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In this chapter, we provide RD Sharma Class 6 ex 4.6 Solutions Chapter 4 Operations on Whole Numbers for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Class 6 ex 4.6 Solutions Chapter 4 Operations on Whole Numbers Maths pdf, Now you will get step by step solution to each question.

RD Sharma Solutions for Class 6 Chapter 4 Operations on Whole Numbers Ex 4.6 Download PDF

Chapter 4: Operations on Whole Numbers – Exercise 4.6

Question: 1

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).

Question: 2

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).

Question: 3

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).

Question: 4

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).

Question: 5

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) × (Predecessor of 1001)

= 1000 × 1000

= 1000000

= one million

Hence, the correct option is (c).

Question: 6

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).

Question: 7

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 …… 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 – 1

= 98

Hence, the correct option is (d).

Question: 8

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 …..

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

Hence, the correct option is (d).

Question: 9

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 — 1 = 99

Hence, the correct option is (b).

Question: 10

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).

Question: 11

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).

Question: 12

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).

Question: 13

The value of 4 x 378 x 25 is:

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

Solution:

By regrouping, we get

4 × 378 × 25 = 4 × 25 × 378

= 100 × 378

= 37800

Hence, the correct option is (a).

Question: 14

The value of 1735 x 1232 – 1735 x 232 is:

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

Solution:

Using distributive law of multiplication over subtraction, we get

1735 × 1232 – 1735 × 232 = 1735 (1232 – 232)

= 1735 × 1000

= 1735000

Hence, the correct option is (c).

Question: 15

The value of 47 × 99 is:

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

Solution:

Since, 99 = 100 — 1

Therefore, 47 × 99 = 47 × (100 — 1)

= 47 × 100 — 47

= 4700 — 47

= 4653

Thus, the value of 47 × 99 is 4653.

Hence, the correct option is (b).

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