In this chapter, we provide RS Aggarwal Solutions for Class 6 Chapter 3 Whole Numbers Exercise 3F for English medium students, Which will very helpful for every student in their exams. Students can download the latest RS Aggarwal Solutions for Class 6 Chapter 3 Whole Numbers Exercise 3F pdf, free RS Aggarwal Solutions Class 6 Chapter 3 Whole Numbers Exercise 3F book pdf download. Now you will get step by step solution to each question.

Textbook | NCERT |

Class | Class 6 |

Subject | Maths |

Chapter | Chapter 3 |

Chapter Name | Whole Numbers |

Exercise | 3F |

Category | RS Aggarwal Solutions |

**RS Aggarwal Class 6 Chapter 3 Whole Numbers Ex 3F Download PDF**

**Question 1.****Solution:**

The smallest whole number is 0 (b)

**Question 2.****Solution:**

The least 4-digit number = 1000

On dividing 1000 by 9, we get

Remainder = 1

Least 4-digit number which is

Divisible by 9 = 1000 – 1 + 9

= 1000 + 8

= 1008 (d)

**Question 3.****Solution:**

The largest 6-digit number = 999999

On dividing by 16, we get

Remainder =15

The greatest 6-digit number divisible by 16

= 999999 – 15

= 999984 (c)

**Question 4.****Solution:**

On dividing 10004 by 12, we get remainder = 8

8 is to be subtracted from 10004 (c)

**Question 5.****Solution:**

On dividing 10056 by 23 We get remainder =12

The least number to be added = 23 – 5

= 18 (b)

**Question 6.****Solution:**

On dividing 457 by 11

We get remainder = 6

Which is greater than half of 11

The number nearest to 457 which is divisible 11 will be = 457 – 6 + 11

= 457 + 5

= 462 (d)

**Question 7.****Solution:**

Whole number between 1018 and 1203 are 1019 to 1202 are 1202 – 1018

= 184 (c)

**Question 8.****Solution:**

Divisor = 46

Quotient =11

Remainder =15

Number = Divisor x Quotient + Remainder

= 46 x 11 + 15

= 506 + 15

= 521 (b)

**Question 9.****Solution:**

Dividend = 199

Quotient =16

Remainder = 7

Divisor = 199−716 = 19216

= 12 (c)

**Question 10.****Solution:**

7589 – ? = 3434

Required number = 7589 – 3434

= 4155 (c)

**Question 11.****Solution:**

587 x 99 = 587 x (100 – 1)

= 587 x 100 – 587 x 1

= 58700 – 587

= 58113 (c)

**Question 12.****Solution:**

4 x 538 x 25 = 538 x 4 x 25

= 538 x 100

= 53800 (c)

**Question 13.****Solution:**

24679 x 92 + 24679 x 8

= 24679 x (92 + 8)

= 24679 x 100

= 2467900 (c)

**Question 14.****Solution:**

1625 x 1625 – 1625 x 625

= 1625 (1625 – 625)

= 1625 x 1000

= 1625000 (a)

**Question 15.****Solution:**

1568 x 185 – 1568 x 85

= 1568 (185 – 85)

= 1568 x 100

= 156800 (c)

**Question 16.****Solution:**

888 + 111 + 555 = 111 x ?

= 11 (8 + 7 + 5)

= 111 x 20 (c)

**Question 17.****Solution:**

The sum of two odd number is an even number (b)

**Question 18.****Solution:**

The product of two odd numbers is an odd number (a)

**Question 19.****Solution:**

If a is a whole number such that

a + a = a, then a = 0

as 0 + 0 = 0

None of these (d)

**Question 20.****Solution:**

The predecessor of 10000 is 10000 – 1

= 9999 (b)

**Question 21.****Solution:**

The successor of 1001 is 1001 + 1

= 1002 (b)

**Question 22.****Solution:**

The smallest even whole number is 2 (b)

**All Chapter RS Aggarwal Solutions For Class 6 Maths**

—————————————————————————–**All Subject NCERT Exemplar Problems Solutions For Class 6**

**All Subject NCERT Solutions For Class 6**

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