Let the number be n. According to the question, 7n + 50 + 40 = 300 ⇒ 7n+90 = 300 ⇒ 7n = 300-90 ⇒ 7n = 210 ⇒ 7n = 210/7 ⇒ n = 30 Thus, the required number is 30. Class 7 Maths Chapter 4 Exercise 4.4 for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-4/
Let the number be n.
According to the question, 7n + 50 + 40 = 300
⇒ 7n+90 = 300
⇒ 7n = 300-90
⇒ 7n = 210
⇒ 7n = 210/7
⇒ n = 30
Thus, the required number is 30.
Let the number of marbles Parmit has be m. According to the question, 5m + 7 = 37 ⇒ 5m = 37 - 7 ⇒ 5m = 30 ⇒ m = 30/5 ⇒ m = 6 Thus, Parmit has 6 marbles. Class 7 Maths Chapter 4 Exercise 4.4 for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-4/
Let the number of marbles Parmit has be m.
According to the question, 5m + 7 = 37
⇒ 5m = 37 – 7
⇒ 5m = 30
⇒ m = 30/5
⇒ m = 6
Thus, Parmit has 6 marbles.
(a) Let the lowest marks be y. According to the question, 2y+7 = 87 ⇒ 2y = 87-7 ⇒ 2y = 80 ⇒ y = 80/2 ⇒ y = 40 Thus, the lowest score is 40. (b) Let the base angle of the triangle be b. Given, a = 40°,b = c Since, a+b+c = 180° [Angle sum property of a triangle] ⇒ 40°+b+b = 180° ⇒ 40° + 2b = 180° ⇒ 2bRead more
(a) Let the lowest marks be y.
According to the question, 2y+7 = 87
⇒ 2y = 87-7
⇒ 2y = 80
⇒ y = 80/2
⇒ y = 40
Thus, the lowest score is 40.
(b) Let the base angle of the triangle be b.
Given, a = 40°,b = c
Since, a+b+c = 180° [Angle sum property of a triangle]
⇒ 40°+b+b = 180°
⇒ 40° + 2b = 180°
⇒ 2b = 180°-40°
⇒ 2b = 140°
⇒ b = 140°/2
⇒ b =70°
Thus, the base angles of the isosceles triangle are 70° each.
(c) Let the score of Rahul be x runs and Sachin’s score is 2 . x
According to the question, x + 2x = 198
⇒ 3x=198
⇒ x = 198/3
⇒ x = 66
Thus, Rahul’s score = 66 runs
And Sachin’s score = 2 x 66 = 132 runs.
(a) Let the number be x. According to the question, 8x + 4 = 60 ⇒ 8x=60-4 ⇒ 8x = 56 ⇒ x = 56/8 ⇒ x = 7 (b) Let the number be y. According to the question, y/5 - 4 = 3 ⇒ y/5 = 3+4 ⇒ y/5 = 7 ⇒ y = 7x5 ⇒ y = 35 Class 7 Maths Chapter 4 Exercise 4.4 for more answers vist to: https://www.tiwariacademy.comRead more
(a) Let the number be x.
According to the question, 8x + 4 = 60
⇒ 8x=60-4
⇒ 8x = 56
⇒ x = 56/8
⇒ x = 7
(b) Let the number be y.
According to the question, y/5 – 4 = 3
⇒ y/5 = 3+4
⇒ y/5 = 7
⇒ y = 7×5
⇒ y = 35
(a) 3 equations starting with x = 2. (i) x = 2 Multiplying both sides by 10, 10x = 20 Adding 2 both sides 10x + 2 = 20+2 = 10x+2 = 22 (ii) x = 2 Multiplying both sides by 5 5x = 10 Subtracting 3 from both sides 5x-3 = 10-3 = 5x-3 = 7 (iii) x = 2 Dividing both sides by 5 x/5 = 2/5 (b) 3 equations staRead more
(a) 3 equations starting with x = 2.
(i) x = 2
Multiplying both sides by 10,
10x = 20
Adding 2 both sides
10x + 2 = 20+2 = 10x+2 = 22
(ii) x = 2
Multiplying both sides by 5
5x = 10
Subtracting 3 from both sides
5x-3 = 10-3 = 5x-3 = 7
(iii) x = 2
Dividing both sides by 5
x/5 = 2/5
(b) 3 equations starting with x = -2.
(i) x = -2
Multiplying both sides by 3
3x=-6
(ii) x = -2
Multiplying both sides by 3
3x=-6
Adding 7 to both sides
3x+7 = -6+7 = 3x+7 =1
2(x+4) = 12 ⇒ x+4 = 12/2 ⇒ x+4 = 6 ⇒ x = 6-4 ⇒ x = 2 Class 7 Maths Chapter 4 Exercise 4.3 for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-4/
2(x+4) = 12
⇒ x+4 = 12/2
⇒ x+4 = 6
⇒ x = 6-4
⇒ x = 2
10p = 100 ⇒ 10p/10 = 100/10 [Dividing both sides by 10] ⇒ p=0 Class 7 Maths Chapter 4 Exercise 4.2 for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-4/
10p = 100
⇒ 10p/10 = 100/10 [Dividing both sides by 10]
⇒ p=0
3n-2 = 46 Step I: 3n-2+2=46+2 ⇒ 3n=48 [Adding 2 both sides] Step II: 3n/3 = 48/3 ⇒ n = 16 [Dividing both sides by 3] Class 7 Maths Chapter 4 Exercise 4.2 for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-4/
3n-2 = 46
Step I: 3n-2+2=46+2
⇒ 3n=48 [Adding 2 both sides]
Step II: 3n/3 = 48/3
⇒ n = 16 [Dividing both sides by 3]
Solve the following riddle: I am a number, Tell my identity! Take me seven times over, And add a fifty! To reach a triple century, You still need forty!
Let the number be n. According to the question, 7n + 50 + 40 = 300 ⇒ 7n+90 = 300 ⇒ 7n = 300-90 ⇒ 7n = 210 ⇒ 7n = 210/7 ⇒ n = 30 Thus, the required number is 30. Class 7 Maths Chapter 4 Exercise 4.4 for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-4/
Let the number be n.
According to the question, 7n + 50 + 40 = 300
⇒ 7n+90 = 300
⇒ 7n = 300-90
⇒ 7n = 210
⇒ 7n = 210/7
⇒ n = 30
Thus, the required number is 30.
Class 7 Maths Chapter 4 Exercise 4.4
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-4/
Solve the following: (i) Irfan says that he has 7 marbles more than five times the marbles Parmit has. Irfan has 37 marbles. How many marbles does Parmit have?
Let the number of marbles Parmit has be m. According to the question, 5m + 7 = 37 ⇒ 5m = 37 - 7 ⇒ 5m = 30 ⇒ m = 30/5 ⇒ m = 6 Thus, Parmit has 6 marbles. Class 7 Maths Chapter 4 Exercise 4.4 for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-4/
Let the number of marbles Parmit has be m.
According to the question, 5m + 7 = 37
⇒ 5m = 37 – 7
⇒ 5m = 30
⇒ m = 30/5
⇒ m = 6
Thus, Parmit has 6 marbles.
Class 7 Maths Chapter 4 Exercise 4.4
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-4/
Solve the following: (a) The teacher tells the class that the highest marks obtained by a student in her class are twice the lowest marks plus 7. The highest score is 87. What is the lowest score? (b) In an isosceles triangle, the base angles are equal. The vertex angle is 40° What are the base angles of the triangle? (Remember,the sum of three angles of a triangle is 180°) (c) Sachin scored twice as many runs as Rahul. Together, their runs fell two short of a double century. How many runs did each one score?
(a) Let the lowest marks be y. According to the question, 2y+7 = 87 ⇒ 2y = 87-7 ⇒ 2y = 80 ⇒ y = 80/2 ⇒ y = 40 Thus, the lowest score is 40. (b) Let the base angle of the triangle be b. Given, a = 40°,b = c Since, a+b+c = 180° [Angle sum property of a triangle] ⇒ 40°+b+b = 180° ⇒ 40° + 2b = 180° ⇒ 2bRead more
(a) Let the lowest marks be y.
According to the question, 2y+7 = 87
⇒ 2y = 87-7
⇒ 2y = 80
⇒ y = 80/2
⇒ y = 40
Thus, the lowest score is 40.
(b) Let the base angle of the triangle be b.
Given, a = 40°,b = c
Since, a+b+c = 180° [Angle sum property of a triangle]
⇒ 40°+b+b = 180°
⇒ 40° + 2b = 180°
⇒ 2b = 180°-40°
⇒ 2b = 140°
⇒ b = 140°/2
⇒ b =70°
Thus, the base angles of the isosceles triangle are 70° each.
(c) Let the score of Rahul be x runs and Sachin’s score is 2 . x
According to the question, x + 2x = 198
⇒ 3x=198
⇒ x = 198/3
⇒ x = 66
Thus, Rahul’s score = 66 runs
And Sachin’s score = 2 x 66 = 132 runs.
Class 7 Maths Chapter 4 Exercise 4.4
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-4/
Set up equations and solve them to find the unknown numbers in the following cases: (a) Add 4 to eight times a number; you get 60. (b) One-fifth of a number minus 4 gives 3.
(a) Let the number be x. According to the question, 8x + 4 = 60 ⇒ 8x=60-4 ⇒ 8x = 56 ⇒ x = 56/8 ⇒ x = 7 (b) Let the number be y. According to the question, y/5 - 4 = 3 ⇒ y/5 = 3+4 ⇒ y/5 = 7 ⇒ y = 7x5 ⇒ y = 35 Class 7 Maths Chapter 4 Exercise 4.4 for more answers vist to: https://www.tiwariacademy.comRead more
(a) Let the number be x.
According to the question, 8x + 4 = 60
⇒ 8x=60-4
⇒ 8x = 56
⇒ x = 56/8
⇒ x = 7
(b) Let the number be y.
According to the question, y/5 – 4 = 3
⇒ y/5 = 3+4
⇒ y/5 = 7
⇒ y = 7×5
⇒ y = 35
Class 7 Maths Chapter 4 Exercise 4.4
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See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-4/
(a) Construct 3 equations starting with x =2. (b) Construct 3 equations starting with x= -2.
(a) 3 equations starting with x = 2. (i) x = 2 Multiplying both sides by 10, 10x = 20 Adding 2 both sides 10x + 2 = 20+2 = 10x+2 = 22 (ii) x = 2 Multiplying both sides by 5 5x = 10 Subtracting 3 from both sides 5x-3 = 10-3 = 5x-3 = 7 (iii) x = 2 Dividing both sides by 5 x/5 = 2/5 (b) 3 equations staRead more
(a) 3 equations starting with x = 2.
(i) x = 2
Multiplying both sides by 10,
10x = 20
Adding 2 both sides
10x + 2 = 20+2 = 10x+2 = 22
(ii) x = 2
Multiplying both sides by 5
5x = 10
Subtracting 3 from both sides
5x-3 = 10-3 = 5x-3 = 7
(iii) x = 2
Dividing both sides by 5
x/5 = 2/5
(b) 3 equations starting with x = -2.
(i) x = -2
Multiplying both sides by 3
3x=-6
(ii) x = -2
Multiplying both sides by 3
3x=-6
Adding 7 to both sides
3x+7 = -6+7 = 3x+7 =1
Class 7 Maths Chapter 4 Exercise 4.3
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Solve the following equations: 4 = 5 (p-2)
4 = 5 (p-2) ⇒ 4 = 5xp-5x2 ⇒ 4 = 5p - 10 ⇒ 5p-10 = 4 ⇒ 5p = 4 + 10 ⇒ 5p = 14 ⇒ p=14/5 Class 7 Maths Chapter 4 Exercise 4.3 for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-4/
4 = 5 (p-2)
⇒ 4 = 5xp-5×2
⇒ 4 = 5p – 10
⇒ 5p-10 = 4
⇒ 5p = 4 + 10
⇒ 5p = 14
⇒ p=14/5
Class 7 Maths Chapter 4 Exercise 4.3
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See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-4/
Solve the following equations: 2(x+4) = 12
2(x+4) = 12 ⇒ x+4 = 12/2 ⇒ x+4 = 6 ⇒ x = 6-4 ⇒ x = 2 Class 7 Maths Chapter 4 Exercise 4.3 for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-4/
2(x+4) = 12
⇒ x+4 = 12/2
⇒ x+4 = 6
⇒ x = 6-4
⇒ x = 2
Class 7 Maths Chapter 4 Exercise 4.3
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Solve the following equations: 2y+5/2 = 37/2
2y+5/2 = 37/2 ⇒ 2y = 37/2 - 5/2 ⇒ 2y = 37-5/2 ⇒ 2y =32/2 ⇒ 2y = 16 ⇒ y = 16/2 ⇒ y= 8 Class 7 Maths Chapter 4 Exercise 4.3 for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-4/
2y+5/2 = 37/2
⇒ 2y = 37/2 – 5/2
⇒ 2y = 37-5/2
⇒ 2y =32/2
⇒ 2y = 16
⇒ y = 16/2
⇒ y= 8
Class 7 Maths Chapter 4 Exercise 4.3
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Solve the following equation: 10p = 100
10p = 100 ⇒ 10p/10 = 100/10 [Dividing both sides by 10] ⇒ p=0 Class 7 Maths Chapter 4 Exercise 4.2 for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-4/
10p = 100
⇒ 10p/10 = 100/10 [Dividing both sides by 10]
⇒ p=0
Class 7 Maths Chapter 4 Exercise 4.2
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See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-4/
Give first the step you will use to separate the variable and then solve the equations: 3n-2 = 46
3n-2 = 46 Step I: 3n-2+2=46+2 ⇒ 3n=48 [Adding 2 both sides] Step II: 3n/3 = 48/3 ⇒ n = 16 [Dividing both sides by 3] Class 7 Maths Chapter 4 Exercise 4.2 for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-4/
3n-2 = 46
Step I: 3n-2+2=46+2
⇒ 3n=48 [Adding 2 both sides]
Step II: 3n/3 = 48/3
⇒ n = 16 [Dividing both sides by 3]
Class 7 Maths Chapter 4 Exercise 4.2
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See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-4/