These NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.1 Questions and Answers are prepared by our highly skilled subject experts.

## NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Exercise 4.1

Question 1.

Complete the last column of the table.

Answer:

Equation | Value | Say, whether the Equation is Satisfied, (Yes/No) |

(i) x + 3 = 0 | x = 3 | No |

(ii) x + 3 = 0 | x = 0 | No |

(iii) x + 3 = 0 | x = -3 | yes |

(iv) x – 7 = 1 | x = 7 | No |

(v) x – 7 = 1 | x = 8 | yes |

(vi) 5x = 25 | x = 0 | No |

(vii) 5x = 25 | x = – 5 | No |

(viii) \(\frac{m}{3}\) = 2 | m = -6 | No |

(ix) \(\frac{m}{3}\) = 2 | m = 0 | No |

(x) \(\frac{m}{3}\) = 2 | m = 6 | yes |

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)

Answer:

(a) n + 5 = 19

1 + 5 =19 (Putting n = 1)

6 ≠ 19

∴ n = 1 is not a solution.

(b) 7n + 5 =19

7(- 2) + 5 =19 (Put n = – 2)

-14 + 5 = 19

-9 ≠ 19

n = -2 is not a solution.

(c) 7n + 5 =19

7(2) + 5=19 (Put n = 2)

14 + 5 = 19

19 = 19

∴ n = 2 is a solution.

(d) 4p – 3 = 13

4(1) – 3 =13 (Put p = 1)

4 – 3 = 13

1 ≠ 13

∴ p = 1 is not a solution.

(e) 4p – 3 = 13

4(- 4) – 3 = 13 (Put p = – 4)

-16 -3 = 13

-19 ≠ 13

∴ p = – 4 is not a solution.

(f) 4p – 3 = 13

4(0) – 3 = 13 (Put p = 0)

0 – 3 = 13

-3 ≠ 13

p = 0 is not a solution.

Question 3.

Solve the following equations by trial and error method:

(i) 5p + 2 = 17

(ii) 3m – 14 = 4

Answer:

(i) 5p + 2 = 17

Put p = 0

L.H.S = 5 (0) + 2

= 0 + 2

= 2 ≠ RHS

Put p = 1

L.H.S = 5(1)+ 2

= 5 + 2

= 7 ≠ RHS

Put p = -1

L.H.S = 5 (-1) + 2

= -5 + 2

= -3 ≠ RHS

Put p = 2

L.H.S = 5 (2) + 2

= 10 + 2

= 12 ≠ RHS

Put p = -2

L.H.S = 5 (-2) + 2

= -10 + 2

= -8 ≠ 17 RHS

Put p = 3

L.H.S = 5 (3) + 2

= 15 + 2

= 17 = RHS

p = 3 is the solution of 5P + 2 = 17

(ii) 3m – 14 =4

Put m = 0

L.H.S = 3m – 14

= 3 (0) – 14

= – 14

– 14 ≠ 4

L.H.S ≠ R.H.S

Put m = 1

L.H.S = 3m -14

= 3(1) – 14

= 3 – 14

= – 11 ≠ R.H.S

L.H.S ≠ R.H.S

Put m = 2

L.H.S = 3m – 14

= 3 (2) – 14

= 6 – 14

= – 8 ≠ R.H.S

L.H.S ≠ R.H.S

Put m = 3

L.H.S = 3(3) – 14

= 9 – 14

= – 5 ≠ R.H.S

L.H.S ≠ R.H.S

Put m = 4

L.H.S = 3(4) – 14

= 12 – 14

= – 2 ≠ R.H.S

L.H.S ≠ R.H.S

Put m = 5

L.H.S = 3(5) – 14

= 15 – 14

= 1 ≠ R.H.S

L.H.S ≠ R.H.S

Put m = 6

L.H.S = 3(6) – 14

= 18 – 14

= 4 = R.H.S

L.H.S = R.H.S

m = 6 is the solution to 3m – 14 = 4

Question 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.

Answer:

(i) x + 4 = 9

(ii) y – 2 = 8

(iii) 10 a = 70

(iv) \(\frac { b }{ 5 }\) = 6

(v) \(\frac { 3t }{ 4 }\) = 15

(vi) 7m + 7 = 77

(vii) \(\frac { x }{ 4 }\) x – 4 = 4

(viii) 6 y – 6 = 60

(ix) \(\frac { z }{ 3 }\) + 3 = 30

Question 5.

Write the following equations in statement forms:

(i) p + 4 = 15

(ii) m – 7 = 3

(iii) 2m = 7

(iv) \(\frac { m }{ 5 }\) = 3

(v) \(\frac { 3m }{ 5 }\) = 6

(vi) 3p + 4 = 25

(vii) 4p – 2 = 18

(viii) \(\frac { p }{ 2 }\) + 2 = 8

Answer:

(i) The sum of p and 4 is 15.

(ii) 7 subtracted from m is 3.

(hi) Twice a number m is 7.

(iv) One-fifth of a number m is 3.

(v) Three-fifth of a number m is 6.

(vi) Three times a number p when added to 4 gives 25.

(vii) 2 subtracted from four times a number p is 18.

(viii) 2 added to half of a number p is 8.

Question 6.

Set up an equation in the following cases:

(i) Irfan says that he has 7 marbles more than live times the marbles Parmit has. Irfan has 37 marbles. (Take 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 is twice the lowest marks plus 7. The highest score is 87. (Take the lowest score to be 1.)

(iv) In an isosceles triangle, the verte x 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).

Answer:

(i) Let Parmit has m marbles.

Then, five time the marbles Parmit has = 5 m.

Irfan has 7 marbles more than five times the marbles parmit has

So, Irfan has (5m + 7) marbles

But it is given that Irfan has 37 marbles.

5m + 7 = 37

(ii) Let Laxmi’s age = y years

3 times Laxmi’s age = 3y years.

Age of Laxmi’s father = 3 times

Laxmi’s age + 4 years

= 3y + 4 years

But Laxmi’s father is 49 years old.

3y + 4 = 49

(iii) Let the lowest score (marks) = 1

Twice the lowest marks = 2 l

Since highest marks = (twice the lowest marks) + 7 = 2l + 7

But the highest marks = 87

2l + 7 = 87

(iv) Let the base angle be b degrees

The base angle of an isosceles triangle are equal.

The other base angle = b degrees

Since the vertex angle = Twice either base angle = 2b degrees.

Also, the sum of three angles of triangle = 180°

b + b + 2b = 180°

(OR)

4b = 180°