In this chapter, we provide RD Sharma Class 6 ex 5.2 Solutions Chapter 5 Negative Numbers and Integers for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Class 6 ex 5.2 Solutions Chapter 5 Negative Numbers and Integers Maths pdf, Now you will get step by step solution to each question.

RD Sharma Solutions for Class 6 Chapter 5 Negative Numbers and Integers Ex 5.2 Download PDF

Chapter 5: Negative Numbers and Integers – Exercise 5.2

Negative Numbers and Integers – Exercise – 5.2 – Q.1

Ans. 

(i) 5 + (-2)

We begin at 0 and first more five units to the right of zero to reach ar a which represents +5. The second number 2 is negative. So, we move 2 units to the left of A to reach at B which represents 3.

Thus, we have = 5 + (-2) = 3.

(ii) (-9) + 4.

We begin at 0 and first move nine units to the left of zero to reach at A. Which represents – 9. The second number 4 is positive. So we over 4 units to the right of A to reach at B. Which represents -5.

Thus, we have.

-9 + 4 = -5.

(iii) (-3) + (-5).

We begin at 0 and first move three units to the left of zero to reach at A which represents -3. The second number +5 is negative. So we move 5 units to the left of A to reach at b which represents -8.

Thus we have = (-3) + (-5) = -3 -5 = -8.

(iv) (-1) + (-2) + 2.

We begin at zero and first move one unit to the left of zero to reach at point A which represents -1. The second number 2 is negative.

So we move 2 units to the left of A to reach at B which represents -3. The Third number is 2 positive. So we move 2 units to the right of B. which is -1.

(-1) + (-2) + 2 = (-1) + 2 -2 = -1.

(v) 6 + (-6)

Thus we have 6 + (-6) = 0

(vi) (-2) + 5 + (-a).

Thus we have = -2 + 5 + (-9)

= 5 – 2 + (-9)

= 3 – 9

= -6.

Negative Numbers and Integers – Exercise – 5.2 – Q.2

Ans. 

(i) – 557 and 488

The integers are to be added are of the unlike signs.Therefore to add them we find the difference of their absolute values and assign the sign of the addend having greater absolute value

(- 557) and 488 = |- 557| – |488|

= 557 – 488

= – 69.

(ii) – 522 + (-160) = – 522 – 160

= – 682

(iii) 2567 and -325

2567 + (-325) = (2567) – (-325)

= 2567 and -325

2567 + (-325) = (2567) – (-325)

= 2567 – 325

= 2242

(iv) – 10025 and 139

-10025 + 139 = [-10025] + [139]

= -10025 + 139

= -9886.

(v) 2567+ (-2578) = 2547 – 2548

= -1.

(vi) 2884 + (-2884) = 2884 – 2884 = 0

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