Integers Class 6 RS Aggarwal Exe-3B Goyal Brothers ICSE Maths Solutions. We provide step by step Solutions of Integers for ICSE Class-6 Foundation RS Aggarwal Mathematics of Goyal Brothers Prakashan . Visit official Website CISCE for detail information about ICSE Board Class-6 Mathematics.
Integers Class 6 RS Aggarwal Exe-3B Operation on Integers Goyal Brothers Prakashan ICSE Foundation Maths Solutions
| Board | ICSE |
| Publications | Goyal brothers Prakashan |
| Subject | Maths |
| Class | 6th |
| Ch-3 | Integers |
| Writer | RS Aggrawal |
| Book Name | Foundation |
| Topics | Solution of Exe-3B Operation on Integers |
| Academic Session | 2023 – 2024 |
Operation on Integers Using Number Lines
Integers Class 6 RS Aggarwal Exe-3B Goyal Brothers Prakashan ICSE Foundation Maths Solutions.
Page- 55,56
Integers Exercise- 3B Operation on Integers Using Number Line
Que-1: Find the sum :
(i) 16 + 21 (ii) (-16) + 21
(iii) 16 + (-21) (iv) (-25) + (-18)
(v) (-32) + (-47) (vi) 54 + (-89)
(vii) (-38) + 45 (viii) 96 + (-103)
(ix) (-150) + (-15)
Solution- (i) 37
(ii) -16 + 21 = 5
(iii) 16 – 21 = -5
(iv) -25 – 18 = -43
(v) -32 – 47 = -79
(vi) 54 – 89 = -35
(vii) -38 + 45 = 7
(viii) 96 – 103 = -7
(ix) -150 – 15 = -165
Que-2: Write the additive inverse of :
(i) 34 (ii) -58 (iii) 0
(iv) -1 (v) 170
Solution- (i) -34
(ii) 58
(iii) 0
(iv) 1
(v) -170
Que-3: Write the successor of :
(i) 101 (ii) -47 (iii) -1
(iv) -80 (v) -301
Solution- (i) 102
(ii) -46
(iii) 0
(iv) -79
(v) -300
Que-4: Write the predecessor of :
(i) 40 (ii) -32 (iii) -70
(iv) -91 (v) 0
Solution- (i) 39
(ii) -33
(iii) -71
(iv) -92
(v) -1
Que-5: Find the difference :
(i) (15) – (21) (ii) (-29) – (9) (iii) 70 – (-8)
(iv) 24 – (-24) (v) (-36) – (64) (vi) 0 – (-20)
(vii) (-63) – (-7) (viii) (-80) – (-20) (ix) (-12) – (-71)
Solution- (i) 15 – 21 = -6
(ii) -29 – 9 = -38
(iii) 70 + 8 = 78
(iv) 24 + 24 = 48
(v) -36 – 64 = -100
(vi) 0 + 20 = 20
(vii) -63 + 7 = -56
(viii) -80 + 20 = -60
(ix) -12 + 71 = 59
Que-6: Subtract :
(i) -180 from 180 (ii) 75 from -75 (iii) -630 from -70
(iv) -95 from 0 (v) -90 from -1 (vi) -16 from -25
Solution- (i) 180 – (-180) = 180 + 180 = 360
(ii) (-75) – 75 = -150
(iii) (-70) – (-630) = 560
(iv) 0 – (-95) = 95
(v) (-1) – (-90) = 89
(vi) (-25) – (-16) = -9
Que-7: Fill in the blanks :
(i) (+23) + (…….) = 0 (ii) (-13) + (……..) = +20
(iii) (-14) + (…….) = -34 (iv) (-20) + (…….) = -8
(v) (-30) – (……..) = -14 (vi) (……..) – (-25) = 2
Solution- (i) -23
(ii) 33
(iii) -20
(iv) 12
(v) -16
(vi) -23
Que-8: Use number line to find :
(i) 5 + 4 (ii) 4 + (-6) (iii) (-4) + 8
(iv) (-5) + 3 (v) (-3) + (-5) (vi) (-6) + (-3)
(vii) 5 – (-2) (viii) (-4) – 5 (ix) 4 – (-4)
Solution- (i) 9
(ii) -2
(iii) 4
(iv) -2
(v) -8
(vi) -9
(vii) 7
(viii) -9
(ix) 8
Que-9: State whether the statement is true or false :
(i) The sum of two integer is always an integer.
(ii) The difference of two integer is always an integer.
(iii) The absolute value of every integer is a positive integer.
(iv) 0 is the smallest positive integer.
(v) Every whole number is an integer.
(vi) The sum of two integers can never be zero.
Solution- (i) True
(ii) True
(iii) True
(iv) False
(v) True
(vi) False
Que-10: What should be added to 15 to get (-15)?
Solution- So, you need to add -30 to 15 to get (-15).
15 + (-30) = (-15)
Que-11: What should be subtracted from (-3) to get +18?
Solution- So, you need to subtract -21 from (-3) to get +18.
(-3) – (-21) = 18
Que-12: The sum of two integers is -23. If one of them is 12, find the other.
Solution- Let’s denote the other integer as x.
Given that the sum of the two integers is -23, we can write the equation as:
12 + x = −23
To find the value of x, we can subtract 12 from both sides:
x = −23 − 12
x = −35
So, the other integer is -35.
Que-13: While playing children’s cards, Amit lost 70 points in the first game, 50 in the second game and 35 in the third game. He gained 60 in the fourth game and 80 in the fifth game. What was his net loss or gain?
Solution- Points lost:
1st game: 70 points lost
2nd game: 50 points lost
3rd game: 35 points lost
Points gained:
4th game: 60 points gained
5th game: 80 points gained
Now, let’s calculate the net loss or gain:
Net loss or gain = (Points gained) – (Points lost)
Net loss or gain = (60+80) − (70+50+35)
Net loss or gain = 140−155
Net loss or gain = −15
So, Amit’s net loss is 15 points.
Que-14: On one day on a hill, the temperature at 8 p.m. was 2°C but at mid-night that day, it fell down to -3°C. By how much degrees did the temperature fall?
Solution-Temperature fall = Temperature at 8 p.m. – Temperature at midnight
Temperature fall = 2°C − (−3°C)
Temperature fall = 2°C + 3°C
Temperature fall = 5°C
So, the temperature fell by 5 degrees Celsius.
Que-15: A car travelled east of Delhi by 100 km and then to the west of it by 130 km. How far from Delhi was the car finally?
Solution- When the car travels east of Delhi by 100 km, it moves away from Delhi. Then, when it travels west of Delhi by 130 km, it moves back toward Delhi. So, we subtract the distance travelled west from the distance travelled east to find the net distance from Delhi.
Net distance from Delhi = Distance east of Delhi – Distance west of Delhi
Net distance from Delhi = 100 km−130 km
Net distance from Delhi = −30 km
The negative sign indicates that the car is 30 km west of Delhi.
So, the car is 30 km away from Delhi, in the west direction
— : end of Integers Class 6 RS Aggarwal Exe-3B Goyal Brothers Prakashan ICSE Foundation Maths Solutions :–
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