ML Aggarwal Whole Number Check Your Progress Class 6 ICSE Maths Solutions. We Provide Step by Step Answer of Check Your Progress Questions for Whole Number as council prescribe guideline for upcoming board exam. Visit official Website CISCE for detail information about ICSE Board Class-6.
ML Aggarwal Whole Number Check Your Progress Class 6 ICSE Maths Solutions
| Board | ICSE |
| Publications | Avichal Publishig Company (APC) |
| Subject | Maths |
| Class | 6th |
| Chapter-2 | Whole Number |
| Writer | ML Aggarwal |
| Book Name | Understanding |
| Topics | Solution of Check Your Progress Questions |
| Edition | 2023-2024 |
Whole Number Check Your Progress
ML Aggarwal Class 6 ICSE Maths Solutions
Page-42
Question 1. Write next three consecutive whole numbers of the number 9998.
Answer :
The next three consecutive number of 9998 are :
9998 + 1 = 9999
9999 + 1 = 10000
10000 + 1 = 10001
so Numbers are = 9999, 10000, 10001
Question 2. Write three consecutive whole numbers occurring just before 567890.
Answer :
The three consecutive whole numbers just before 567890 are:
567890 – 1 = 567889 – 1 = 567888 – 1
= 567887
∴ These are : 567889, 567888, ,567887
Question 3. Find the product of the successor and the predecessor of the smallest number of 3-digits.
Answer :
Smallest number of 3-digits = 100
Successor = 100 + 1
Predecessor = 100 -1
∴ Product = 100 + 1 × 100 – 1
= 101 × 99 = 9999
Question 4. Find the number of whole numbers between the smallest and the greatest numbers of 2-digits.
Answer :
Smallest number of 2-digits = 10
Greatest number of 2-digits = 99
Numbers between 10 and 99
11, 12, ………, 98
= 98 – 10 = 88
Question 5. Find the following sum by suitable arrangements:
(i) 678 + 1319 + 322 + 5681
(ii) 777 + 546 + 1463 + 223 + 537
Answer :
(i) 678 + 1319 + 322 + 5681
= ( 678 + 322 ) + ( 5681 + 1319 )
= 1000 + 7000 = 8000
(ii) 777 + 546 + 1463 + 223 + 537
= (777 + 223) + ( 1463 + 537) + 546
= 1000 + 2000 + 546 = 3546
Question 6. Determine the following products by suitable arrangements:
(i) 625 × 437 × 16
(ii) 309 × 25 × 7 × 8
Answer :
(i) 625 × 437 × 16
= 437 × ( 625 × 16)
= 437 × 10000 = 4370000
(ii) 309 × 25 × 7 × 8
= (309 × 7) × (25 × 8)
= 2163 × 200 = 432600
Question 7. Find the value of the following by using suitable properties:
(i) 236 × 414 + 236 × 563 + 236 × 23
(ii) 370 × 1587 – 37 × 10 × 587
Answer :
(i) 236 × 414 + 236 × 563 + 236 × 23
= 236 × ( 414 + 563 + 23)
= 236 × ( 1000) = 236000
(ii) 370 × 1587 – 37 ×10 × 587
= 37 × 10 ( 1587 – 587)
= 370 × 1000 = 370000
Question 8. Divide 6528 by 29 and check the result by division algorithm.
Answer :
29 ) 6528 ( 225
Reminder 3
Quotient = 225
Question 9. Find the greatest 4-digit number which is exactly divisible by 357.
Answer :
Largest 4 digit number is 9999
357 ) 9999( 28
Reminder =3,
Question 10. Find the smallest 5-digit number which is exactly divisible by 279.
Answer :
Smallest 5 digit number is 10000
279 ) 10000 ( 35
Reminder = 235
Dividing it by 279, we get remainder = 235
To make the smallest 5-digit number exactly divisible by 279, we have
to add 279 – 235 = 44 in 10000
∴ 10000 + 44 = 10044.
— : End of ML Aggarwal Whole Number Check Your Progress Class 6 ICSE Maths Solutions :–
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