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Please find the detailed Solutions for class 7th Maths Chapter 4 Simple Equations Exercise 4.1 Solutions, you can check all the solutions from here chapter wise
NCERT Class 7th Mathematics Chapter 4 Simple Equations Exercise 4.1 Solutions
1. Complete the last column of the table.
Say, whether the equation is satisfied. (Yes/ No)
(i) x + 3 = 0, x = 3
Sol: Let us put x=3 in the equation
L.H.S. = 3+3 = 6 ≠ R.H.S.
No, the equation is not satisfied.
(ii) x + 3 = 0 x = 0
Sol: Let us put x=0 in the equation
L.H.S. = 0+3 = 3 ≠ R.H.S.
No, the equation is not satisfied.
(iii) x + 3 = 0 x = – 3
Sol: Let us put x= -3 in the equation
L.H.S. = -3+3 = 0 = R.H.S.
Yes, the equation is satisfied.
(iv) x – 7 = 1 x = 7
Sol: Let us put x=7 in the equation
L.H.S. = 7-7 = 0 ≠ R.H.S.
No, the equation is not satisfied.
(v) x – 7 = 1 x = 8
Sol: Let us put x=8 in the equation
L.H.S. = 8-7 = 1 = R.H.S.
Yes, the equation is satisfied.
(vi) 5x = 25 x = 0
Sol: Let us put x=0 in the equation
L.H.S. = 5*0 = 0 ≠ R.H.S.
No, the equation is not satisfied.
(vii) 5x = 25 x = 5
Sol: Let us put x=5 in the equation
L.H.S. = 5*5 = 25 = R.H.S.
Yes, the equation is satisfied.
(viii) 5x = 25 x = – 5
Sol: Let us put x=-5 in the equation
L.H.S. = 5*-5 = -25 ≠ R.H.S.
No, the equation is not satisfied.
(ix) m/3 = 2 m = – 6
Sol: Let us put m=-6 in the equation
L.H.S. = -6/3 = -2 ≠ R.H.S.
No, the equation is not satisfied.
(x) m/3 = 2 m = 0
Sol: Let us put m=0 in the equation
L.H.S. = 0/3 = 0 ≠ R.H.S.
No, the equation is not satisfied.
(xi) m/3 = 2 m = 6
Sol: Let us put m=6 in the equation
L.H.S. = 6/3 = 2 = R.H.S.
Yes, the equation is satisfied.
2. Check whether the value given in the brackets is a solution to the given equation or
not:
(a) n + 5 = 19 (n = 1)
Sol: Let us put n=1 in the equation
L.H.S. = 1+5 = 6 ≠ R.H.S.
n=1, is not the solution of the given equation.
(b) 7n + 5 = 19 (n = – 2)
Sol: Let us put n=-2 in the equation
L.H.S. = 7*(-2) + 5 = -14+5 = -9 ≠ R.H.S.
n=-2, is not the solution of the given equation.
(c) 7n + 5 = 19 (n = 2)
Sol: Let us put n=2 in the equation
L.H.S. = 7*2+5 = 14+5 =19 = R.H.S.
n=2, is the solution of the given equation.
(d) 4p – 3 = 13 (p = 1)
Sol: Let us put p=1 in the equation
L.H.S. = (4*1)-3 = 4-3 = 1 ≠ R.H.S.
p=1, is not the solution of the given equation.
(e) 4p – 3 = 13 (p = – 4)
Sol: Let us put p=-4 in the equation
L.H.S. = 4*(-4)-3 = -16-3 = -19 ≠ R.H.S.
n=1, is not the solution of the given equation.
(f) 4p – 3 = 13 (p = 0)
Sol: Let us put p=0 in the equation
L.H.S. = 4*0-3 = 0-3 = -3 ≠ R.H.S.
p=0, is not the solution of the given equation.
3. Solve the following equations by trial and error method:
(i) 5p + 2 = 17
Sol: Putting p=1
5*1 + 2 = 5+2= 7 ≠ R.H.S.
Putting p=2
5*2 + 2 = 10+2= 12 ≠ R.H.S.
Putting p=3
5*3 + 2 = 15+2= 17 = R.H.S.
So, p=3 is the solution of the equation.
(ii) 3m – 14 = 4
Sol: Putting m=4
3*4 – 14 = 12-14 = -2 ≠ R.H.S.
Putting m=5
3*5 – 14 = 15-14 = 1 ≠ R.H.S.
Putting m=6
3*6 – 14 = 18-14 = 4 = R.H.S.
So, m=6 is the solution of the given equation.
4. Write equations for the following statements:
(i) The sum of numbers x and 4 is 9.
Ans: x+4 = 9
(ii) The difference between y and 2 is 8.
Ans: y-2 = 8
(iii) Ten times a is 70.
Ans: 10a = 70
(iv) The number b divided by 5 gives 6.
Ans: b/5 = 6
(v) Three fourth of t is 15.
Ans: (3/4)*t = 15
(vi) Seven times m plus 7 gets you 77.
Ans: 7m+7 = 77
(vii) One fourth of a number minus 4 gives 4.
Ans: (1/4)x-4 = 4
(viii) If you take away 6 from 6 times y, you get 60.
Ans: 6y-6 = 60
(ix) If you add 3 to one third of z, you get 30.
Ans: (1/3)z+3 = 30
5. Write the following equations in statement forms:
(i) p + 4 = 15
Ans: The sum of p and 4 is 15
(ii) m – 7 = 3
Ans: 7 subtracted from m is 3
(iii) 2m = 7
Ans: Twice of a number m is 7
(iv) m/5 = 3
Ans: One fifth of a number is 3
(v) 3m/5 = 6
Ans: Three fifth of a number is 6
(vi) 3p + 4 = 25
Ans: Three times of p when added to 4 is 25
(vii) 4p – 2 = 18
Ans: 4 subtracted from Four times of p is 18
(viii) p/2 + 2 = 8
Ans: 2 added to half of a number p is 8
6. 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. (Take m to be the number of Parmit’s marbles.)
Sol: Let Parmit has m marbes
Then Irfan has 7 marbles more than five times the marbles Parmit has
So Irfan has 5m+7 marbles.
But Irfan Has 37 marbles
So, 5m+7 = 37
(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.)
Sol: Let the age of Laxmi = y
Then his father’s age = 3y+4
But his father’s age is 49 years
So, 3y+4 = 49
(iii) The teacher tells the class that the highest marks obtained by a student in her
class is twice the lowest marks plus 7. The highest score is 87. (Take the lowest
score to be l.)
Sol: Let the lowest marks be l
2*Lowest marks + 7 = Highest marks
2*l + 7 = 87
So, 2l+7 = 87
(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
degrees).
Sol: In an isosceles triangle 2 angles are of equal measure
Let base angle = b
Vertex angle = 2* base angle = 2b
Now, in a triangle sum of all angles = 180
b+b+2b = 180 degrees
4b = 180 degrees
You can find the solutions for previous exercise from here
- Class 7th Maths – Exercise 3.4 Solutions
- Class 7th Maths – Exercise 3.3 Solutions
- Class 7th Maths – Exercise 3.2 Solutions
- Class 7th Maths – Exercise 3.1 Solutions
- Class 7th Maths – Exercise 2.7 Solutions
- Class 7th Maths – Exercise 2.6 Solutions
- Class 7th Maths – Exercise 2.5 Solutions
- Class 7th Maths – Exercise 2.4 Solutions
- Class 7th Maths – Exercise 2.3 Solutions
- Class 7th Maths – Exercise 2.2 Solutions
- Class 7th Maths – Exercise 2.1 Solutions
- Class 7th Maths – Exercise 1.4 Solutions
- Class 7th Maths – Exercise 1.3 Solutions
