Objective Type Questions present at the end of Chapter 4 help students self-analyse their knowledge about the concepts covered in the entire chapter. These questions are answered in RD Sharma Solutions for Class 6 by expert Mathematicians to help students perform well in the exam. The important formulae and shortcuts used in solving problems are highlighted, which help students grasp them faster. RD Sharma Solutions for Class 6 Maths Chapter 4 Operations on Whole Numbers Objective Type Questions PDF is available here.
RD Sharma Solutions for Class 6 Maths Chapter 4: Operations on Whole Numbers Objective Type Questions
Access answers to Maths RD Sharma Solutions for Class 6 Chapter 4: Operations on Whole Numbers Objective Type Questions
Mark the correct alternative in each of the following:
1. Which one of the following is the smallest whole number?
(a) 1 (b) 2 (c) 0 (d) None of these
Solution:
Option (c) is the correct answer.
We know that the set of whole numbers is {0, 1, 2, 3, 4 …}.
Hence, the smallest whole number is 0.
2. Which one of the following is the smallest even whole number?
(a) 0 (b) 1 (c) 2 (d) None of these
Solution:
Option (c) is the correct answer.
We know that the natural numbers, along with 0, form the collection of whole numbers.
Hence, the numbers 0, 1, 2, 3, 4 … form the collection of whole numbers.
So the number which is divisible by 2 is an even number, and 2 is the smallest even number.
3. Which one of the following is the smallest odd whole number?
(a) 0 (b) 1 (c) 3 (d) 5
Solution:
Option (b) is the correct answer.
We know that the natural numbers, along with 0, form the collection of whole numbers.
Hence, the numbers 0, 1, 2, 3, and 4 … form the collection of whole numbers.
So the natural number which is not divisible by 2 is called an odd whole number, and 1 is the smallest odd whole number.
4. How many whole numbers are between 437 and 487?
(a) 50 (b) 49 (c) 51 (d) None of these
Solution:
Option (b) is the correct answer.
We know that the whole numbers between 437 and 487 are 438, 439, 440, 441, …, 484, 485 and 486.
In order to find the required number of whole numbers, subtract 437 from 487 and then subtract 1 again.
Hence, there are (487 − 437) − 1 whole number lying between 437 and 487.
So we get (487 − 437) − 1 = 50 − 1 = 49
5. The product of the successor of 999 and the predecessor of 1001 is
(a) one lakh (b) one billion (c) one million (d) one crore
Solution:
Option (c) is the correct answer.
We know that the successor of 999 = 999 + 1 = 1000
So the predecessor of 1001 = 1001 − 1 = 1000
It can be written as
Product of them = (Successor of 999) × (Predecessor of 1001)
By substituting the values
Product of them = 1000 × 1000 = 1000000 = one million
6. Which one of the following whole numbers does not have a predecessor?
(a) 1 (b) 0 (c) 2 (d) None of these
Solution:
Option (b) is the correct answer.
We know that the numbers 0, 1, 2, 3, 4 … form the collection of whole numbers.
Hence, the smallest whole number is 0, which does not have a predecessor.
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:
Option (d) is the correct answer.
We know that the smallest whole number = 0
So the greatest 2-digit whole number = 99
Whole numbers which lie between 0 and 99 are 1, 2, 3, 4,…, 97, 98.
In order to find the number of whole numbers between 0 and 99, first subtract 1 from the difference of 0 and 99.
So the number of whole numbers between 0 and 99 = (99 − 0) − 1 = 99 − 1 = 98
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:
Option (d) is the correct answer.
We know that 0 + 0 = 0, 1 + 1 = 2, 2 + 2 = 4….
Hence, the statement n + n = n is true only when n = 0.
9. The predecessor of the smallest 3-digit number is
(a) 999 (b) 99 (c) 100 (d) 101
Solution:
Option (b) is the correct answer.
We know that the smallest 3-digit number = 100
So the predecessor of 3 digit number = 100 − 1 = 99
10. The least number of 4-digits which is exactly divisible by 9 is
(a) 1008 (b) 1009 (c) 1026 (d) 1018
Solution:
Option (a) is the correct answer.
We know that the least 4-digit number = 1000
Hence, the least 4-digits which is exactly divisible by 9 is 1000 + (9 − 1) = 1008
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:
Option (c) is the correct answer.
It is given that
Divisor = 53, Quotient = 8 and Remainder = 5.
By using the relation, we get
Dividend = Divisor × Quotient + Remainder
By substituting the values
Dividend = 53 × 8 + 5 = 424 + 5 =429
Hence, the required number is 429.
12. The whole number n satisfying n + 35 = 101 is
(a) 65 (b) 67 (c) 64 (d) 66
Solution:
Option (d) is the correct answer.
It is given that
n + 35 = 101
By adding − 35 on both sides
n + 35 + (− 35) = 101 + (− 35)
On further calculation
n + 0 = 66
So we get
n = 66
13. The 4 × 378 × 25 is
(a) 37800 (b) 3780 (c) 9450 (d) 30078
Solution:
Option (a) is the correct answer.
We can write it as
4 × 378 × 25 = 4 × 25 × 378
On further calculation
4 × 378 × 25 = 100 × 378 = 37800
14. The value of 1735 × 1232 − 1735 × 232 is
(a) 17350 (b) 173500 (c) 1735000 (d) 173505
Solution:
Option (c) is the correct answer.
By using the distributive law of multiplication over subtraction
1735 × 1232 − 1735 × 232 = 1735(1232 − 232)
On further calculation
1735 × 1232 − 1735 × 232 = 1735 × 1000 = 1735000
15. The value of 47 × 99 is
(a) 4635 (b) 4653 (c) 4563 (d) 6453
Solution:
Option (b) is the correct answer.
It can be written as
99 = 100 − 1
So we get
47 × 99 = 47 × (100 − 1)
On further calculation
47 × 99 = 47× 100 – 47 = 4700 – 47 = 4653
Hence, the value of 47 × 99 is 4653.
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