May 30th, 2007 — CAT, CAT Preparation, MBA, MBA Study Material, Prep Material

When you take CAT, what makes a crucial difference in determining whether you crack CAT and get into the IIMs or not, is how quick you can attempt questions.

One of the key observations that might throw light on CAT cracking methodologies is that - in most cases students spend about 60% of the time in quant and DI section on calculations. If you can **make a difference to your calculation speed,** even if it is marginal, it will make a substantial difference to your getting into an IIM. After all people make it to the cut off by a margin of 0.25 marks.

Practice speed calculation whenever you have to calculate anything. Engineers particularly, should try and throw their fx 100 calculators away, at least, for the time being. And commerce graduates do not go anywhere near your calculators for the next 6 months.

Number System Tips

If you have to find the square of numbers ending with ‘5′.

e.g 25 * 25. Find the square of the units digit (which is 5) = 25. Write this down.

Then take the tenths digit (2 in this case) and increment it by 1 (therefore, 2 becomes 3). Now multiply 2 with 3 = 6.

Write ‘6′ before 25 and you get the answer = 625.

For instance, if you have to find the square of 45.

45 * 45. The square of the units digit = 25

Increment 4 by 1. It will give you ‘5′. Now multiply 4 * 5 = 20.

Write 20 before 25. The answer is 2025.

In the case of numbers like 125. The rule applies without an iota of difference.

The square of the units digit = 25.

Increment 12 by 1. It will give you 13. Now multiply 12*13 = 156.

Write 156 before 25. The answer is 15625.

Try out the above technique with the following numbers.

Questions:

1. 225 * 225

2. 75 * 75

3. 145 * 145

4. 35 * 35

5. 375 * 375.

Answers and Solution:

1. 225 * 225. Square of units digits is 25. Increment 22 by 1 to get 23. The product of 22 and 23 is 506. Therefore, the answer is 50625.

2. 75 * 75. Square of the units digits is 25. Increment 7 by 1 to get 8. The product of 7*8 = 56. Therefore, the answer is 5625.

3. 145 * 145. Square of the units digits is 25. Increment 14 by 1 to get 15. The product of 14*15 = 210. Therefore, the answer is 21025.

4. 35 * 35. Square of the units digits is 25. Increment 3 by 1 to get 4. The product of 3*4=12. Therefore, the answer is 1225.

5. 375 * 375. Square of the units digits is 25. Increment 37 by 1 to get 38. The product of 37*38=1406. Therefore, the answer is 140625.

May 30th, 2007 — CAT, CAT Preparation, Entrance Exams, GRE, GRE Preparation, MBA, MBA Study Material, Prep Material

**Exercise1 + Answers + Explanations**

Question:

The sum of the first 100 numbers, 1 to 100 is divisible by

(1) 2, 4 and 8

(2) 2 and 4

(3) 2 only

(4) None of these

Correct Answer - (3)

Explanation:

The sum of the first 100 natural numbers is given by (n(n + 1))/2 = (100(101))/2 = 50(101).

101 is an odd number and 50 is divisible by 2. Hence, 50 (101) will be divisible by 2.

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Question:

What is the minimum number of square marbles required to tile a floor of length 5 metres 78 cm and width 3 metres 74 cm?

(1) 176

(2) 187

(3) 54043

(4) 748

Correct Answer - (2)

Explanation:

The marbles used to tile the floor are square marbles. Therefore, the length of the marble = width of the marble.As we have to use whole number of marbles, the side of the square should a factor of both 5 m 78 cm and 3m 74. And it should be the highest factor of 5 m 78 cm and 3m 74. 5 m 78 cm = 578 cm and 3 m 74 cm = 374 cm.

The HCF of 578 and 374 = 34.

Hence, the side of the square is 34.

The number of such square marbles required = = 187 marbles.

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

(1) 3

(2) 2

(3) 0

(4) 5

Correct Answer - (3)

Explanation:

6 is an even multiple of 3. When any even multiple of 3 is divided by 6, it will leave a remainder of 0. Or in other words it is perfectly divisible by 6.

On the contrary, when any odd multiple of 3 is divided by 6, it will leave a remainder of 3. For e.g when 9 an odd multiple of 3 is divided by 6, you will get a remainder of 3.

9 is an odd multiple of 3. And all powers of 9 are odd multiples of 3. Therefore, when each of the 8 powers of 9 listed above are divided by 6, each of them will leave a remainder of 3.

The total value of the remainder = 3 + 3 + …. + 3 (8 remainders) = 24. 24 is divisible by 6. Hence, it will leave no remainder.

Hence the final remainder when the expression 9^1 + 9^2 + 9^3 + ….. + 9^8 is divided by 6 will be equal to ‘0′.

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Question:

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

(1) 0

(2) 3

(3) 4

(4) None of these

Correct Answer - (2)

Explanation:

Any number that is divisible by ‘6′ will be a number that is divisible by both ‘2′ and ‘3′. i.e. the number should an even number and should be divisible by three (sum of its digits should add to a multiple of ‘3′). Any number that is divisible by ‘9′ is also divisible by ‘3′ but unless it is an even number it will not be divisibly by ‘6′. In the above case, ‘9′ is an odd number. Any power of ‘9′ which is an odd number will be an odd number which is divisible by ‘9′.

Therefore, each of the terms 91, 92 etc are all divisibly by ‘9′ and hence by ‘3′ but are odd numbers.

Any multiple of ‘3′ which is odd when divided by ‘6′ will leave a reminder of ‘3′. For example 27 is a multiple of ‘3′ which is odd. 27/6 will leave a reminder of ‘3′. Or take 45 which again is a multiple of ‘3′ which is odd. 45/6 will also leave a reminder of ‘3′.

Each of the individual terms of the given expression 91 + 92 + 93 + …… + 99 when divided by 6 will leave a reminder of ‘3′. There are 9 such terms. The sum of all the reminders will therefore be equal to 9*3 = 27.

27/6 will leave a reminder of ‘3′.

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Question:

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

(1) n! +1

(2) (n+1)!

(3) (n+1)!-1

(4) (n+1)!+1

Correct Answer - (3)

Explanation:

1.1! can be written as (2!-1!), likewise 2.2! = (3!-2!) , 3.3! = (4!-3!) and n.n! = (n+1)!-n!.

Thus by summing it up we get the result as given below.

1.1! + 2.2! + 3.3! + ……+n.n! = (2!-1!) + (3!-2!) + (4!-3!) +……+ (n+1)!-n! = (n+1)!-1.

May 30th, 2007 — CAT, CAT Preparation, MBA, MBA Study Material, Prep Material

**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

May 30th, 2007 — CAT, CAT Preparation, Entrance Exams, MBA, MBA Study Material, Prep Material

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?

May 30th, 2007 — CAT, CAT Preparation, Entrance Exams, MBA, MBA Study Material, Prep Material

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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