NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.1, are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.1.
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.1
Question 1.
Complete the last column of the table.
Solution:
Question 2.
Check whether the value given in the brackets is a solution to the given equation or not.
(a) n + 5 = 19 (n = 1)
(b) 7n + 5 = 19 (n = – 2)
(c) 7n + 5 = 19 (n = 2)
(d) 4p – 3 = 13 (p = 1)
(e) 4p – 3 = 13 (p = – 4)
(f) 4p – 3 = 13 (p = 0).
Solution:
(a) n + 5 = 19 (n = 1)
L.H.S. = n + 5 = 1 + 5 | when n = 1 = 5
R.H.S. = 19
∵ L.H.S. ≠ R.H.S.
∴ n = 1 is not a solution to the given equation n + 5 = 19.
(b) 7n + 5 = 19 (n = – 2)
L.H.S. = 7n + 5 = 7(- 2) + 5 | when n = – 2 = – 14 + 5 = – 9
R.H.S. = 19
∵ L.H.S. ≠ R.H.S.
∴ n = – 2 is not a solution to the given equation 7n + 5 = 19.
(c) 7n + 5 = 19 (n = 2)
L.H.S. = In + 5 = 7(2) + 5 | when n = 2 = 14 + 5 = 19 = R.H.S.
∴ n = 2 is a solution to the given equation 7n + 5 = 19.
(d) 4p – 3 = 13 (p = 1)
L.H.S. = 4p – 3 = 4(1) – 3 | when p = 1 = 4 – 3 = 1
R.H.S. = 13
∵ L.H.S. ≠ R.H.S.
∴ p = 1 is not a solution to the given equation 4p – 3 = 13.
(e) 4p – 3 = 13 (p = – 4)
L.H.S. = 4p – 3 = 4(- 4) – 3 , | when p = – 4 = – 16 – 3 = – 19
R.H.S. = 13
∵ L.H.S. ≠ R.H.S.
∴ p = – 4 is not a solution to the given equation
4p – 3 = 13.
(f) 4p – 3 = 13 (p = 0)
L.H.S. = 4 (p) – 3 = 4(0) – 3 | when p = 0 = 0 – 3 = – 3
R.H.S. = 13
∵ L.H.S. ≠ R.H.S.
∴ p = 0 is not a solution to the given equation 4p – 3 = 13.
Question 3.
Solve the following equations by trial and error method.
- 5p + 2 = 17
- 3m – 14 = 4.
Solution:
(i) 5p + 2 = 17
So, p = 3 is the solution of the given equation 5p + 2 = 17.
(ii) 3m – 14 = 4
So, m = 6 is the solution of the given equation 3m – 14 = 4.
Question 4.
Write equations for the following statements.
- The sum of numbers x and 4 is 9.
- 2 subtracted from y is 8.
- Ten times a is 70.
- The number b divided by 5 gives 6.
- Three-fourth oft is 15.
- Seven times m plus 7 gets you 77.
- One-fourth of a number x minus 4 gives 4.
- If you take away 6 from 6 times y, you get 60.
- If you add 3 to one-third of z, you get 30.
Solution:
- x + 4 = 9
- y – 2 = 8
- 10 α = 70
- \(\frac { b }{ 5 } \) = 6
- \(\frac { 3 }{ 4 } \) t = 15
- 7m + 7 = 77
- \(\frac { 1 }{ 4 } \) x – 4 = 4
- 6y – 6 = 60
- \(\frac { 1 }{ 3 } \) z + 3 = 30
Question 5.
Write the following equations in statement forms:
- p + 4 = 15
- m – 7 = 3
- 2m = 7
- \(\frac { m }{ 5 } \) = 3
- \(\frac { 3m }{ 5 } \) = 6
- 3p + 4 = 25
- 4p – 2 = 18
- \(\frac { p }{ 2 } \) + 2 = 8.
Solution:
- The sum of p and 4 is 15.
- 7 subtracted from m is 3.
- Twice a number m is 7.
- One-fifth of a number m is 3.
- Three-fifth of a number m is 6.
- Three times a numberp when added to 4 gives 25.
- 2 subtracted from four times a number p is 18.
- Add 2 to half of a number p to get 8.
Question 6.
Set up an equation in the following cases:
- 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.)
- 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.)
- 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.)
- 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 triangle is 180 degrees).
Solution:
(i) Let Parmit has m marbles. Then, five times the marbles Parmit has = 5m
Irfan has 7 marbles more than five times the marbles Parmit has = 5m + 7,
i.e., Irfan has (5m + 7) marbles
But it is given that Irfan has 37 marbles. Therefore, 5m + 7 = 37
(ii) Let the age of Laxmi be y years.
Then, three times Laxmi’s age = 3y years Laxmi’s father is 4 years older than three times Laxmi’s age,
i.e., Age of Laxmi’s father = (3y + 4) years But it is given that Laxmi’s father is 49 years old Therefore, 3y + 4 = 49
(iii) Let the lowest marks be l.
Then, twice the lowest marks = 21
Highest score obtained by a student in her class is twice the lowest marks plus 7, i.e.,
Highest score = 2l + 7
But this is given to be 87
Therefore, 2l + 7 = 87
(iv) Let the base angle be b in degrees. Then the vertex angle is 2b in degrees.
∵ Sum of the angles of a triangle is 180 degrees.
∴ 2 b + b + b = 180° or 4 b = 180°
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