26 x (-48) + (-48) x (-36) ⇒ (-48)x[26+(-36)] [Distributive property] ⇒ (-48) x (-10) ⇒ 480 Class 7 Maths Exercise 1.3 Questions 4, 5 for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-1/
26 x (-48) + (-48) x (-36)
⇒ (-48)x[26+(-36)] [Distributive property]
⇒ (-48) x (-10)
⇒ 480
(i) (-1) x a = -a, where a is an integer. (ii) (a) (–1)x(-22) = 22 (b) (–1)x37=-37 (c) (–1)x0=0 Class 7 Maths Exercise 1.3 Questions 1, 2, 3 for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-1/
(i) (-1) x a = -a, where a is an integer.
(ii) (a) (–1)x(-22) = 22
(b) (–1)x37=-37
(c) (–1)x0=0
(a) 3 x (–1) = -3 (b) (–1) x 225 = -225 Class 7 Maths Exercise 1.3 Questions 1, 2, 3 for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-1/
(-5) + (-8) = (-8) + (-5) [Commutative property] Class 7 Maths Exercise 1.2 Questions 1, 2, 3, 4 for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-1/
Team A scored -40,10,0 Total score of Team A = -40+10+0 =-30 Team B scored 10,0,-40 Total score of Team B = 10+0+(-40) = 10 + 0 - 40 = -30 Thus, scores of both teams are same. Yes, we can add integers in any order due to commutative property. Class 7 Maths Exercise 1.2 Questions 1, 2, 3, 4 for moreRead more
Team A scored -40,10,0
Total score of Team A = -40+10+0 =-30
Team B scored 10,0,-40
Total score of Team B = 10+0+(-40) = 10 + 0 – 40 = -30
Thus, scores of both teams are same.
Yes, we can add integers in any order due to commutative property.
(a) -2-(-10)-2+10=8 (b) (-7)+2=-5 (c) (-2)-1=-2-1=-3 Class 7 Maths Exercise 1.2 Questions 1, 2, 3, 4 for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-1/
(a) One such pair whose sum is -7: -5+(-2)=-7 (b) One such pair whose difference is -10: -2-8=-10 (c) One such pair whose sum is 0: -5+5 = 0 Class 7 Maths Exercise 1.2 Questions 1, 2, 3, 4 for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-1/
(a) One such pair whose sum is -7: -5+(-2)=-7
(b) One such pair whose difference is -10: -2-8=-10
(c) One such pair whose sum is 0: -5+5 = 0
(i) a=21,b=18 Given: a=21,b=18 We have a-(-b) = a + b Putting the values in L.H.S. = a-(-b)= 21-(-18) = 21+18 =39 Putting the values in R.H.S. = a + b = 21 + 19 = 39 Since, L.H.S. = R.H.S Hence, verified. (ii) Given: a = 118,b = 125 We have a-(-b)=a+b Putting the values in L.H.S. = a-(-b) = 118 = (-Read more
(i) a=21,b=18
Given: a=21,b=18
We have a-(-b) = a + b
Putting the values in L.H.S. = a-(-b)= 21-(-18) = 21+18 =39
Putting the values in R.H.S. = a + b = 21 + 19 = 39
Since, L.H.S. = R.H.S
Hence, verified.
(ii) Given: a = 118,b = 125
We have a-(-b)=a+b
Putting the values in L.H.S. = a-(-b) = 118 = (-125) = 118+125 =243
Putting the values in R.H.S. = a+b = 118 + 125 = 243
Since, L.H.S. = R.H.S
Hence, verified.
Find the product, using suitable properties: 26 x (-48) + (-48) x (-36)
26 x (-48) + (-48) x (-36) ⇒ (-48)x[26+(-36)] [Distributive property] ⇒ (-48) x (-10) ⇒ 480 Class 7 Maths Exercise 1.3 Questions 4, 5 for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-1/
26 x (-48) + (-48) x (-36)
⇒ (-48)x[26+(-36)] [Distributive property]
⇒ (-48) x (-10)
⇒ 480
Class 7 Maths Exercise 1.3 Questions 4, 5
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-1/
Starting from (-1) x5, write various products showing some patterns to show (-1) x (-1) = 1.
(-1)x5 = -5 (-1)x4 = -4 (-1)x3 = -3 (-1)x2 = -2 (-1)x1 = -1 (-1)x0 = 0 (-1)x(-1)=1 Class 7 Maths Exercise 1.3 Questions 4, 5 for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-1/
(-1)x5 = -5 (-1)x4 = -4
(-1)x3 = -3 (-1)x2 = -2
(-1)x1 = -1 (-1)x0 = 0
(-1)x(-1)=1
Class 7 Maths Exercise 1.3 Questions 4, 5
for more answers vist to:
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(i) For any integer a, what is (-1) x a equal to? (ii) Determine the integer whose product with(-1) is: (a) –22 (b) 37 (c) 0
(i) (-1) x a = -a, where a is an integer. (ii) (a) (–1)x(-22) = 22 (b) (–1)x37=-37 (c) (–1)x0=0 Class 7 Maths Exercise 1.3 Questions 1, 2, 3 for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-1/
(i) (-1) x a = -a, where a is an integer.
(ii) (a) (–1)x(-22) = 22
(b) (–1)x37=-37
(c) (–1)x0=0
Class 7 Maths Exercise 1.3 Questions 1, 2, 3
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-1/
Verify the following: (a) 18 x [7 + (–3)] = [18 x 7] + [18 x (–3)] (b) (–21) x [(–4) + (–6)] = [(–21) x (-4)] + [(–21) x (–6)]
(a) 18 x [7 + (–3)] = [18 x 7] + [18 x (–3)] ⇒ 18 X 4 = 126+(-54) ⇒ 72 = 72 ⇒ L.H.S = R.H.S Hence verified. (b) (–21) x [(–4) + (–6)] = [(–21) x (-4)] + [(–21) x (–6)] ⇒ (–21) x (-10) = 84 + 126 ⇒ 210 = 210 ⇒ L.H.S = R.H.S Hence verified. Class 7 Maths Exercise 1.3 Questions 1, 2, 3 for more answersRead more
(a) 18 x [7 + (–3)] = [18 x 7] + [18 x (–3)]
⇒ 18 X 4 = 126+(-54)
⇒ 72 = 72
⇒ L.H.S = R.H.S
Hence verified.
(b) (–21) x [(–4) + (–6)] = [(–21) x (-4)] + [(–21) x (–6)]
⇒ (–21) x (-10) = 84 + 126
⇒ 210 = 210
⇒ L.H.S = R.H.S
Hence verified.
Class 7 Maths Exercise 1.3 Questions 1, 2, 3
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-1/
Find the each of the following products: (a) 3 x (–1) (b) (–1) x 225
(a) 3 x (–1) = -3 (b) (–1) x 225 = -225 Class 7 Maths Exercise 1.3 Questions 1, 2, 3 for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-1/
(a) 3 x (–1) = -3
(b) (–1) x 225 = -225
Class 7 Maths Exercise 1.3 Questions 1, 2, 3
for more answers vist to:
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Fill in the blanks to make the following statements true: (-5) + (-8) = (-8) + (…….)
(-5) + (-8) = (-8) + (-5) [Commutative property] Class 7 Maths Exercise 1.2 Questions 1, 2, 3, 4 for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-1/
(-5) + (-8) = (-8) + (-5) [Commutative property]
Class 7 Maths Exercise 1.2 Questions 1, 2, 3, 4
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-1/
In a quiz, team A scored – 40,10,0 and team B scores 10, 0, -40 in three successive rounds. Which team scored more? Can we say that we can add integers in any order?
Team A scored -40,10,0 Total score of Team A = -40+10+0 =-30 Team B scored 10,0,-40 Total score of Team B = 10+0+(-40) = 10 + 0 - 40 = -30 Thus, scores of both teams are same. Yes, we can add integers in any order due to commutative property. Class 7 Maths Exercise 1.2 Questions 1, 2, 3, 4 for moreRead more
Team A scored -40,10,0
Total score of Team A = -40+10+0 =-30
Team B scored 10,0,-40
Total score of Team B = 10+0+(-40) = 10 + 0 – 40 = -30
Thus, scores of both teams are same.
Yes, we can add integers in any order due to commutative property.
Class 7 Maths Exercise 1.2 Questions 1, 2, 3, 4
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-1/
(a) Write a pair of negative integers whose difference gives 8. (b) Write a negative integer and a positive integer whose is -5. (c) Write a negative integer and a positive integer whose difference is -3.
(a) -2-(-10)-2+10=8 (b) (-7)+2=-5 (c) (-2)-1=-2-1=-3 Class 7 Maths Exercise 1.2 Questions 1, 2, 3, 4 for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-1/
(a) -2-(-10)-2+10=8
(b) (-7)+2=-5
(c) (-2)-1=-2-1=-3
Class 7 Maths Exercise 1.2 Questions 1, 2, 3, 4
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-1/
Write down a pair of integers whose: (a) sum is -7 (b) difference is -10 (a) sum is 0
(a) One such pair whose sum is -7: -5+(-2)=-7 (b) One such pair whose difference is -10: -2-8=-10 (c) One such pair whose sum is 0: -5+5 = 0 Class 7 Maths Exercise 1.2 Questions 1, 2, 3, 4 for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-1/
(a) One such pair whose sum is -7: -5+(-2)=-7
(b) One such pair whose difference is -10: -2-8=-10
(c) One such pair whose sum is 0: -5+5 = 0
Class 7 Maths Exercise 1.2 Questions 1, 2, 3, 4
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-1/
Verify a-(-b) = a+b for the following values of and a : b (i) a=21,b=18 (ii) a=118,b=125
(i) a=21,b=18 Given: a=21,b=18 We have a-(-b) = a + b Putting the values in L.H.S. = a-(-b)= 21-(-18) = 21+18 =39 Putting the values in R.H.S. = a + b = 21 + 19 = 39 Since, L.H.S. = R.H.S Hence, verified. (ii) Given: a = 118,b = 125 We have a-(-b)=a+b Putting the values in L.H.S. = a-(-b) = 118 = (-Read more
(i) a=21,b=18
Given: a=21,b=18
We have a-(-b) = a + b
Putting the values in L.H.S. = a-(-b)= 21-(-18) = 21+18 =39
Putting the values in R.H.S. = a + b = 21 + 19 = 39
Since, L.H.S. = R.H.S
Hence, verified.
(ii) Given: a = 118,b = 125
We have a-(-b)=a+b
Putting the values in L.H.S. = a-(-b) = 118 = (-125) = 118+125 =243
Putting the values in R.H.S. = a+b = 118 + 125 = 243
Since, L.H.S. = R.H.S
Hence, verified.
7 Maths Chapter 1 Exercise 1.1 Question 7, 8
for more answers vist to:
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