(a) 3l = 42 ⇒ 3l/3 = 42/3 [Dividing both sides by 3] ⇒ l = 14 (b) b/2 = 6 ⇒ b/2 x 2 = 6x2 [Multiplying both sides by 2] ⇒ b = 12 Class 7 Maths Chapter 4 Exercise 4.2 for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-4/
(a) 3l = 42
⇒ 3l/3 = 42/3 [Dividing both sides by 3]
⇒ l = 14
(b) b/2 = 6
⇒ b/2 x 2 = 6×2 [Multiplying both sides by 2]
⇒ b = 12
(a) x-1 = 0 ⇒ x-1+1 = 0+1 [Adding 1 both sides] ⇒ x = 1 (b) x+1 = 0 ⇒ x+1-1 = 0-1 [Subtracting 1 both sides] ⇒ x =-1 Class 7 Maths Chapter 4 Exercise 4.2 for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-4/
(a) x-1 = 0
⇒ x-1+1 = 0+1 [Adding 1 both sides]
⇒ x = 1
(b) x+1 = 0
⇒ x+1-1 = 0-1 [Subtracting 1 both sides]
⇒ x =-1
(i) Let m be the number of Parmit’s marbles. ∴ 5m+7 = 37 (ii) Let the age of Laxmi be y years. ∴ 3y+4 = 49 (iii) Let the lowest score be l. ∴ 2l+7=87 (iv) Let the base angle of the isosceles triangle be b, so vertex angle = 2b. ∴ 2b+b+b = 180° ⇒ 4b =180° [Angle sum property of a∆] Class 7 Maths ChapRead more
(i) Let m be the number of Parmit’s marbles.
∴ 5m+7 = 37
(ii) Let the age of Laxmi be y years.
∴ 3y+4 = 49
(iii) Let the lowest score be l.
∴ 2l+7=87
(iv) Let the base angle of the isosceles triangle be b, so vertex angle = 2b.
∴ 2b+b+b = 180°
⇒ 4b =180° [Angle sum property of a∆]
(i) The sum of numbers p and 4 is 15. (ii) 7 subtracted from m is 3. (iii) Two times m is 7. (iv) The number m is divided by 5 gives 3. (v) Three-fifth of the number m is 6. (vi) Three times p plus 4 gets 25. (vii) If you take away 2 from 4 times p, you get 18. (viii) If you added 2 to half is p, yoRead more
(i) The sum of numbers p and 4 is 15.
(ii) 7 subtracted from m is 3.
(iii) Two times m is 7.
(iv) The number m is divided by 5 gives 3.
(v) Three-fifth of the number m is 6.
(vi) Three times p plus 4 gets 25.
(vii) If you take away 2 from 4 times p, you get 18.
(viii) If you added 2 to half is p, you get 8.
5p+2 =17 Putting p = -3 in L.H.S. 5(-3)+2 = -15+2 =-13 ∵ -13 ≠ 17 Therefore, p = -3 is not the solution. Putting p = -2 in L.H.S. 5(-2)+2 = -10+2 =-8 ∵ -8 ≠ 17 Therefore, p = -2 is not the solution. Putting p = -1 in L.H.S. 5(-1)+2 = -5+2 =-3 ∵ -3 ≠ 17 Therefore, p = -1 is not the solution. PuttingRead more
5p+2 =17
Putting p = -3 in L.H.S. 5(-3)+2 = -15+2 =-13
∵ -13 ≠ 17 Therefore, p = -3 is not the solution.
Putting p = -2 in L.H.S. 5(-2)+2 = -10+2 =-8
∵ -8 ≠ 17 Therefore, p = -2 is not the solution.
Putting p = -1 in L.H.S. 5(-1)+2 = -5+2 =-3
∵ -3 ≠ 17 Therefore, p = -1 is not the solution.
Putting p = 0 in L.H.S. 5(0)+2 = 0+2 = 2
∵ 2 ≠ 17 Therefore, p = 0 is not the solution.
Putting p = 1 in L.H.S. 5(1)+2 = 5+2 = 7
∵ 7 ≠ 17 Therefore, p = 1 is not the solution.
Putting p = 2 in L.H.S. 5(2)+2 = 10+2 = 12
∵ 12 ≠ 17 Therefore, p = 2 is not the solution.
Putting p = 3 in L.H.S. 5(3)+2 = 15+2 = 17
∵ 17 ≠ 17 Therefore, p = 3 is the solution.
n+5=19(n=1) Putting n = 1 in L.H.S., 1 + 5 = 6 ∵ L.H.S. ≠ R.H.S., ∴ n=1 is not the solution of given equation. Class 7 Maths Chapter 4 Exercise 4.1 for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-4/
n+5=19(n=1)
Putting n = 1 in L.H.S.,
1 + 5 = 6
∵ L.H.S. ≠ R.H.S.,
∴ n=1 is not the solution of given equation.
Give first the step you will use to separate the variable and then solve the equations: (a) 3l =42 (b) b/2=6
(a) 3l = 42 ⇒ 3l/3 = 42/3 [Dividing both sides by 3] ⇒ l = 14 (b) b/2 = 6 ⇒ b/2 x 2 = 6x2 [Multiplying both sides by 2] ⇒ b = 12 Class 7 Maths Chapter 4 Exercise 4.2 for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-4/
(a) 3l = 42
⇒ 3l/3 = 42/3 [Dividing both sides by 3]
⇒ l = 14
(b) b/2 = 6
⇒ b/2 x 2 = 6×2 [Multiplying both sides by 2]
⇒ b = 12
Class 7 Maths Chapter 4 Exercise 4.2
for more answers vist to:
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: (a) x-1 = 0 (b) x+1 = 0
(a) x-1 = 0 ⇒ x-1+1 = 0+1 [Adding 1 both sides] ⇒ x = 1 (b) x+1 = 0 ⇒ x+1-1 = 0-1 [Subtracting 1 both sides] ⇒ x =-1 Class 7 Maths Chapter 4 Exercise 4.2 for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-4/
(a) x-1 = 0
⇒ x-1+1 = 0+1 [Adding 1 both sides]
⇒ x = 1
(b) x+1 = 0
⇒ x+1-1 = 0-1 [Subtracting 1 both sides]
⇒ x =-1
Class 7 Maths Chapter 4 Exercise 4.2
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-4/
Set up an equation in the following cases: (i) Irfan says that he has 7 marbles more than five times the marbles Parmit has. Irfan has 37 marbles. (Tale m to be the number of Parmit’s marbles.) (ii) Laxmi’s father is 49 years old. He is 4 years older than three times Laxmi’s age. (Take Laxmi’s age to be y years.) (iii) 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. (Take the lowest score to be l ) . (iv) In an isosceles triangle, the vertex angle is twice either base angle. (Let the base angle be b in degrees. Remember that the sum of angles of a triangle is 180°.)
(i) Let m be the number of Parmit’s marbles. ∴ 5m+7 = 37 (ii) Let the age of Laxmi be y years. ∴ 3y+4 = 49 (iii) Let the lowest score be l. ∴ 2l+7=87 (iv) Let the base angle of the isosceles triangle be b, so vertex angle = 2b. ∴ 2b+b+b = 180° ⇒ 4b =180° [Angle sum property of a∆] Class 7 Maths ChapRead more
(i) Let m be the number of Parmit’s marbles.
∴ 5m+7 = 37
(ii) Let the age of Laxmi be y years.
∴ 3y+4 = 49
(iii) Let the lowest score be l.
∴ 2l+7=87
(iv) Let the base angle of the isosceles triangle be b, so vertex angle = 2b.
∴ 2b+b+b = 180°
⇒ 4b =180° [Angle sum property of a∆]
Class 7 Maths Chapter 4 Exercise 4.1
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-4/
Write the following equations in statement form: (i) p+4=15 (ii) m -7 = 3 (iii) 2m = 7 (iv) m/5 =3
(i) The sum of numbers p and 4 is 15. (ii) 7 subtracted from m is 3. (iii) Two times m is 7. (iv) The number m is divided by 5 gives 3. (v) Three-fifth of the number m is 6. (vi) Three times p plus 4 gets 25. (vii) If you take away 2 from 4 times p, you get 18. (viii) If you added 2 to half is p, yoRead more
(i) The sum of numbers p and 4 is 15.
(ii) 7 subtracted from m is 3.
(iii) Two times m is 7.
(iv) The number m is divided by 5 gives 3.
(v) Three-fifth of the number m is 6.
(vi) Three times p plus 4 gets 25.
(vii) If you take away 2 from 4 times p, you get 18.
(viii) If you added 2 to half is p, you get 8.
Class 7 Maths Chapter 4 Exercise 4.1
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-4/
Write equations for the following statements: (i) The sum of numbers x and 4 is 9. (ii) 2 subtracted from y is 8. (iii) Ten times a is 70. (iv) The number b divided by 5 gives 6. (v) Three-fourth of t is 15. (vi) Seven times m plus 7 gets you 77. (vii) One-fourth of a number x minus 4 gives 4. (viii) If you take away 6 from 6 times , y you get 60. (ix) If you add 3 to one-third of , z you get 30.
(i) x + 4 = 9 (ii) y - 2 = 8 (iii) 10a = 70 (iv) b/5=6 (v) 3/4(t) = 15 (vi) 7m+7 = 77 (vii) x/4 - 4 = 4 (viii) 6y-6=60 (ix) z/3+3=30 Class 7 Maths Chapter 4 Exercise 4.1 for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-4/
(i) x + 4 = 9 (ii) y – 2 = 8
(iii) 10a = 70 (iv) b/5=6
(v) 3/4(t) = 15 (vi) 7m+7 = 77
(vii) x/4 – 4 = 4 (viii) 6y-6=60
(ix) z/3+3=30
Class 7 Maths Chapter 4 Exercise 4.1
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-4/
Solve the following equations by trial and error method: 5p+2 =17
5p+2 =17 Putting p = -3 in L.H.S. 5(-3)+2 = -15+2 =-13 ∵ -13 ≠ 17 Therefore, p = -3 is not the solution. Putting p = -2 in L.H.S. 5(-2)+2 = -10+2 =-8 ∵ -8 ≠ 17 Therefore, p = -2 is not the solution. Putting p = -1 in L.H.S. 5(-1)+2 = -5+2 =-3 ∵ -3 ≠ 17 Therefore, p = -1 is not the solution. PuttingRead more
5p+2 =17
Putting p = -3 in L.H.S. 5(-3)+2 = -15+2 =-13
∵ -13 ≠ 17 Therefore, p = -3 is not the solution.
Putting p = -2 in L.H.S. 5(-2)+2 = -10+2 =-8
∵ -8 ≠ 17 Therefore, p = -2 is not the solution.
Putting p = -1 in L.H.S. 5(-1)+2 = -5+2 =-3
∵ -3 ≠ 17 Therefore, p = -1 is not the solution.
Putting p = 0 in L.H.S. 5(0)+2 = 0+2 = 2
∵ 2 ≠ 17 Therefore, p = 0 is not the solution.
Putting p = 1 in L.H.S. 5(1)+2 = 5+2 = 7
∵ 7 ≠ 17 Therefore, p = 1 is not the solution.
Putting p = 2 in L.H.S. 5(2)+2 = 10+2 = 12
∵ 12 ≠ 17 Therefore, p = 2 is not the solution.
Putting p = 3 in L.H.S. 5(3)+2 = 15+2 = 17
∵ 17 ≠ 17 Therefore, p = 3 is the solution.
Class 7 Maths Chapter 4 Exercise 4.1
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-4/
Check whether the value given in the brackets is a solution to the given equation or not: n+5=19(n=1)
n+5=19(n=1) Putting n = 1 in L.H.S., 1 + 5 = 6 ∵ L.H.S. ≠ R.H.S., ∴ n=1 is not the solution of given equation. Class 7 Maths Chapter 4 Exercise 4.1 for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-4/
n+5=19(n=1)
Putting n = 1 in L.H.S.,
1 + 5 = 6
∵ L.H.S. ≠ R.H.S.,
∴ n=1 is not the solution of given equation.
Class 7 Maths Chapter 4 Exercise 4.1
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-4/