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

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

**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

**Objective questions**

**Mark against the correct answer in each of the following :**

**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 = \(\\ \frac { 199-7 }{ 16 } \) = \(\\ \frac { 192 }{ 16 } \)

= 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)

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

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

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