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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.
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.
(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
= 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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Chapter 5: Negative Numbers and Integers –...
