Open App

Class 6 Exam > Class 6 Notes > Mathematics (Maths) Class 6 > RD Sharma Solutions: Negative Numbers and Integers (Exercise 5.4)

Negative Numbers and Integers (Exercise 5.4) RD Sharma Solutions | Mathematics (Maths) Class 6 PDF Download

Download, print and study this document offline

Please wait while the PDF view is loading

 Page 1


 
 
 
 
 
 
Exercise 5.4                                                                             page: 5.17 
1. Subtract the first integer from the second in each of the following: 
(i) 12, -5 
(ii) – 12, 8 
(iii) – 225, - 135 
(iv) 1001, 101 
(v) – 812, 3126 
(vi) 7560, - 8 
(vii) – 3978, - 4109 
(viii) 0, - 1005 
Solution: 
 
(i) 12, -5 
So by subtracting the first integer from the second 
-5 – 12 = - 17 
 
(ii) – 12, 8 
So by subtracting the first integer from the second 
8 – (-12) = 8 + 12 = 20 
 
(iii) – 225, - 135 
So by subtracting the first integer from the second 
-135 – (-225) = 225 – 135 = 90 
 
(iv) 1001, 101 
So by subtracting the first integer from the second 
101 – 1001 = - 900 
 
(v) – 812, 3126 
So by subtracting the first integer from the second 
3126 – (-812) = 3126 + 812 = 3938 
 
(vi) 7560, - 8 
So by subtracting the first integer from the second 
-8 – 7560 = - 7568 
 
(vii) – 3978, - 4109 
So by subtracting the first integer from the second 
-4109 – (-3978) = - 4109 + 3978 = -131 
 
(viii) 0, - 1005 
So by subtracting the first integer from the second 
-1005 – 0 = - 1005 
 
2. Find the value of: 
(i) – 27 – (- 23) 
(ii) – 17 – 18 – (-35) 
(iii) – 12 – (-5) – (-125) + 270 
(iv) 373 + (-245) + (-373) + 145 + 3000 
Page 2


 
 
 
 
 
 
Exercise 5.4                                                                             page: 5.17 
1. Subtract the first integer from the second in each of the following: 
(i) 12, -5 
(ii) – 12, 8 
(iii) – 225, - 135 
(iv) 1001, 101 
(v) – 812, 3126 
(vi) 7560, - 8 
(vii) – 3978, - 4109 
(viii) 0, - 1005 
Solution: 
 
(i) 12, -5 
So by subtracting the first integer from the second 
-5 – 12 = - 17 
 
(ii) – 12, 8 
So by subtracting the first integer from the second 
8 – (-12) = 8 + 12 = 20 
 
(iii) – 225, - 135 
So by subtracting the first integer from the second 
-135 – (-225) = 225 – 135 = 90 
 
(iv) 1001, 101 
So by subtracting the first integer from the second 
101 – 1001 = - 900 
 
(v) – 812, 3126 
So by subtracting the first integer from the second 
3126 – (-812) = 3126 + 812 = 3938 
 
(vi) 7560, - 8 
So by subtracting the first integer from the second 
-8 – 7560 = - 7568 
 
(vii) – 3978, - 4109 
So by subtracting the first integer from the second 
-4109 – (-3978) = - 4109 + 3978 = -131 
 
(viii) 0, - 1005 
So by subtracting the first integer from the second 
-1005 – 0 = - 1005 
 
2. Find the value of: 
(i) – 27 – (- 23) 
(ii) – 17 – 18 – (-35) 
(iii) – 12 – (-5) – (-125) + 270 
(iv) 373 + (-245) + (-373) + 145 + 3000 
 
 
 
 
 
 
(v) 1 + (-475) + (-475) + (-475) + (-475) + 1900 
(vi) (-1) + (-304) + 304 + 304 + (-304) + 1 
Solution: 
 
(i) – 27 – (- 23) 
So we get 
= - 27 + 23 
On further calculation 
= 23 – 27  
We get 
= - 4 
 
(ii) – 17 – 18 – (-35) 
So we get 
= - 35 + 35 
On further calculation 
= 0  
 
(iii) – 12 – (-5) – (-125) + 270 
So we get 
= - 12 + 5 + 125 + 270 
On further calculation 
= 400 – 12 
We get 
= 388 
 
(iv) 373 + (-245) + (-373) + 145 + 3000 
So we get 
= 373 – 245 – 373 + 145 + 3000 
On further calculation 
= 3145 + 373 – 373 – 245 
We get 
= 3145 – 245 
By subtraction 
= 2900 
 
(v) 1 + (-475) + (-475) + (-475) + (-475) + 1900 
So we get 
= 1 – 950 – 950 + 1900 
On further calculation 
= 1900 + 1 – 1900 
We get 
= 1 
 
(vi) (-1) + (-304) + 304 + 304 + (-304) + 1 
So we get 
= - 1 + 1 – 304 + 304 – 304 + 304  
On further calculation  
= 0 
 
Page 3


 
 
 
 
 
 
Exercise 5.4                                                                             page: 5.17 
1. Subtract the first integer from the second in each of the following: 
(i) 12, -5 
(ii) – 12, 8 
(iii) – 225, - 135 
(iv) 1001, 101 
(v) – 812, 3126 
(vi) 7560, - 8 
(vii) – 3978, - 4109 
(viii) 0, - 1005 
Solution: 
 
(i) 12, -5 
So by subtracting the first integer from the second 
-5 – 12 = - 17 
 
(ii) – 12, 8 
So by subtracting the first integer from the second 
8 – (-12) = 8 + 12 = 20 
 
(iii) – 225, - 135 
So by subtracting the first integer from the second 
-135 – (-225) = 225 – 135 = 90 
 
(iv) 1001, 101 
So by subtracting the first integer from the second 
101 – 1001 = - 900 
 
(v) – 812, 3126 
So by subtracting the first integer from the second 
3126 – (-812) = 3126 + 812 = 3938 
 
(vi) 7560, - 8 
So by subtracting the first integer from the second 
-8 – 7560 = - 7568 
 
(vii) – 3978, - 4109 
So by subtracting the first integer from the second 
-4109 – (-3978) = - 4109 + 3978 = -131 
 
(viii) 0, - 1005 
So by subtracting the first integer from the second 
-1005 – 0 = - 1005 
 
2. Find the value of: 
(i) – 27 – (- 23) 
(ii) – 17 – 18 – (-35) 
(iii) – 12 – (-5) – (-125) + 270 
(iv) 373 + (-245) + (-373) + 145 + 3000 
 
 
 
 
 
 
(v) 1 + (-475) + (-475) + (-475) + (-475) + 1900 
(vi) (-1) + (-304) + 304 + 304 + (-304) + 1 
Solution: 
 
(i) – 27 – (- 23) 
So we get 
= - 27 + 23 
On further calculation 
= 23 – 27  
We get 
= - 4 
 
(ii) – 17 – 18 – (-35) 
So we get 
= - 35 + 35 
On further calculation 
= 0  
 
(iii) – 12 – (-5) – (-125) + 270 
So we get 
= - 12 + 5 + 125 + 270 
On further calculation 
= 400 – 12 
We get 
= 388 
 
(iv) 373 + (-245) + (-373) + 145 + 3000 
So we get 
= 373 – 245 – 373 + 145 + 3000 
On further calculation 
= 3145 + 373 – 373 – 245 
We get 
= 3145 – 245 
By subtraction 
= 2900 
 
(v) 1 + (-475) + (-475) + (-475) + (-475) + 1900 
So we get 
= 1 – 950 – 950 + 1900 
On further calculation 
= 1900 + 1 – 1900 
We get 
= 1 
 
(vi) (-1) + (-304) + 304 + 304 + (-304) + 1 
So we get 
= - 1 + 1 – 304 + 304 – 304 + 304  
On further calculation  
= 0 
 
 
 
 
 
 
 
3. Subtract the sum of – 5020 and 2320 from – 709. 
Solution: 
 
We know that the sum of -5020 and 2320 is 
-5020 + 2320 
It can be written as 
= 2320 – 5020 
So we get 
= - 2700 
Subtracting from – 709 we get 
= - 709 – (-2700) 
We get 
= - 709 + 2700 
By subtraction 
= 1991 
 
4. Subtract the sum of – 1250 and 1138 from the sum of 1136 and - 1272. 
Solution: 
 
We know that the sum of – 1250 and 1138 is 
-1250 + 1138 
It can be written as 
= 1138 – 1250 
So we get 
= - 112 
 
We know that the sum of 1136 and – 1272 is 
1136 – 1272 = - 136 
So we get 
-136 – (-112) = - 136 + 112 = -24 
 
5. From the sum of 233 and – 147, subtract – 284. 
Solution: 
 
We know that the sum of 233 and – 147 is 
233 – 147 = 86 
Subtracting – 284 we get 
86 – (-284) = 86 + 284 = 370 
 
6. The sum of two integers is 238. If one of the integers is – 122, determine the other. 
Solution: 
 
It is given that 
Sum of two integers = 238 
One of the integers = - 122 
So the other integer = - (-122) + 238 
On further calculation 
Other integer = 238 + 122 = 360 
 
7. The sum of two integers is – 223. If one of the integers is 172, find the other. 
Page 4


 
 
 
 
 
 
Exercise 5.4                                                                             page: 5.17 
1. Subtract the first integer from the second in each of the following: 
(i) 12, -5 
(ii) – 12, 8 
(iii) – 225, - 135 
(iv) 1001, 101 
(v) – 812, 3126 
(vi) 7560, - 8 
(vii) – 3978, - 4109 
(viii) 0, - 1005 
Solution: 
 
(i) 12, -5 
So by subtracting the first integer from the second 
-5 – 12 = - 17 
 
(ii) – 12, 8 
So by subtracting the first integer from the second 
8 – (-12) = 8 + 12 = 20 
 
(iii) – 225, - 135 
So by subtracting the first integer from the second 
-135 – (-225) = 225 – 135 = 90 
 
(iv) 1001, 101 
So by subtracting the first integer from the second 
101 – 1001 = - 900 
 
(v) – 812, 3126 
So by subtracting the first integer from the second 
3126 – (-812) = 3126 + 812 = 3938 
 
(vi) 7560, - 8 
So by subtracting the first integer from the second 
-8 – 7560 = - 7568 
 
(vii) – 3978, - 4109 
So by subtracting the first integer from the second 
-4109 – (-3978) = - 4109 + 3978 = -131 
 
(viii) 0, - 1005 
So by subtracting the first integer from the second 
-1005 – 0 = - 1005 
 
2. Find the value of: 
(i) – 27 – (- 23) 
(ii) – 17 – 18 – (-35) 
(iii) – 12 – (-5) – (-125) + 270 
(iv) 373 + (-245) + (-373) + 145 + 3000 
 
 
 
 
 
 
(v) 1 + (-475) + (-475) + (-475) + (-475) + 1900 
(vi) (-1) + (-304) + 304 + 304 + (-304) + 1 
Solution: 
 
(i) – 27 – (- 23) 
So we get 
= - 27 + 23 
On further calculation 
= 23 – 27  
We get 
= - 4 
 
(ii) – 17 – 18 – (-35) 
So we get 
= - 35 + 35 
On further calculation 
= 0  
 
(iii) – 12 – (-5) – (-125) + 270 
So we get 
= - 12 + 5 + 125 + 270 
On further calculation 
= 400 – 12 
We get 
= 388 
 
(iv) 373 + (-245) + (-373) + 145 + 3000 
So we get 
= 373 – 245 – 373 + 145 + 3000 
On further calculation 
= 3145 + 373 – 373 – 245 
We get 
= 3145 – 245 
By subtraction 
= 2900 
 
(v) 1 + (-475) + (-475) + (-475) + (-475) + 1900 
So we get 
= 1 – 950 – 950 + 1900 
On further calculation 
= 1900 + 1 – 1900 
We get 
= 1 
 
(vi) (-1) + (-304) + 304 + 304 + (-304) + 1 
So we get 
= - 1 + 1 – 304 + 304 – 304 + 304  
On further calculation  
= 0 
 
 
 
 
 
 
 
3. Subtract the sum of – 5020 and 2320 from – 709. 
Solution: 
 
We know that the sum of -5020 and 2320 is 
-5020 + 2320 
It can be written as 
= 2320 – 5020 
So we get 
= - 2700 
Subtracting from – 709 we get 
= - 709 – (-2700) 
We get 
= - 709 + 2700 
By subtraction 
= 1991 
 
4. Subtract the sum of – 1250 and 1138 from the sum of 1136 and - 1272. 
Solution: 
 
We know that the sum of – 1250 and 1138 is 
-1250 + 1138 
It can be written as 
= 1138 – 1250 
So we get 
= - 112 
 
We know that the sum of 1136 and – 1272 is 
1136 – 1272 = - 136 
So we get 
-136 – (-112) = - 136 + 112 = -24 
 
5. From the sum of 233 and – 147, subtract – 284. 
Solution: 
 
We know that the sum of 233 and – 147 is 
233 – 147 = 86 
Subtracting – 284 we get 
86 – (-284) = 86 + 284 = 370 
 
6. The sum of two integers is 238. If one of the integers is – 122, determine the other. 
Solution: 
 
It is given that 
Sum of two integers = 238 
One of the integers = - 122 
So the other integer = - (-122) + 238 
On further calculation 
Other integer = 238 + 122 = 360 
 
7. The sum of two integers is – 223. If one of the integers is 172, find the other. 
 
 
 
 
 
 
Solution: 
 
It is given that 
Sum of two integers = - 223 
One of the integers = 172 
So the other integer = - 223 – 172 = - 395 
 
8. Evaluate the following: 
(i) – 8 – 24 + 31 – 26 – 28 + 7 + 19 – 18 – 8 + 33 
(ii) – 26 – 20 + 33 – (-33) + 21 + 24 – (-25) – 26 – 14 – 34 
Solution: 
 
(i) – 8 – 24 + 31 – 26 – 28 + 7 + 19 – 18 – 8 + 33 
We get 
= - 8 – 24 – 26 – 28 – 18 – 8 + 31 + 7 + 19 + 33 
On further calculation 
= - 32 – 26 – 28 – 26 + 38 + 19 + 33 
It can be written as 
= 38 – 32 – 26 – 28 + 33 – 26 + 19 
So we get 
= 6 – 26 – 28 + 7 + 19 
By calculation 
= 6 – 28 – 26 + 26 
= 6 – 28 
By subtraction 
= - 22 
 
(ii) – 26 – 20 + 33 – (-33) + 21 + 24 – (-25) – 26 – 14 – 34 
We get 
= - 46 + 33 + 33 + 21 + 24 + 25 – 26 – 14 – 34 
On further calculation 
= - 46 + 66 + 21 + 24 + 25 + (-74) 
It can be written as 
= - 46 + 66 + 70 – 74 
So we get 
= - 46 – 4 + 66 
By calculation 
= - 50 + 66 
= 66 – 50 
By subtraction 
= 16 
 
9. Calculate  
1 – 2 + 3 – 4 + 5 – 6 + ……… + 15 – 16 
Solution: 
 
It can be written as 
1 – 2 + 3 – 4 + 5 – 6 + 7 – 8 + 9 – 10 + 11 – 12 + 13 – 14 + 15 – 16 
We get 
= - 1 – 1 – 1 – 1 – 1 – 1 – 1 – 1  
Page 5


 
 
 
 
 
 
Exercise 5.4                                                                             page: 5.17 
1. Subtract the first integer from the second in each of the following: 
(i) 12, -5 
(ii) – 12, 8 
(iii) – 225, - 135 
(iv) 1001, 101 
(v) – 812, 3126 
(vi) 7560, - 8 
(vii) – 3978, - 4109 
(viii) 0, - 1005 
Solution: 
 
(i) 12, -5 
So by subtracting the first integer from the second 
-5 – 12 = - 17 
 
(ii) – 12, 8 
So by subtracting the first integer from the second 
8 – (-12) = 8 + 12 = 20 
 
(iii) – 225, - 135 
So by subtracting the first integer from the second 
-135 – (-225) = 225 – 135 = 90 
 
(iv) 1001, 101 
So by subtracting the first integer from the second 
101 – 1001 = - 900 
 
(v) – 812, 3126 
So by subtracting the first integer from the second 
3126 – (-812) = 3126 + 812 = 3938 
 
(vi) 7560, - 8 
So by subtracting the first integer from the second 
-8 – 7560 = - 7568 
 
(vii) – 3978, - 4109 
So by subtracting the first integer from the second 
-4109 – (-3978) = - 4109 + 3978 = -131 
 
(viii) 0, - 1005 
So by subtracting the first integer from the second 
-1005 – 0 = - 1005 
 
2. Find the value of: 
(i) – 27 – (- 23) 
(ii) – 17 – 18 – (-35) 
(iii) – 12 – (-5) – (-125) + 270 
(iv) 373 + (-245) + (-373) + 145 + 3000 
 
 
 
 
 
 
(v) 1 + (-475) + (-475) + (-475) + (-475) + 1900 
(vi) (-1) + (-304) + 304 + 304 + (-304) + 1 
Solution: 
 
(i) – 27 – (- 23) 
So we get 
= - 27 + 23 
On further calculation 
= 23 – 27  
We get 
= - 4 
 
(ii) – 17 – 18 – (-35) 
So we get 
= - 35 + 35 
On further calculation 
= 0  
 
(iii) – 12 – (-5) – (-125) + 270 
So we get 
= - 12 + 5 + 125 + 270 
On further calculation 
= 400 – 12 
We get 
= 388 
 
(iv) 373 + (-245) + (-373) + 145 + 3000 
So we get 
= 373 – 245 – 373 + 145 + 3000 
On further calculation 
= 3145 + 373 – 373 – 245 
We get 
= 3145 – 245 
By subtraction 
= 2900 
 
(v) 1 + (-475) + (-475) + (-475) + (-475) + 1900 
So we get 
= 1 – 950 – 950 + 1900 
On further calculation 
= 1900 + 1 – 1900 
We get 
= 1 
 
(vi) (-1) + (-304) + 304 + 304 + (-304) + 1 
So we get 
= - 1 + 1 – 304 + 304 – 304 + 304  
On further calculation  
= 0 
 
 
 
 
 
 
 
3. Subtract the sum of – 5020 and 2320 from – 709. 
Solution: 
 
We know that the sum of -5020 and 2320 is 
-5020 + 2320 
It can be written as 
= 2320 – 5020 
So we get 
= - 2700 
Subtracting from – 709 we get 
= - 709 – (-2700) 
We get 
= - 709 + 2700 
By subtraction 
= 1991 
 
4. Subtract the sum of – 1250 and 1138 from the sum of 1136 and - 1272. 
Solution: 
 
We know that the sum of – 1250 and 1138 is 
-1250 + 1138 
It can be written as 
= 1138 – 1250 
So we get 
= - 112 
 
We know that the sum of 1136 and – 1272 is 
1136 – 1272 = - 136 
So we get 
-136 – (-112) = - 136 + 112 = -24 
 
5. From the sum of 233 and – 147, subtract – 284. 
Solution: 
 
We know that the sum of 233 and – 147 is 
233 – 147 = 86 
Subtracting – 284 we get 
86 – (-284) = 86 + 284 = 370 
 
6. The sum of two integers is 238. If one of the integers is – 122, determine the other. 
Solution: 
 
It is given that 
Sum of two integers = 238 
One of the integers = - 122 
So the other integer = - (-122) + 238 
On further calculation 
Other integer = 238 + 122 = 360 
 
7. The sum of two integers is – 223. If one of the integers is 172, find the other. 
 
 
 
 
 
 
Solution: 
 
It is given that 
Sum of two integers = - 223 
One of the integers = 172 
So the other integer = - 223 – 172 = - 395 
 
8. Evaluate the following: 
(i) – 8 – 24 + 31 – 26 – 28 + 7 + 19 – 18 – 8 + 33 
(ii) – 26 – 20 + 33 – (-33) + 21 + 24 – (-25) – 26 – 14 – 34 
Solution: 
 
(i) – 8 – 24 + 31 – 26 – 28 + 7 + 19 – 18 – 8 + 33 
We get 
= - 8 – 24 – 26 – 28 – 18 – 8 + 31 + 7 + 19 + 33 
On further calculation 
= - 32 – 26 – 28 – 26 + 38 + 19 + 33 
It can be written as 
= 38 – 32 – 26 – 28 + 33 – 26 + 19 
So we get 
= 6 – 26 – 28 + 7 + 19 
By calculation 
= 6 – 28 – 26 + 26 
= 6 – 28 
By subtraction 
= - 22 
 
(ii) – 26 – 20 + 33 – (-33) + 21 + 24 – (-25) – 26 – 14 – 34 
We get 
= - 46 + 33 + 33 + 21 + 24 + 25 – 26 – 14 – 34 
On further calculation 
= - 46 + 66 + 21 + 24 + 25 + (-74) 
It can be written as 
= - 46 + 66 + 70 – 74 
So we get 
= - 46 – 4 + 66 
By calculation 
= - 50 + 66 
= 66 – 50 
By subtraction 
= 16 
 
9. Calculate  
1 – 2 + 3 – 4 + 5 – 6 + ……… + 15 – 16 
Solution: 
 
It can be written as 
1 – 2 + 3 – 4 + 5 – 6 + 7 – 8 + 9 – 10 + 11 – 12 + 13 – 14 + 15 – 16 
We get 
= - 1 – 1 – 1 – 1 – 1 – 1 – 1 – 1  
 
 
 
 
 
 
By calculation 
= - 8 
 
10. Calculate the sum: 
5 + (-5) + 5 + (-5) + ….. 
(i) if the number of terms is 10. 
(ii) if the number of terms is 11. 
Solution: 
 
(i) if the number of terms is 10 
We get 
5 + (-5) + 5 + (-5) + 5 + (-5) + 5 + (-5) + 5 + (-5) 
On further calculation 
= 5 – 5 + 5 – 5 + 5 – 5 + 5 – 5 + 5 – 5 = 0 
 
(ii) if the number of terms is 11 
We get 
5 + (-5) + 5 + (-5) + 5 + (-5) + 5 + (-5) + 5 + (-5) + 5 
On further calculation 
= 5 – 5 + 5 – 5 + 5 – 5 + 5 – 5 + 5 – 5 + 5 = 5 
 
11. Replace * by < or > in each of the following to make the statement true: 
(i) (-6) + (-9) * (-6) – (-9) 
(ii) (-12) – (-12) * (-12) + (-12) 
(iii) (-20) – (-20) * 20 - (65) 
(iv) 28 – (-10) * (-16) – (-76) 
Solution: 
 
(i) (-6) + (-9) < (-6) – (-9) 
 
(ii) (-12) – (-12) > (-12) + (-12) 
 
(iii) (-20) – (-20) > 20 - (65) 
 
(iv) 28 – (-10) < (-16) – (-76) 
 
12. If ? is an operation on integers such that a ? b = - a + b – (-2) for all integers a, b. Find the value of  
(i) 4 ? 3 
(ii) (-2) ? (-3) 
(iii) 6 ? (-5) 
(iv) (-5) ? 6 
Solution: 
 
(i) 4 ? 3 
By substituting values in a ? b = - a + b – (-2) 
We get 
4 ? 3 = - 4 + 3 – (-2) = 1 
 
(ii) (-2) ? (-3) 
By substituting values in a ? b = - a + b – (-2)

	Mathematics (Maths) Class 6 94 videos\|348 docs\|54 tests

Mathematics (Maths) Class 6

94 videos|348 docs|54 tests

Join Course for Free

Top Courses for Class 6

View all

Related Exams

Class 6

About this Document

	4.76/5 Rating
	Nov 24, 2024 Last updated

Document Description: RD Sharma Solutions: Negative Numbers and Integers (Exercise 5.4) for Class 6 2024 is part of Mathematics (Maths) Class 6 preparation. The notes and questions for RD Sharma Solutions: Negative Numbers and Integers (Exercise 5.4) have been prepared according to the Class 6 exam syllabus. Information about RD Sharma Solutions: Negative Numbers and Integers (Exercise 5.4) covers topics like and RD Sharma Solutions: Negative Numbers and Integers (Exercise 5.4) Example, for Class 6 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for RD Sharma Solutions: Negative Numbers and Integers (Exercise 5.4).

Introduction of RD Sharma Solutions: Negative Numbers and Integers (Exercise 5.4) in English is available as part of our Mathematics (Maths) Class 6 for Class 6 & RD Sharma Solutions: Negative Numbers and Integers (Exercise 5.4) in Hindi for Mathematics (Maths) Class 6 course. Download more important topics related with notes, lectures and mock test series for Class 6 Exam by signing up for free. Class 6: Negative Numbers and Integers (Exercise 5.4) RD Sharma Solutions | Mathematics (Maths) Class 6

Description

Full syllabus notes, lecture & questions for Negative Numbers and Integers (Exercise 5.4) RD Sharma Solutions | Mathematics (Maths) Class 6 - Class 6 | Plus excerises question with solution to help you revise complete syllabus for Mathematics (Maths) Class 6 | Best notes, free PDF download

Information about RD Sharma Solutions: Negative Numbers and Integers (Exercise 5.4)

In this doc you can find the meaning of RD Sharma Solutions: Negative Numbers and Integers (Exercise 5.4) defined & explained in the simplest way possible. Besides explaining types of RD Sharma Solutions: Negative Numbers and Integers (Exercise 5.4) theory, EduRev gives you an ample number of questions to practice RD Sharma Solutions: Negative Numbers and Integers (Exercise 5.4) tests, examples and also practice Class 6 tests

	Mathematics (Maths) Class 6 94 videos\|348 docs\|54 tests

Mathematics (Maths) Class 6

94 videos|348 docs|54 tests

Join Course for Free

Download as PDF

Explore Courses for Class 6 exam

Top Courses for Class 6

Explore Courses

Signup for Free!

Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.

Start learning for Free

10M+ students study on EduRev

mock tests for examination

practice quizzes

Negative Numbers and Integers (Exercise 5.4) RD Sharma Solutions | Mathematics (Maths) Class 6

study material

Negative Numbers and Integers (Exercise 5.4) RD Sharma Solutions | Mathematics (Maths) Class 6

Extra Questions

MCQs

Semester Notes

video lectures

pdf

Negative Numbers and Integers (Exercise 5.4) RD Sharma Solutions | Mathematics (Maths) Class 6

Sample Paper

Objective type Questions

Summary

Previous Year Questions with Solutions

shortcuts and tricks

ppt

Viva Questions

Free

Exam

past year papers

Important questions

;

Additional Information about RD Sharma Solutions: Negative Numbers and Integers (Exercise 5.4) for Class 6 Preparation

RD Sharma Solutions: Negative Numbers and Integers (Exercise 5.4) Free PDF Download

The RD Sharma Solutions: Negative Numbers and Integers (Exercise 5.4) is an invaluable resource that delves deep into the core of the Class 6 exam. These study notes are curated by experts and cover all the essential topics and concepts, making your preparation more efficient and effective. With the help of these notes, you can grasp complex subjects quickly, revise important points easily, and reinforce your understanding of key concepts. The study notes are presented in a concise and easy-to-understand manner, allowing you to optimize your learning process. Whether you're looking for best-recommended books, sample papers, study material, or toppers' notes, this PDF has got you covered. Download the RD Sharma Solutions: Negative Numbers and Integers (Exercise 5.4) now and kickstart your journey towards success in the Class 6 exam.

Importance of RD Sharma Solutions: Negative Numbers and Integers (Exercise 5.4)

The importance of RD Sharma Solutions: Negative Numbers and Integers (Exercise 5.4) cannot be overstated, especially for Class 6 aspirants. This document holds the key to success in the Class 6 exam. It offers a detailed understanding of the concept, providing invaluable insights into the topic. By knowing the concepts well in advance, students can plan their preparation effectively. Utilize this indispensable guide for a well-rounded preparation and achieve your desired results.

RD Sharma Solutions: Negative Numbers and Integers (Exercise 5.4) Notes

RD Sharma Solutions: Negative Numbers and Integers (Exercise 5.4) Notes offer in-depth insights into the specific topic to help you master it with ease. This comprehensive document covers all aspects related to RD Sharma Solutions: Negative Numbers and Integers (Exercise 5.4). It includes detailed information about the exam syllabus, recommended books, and study materials for a well-rounded preparation. Practice papers and question papers enable you to assess your progress effectively. Additionally, the paper analysis provides valuable tips for tackling the exam strategically. Access to Toppers' notes gives you an edge in understanding complex concepts. Whether you're a beginner or aiming for advanced proficiency, RD Sharma Solutions: Negative Numbers and Integers (Exercise 5.4) Notes on EduRev are your ultimate resource for success.

RD Sharma Solutions: Negative Numbers and Integers (Exercise 5.4) Class 6 Questions

The "RD Sharma Solutions: Negative Numbers and Integers (Exercise 5.4) Class 6 Questions" guide is a valuable resource for all aspiring students preparing for the Class 6 exam. It focuses on providing a wide range of practice questions to help students gauge their understanding of the exam topics. These questions cover the entire syllabus, ensuring comprehensive preparation. The guide includes previous years' question papers for students to familiarize themselves with the exam's format and difficulty level. Additionally, it offers subject-specific question banks, allowing students to focus on weak areas and improve their performance.

Study RD Sharma Solutions: Negative Numbers and Integers (Exercise 5.4) on the App

Students of Class 6 can study RD Sharma Solutions: Negative Numbers and Integers (Exercise 5.4) alongwith tests & analysis from the EduRev app, which will help them while preparing for their exam. Apart from the RD Sharma Solutions: Negative Numbers and Integers (Exercise 5.4), students can also utilize the EduRev App for other study materials such as previous year question papers, syllabus, important questions, etc. The EduRev App will make your learning easier as you can access it from anywhere you want. The content of RD Sharma Solutions: Negative Numbers and Integers (Exercise 5.4) is prepared as per the latest Class 6 syllabus.

Education Revolution