# ML Aggarwal CBSE Solutions Class 6 Math First Chapter Knowing Our Numbers Exercise 1.2

### ML Aggarwal CBSE Solutions Class 6 Math First Chapter Knowing Our Numbers Exercise 1.2

(1) Use the appropriate symbol < or > to fill in the blanks:

(i) 173 < 189

(ii) 1058 < 1074

(iii) 8315 > 8017

(2) In each of the following pairs of numbers, state which number is smaller :

(i) Smallest Number: 503

(ii) 1139 is smallest number.

(iii) 24531 is smallest number.

(3) Find the greatest and the smallest numbers in each row:

(i) Smallest number: 571

Greatest number: 71834

(ii) Smallest number: 7691

Greatest number: 12002

(4) Arrange the following numbers in ascending order:

Solution: 34 < 43 < 304 < 340 < 430.

(5) Arrange the following numbers in descending order:

Solution: 7529 > 7333 > 553 > 335 > 53

(6) Write all possible 2-digit numbers that can be formed by using the digits 2, 3 and 4. Repetition of digits is not allowed. Also find their sum.

Solution : All possible two digit numbers are – 23, 34, 43, 42, 32, 24 .

(7) Write all possible 3-digit numbers using the digits 3, 1 and 5. Repetition of digits is not allowed.

Solution: 315, 353, 135, 153, 531, 513.

(8) Write all possible 3 digit numbers using the digits 7, 0 and 6. Repetition of digits is not allowed . Also find their sum.

Solution: Possible 3 digit numbers are:   706 and 607.

Their sum = 706 + 607 = 1313

(9) Write all possible 2 digits numbers using the digits 4, 0 and 9.l Repetition of digits is not allowed. Also find their sum.

Solution: 40 + 90 + 49 + 94 = 273

(10) Write all possible 2-digit numbers that can be formed by using the digits 3, 7 and 9. Repetition of digits is allowed.

Solution: all possible two digit numbers are = 37,39, 73, 79, 93, 97.

(11) Write all possible numbers using the digits 3, 1 and 5. Repetition of digits is not allowed.

Solution: All possible numbers are: 1, 3, 5, 13, 15, 31, 35, 53, 51, 135, 153, 315, 351, 531 & 513.

(12)  How many 6 digits numbers are there in all?

Solution: The lowest 6-digit number is 100,000 and the highest is 999,999.

Subtract 999,999−100,000=899,999, but then add 1, because we want to include both 100,000 and 999,999.

(13) Write down the greatest number and the smallest number of 4 digits that can be formed by the digits 7, 5, 0 and 4 using each digit only once.

Solution: The greatest 4 digit number are – 7540

The smallest 4 digits number are – 4057

(14) Rearrange the digits of the number 5701024 to get the largest number and the smallest number of 7 digits.

Solution: The greatest 7 digit numbers is – 7542100

The smallest 7 digit number is – 1002457

(15) Keeping the place value of digit 3 in the number 730265 same, rearrange the digits of the given number to get the largest number and smallest number of 6 digits.

Solution: 736520;230567

(16) Form the smallest and greatest 4-digit numbers by using any one digit twice form the digits:

Solution: (i) Greatest number: 9953

Smallest number: 2235

(ii) Greatest number:

Smallest number: 6641

Smallest number: 1004

(iii) Greatest number: 8865

Smallest number: 1145

(17) Write (i) the greatest number of 6 digits (ii) the smallest number of 7 digits. Also find their difference.

Solution: The greatest number  of 6 digits : 999999

The smallest number of 7 digits: 1000000.

Their differences are = 1000000 – 999999 = 1.

(18) Write the greatest 4-digit number having distinct digits.

Solution: 9876

(19) Write the smallest 4-digit number having distinct digits.

Solution: 1234

(20) Write the greatest 6 digit number using three different digits.

Solution: 999987

(21) Write the smallest 7 digit number using four different digits.

Solution: 1000234

(22) Write the greatest and the smallest 4-dit members using four different digits with the conditions as given:

Solution: (i) Digit 7 is always at unit place.

(ii) Digit 4 is always at tens place

(iii) Digit 9 is always at hundred place

(iv) Digit 2 is always at thousand place.

Solution: (i) 9867; 1027

(ii) 9847; 1042

(iii) 8976; 1902

(iv) 2987; 2013

Updated: September 21, 2019 — 10:02 am