RD Sharma Solutions for Class 6 Maths Exercise 4.4 PDF
RD Sharma Solutions consists of problems solved by subject experts after conducting wide research on each concept. They mainly focus on the concepts which are important for the final exam. The methods of solving problems are based on the CBSE syllabus and guidelines. The step-wise detailed structure of the solutions improves knowledge about the concepts among students. PDFs of solutions can be downloaded from BYJU’S website, which are important for exam preparation. RD Sharma Solutions for Class 6 Maths Chapter 4 Operations on Whole Numbers Exercise 4.4 are provided here.
RD Sharma Solutions for Class 6 Maths Chapter 4: Operations on Whole Numbers Exercise 4.4
Access answers to Maths RD Sharma Solutions for Class 6 Chapter 4: Operations on Whole Numbers Exercise 4.4
1. Does there exist a whole number a such that a ÷ a = a?
Solution:
Yes. There exists a whole number ‘a’ such that a ÷ a = a.
We know that the whole number is 1 where 1 ÷ 1 = 1.
2. Find the value of:
(i) 23457 ÷ 1
(ii) 0 ÷ 97
(iii) 476 + (840 ÷ 84)
(iv) 964 – (425 ÷ 425)
(v) (2758 ÷ 2758) – (2758 ÷ 2758)
(vi) 72450 ÷ (583 – 58)
Solution:
(i) 23457 ÷ 1
By division
23457 ÷ 1 = 23457
(ii) 0 ÷ 97
By division
0 ÷ 97 = 0
(iii) 476 + (840 ÷ 84)
On further calculation
476 + (840 ÷ 84) = 476 + 10
= 486
(iv) 964 – (425 ÷ 425)
On further calculation
964 – (425 ÷ 425) = 964 – 1
= 963
(v) (2758 ÷ 2758) – (2758 ÷ 2758)
On further calculation
(2758 ÷ 2758) – (2758 ÷ 2758) = 1 – 1
= 0
(vi) 72450 ÷ (583 – 58)
On further calculation
72450 ÷ (583 – 58) = 72450 ÷ 525
= 138
3. Which of the following statements are true:
(i) 10 ÷ (5 × 2) = (10 ÷ 5) × (10 ÷ 2)
(ii) (35 – 14) ÷ 7 = 35 ÷ 7 – 14 ÷ 7
(iii) 35 – 14 ÷ 7 = 35 ÷ 7 – 14 ÷ 7
(iv) (20 – 5) ÷ 5 = 20 ÷ 5 – 5
(v) 12 × (14 ÷ 7) = (12 × 14) ÷ (12 × 7)
(vi) (20 ÷ 5) ÷ 2 = (20 ÷ 2) ÷ 5
Solution:
(i) False.
We know that
LHS = 10 ÷ (5 × 2)
So we get
= 10 ÷ 10
= 1
RHS = (10 ÷ 5) × (10 ÷ 2)
So we get
= 2 × 5
= 10
(ii) True.
We know that
LHS = (35 – 14) ÷ 7
So we get
= 21 ÷ 7
= 3
RHS = 35 ÷ 7 – 14 ÷ 7
So we get
= 5 – 2
= 3
(iii) False.
We know that
LHS = 35 – 14 ÷ 7
So we get
= 35 – 2
= 33
RHS = 35 ÷ 7 – 14 ÷ 7
So we get
= 5 – 2
= 3
(iv) False.
We know that
LHS = (20 – 5) ÷ 5
So we get
= 15 ÷ 5
= 3
RHS = 20 ÷ 5 – 5
So we get
= 4 – 5
= – 1
(v) False.
We know that
LHS = 12 × (14 ÷ 7)
So we get
= 12 × 2
= 24
RHS = (12 × 14) ÷ (12 × 7)
So we get
= 168 ÷ 84
= 2
(vi) True.
We know that
LHS = (20 ÷ 5) ÷ 2
So we get
= 4 ÷ 2
= 2
RHS = (20 ÷ 2) ÷ 5
So we get
= 10 ÷ 5
= 2
4. Divide and check the quotient and remainder:
(i) 7772 ÷ 58
(ii) 6906 ÷ 35
(iii) 16135 ÷ 875
(iv) 16025 ÷ 1000
Solution:
(i) 7772 ÷ 58
So we get 7772 ÷ 58 = 134
By verifying
We know that
Dividend = Divisor × Quotient + Remainder
By substituting values
7772 = 58 × 134 + 0
So we get
7772 = 7772
LHS = RHS
(ii) 6906 ÷ 35
So we get quotient = 197 and remainder = 11
By verifying
We know that
Dividend = Divisor × Quotient + Remainder
By substituting values
6906 = 35 × 197 + 11
On further calculation
6906 = 6895 + 11
We get
6906 = 6906
LHS = RHS
(iii) 16135 ÷ 875
So we get quotient = 18 and remainder = 385
By verifying
We know that
Dividend = Divisor × Quotient + Remainder
By substituting values
16135 = 875 × 18 + 385
On further calculation
16135 = 15750 + 385
We get
16135 = 16135
LHS = RHS
(iv) 16025 ÷ 1000
So we get quotient = 16 and remainder = 25
By verifying
We know that
Dividend = Divisor × Quotient + Remainder
By substituting values
16025 = 1000 × 16 + 25
On further calculation
16025 = 16000 + 25
We get
16025 = 16025
LHS = RHS
5. Find a number which when divided by 35 gives the quotient 20 and remainder 18.
Solution:
We know that
Dividend = Divisor × Quotient + Remainder
By substituting values
Dividend = 35 × 20 + 18
On further calculation
Dividend = 700 + 18
So we get
Dividend = 718
6. Find the number which when divided by 58 gives a quotient 40 and remainder 31.
Solution:
We know that
Dividend = Divisor × Quotient + Remainder
By substituting values
Dividend = 58 × 40 + 31
On further calculation
Dividend = 2320 + 31
So we get
Dividend = 2351
7. The product of two numbers is 504347. If one of the numbers is 1591, find the other.
Solution:
The product of two numbers = 504347
One of the numbers = 1591
Consider A as the number
A × 1591 = 504347
So by division
A = 317
8. On dividing 59761 by a certain number, the quotient is 189 and the remainder is 37. Find the divisor.
Solution:
It is given that
Dividend = 59761
Quotient = 189
Remainder = 37
Consider Divisor = A
We know that
Dividend = Divisor × Quotient + Remainder
By substituting values
59761 = A × 189 + 37
On further calculation
59761 – 37 = A × 189
So we get
59724 = A × 189
By division
A = 316
9. On dividing 55390 by 299, the remainder is 75. Find the quotient.
Solution:
It is given that
Dividend = 55390
Divisor = 299
Remainder = 75
Consider Quotient = A
We know that
Dividend = Divisor × Quotient + Remainder
By substituting values
55390 = 299 × A + 75
On further calculation
55390 – 75 = A × 299
So we get
55315 = A × 299
By division
A = 185
RD Sharma Solutions for Class 6 Maths Chapter 4 – Operations on Whole Numbers Exercise 4.4
RD Sharma Solutions Class 6 Maths Chapter 4 Operations on Whole Numbers Exercise 4.4 help students gain knowledge about division and the important properties which are used in solving exercise-wise problems.
Key Features of RD Sharma Solutions for Class 6 Maths Chapter 4: Operations on Whole Numbers Exercise 4.4
- Problem-solving and logical thinking abilities are improved among students.
- The solutions help students solve problems more accurately and in an efficient manner.
- They aid students in grasping the important methods which are used in solving problems based on the CBSE guidelines.
- The solutions can be referred to while solving exercise-wise problems in the RD Sharma textbook for the latest syllabus.
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