The document RD Sharma Solutions -Ex-4.6, Operations On Whole Numbers, Class 6, Maths Class 6 Notes | EduRev is a part of the Class 6 Course RD Sharma Solutions for Class 6 Mathematics.

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**Q1. Which one of the following is the smallest whole number?**

**(a) 1 (b) 2 (c) 0 (d) None of these**

**Solution:** The set of whole numbers is { 0 , 1, 2, 3, 4, ...}.

So, the smallest whole number is 0.

Hence, the correct option is (c).

**Q2. Which one of the following is the smallest even whole number?**

**(a) 0 (b) 1 (c) 2 (d) None of these**

**Solution: **The natural numbers along with 0 form the collection of whole numbers.

So, the numbers 0, 1, 2, 3, 4, ... form the collection of whole numbers.

The number which is divisible by 2 is an even number.

So, in the collection "0, 1, 2, 3, 4, ...", 2 is the smallest even number.

Hence, the correct option is (c).

**Q3. Which one of the following is the smallest odd whole number?**

**(a) 0 (b) 1 (c) 3 (d) 5**

**Solution: **The natural numbers along with 0 form the collection of whole numbers.

So, the numbers 0, 1, 2, 3, 4, ... form the collection of whole numbers.

A natural number which is not divisible by 2 is called an odd whole number.

So, in the collection "0, 1, 2, 3, 4, ...", 1 is the smallest odd whole number.

Hence, the correct option is (b).

**Q4. How many whole numbers are between 437 and 487?**

**(a) 50 (b) 49 (c) 51 (d) None of these**

**Solution: **The whole numbers between 437 and 487 are 438, 439, 440, 441, ... , 484, 485 and 486. To find the required number of whole numbers,

We need to subtract 437 from 487 and then subtract again 1 from the result.

Thus, there are (487 - 437) - 1 whole numbers between 437 and 487.

Now, (487 - 437) - 1 = 50 - 1 = 49

Hence, the correct option is (b).

**Q5. The product of the successor 999 and predecessor of 1001 is:**

**(a) one lakh (b) one billion (c) one million (d) one crore**

**Solution:** Successor of 999 = 999 + 1 = 1000

Predecessor of 1001 = 1001 - 1 = 1000

Now,

Product = (Successor of 999) x (Predecessor of 1001)

= 1000 x 1000

= 1000000

= one million

Hence, the correct option is (c).

**Q6. Which one of the following whole numbers does not have a predecessor?**

**(a) 1 (b) 0 (c) 2 (d) None of these**

**Solution:** The numbers 0, 1, 2, 3, 4, .... form the collection of whole numbers.

The smallest whole number is 0.

So, 0 does not have a predecessor.

Hence, the correct option is (b).

**Q7. The number of whole numbers between the smallest whole number and the greatest 2 digit number is:**

**(a) 101 (b) 100 (c)99 (d)98**

**Solution: **Smallest whole number = 0

Greatest 2-digit whole number = 99

The whole numbers between 0 and 99 are 1, 2, 3, 4 â€¦â€¦ 97, 98.

To find the number of whole numbers between 0 and 99,

Subtract 1 from the difference of 0 and 99.

Therefore, Number of whole numbers between 0 and 99 = (99 - 0) - 1

= 99 â€“ 1

= 98

Hence, the correct option is (d).

**Q8. If n is a whole number such that n + n = n, then n =?**

**(a) 1 (b)2 (c)3 (d) None of these**

**Solution:** Here, 0 + 0 = 0, 1 + 1 = 2 , 2 + 2 = 4 â€¦..

So, the statement n + n = n is true only when n = 0.

Hence, the correct option is (d).

**Q9. The predecessor of the smallest 3 digit number is:**

**(a) 999 (b)99 (c) 100 (d)101**

**Solution:** Smallest 3-digit number = 100

Predecessor of 3-digit number = 100 â€” 1 = 99

Hence, the correct option is (b).

**Q10. The least number of 4 digits which is exactly divisible by 9 is:**

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

**Solution:** Least 4-digit number = 1000

The least 4-digit number exactly divisible by 9 is 1000 + (9 - 1) = 1008.

Hence, the correct option is (a).

**Q11. The number which when divided by 53 gives 8 as quotient and 5 as remainder is:**

**(a) 424 (b)419 (c)429 (d)None of these**

**Solution**: Here, Divisor = 53, Quotient = 8 and Remainder = 5.

Now, using the relation Dividend = Divisor x Quotient + Remainder

We get

Dividend = 53 x 8 + 5

= 424 + 5

=429

Thus, the required number is 429.

Hence, the correct option is (c).

**Q12. The whole number n satisfying n + 35 = 101 is:**

**(a)65 (b)67 (c)64 (d)66**

**Solution:** Here, n+ 35 = 101.

Adding - 35 on both sides, we get

n + 35 + (- 35)= 101 + (- 35)

n + 0 = 66

n=66

Hence, the correct option is (d).

**Q13. The value of 4 x 378 x 25 is :**

**(a)37800 (b)3780 (c)9450 (d)30078**

**Solution:** By regrouping, we get

4 x 378 x 25 = 4 x 25 x 378

= 100 x 378

= 37800

Hence, the correct option is (a).

**Q14. The value of 1735 x 1232 â€“ 1735 x 232 is:**

**(a)17350 (b)173500 (c)1735000 (d)173505**

**Solution:** Using distributive law of multiplication over subtraction, we get

1735 x 1232 â€“ 1735 x 232 = 1735 (1232 â€“ 232)

= 1735 x 1000

= 1735000

Hence, the correct option is (c).

**Q15. The value of 47 x 99 is :**

**(a)4635 (b)4653 (c)4563 (d)6453**

**Solution:** Since, 99 = 100 â€” 1

Therefore, 47 x 99 = 47 x (100 â€” 1)

= 47x 100 â€” 47

= 4700 â€” 47

= 4653

Thus, the value of 47 x 99 is 4653.

Hence, the correct option is (b).

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