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  1. Asked: In:

    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!

    Best Answer
    Aahna

    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:
    https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-4/

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    • 5
  2. Asked: In:

    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?

    Best Answer
    Aahna

    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:
    https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-4/

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    • 4
  3. Asked: In:

    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?

    Best Answer
    Aahna

    (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:
    https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-4/

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    • 4
  4. Asked: In:

    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.

    Best Answer
    Aahna

    (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:
    https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-4/

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    • 3
  6. Asked: In:

    (a) Construct 3 equations starting with x =2. (b) Construct 3 equations starting with x= -2.

    Aahna

    (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
    Multiplying both sides by 3
    3x=-6
    Adding 10 to both sides
    3x+10 = -6+10=3x+10=4

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    • 4
  7. Asked: In:

    (a) Construct 3 equations starting with x =2. (b) Construct 3 equations starting with x= -2.

    Best Answer
    Aahna

    (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

    for more answers vist to:
    https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-4/

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    • 4
  8. Asked: In:

    Solve the following equations: 4 = 5 (p-2)

    Best Answer
    Aahna

    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

    for more answers vist to:
    https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-4/

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    • 4
  9. Asked: In:

    Solve the following equations: 2(x+4) = 12

    Best Answer
    Aahna

    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

    for more answers vist to:
    https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-4/

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    • 4
  11. Asked: In:

    Solve the following equations: 2y+5/2 = 37/2

    Best Answer
    Aahna

    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

    for more answers vist to:
    https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-4/

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    • 4
  12. Asked: In:

    Solve the following equation: 10p = 100

    Best Answer
    Aahna

    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

    for more answers vist to:
    https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-4/

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    • 5