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 :–

Return to **–** ML Aggarwal Maths Solutions for ICSE Class -6

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