RD Sharma Solutions for Class 6 Maths Exercise 5.2 PDF
Mathematics is one of the challenging subjects which requires regular practice and strong conceptual knowledge. The students can improve their logical thinking and problem-solving abilities by using RD Sharma Solutions. The experts at BYJU’S solve the problems from the RD Sharma textbook following the latest CBSE syllabus. The primary objective of creating solutions is to help students perform well in the final exam. The solutions created by the subject-matter experts are accurate. RD Sharma Solutions for Class 6 Maths Chapter 5 Negative Numbers and Integers Exercise 5.2 are provided here.
RD Sharma Solutions for Class 6 Maths Chapter 5: Negative Numbers and Integers Exercise 5.2
Access answers to Maths RD Sharma Solutions for Class 6 Chapter 5: Negative Numbers and Integers Exercise 5.2
1. Draw a number line and represent each of the following on it:
(i) 5 + (-2)
(ii) (-9) + 4
(iii) (-3) + (-5)
(iv) 6 + (-6)
(v) (-1) + (-2) + 2
(vi) (-2) + 7 + (-9)
Solution:
(i) 5 + (-2)
From 0 move towards right of first five units to obtain + 5
So the second number is – 2 so move 2 units towards left of + 5 we get + 3
Therefore, 5 + (-2) = 3.
(ii) (-9) + 4
From 0 move towards left of nine units to obtain – 9
So the second number is 4 so move 4 units towards right of – 9 we get – 5
Therefore, (-9) + 4 = – 5.
(iii) (-3) + (-5)
From 0 move towards left of three units to obtain – 3
So the second number is – 5 so move 5 units towards left of – 3 we get – 8
Therefore, (-3) + (-5) = – 8.
(iv) 6 + (-6)
From zero move towards right of six units to obtain 6
So the second number is – 6 so move 6 units towards left of 6 we get 0
Therefore, 6 + (-6) = 0.
(v) (-1) + (-2) + 2
From zero move towards left of one unit to obtain – 1
So the second number is – 2 so move 2 units towards left of – 1 we get – 3
The third number is 2 so move 2 units towards right of – 3 we get – 1
Therefore, (-1) + (-2) + 2 = – 1.
(vi) (-2) + 7 + (-9)
From zero move towards left of two units to obtain – 2
So the second number is 7 so move 7 units towards right of – 2 we get 5
The third number is – 9 so move 9 units towards left of 5 we get – 4
Therefore, (-2) + 7 + (-9) = – 4.
2. Find the sum of
(i) -557 and 488
(ii) -522 and -160
(iii) 2567 and – 325
(iv) -10025 and 139
(v) 2547 and -2548
(vi) 2884 and -2884
Solution:
(i) -557 and 488
We get
-557 + 488
It can be written as
|-557| – |488| = 557 – 488 = 69.
(ii) -522 and -160
We get
-522 + (-160)
It can be written as
-522 – 160 = – 682
(iii) 2567 and – 325
We get
2567 + (-325)
It can be written as
2567 – 325 = 2242
(iv) -10025 and 139
We get
-10025 + 139
It can be written as
-10025 + 139 = -9886
(v) 2547 and -2548
We get
2547 + (-2548)
It can be written as
2547 – 2548 = -1
(vi) 2884 and -2884
We get
2884 + (-2884)
It can be written as
2884 – 2884 = 0
RD Sharma Solutions for Class 6 Maths Chapter 5 – Negative Numbers and Integers Exercise 5.2
RD Sharma Solutions Class 6 Maths Chapter 5 Negative Numbers and Integers Exercise 5.2 consist of important concepts like the addition of integers and the procedure followed in solving problems.
Key Features of RD Sharma Solutions for Class 6 Maths Chapter 5: Negative Numbers and Integers Exercise 5.2
- RD Sharma Solutions help students to revise important concepts for the final exam.
- The students can self-assess their knowledge about the concepts which are covered in each exercise.
- Solving a huge number of problems improves efficiency in answering questions.
- They help students to know about the various types of problems that would appear in the annual exam.
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