# Problems on Numbers for Exam

Q31. The smallest whole number is

(a) 0

(b) 1

(c) 2

(d) None of the above

**Answer:** (a) 0

Q32. The smallest natural number is

(a) 0

(b) 1

(c) 2

(d) None of the above

**Answer:** (b) 1

Q33. The sum of first 6 prime number is

(a) 38

(b) 39

(c) 40

(d) 41

**Answer:** (d) 41

2+3+5+7+11+13

41

Q34. Which of the following is not a prime number

(a) 11

(b) 21

(c) 31

(d) 41

**Answer:** (b) 21

Q35. The difference between the square of two consecutive even integers is always divisible by

(a) 2

(b) 3

(c) 4

(d) 7

**Answer:** (c) 4

Let the two consucative even integers are 2x and 2x+2

Now (2x+2)^{2} - (2x)^{2} = 4x^{2} + 8x + 4 - 4x^{2}

=8x + 4

= 4 (2x + 1) which is divisible by 4

Q36. There are four prime numbers written in ascending order. The product of first three is 385 and that of the last three is 1001. The first number is

(a) 3

(b) 4

(c) 5

(d) 10

**Answer:** (c) 5

Let the prime numbers be a, b, c, d

Now abc = 385

bcd = 1001

∴ | abc | = | 385 |

bcd | 1001 |

∴ | a | = | 5 |

d | 13 |

∴ a = 5 d = 13

∴ First number is 5

Q37. A number when divided by 6 leaves a remainder 3. When the square of the same number is divided by 6, the remainder is

(a) 0

(b) 1

(c) 2

(d) 3

**Answer:** (d) 3

Let the number is x

∴ x = 6k + 3

∴ x^{2} = (6k + 3)^{2} = 36k^{2} + 36k + 9

= 36k^{2} + 36k + 6 + 3

= 6 (6k^{2} + 6k + 1) + 3

∴ Remainder = 3

Q38. A number is divided by 56 gives 28 as remainder. If the same number is divided by 8, the remainder will be

(a) 2

(b) 3

(c) 4

(d) 5

**Answer:** (c) 4

Let the number is x

∴ x = 56k + 28

x = (8 X 7k) + (8 X 3) + 4

x = 8(7k + 3) + 4

Remainder = 4

Q39. In a question on division with zero remainder, a candidate took 21 as divisor instead of 12. The quotient obtained by him was 20. The correct quotient is

(a) 35

(b) 48

(c) 56

(d) 66

**Answer:** (a) 35

Divisor = 21

Quotient = 20

Remainder = 0

∴ Number = Divisor X Quotient + Remainder

= 21 X 20 + 0

= 420

Correct quotient = 420 ÷ 12

=35

Q40. If 27 * 324 is divisible by 3, then what is the smallest digit is there in place of *

(a) 0

(b) 3

(c) 5

(d) 7

**Answer:** (a) 0

2+7+*+3+2+4 = 18 + * which is divisible by 3

=18 + 0 which is divisible by 3

18

∴ * = 0