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
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
for more answers vist to:
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.
(iii) x =-2 Multiplying both sides by 3 3x=-6 Adding 10 to both sides 3x+10 = -6+10=3x+10=4
(iii) x =-2
See lessMultiplying both sides by 3
3x=-6
Adding 10 to both sides
3x+10 = -6+10=3x+10=4