# RS Aggarwal Class 6 Solutions Chapter 3 - Whole Numbers Ex 3B (3.2)

## RS Aggarwal Class 6 Chapter 3 - Whole Numbers Ex 3B (3.2) Solutions Free PDF

RS Aggarwal solutions are extremely helpful to solve all your problems and pass your examination with flying colours. These solutions are considered as one of the preferable study material as they are closely aligned with the latest syllabus. We at BYJUâ€™S provide you with the comprehensive solutions of RS Aggarwal Class 6 Solutions Chapter 3- Whole Numbers Ex 3B (3.2) in downloadable pdf format created by our subject experts.

The solutions we provide here cover all the important topics in depth. It provides detailed, step-by-step solutions to all questions. These solutions serve as an invaluable aid to students while preparing for the exams. It covers the latest CBSE syllabus based on CCE guidelines. Practicing these solutions will help you to revise the entire syllabus before the exam and score more marks. Solve all the questions mentioned in the exercise and if you are stuck in between take help from the RS Aggarwal Maths Solutions.

## Download PDF of RS Aggarwal Class 6 Solutions Chapter 3 – Whole Numbers System Ex 3B (3.2)

Find the correct answer for each of the following:

Q1: The smallest whole number is

(a) 1 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (b) 0 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (c) 2 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (d) none of these

Sol:

(b) 0

The smallest whole number is 0.

Q2: The least number of 4 digit which is exactly divisible by 9 is

(a) 1018 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (b) 1026 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (c) 1009 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (d) 1008

Sol:

(d) 1008

(a)

Hence, 1018 is not exactly divisible by 9.

(b)

Here, 1026 is exactly divisible by 9.

(c)

Here, 1008 is not exactly divisible by 9.

(d)

Hence, 1008 is exactly divisible by 9.

(b) and (d) are exactly divisible by 9, but (d) is the least number which is exactly divisible by 9.

Q3: The largest number of 6 digits which is exactly divisible by 16 is

(a) 999980 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (b) 999982 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (c) 999984 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (d) 999964

Sol:

(c) 999984

(a)

Hence, 999980 is not exactly divisible by 16.

(b)

Hence, 999982 is not exactly divisible by 16.

(c)

Hence, 999984 is exactly divisible by 16.

Hence, 999964 is not exactly divisible by 16.

The largest six digit number which is exactly divisible by 16 is 999984.

Q4: What least number should be subtracted from 10004 to get a number exactly divisible by 12?

(a) 4 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (b) 6 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (c) 8 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (d) 20

Sol:

(c) 8

Here we have to tell what least number should be subtracted from 1004 to get a number exactly divisible by 12.

So, we will first divide 1004 by 12.

Remainder = 8

So, 8 should be subtracted from 1004 to get the number exactly divisible by 12.

i.e. 1004 – 8 = 9996

Hence, 9996 is exactly divisible by 12.

Q5: What least number should be added to 10056 to get a number exactly divisible by 23?

(a) 5 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (b) 18 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (c) 13 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (d) 10

Sol:

(a) 18

Here, we have to tell what least number must be added to 1056 to get a number exactly divisible by 23.

So, first we will divide 1056 by 23.

Remainder = 5

Required number = 23 – 5 = 18

So, 18 must be added to 1056 to get a number exactly divisible by 23.

Hence, 10074 is exactly divisible by 23.

Q6: What least number is nearest to 457 which is divisible by 11?

(a) 450 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (b) 451 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (c) 460 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (d) 462

Sol:

(d) 462

(a)

Hence, 450 is not divisible by 11.

(b)

Hence, 451 is divisible by 11.

(c)

Hence, 460 is not divisible by 11.

(d)

Hence, 462 is divisible by 11.

Here, the number given in the options (b) and (d) are divisible by 11. However, we want a whole number nearest to 457 which is divisible by 11.

So, 462 is whole number nearest to 457 and divisible by 11.

Q7: How many whole numbers are there between 1018 and 1203?

(a) 185 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (b) 186 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (c) 184 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (d) none of these

Sol:

(c) 184

Number of whole numbers = (1203 – 1018) – 1

= 185 – 1

= 184

Q8 : A number when divided by 46 gives 11 as quotient and 15 as remainder. The number is

(a) 491 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (b) 521 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (c) 701 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (d) 679

Sol:

(b) 521

Divisor = 46

Quotient = 11

Remainder = 15

Dividend = Divisor $\times$ quotient + remainder

= 46 $\times$ 11 + 15

= 506 + 15

= 521

Q9: In a division sum, we have dividend = 199, quotient = 16 and remainder = 7. The divisor is

(a) 11 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (b) 23 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (c) 12 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (d) none of these

Sol:

(c) 12

Dividend = 199

Quotient = 16

Remainder = 7

According to the division algorithm, we have:

Dividend = Divisor $\times$ quotient + remainder

$\Rightarrow$ 199 = Divisor $\times$ 16 + 7

$\Rightarrow$ 199 – 7 = Divisor $\times$ 16

$\Rightarrow$ Divisor = 192 $\div$ 16

Q10: 7589 – ? = 3434

(a) 11023 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (b) 4245 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (c) 4155 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (d) none of these

Sol:

(a) 11023

7589 – ? = 3434

$\Rightarrow$ 7589 – x = 3434

$\Rightarrow$ x = 7589 + 3434

$\Rightarrow$ x = 11023

Q11: 587 $\times$ 99 = ?

(a) 57213 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (b) 58513 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (c) 58113 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (d) 56413

Sol:

(c) 58113

587 $\times$ 99

= 587 $\times$ (100 – 1)

= 587 $\times$ 100 – 587 $\times$ 1

= 58700 – 587

= 58113

Q12: Â 4 $\times$ 538 $\times$ 25 = ?

(a) 32280 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (b) 26900 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (c) 53800 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (d) 10760

Sol:

(c) 53800

4 $\times$ 538 $\times$ 25

= 100 $\times$ 538

= 53800

Q13: 24679 $\times$ 92 + 24679 $\times$ 8 = ?

(a) 493580 Â Â Â Â Â Â Â Â Â Â Â (b) 1233950 Â Â Â Â Â Â Â Â Â Â Â (c) 2467900 Â Â Â Â Â Â Â Â Â Â Â Â Â (d) none of these

Sol:

(c) 2467900

By using the distributive property, we have:

24679 $\times$ 92 + 24679 $\times$ 8

= 24679 $\times$ (92 + 8)

= 24679 $\times$ 100

= 2467900

Q14: 1625 $\times$ 1625 – 1625 $\times$ 625

(a) 1625000 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (b) 162500 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (c) 325000 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (d) 812500

Sol:

(a) 1625000

By using the distributive property, we have:

1625 $\times$ 1625 – 1625 $\times$ 625

= 1625 $\times$ (1625 – 625)

= 1625 $\times$ 1000

= 1625000

Q15: 1568 $\times$ 185 – 1568 $\times$ 85 = ?

(a) 7840 Â Â Â Â Â Â Â Â Â Â Â Â Â Â (b) 15680 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (c) 156800 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (d) none of these

Sol:

(c) 156800

By using the distributive property, we have;

1568 $\times$ 185 – 1568 $\times$ 85

= 1568 $\times$ (185 – 85)

= 1568 $\times$ 100

= 156800

Q16: (888 + 777 + 555) = (111 $\times$ ?)

(a) 120 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (b) 280 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (c) 20 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (d) 140

Sol:

(c) 20

(888 + 777 + 555) = (111 $\times$ ?)

$\Rightarrow$ (888 + 777 + 555) = 111 $\times$ (8 + 7 + 5) Â Â Â Â Â [by taking 111 common]

= 111 $\times$ 20 = 2220

Q17: The sum of two odd numbers is

(a) an odd number Â (b) an even number Â (c) a prime number Â (d) a multiple of 3

Sol:

(b) An even number

The sum of two odd numbers is an even number.

Example: 5 + 3 = 8

Q18:The product of two odd numbers is

(a) an odd number Â (b) an even number Â (c) a prime number Â (d) none of these

Sol:

(a) an odd number

The product of two odd numbers is an odd number.

Example: 5 $\times$ 3 = 15

Q19: If a is a whole number such that a + a = a, then a = ?

(a) 1 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (b) 2 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (c) 3 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (d) none of these

Sol:

(d) None of these

Given: a is a whole number such that a + a = a

If a = 1, then 1 + 1 = 2 $\neq$ 1

If a = 2, then 2 + 2 = 4 $\neq$ 1

If a = 3, then 3 + 3 = 6 $\neq$ 1

Q20: The predecessor of 10000 is

(a) 10001 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (b) 9999 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (c) none of these

Sol:

(b) 9999

Predecessor of 10000 = 10000 – 1 = 9999

Q21: The successor of 1001 is

(a) 1000 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (b) 1002 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (c) none of these

Sol:

(b) 1002

Successor of 1001 = 1001 + 1 = 1002

Q22: The smallest even whole number is

(a) 0 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (b) 2 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (c) none of these

Sol:

(b) 2

The smallest even whole number is 2. Zero is neither an even number nor an odd number.

### Key Features of RS Aggarwal Class 6 Solutions Chapter 3 – Whole Numbers Ex 3B (3.2)

• Each and every step is properly explained with proper reasons to understand the solution.
• All the solutions are easy to digest and can be comprehended pretty easily.
• The R.S.Agarwal Class 6 Solutions helps the students to create a strong base for mathematics and prepare for final
• examinations.
• The solution contains ample no of questions for students to practice before the exam.

#### Practise This Question

Solve the equation  3x+1=65x1