NCTB Class 6 Math Chapter One Exercise 1.2 Solution

NCTB Class 6 Math Chapter One Exercise 1.2 Solution by Math Expert. Bangladesh Board Class 6 Math Solution Chapter 1 Natural Numbers and Fractions Exercise 1.2 Solution.

Board

NCTB

Class

6

Subject

Mathematics

Chapter

1

Chapter Name

Natural Numbers and Fractions

Exercise

1.2 Solution

Exercise – 1.2

(1) Write down the prime numbers between 30 to 70:

Solution:-  The prime numbers between 30 to 70 are —-31, 37, 41, 43, 47, 53, 59, 61, 67.

Because, prime No is that number which is dividable by 1 & the number itself.

(2) Determine which of the following pairs are co-prime:

(2)(a) 27, 54

Solution:- 27 = 3x3x3x1

54 = 3x3x3x2

This is not co –prime.

(2)(b) 63, 91

Solution:- 63 = 3x3x7

91 = 7×13

This is not co –prime

(2)(c) 189, 210

Solution:- 3x3x3x7

210 = 3x2x5x7

This pair is not a co- prime number

(2)(d) 52, 97

Solution:- 52 = 1x2x2x13

97 = 1×97

This pair (52, 97) is a co-prime number.

Because there is a only one common factor that is 1.

(3) Which of the following numbers are divisible by the numbers as indicated?

 (a) 545, 6774, 8535 by 3

Solution:- In these all numbers, the numbers 6774 and 8553 is divisible by 3.

6774 =(6+7+7+4) ÷ 3

= 24÷ 3

= 8

8553 (8+5+5+3) ÷ 3

= 21÷3

= 7

(3)(b) 8542, 2184, 5274 by 4

Solution:- From these numbers, only 2184 is divisible by 4.

(3)(c) 2184, 1074, 7832 by 6

Solution:- In these numbers, the numbers 2184 and 1074 is divisible by 6.

Because, 84 ÷ 4 = 21

8= ten’s place and 4= one’s place

Because, these numbers are divided by 2 & 3

So, these divisible by 6.

2184÷ 2 =1092

÷ 3 = 728

1074÷2 = 537

÷ 3 = 358

(3)(d) 5075, 1737, 2193 by 9

Solution:- From these numbers, only 1737 is divisible by 9

1737 = (1+7+3+7) ÷ 9

= 18 ÷ 9

= 2

(4) For divisible by 9;

(4) Which digits are to be put in the blanks os that each of the following numbers is divisible by 9?

(4)(a) 5 – 4723

Solution:- 5 6 4723

Therefore, (5+6+4+7+2+3) ÷ 9

= 27 ÷ 9

= 3.

(4) (b) 812 — 74

Solution:- 812 5 74

Therefore, (8+1+2+5+7+4) ÷ 9

= 27÷9

= 3

(4)(c) — 41578

Solution:- 2 41578

(2+4+1+5+7+8) ÷ 9

= 27 ÷ 9

= 3

(4)(d) 5742 —

Solution:- 5742 9

5+7+4+2+9 ÷ 9

= 27÷ 9

= 3

(5) Determine the smallest number of 5 digits which is divisible by 3.

Solution:- The smallest number of 5 digits are = 10000 (Not divisible by 3)

So, the number is = (10000+2)

= 10002

Therefore, (1+0+0+0+2) ÷ 3

= 3÷3

= 1

So, the number is 10002 (Answer).

(6) Determine the greatest number of 7 digits which id divisible by 6.

Solution:-The greatest number of 7 digits which is divisible by 6 that is 9999996 (Answer).

9999996÷ 2               [6=2×3]

= 4999998

9999996÷3

= 3333332

(7) Determine whether the greatest and the smallest number formed by the digits 3, 0, 5, 2, 7 are divisible by 4 and 5.

Solution:- The number is 75320 and 35720 that’s number are divisible 4 8 5 (Answer).

Updated: November 26, 2020 — 6:15 am

2 Comments

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  1. Nowfat Jahan Zimha ( class six)

    Thanks for shareing
    Can you please do the maths descriptively? You are doing in shorcut so I can’t understand 😔😔

    1. Sure we will do it in upcoming days

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