RS Aggarwal 2021-2022 for Class 6 Maths Chapter 3 - Integer
RS Aggarwal Class 6 Math Solution Chapter 4 Integer Exercise 4C is available here. RS Aggarwal Class 6 Math Solutions are solved by expert teachers in step by step, which help the students to understand easily. RS Aggarwal textbooks are responsible for a strong foundation in Maths. These textbook solutions help students in exams as well as their daily homework routine. The solutions included are easy to understand, and each step in the solution is described to match the students’ understanding.
Rs Aggarwal Class 6 Math Solution Chapter 4 Integer
Exercise 4C
1: Subtract:
(i) 18 from − 34
(ii) − 15 from 25
(iii) − 28 from − 43
(iv) 68 from − 37
(v) 219 from 0
(vi) − 92 from 0
(vii) − 135 from − 250
(viii) − 2768 from − 287
(ix) 6240 from − 271
(x) − 3012 from 6250
Solution:
(i) −34 − 18
= −52
(ii) 25 − ( −15)
= 25 + 15
= 40
(iii) −28 from −43
= −43 − ( −28)
= −43 + 28
= −15
(iv) 68 from −37
= −37 − 68
= −105
(v) 219 from 0
= 0 − 219
= −219
(vi) −92 from 0
= 0 − ( −92)
= 0 + 92
= 92
(vii) −135 from −250
= −250 − ( −135)
= −250 + 135
= −115
(viii) −2768 from −287
= −287 − ( −2768)
= −287 + 2768
= 2481
(ix) 6240 from −271
= −271 − (6240)
= −271 − 6240
= −6511
(x) −3012 from 6250
= 6250 − (−3012)
= 6250 + 3012
= 9262
2: Subtract the sum of − 1050 and 813 from 23.
Solution:
First, we find the Sum of −1050 and 813:
= −1050 + 813
= −237
Now, Subtracting the sum of −1050 and 813 from −23:
= −23 − (−237)
= −23 +237
= 214
3: Subtract the sum of − 250 and 138 from the sum of 136 and − 272
Solution:
First, we find the Sum of 136 and − 272
= 138 + (−250)
= 138 − 250
= −112
Then, we find the Sum of − 250 and 138
= 136 + (−272)
= 136 − 272
= −136
Now, Subtracting the sum of −250 and 138 from the sum of 136 and −272:
= −136 − (−112)
= −136 + 112
= −24
4: From the sum of 33 and − 47, subtract − 84.
Solution:
The sum of 33 and −47:
= 33 + (−47)
= 33 − 47
= −14
Now, Subtracting −84 from −14:
= −14 − (−84)
= −14 + 84
= 70
5: Add − 36 to the difference of − 8 and − 68
Solution:
First, we find the difference of − 8 and − 68
= − 8 − (−68)
= − 8 + 68
= 60
Now, Adding − 36 to 60
= −36 + 60
= 24
6: Simplify:
(i) [37 − (− 8)] + [11 − ( − 30)]
(ii) − 13 − (− 17)] + [− 22 − ( − 40)]
Solution:
(i) [37 − (−8)] + [11 − (−30)]
= (37 + 8) + (11 + 30)
= 45 + 41
= 86
(ii) [−13 − (−17) + [−22 − (−40)]
= (−13 +17) + ( − 22 + 40)
= 4 + 18
= 22
7: Find 34 − ( − 72) and ( − 72) − 34. Are they equal?
Solution:
No, they are not equal.
⇒ 34 − (−72) = (−72) − 34
⇒ 34 + 72 = −72 − 34
⇒ 106 ≠ −106. So, they are not equal.
8: The sum of two integers is − 13. If one of the numbers is 170, find the other.
Solution:
The sum of two integers = – 13
One number =170
The other number = ( The sum of two integers ) – ( One number )
= – 13 – 170
= ( – 13) + ( – 170)
= – 183
Solution:
The sum of two integers = 65
One number = – 47
The other number = ( The sum of two integers ) – ( One number )
= 65 – (– 47 )
= 65 + 47
= 112
10: Which of the following statements are true and which are false?
(i) The sum of two integers is always an integer
(ii) The difference of two integers is always an integer.
(iii) − 14 > − 8 − ( − 7).
(iv) − 5 − 2 > 8.
(v) ( − 7) − 3 = ( − 3) − ( − 7)
Solution:
(i) True :
(ii) True
(iii) False
– 14 > – 8 – ( – 7)
– 14 > – 8 + 7 .
– 14 > – 1 which is not true.
Smaller negative number is greater than larger negative number.
(iv) True
− 5 − 2 > – 8
– 7 > – 8
Smaller negative number is greater than larger negative number.
(v) False
( − 7) − 3 = ( − 3) − ( − 7)
− 10 = − 3 + 7
− 10 = 4, which is not true.
Positive number is greater than negative number.
11: Point A is on a mountain, which is 5200 metres above sea level and point B is in a mine which is 39600 metres below sea level. Find the vertical distance between A and B.
Solution:
We know that height above the sea level count as positive and depth below the sea level count as negative.
∴ Height of point A from sea level = 5700 m
Depth of point B from sea level = − 39600 m
Distance between A and B = Distance of point A from sea level − Distance of point B from sea level
= 5700 − ( − 39600 )
= 5700 + 39600
= 45300 m
12: On a day in Srinagar, the temperature at 6 pm. was 1°C but at midnight that day, it dropped to − 4°C. By how many degrees Celsius did the temperature fall?
Solution:
Temperature at 6 p.m. = 1°C
Temperature at mid-night = − 4°C
Required temperature fall = 1°C − ( − 4°C)
= 1°C + 4°C
= 5°C.
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