hardworking

Conclusions – (i) some hardworking students are smart
(ii) some smart students are hardworking
Solution – Both the conclusions are correct, as some students are definitely both smart and hardworking.

2. Statement–
tennis players get married only to models

Conclusions – (i) Madhu is married to a tennis player
(ii) Madhu is not married to a tennis player
Solution – Here either Madhu can be married to a tennis player or to someone who is not a tennis player. So either conclusion (i) or (ii) follow but not both.

3. Statements –
some roses are red
some red are black.

Conclusions – (i) no black is a rose.
(ii) no rose is a black
Solution – Here both the statements are possibilities. It may be that some roses are black or no roses are black. Hence neither of the conclusions follows.

by Jasveen Grewal *

If you are one of the aspirants looking forward to “bell” the CAT this year, I’m sure you’ve already started with the preparations of brushing-up your “sectional” skills. As a teacher of verbal ability, I often come across students who are befuddled about why this particular examination has questions on ENGLISH USAGE! Through this article I hope to satisfy such queries.

An MBA from the IIMs or other top-rung B-schools is a dream worth pursuing. A couple of years of rigorous training and you emerge dynamic managers with strong personalities - ready to conquer the world!

And today, as aspiring MBAs, you are awestruck by the corporate leaders. But, what if you are listening to the speech by a manager (someone with an aura, someone you have admired for his persona & are glad to be able to listen to) and he makes the mistake of saying ‘oversee’ in place of ‘overlook’. (To ‘oversee’ & to ‘overlook’ - two seemingly similar words, aren’t they? But the meaning of these two words is totally different. The two words are almost antonyms.)

The demi-god image of the manager shatters right in front of you, does it not? The situation analysis, I guess, answers the most vital question of most of the students taking various competitive exams, that is, why do we have the English Usage as a part of any competitive examination? The answer simply lies in the fact that competitive exams, especially CAT, select or reject a candidate by gauging him upon parameters, which are necessary for the candidate to be successful in the field he wishes to pursue.

Believe me, no one would like to opt for you as a student in one’s college if one has to build you up right from the scratch. One would prefer to shape you up & galvanize your skills. The drive should be present in you already! Therefore, to check your potential for the future you plan to have for yourself, you get various questions in the examination, including the questions on English Usage.

You need to adopt different approaches for different passages.
These are as follows:

Read Passage –> Answer Questions:

This is the traditional way or reading the passage first and then approaching the questions.

Answer Questions -> Read Passage –> Answer Questions:

This approach most commonly used is to have a look at the questions first and then read the passage, underlining part of the passage where an answer to the questions read may be found. After reading the passage, one can go to the questions and if need be just revisit the underlined part. However, while one is reading the questions, one must not waste a lot of time reading any inference-based questions and just keep a look out for names, events and terms that can be found in the passage.

Answer Questions -> Read Passage -> Answer Questions -> Read Passage:

In this approach, one never ends up reading the passage. A question is taken one at a time and its answer is searched for in the passage. Obviously not many inference-based questions can be answered in this way, so these questions will have to be left. But if it were a fact-based passage, these types of questions would be minimal. Also one can take advantage that usually the answers to the questions would lie sequentially in the passage.

You could also follow this approach. Around one-third of the passage is read, which gives one a fair idea of the structure of the passage. Then questions are looked at and all those questions whose answers lie in the part of passage read are attempted. Then for the remaining passage the same approach is followed. This also helps in saving time searching for the answers as the amount of passage unread is less and also one understands the flow of the passage. A few inferential questions can also be answered using this approach.

The above approaches coupled with a proper selection of the passages to be attempted should help one comfortably clear the cut-offs of the Reading Comprehension section.

Selecting the correct passages:

The number of passages asked will be minimum five. If 3 are attempted it is a fair attempt. Passages should be selected based on the following.
1. The Comfort level with the topic of the passage
2. Degree of Difficulty level of the language used
3. Trying Understanding the structure of the passage
4. Knowing Length of the passage and number of the questions asked

Management:

1. Any Edition of Harvard Business Review
2. Case Studies of Businessworld Magazine
3. What They Don’t Teach You At Harvard Business School - Mark H. McCormack

Physics/Science:

1. Brief History of Time - Stephen Hawking
2. Any work of Isaac Asimov
3. The Tao of Physics - Fritjof Capra

Philosophy:

1. Zen And The Art of Motor Cycle Maintenance
2. The Fountainhead - Ayn Rand

Abstract Topic:

Futureshock, Power Shift, Third Wave - Alvin Toffler

Literary Works:

1. God of Small Things - Arundhati Roy
2. Midnight’s Children - Salman Rushdie

Business:

Made In Japan - Akio Morito

Religion:

Any book written by Swami Vivekanand

Sports:

Idols - Sunil Gavaskar

General:

Uncommon Wisdom - Fritjof Capra

Law:

Any of John Grisham’s works

Medical:

Doctors - Erich Segal

Anthropology/Sociology:

1. Manwatching
2. The Naked Ape - Desmond Morris

Note: The above list is just an indicative list.