## RS Aggarwal Class 6 Solutions Chapter 3 Whole Numbers Ex 3E

These Solutions are part of RS Aggarwal Solutions Class 6. Here we have given RS Aggarwal Solutions Class 6 Chapter 3 Whole Numbers Ex 3E.

**Other Exercises**

- RS Aggarwal Solutions Class 6 Chapter 3 Whole Numbers Ex 3A
- RS Aggarwal Solutions Class 6 Chapter 3 Whole Numbers Ex 3B
- RS Aggarwal Solutions Class 6 Chapter 3 Whole Numbers Ex 3C
- RS Aggarwal Solutions Class 6 Chapter 3 Whole Numbers Ex 3D
- RS Aggarwal Solutions Class 6 Chapter 3 Whole Numbers Ex 3E
- RS Aggarwal Solutions Class 6 Chapter 3 Whole Numbers Ex 3F

**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

Hope given RS Aggarwal Solutions Class 6 Chapter 3 Whole Numbers Ex 3E are helpful to complete your math homework.

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