In this chapter, we provide RS Aggarwal Solutions for Class 6 Chapter 3 Whole Numbers Exercise 3E 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 3E pdf, free RS Aggarwal Solutions Class 6 Chapter 3 Whole Numbers Exercise 3E 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 | 3E |

Category | RS Aggarwal Solutions |

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

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

(i) By actual division, we have

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

(i) By actual division, we have

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

(i) We know that any number (non-zero) divided by 1 gives the number itself

65007 ÷ 1 = 65007

(ii) We know that 0 divided by any natural number gives 0

0 ÷ 879 = 0

(iii) 981 + 5720 ÷ 10 = 981 + (5720 ÷ 10)

= 981 + 572

= 1553

(iv) 1507 – 625 ÷ 25 = 1507 – (625 ÷ 25)

= 1507 – 25

= 1482

(v) 32277 ÷ (648 – 39)

= 32277 ÷ 609

32277 ÷ (648 – 39) = 53

(vi) 1573 ÷ 1573 – 1573 ÷ 1573

= (1573 ÷ 1573) – (1573 ÷ 1573)

= 1 – 1

= 0

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

We have n ÷ n = n

let n = 1, 1 ÷ 1 = 1

1 = 1

which is true

∴ Hence 1 is the required whole number.

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

Product of two numbers = 504347

One number = 317

Other number = 504347 ÷ 317

∴ Other number = 1591

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

Here Dividend = 59761, Quotient = 189

∴ Remainder = 37

We know that Dividend = Divisor x Quotient + Remainder

59761 = Divisor x 189 + 37

59761 – 37 = Divisor x 189

59724 = Divisor x 189

Divisor x 189 = 59724

∴ Divisor = 59724 ÷ 189

∴ Divisor = 59724 ÷ 189 = 316

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

Here dividend = 55390,

Divisor = 299 and Remainder = 75

By division algorithm, we have

Dividend = Quotient x Divisor + Remainder

55390 = Quotient x 299 + 75

55390 – 75 = Quotient x 299

55315 = Quotient x 299

Quotient x 299 = 55315

Quotient = 55315 ÷ 299

∴ Required quotient = 185

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

On dividing 13601 by 87, we have

It is clear that if we subtract 29 from 13601, the resulting number will be exactly divisible by 87.

∴ The required least number = 29.

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

Here dividend = 1056, Divisor = 23

By actual division, we have

It is clear that if we add 2 to 21, it will become 23 which is divisible by 23.

∴ Required least number = 2.

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

Greatest 4-digit number = 9999

On, dividing by 16, we get remainder as 15

∴ The required largest 4-digit number = 9999 – 15

= 9984

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

Largest number of 5-digits = 99999

On dividing 99999 by 653, we have

∴ Quotient = 153, Remainder = 90

Check : By division algorithm Dividend = Divisor x Quotient + Remainder

= 653 x 153 + 90

= 99909 + 90

= 99999

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

The least 6-digit number = 100000

On dividing 100000 by 83, we have

It is clear that if we add 15 to 68, it will become 83 which is divisible by 83.

∴ Required least 6-digit number = 100000 + 15

= 100015

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

Cost of 1 dozen bananas = Rs. 29

Bananas can be purchase in Rs. 1392

1392 ÷ 29

= 48 dozens

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

Total number of trees = 19625

Total number of rows = 157

Number of trees in each row = 19625 ÷ 157

∴ Number of trees in each row = 125

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

Total population of the town = 517530

Since there is one educated person out of 15

Total number of educated persons in the town = 517530 ÷ 15

∴ Total number of educated persons in the town = 34502.

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

Cost of 23 colour TV sets = Rs. 570055

Cost of 1 colour TV set

∴ Cost of 1 color TV set

= Rs. 570055 ÷ 23

= Rs. 24785

**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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