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

13 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

    2. I also have the same question

    3. Amit rakshat I support this girl as a student

      1. Moumita Sure we are also thinking on that. Please tell us question which you guys cant understand..

  2. Super helpful

    1. Thanks for sharing.
      We are really got a super help.????

      1. Sabiha tell us more subject so that we can help

  3. Leyana Chowdhury

    Hi

  4. 3)=C mistake because 84 in ones and tens place this formula is for 4 not for 6. Too much useful all maths but please try to explainn it descriptively so that we can under stand and please answer all maths don’t skip any of them

  5. Sabiha Rahman Chowdhury

    Thanks for sharing.
    We are really got a super help.????

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