**TYPES OF NUMBERS**

- Integers : All numbers whose fractional part is 0 (zero) like -3, -2, 1, 0, 10, 100 are integers.
- Natural Numbers : Counting numbers like 1, 2, 3, 4, 5, 6 … Basically, all integers greater than 0 are natural numbers … More on Numbers

Question 1 |

Which is not the prime number?

43 | |

57 | |

73 | |

101 |

**Arithmetic Aptitude**

**Numbers**

**Discuss it**

Question 1 Explanation:

A positive natural number is called prime number if nothing divides it except the number itself and 1. 57 is not a prime number as it is divisible by 3 and 19.

Question 2 |

How many terms are there in 3,9,27,81........531441?

25 | |

12 | |

13 | |

14 |

**Arithmetic Aptitude 4**

**Numbers**

**Discuss it**

Question 2 Explanation:

3, 9, 27, 81..............531441 form a G.P. with a = 3 and r = 9/3 = 3 Let the number of terms be n Then 3 x 3^{n-1}= 531441 ∴ 3^{n}= 3^{12}∴ n = 12

Question 3 |

If the average of four consecutive odd numbers is 12, find the smallest of these numbers?

5 | |

7 | |

9 | |

11 |

**Arithmetic Aptitude 4**

**Numbers**

**Discuss it**

Question 3 Explanation:

Let the numbers be x, x+2, x+4 and x+6 Then (x + x + 2 + x + 4 + x + 6)/4 = 12 ∴ 4x + 12 = 48 ∴ x = 9

Question 4 |

If the sum of two numbers is 13 and the sum of their square is 85. Find the numbers?

6 & 7 | |

5 & 8 | |

4 & 9 | |

3 & 10 |

**Arithmetic Aptitude 4**

**Numbers**

**Discuss it**

Question 4 Explanation:

Let the numbers be x and 13-x
Then x

^{2}+ (13 – x)^{2}= 85 ∴ x^{2}+ 169 + x^{2}– 26x = 85 ∴ 2 x^{2}– 26x + 84 = 0 ∴ x^{2}– 13x + 42 = 0 ∴ (x-6)(x-7)=0 Hence numbers are 6 & 7Question 5 |

The difference between a two-digit number and the number obtained by interchanging the positions of its digits is 45. What is the difference between the two digits of that number?

5 | |

7 | |

6 | |

None of these |

**Arithmetic Aptitude 5**

**Numbers**

**Discuss it**

Question 5 Explanation:

Let the ten’s digit be x and unit’s digit be y Then (10x + y) – (10y + x) = 45 9(x – y) = 45 x – y = 5

Question 6 |

A two-digit number is such that the product of the digits is 12. When 9 is subtracted from the number, the digits are reversed. The number is:

34 | |

62 | |

43 | |

26 |

**Arithmetic Aptitude 5**

**Numbers**

**Discuss it**

Question 6 Explanation:

Let the ten’s and unit’s digit be x and y. Then 10x + - 9 = 10 x + x 10x2 + 12 -9x = 120 + x2 9x2 – 9x – 108 = 0 x2 –x – 12 = 0 x2 –4x + 3x – 12 = 0 (x – 4) (x + 3) = 0 Therefore x = 4 Hence the required no. is 43

Question 7 |

Find a positive number which when increased by 16 is equal to 80 times the reciprocal of the number

20 | |

-4 | |

-10 | |

4 |

**Arithmetic Aptitude 5**

**Numbers**

**Discuss it**

Question 7 Explanation:

Let the number be x. Then x + 16 = 80 * (1/x) x^{2}+ 16x – 80 = 0 x^{2}+ 20x – 4x – 80 =0 (x + 20) (x -4) Therefore x = 4

Question 8 |

What is the sum of two consecutive odd numbers, the difference of whose squares is 56?

30 | |

28 | |

34 | |

32 |

**Arithmetic Aptitude 5**

**Numbers**

**Discuss it**

Question 8 Explanation:

Let the no. be x and (x +2). Then (x +2)2 – x2 = 56 4x + 4 = 56 x + 1 = 14 x = 13 Sum of numbers = x + (x +2) = 28

Question 9 |

The product of two numbers is 108 and the sum of their squares is 225. The difference of the number is:

5 | |

4 | |

3 | |

None of these |

**Arithmetic Aptitude 5**

**Numbers**

**Discuss it**

Question 9 Explanation:

Let the numbers be x and y. Then xy = 108 and x^{2}+ y^{2}= 225 (x –y)^{2}= x^{2}+ y^{2}– 2xy (x –y)^{2}= 225 – 216 (x –y)^{2}= 9 Therefore (x –y) = 3

Question 10 |

The average of 21 results is 20. Average of 1

^{st}10 of them is 24 that of last 10 is 14. the result of 11'th is :42 | |

44 | |

46 | |

40 |

**Arithmetic Aptitude 6**

**Numbers**

**Discuss it**

Question 10 Explanation:

11'th result = sum of 21 results – sum of 20 results = 21 x 20 – (24 x 10 + 14 x 10) = 420 – (240 + 140) = 420- 380 = 40

Question 11 |

What could be the maximum value of Y in the following equation given that neither of X, Y, Z is zero?
5X8 + 3Y4 + 2Z1 = 1103

0 | |

7 | |

8 | |

9 |

**Numbers**

**Simplification and Approximation**

**Discuss it**

Question 11 Explanation:

1 1 <- CARRY 5 X 8 + 3 Y 4 + 2 Z 1 -------- 11 0 3 --------Clearly, X + Y + Z + 1 = 10 => X + Y + Z = 9 Now, since neither of X, Y, Z can be zero, the value of Y will be maximum when X = Z = 1. => Max Y = 7

Question 12 |

Which of the following are prime numbers ?
(i) 147
(ii) 327
(iii) 547
(iv) 637

147 | |

327 | |

547 | |

637 |

**Numbers**

**Discuss it**

Question 12 Explanation:

(i) 147
13

^{2}= 169 > 147. Prime numbers less than 13 are 2, 3, 5, 7, 11. 147 is divisible by 3. Therefore, 147 is not a prime number. It is a composite number. (ii) 327 19^{2}= 361 > 327. Prime numbers less than 19 are 2, 3, 5, 7, 11, 13, 17. 327 is divisible by 3. Therefore, 327 is not a prime number. It is a composite number. (iii) 547 24^{2}= 576 > 547. Prime numbers less than 24 are 2, 3, 5, 7, 11, 13, 17, 19, 23. 547 is not divisible by any of the above prime numbers. Therefore, it is a prime number. (iv) 637 26^{2}= 676 > 637. Prime numbers less than 26 are 2, 3, 5, 7, 11, 13, 17, 19, 23. 637 is divisible by 7. Therefore, 637 is not a prime number. It is a composite number. Hence, 547 is the only prime number among the given numbers.Question 13 |

What is the unit's digit in the product (267)

^{153}x (66666)^{72}?7 | |

6 | |

1 | |

2 |

**Numbers**

**Simplification and Approximation**

**Discuss it**

Question 13 Explanation:

The unit's digit in (267)

^{153}x (66666)^{72}is same as unit's digit in (7)^{153}x (6)^{72}. Now, unit's digit in 7^{4}is 1. => Unit's digit in 7^{z}is 1, where 'z' is a multiple of 4. 152 is the nearest number to 153, which is a multiple of 4. => Unit's digit in 7^{152}is 1 => Unit's digit in 7^{153}is 1 x 7 = 7 Also, unit's digit when 6 is raised to any power remains 6, i.e., for all 'n', 6^{n}will have 6 in the unit's place. Therefore, unit's digit in (267)^{153}x (66666)^{72}= unit's digit in (7)^{153}x (6)^{72}= unit's digit in 7 x 6 = 2 Hence, D (2) is the correct answer.Question 14 |

What is the total number of prime factors in the expression 12

^{12}x 16^{16}x 18^{18}?46 | |

154 | |

3456 | |

2 |

**Numbers**

**Discuss it**

Question 14 Explanation:

12

^{12}= (2 x 2 x 3)^{12}= 2^{24}x 3^{12}16^{16}= (2 x 2 x 2 x 2)^{16}= 2^{64}18^{18}= (2 x 3 x 3)^{18}= 2^{18}x 3^{36}Therefore, total number of prime factors = 24 + 12 + 64 + 18 + 36 = 154Question 15 |

What should be assigned to # so that 2582#724 is divisible by 11 ?

4 | |

5 | |

6 | |

7 |

**Numbers**

**Number Divisibility**

**Discuss it**

Question 15 Explanation:

For a number to be divisible by 11, the difference of the sum of numbers at even and odd places should be either 0 or a multiple of 11.
We assume the leftmost digit at position 1.
So, Sum of numbers at even places = 5 + 2 + 7 + 4 = 18
Sum of numbers at odd places = 2 + 8 + # + 2 = 12 + #
Now, to make the number divisible by 11, we equate the sums obtained in the above step.
18 = 12 + #
=> # = 6
Therefore, C (6) is the correct choice.

Question 16 |

What must be subtracted from the greatest five digit number so as to find the greatest five digit number divisible by 23 ?

22 | |

20 | |

18 | |

16 |

**Numbers**

**Number Divisibility**

**Discuss it**

Question 16 Explanation:

99999 is the greatest five digit number.
23 x 4347 = 99981, which is the largest five digit number divisible by 23.
=> 99999 - 99981 = 18
Thus, C (18) is the correct choice.

Question 17 |

On dividing 201098 by a certain number, the quotient is 67 and the remainder is 31. Find the divisor.

3011 | |

3001 | |

3021 | |

2991 |

**Numbers**

**Discuss it**

Question 17 Explanation:

Dividend = (Divisor x Quotient) + Remainder
=> Divisor = (Dividend - Remainder) / Quotient
=> Divisor = (201098 - 31) / 67 = 3001
Thus, B (3001) is the correct choice.

Question 18 |

A number leaves remainders 1, 4 and 7 on being divided successively by 3, 5 and 8 respectively. What is the product of the remainders when the order of divisors is reversed ?

28 | |

48 | |

24 | |

36 |

**Numbers**

**Discuss it**

Question 18 Explanation:

8 x 1 + 7 = 15
5 x 15 + 4 = 79
3 x 79 + 1 = 238
Thus, 238 is the number.
Now, we divide 238 successively by 8, 5, 3 in that particular order.
238 / 8 => Q = 29, R = 6
29 / 5 => Q = 5, R = 4
5 / 3 => Q = 1, R = 2
Therefore, product of remainders = 6 x 4 x 2 = 48

Question 19 |

What is the smallest number which when divided by 6, 9, 11, 16 and 22 leaves remainder 3 in each case ?

3267 | |

1584 | |

1587 | |

9504 |

**Numbers**

**Discuss it**

Question 19 Explanation:

LCM (6,9,11,16,22) = 1584
The required number = 1584 + 3 = 1587

Question 20 |

What is the largest number which divides 26, 47, 84 to leave remainders 2, 3, 0 respectively ?

4 | |

6 | |

8 | |

1 |

**Numbers**

**Discuss it**

Question 20 Explanation:

26 - 2 = 24
47 - 3 = 44
84 - 0 = 84
The number divides 26, 47, 84 to leave remainders 2, 3, 0 respectively.
=> The number should divide 24, 44, 84 completely.
HCF (24, 44, 84) = 4
Hence, 4 is the required number.

Question 21 |

Three friends started running together on a circular track at 8:00:00 am. Time taken by them to complete one round of the track is 15 min, 20 min, 30 min respectively. If they run continuously without any halts, then at what time will they meet again at the starting point for the fourth time ?

8:30:00 am | |

9:00:00 pm | |

12:00:00 pm | |

12:00:00 am |

**Numbers**

**LCM**

**Discuss it**

Question 21 Explanation:

LCM (15, 20, 30) = 60
=> They meet at the starting point after every 60 min, i.e., after every 1 hour.
Therefore, they will meet at the starting point for the fourth time after 4 hours, i.e., at 12:00:00 pm.

Question 22 |

The LCM of two co-prime numbers is 117. What is the sum of squares of the numbers ?

220 | |

1530 | |

250 | |

22 |

**Numbers**

**LCM**

**Discuss it**

Question 22 Explanation:

117 = 3 x 3 x 13
As the numbers are co-prime, HCF = 1.
So, the numbers have to be 9 and 13.
9

^{2}= 81 13^{2}= 169 Therefore, required answer = 250Question 23 |

HCF of two numbers is 11 and their LCM is 385. If the numbers do not differ by more than 50, what is the sum of the two numbers ?

132 | |

35 | |

12 | |

36 |

**Numbers**

**LCM**

**HCF**

**Discuss it**

Question 23 Explanation:

Product of numbers = LCM x HCF = 11 x 385 = 4235
Let the numbers be of the form 11m and 11n, such that 'm' and 'n' are co-primes.
=> 11m x 11n = 4235
=> m x n = 35
=> (m,n) can be either of (1, 35), (35, 1), (5, 7), (7, 5).
=> The numbers can be (11, 385), (385, 11), (55, 77), (77, 55).
But it is given that the numbers cannot differ by more than 50.
Hence, the numbers are 55 and 77.
Therefore, sum of the two numbers = 55 + 77 = 132

Question 24 |

Two numbers are in the ratio of 5:7. If their LCM is 105, what is the difference between their squares ?

216 | |

210 | |

72 | |

840 |

**Numbers**

**LCM**

**HCF**

**Discuss it**

Question 24 Explanation:

Let 'h' be the HCF of the two numbers.
=> The numbers are 5h and 7h.
We know that Product of Numbers = LCM x HCF
=> 5h x 7h = 105 x h
=> h = 3
So, the numbers are 15 and 21.
Therefore, difference of their squares = 21

^{2}- 15^{2}= 441 - 225 = 216Question 25 |

Which is the largest number that divides 17, 23, 35, 59 to leave the same remainder in each case ?

2 | |

3 | |

6 | |

12 |

**Numbers**

**HCF**

**Discuss it**

Question 25 Explanation:

Required Number = HCF (23-17, 35-23, 59-35, 59-17)
Required Number = HCF (6, 12, 24, 42) = 6

Question 26 |

Which of the following is the largest of all ?
(i) 7/8
(ii) 15/16
(iii) 23/24
(iv) 31/32

(i) | |

(ii) | |

(iii) | |

(iv) |

**Numbers**

**LCM**

**Number Divisibility**

**Discuss it**

Question 26 Explanation:

LCM (8, 16, 24, 32) = 96
7/8 = 84/96
15/16 = 90/96
23/24 = 92/96
31/32 = 93/96
Hence, 31/32 is the largest of all.

Question 27 |

A partial order P is defined on the set of natural numbers as follows. Here x/y denotes integer division.

i. (0, 0) ∊ P.

ii. (a, b) ∊ P if and only if a % 10 ≤ b % 10 and (a/10, b/10) ∊ P.

Consider the following ordered pairs:

i. (101, 22)

ii. (22, 101)

iii. (145, 265)

iv. (0, 153)

Which of these ordered pairs of natural numbers are contained in P?

i. (0, 0) ∊ P.

ii. (a, b) ∊ P if and only if a % 10 ≤ b % 10 and (a/10, b/10) ∊ P.

Consider the following ordered pairs:

i. (101, 22)

ii. (22, 101)

iii. (145, 265)

iv. (0, 153)

Which of these ordered pairs of natural numbers are contained in P?

(i) and (iii) | |

(ii) and (iv) | |

(i) and (iv) | |

(iii) and (iv) |

**Numbers**

**Gate IT 2007**

**Discuss it**

There are 27 questions to complete.