NCERT Class 7 Maths Chapter 4 Simple Equations Exercise 4.1

Chapter 4 Simple Equations Exercise 4.1

Question 1: Complete the last column of the table.


S. No.

Equation

Value

Say, whether the equation is satisfied. (Yes/No)

1

x + 3 = 0

x = 3

= 3 + 3 ≠ 0

Therefore, the equation is not satisfied.

2

x + 3 = 0

x = 0

= 0 + 3 ≠ 0

Therefore, the equation is not satisfied.

3

x + 3 = 0

x = -3

= -3 + 3 = 0

Therefore, the equation is satisfied.

4

x – 7 = 1

x = 7

= 7 - 7 ≠ 1

Therefore, the equation is not satisfied.

5

x – 7 = 1

x = 8

= 8 - 7 = 1

Therefore, the equation is satisfied.

6

5x = 25

x = 0

= 5 x 0 ≠ 25

Therefore, the equation is not satisfied.

7

5x = 25

x = 5

= 5 x 5 = 25

Therefore, the equation is satisfied.

8

5x = 25

x = -5

= 5 x -5 ≠ 25

Therefore, the equation is not satisfied.

9

(m/3) = 2

m = – 6

= -6/2 ≠ 2

Therefore, the equation is not satisfied.

10

(m/3) = 2

m = 0

= 0/3 ≠ 2

Therefore, the equation is not satisfied.

11

(m/3) = 2

m = 6

= 6/2 = 3

Therefore, the equation is satisfied.


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)
Answer:

LHS

RHS

1 + 5

19

6

19


6 ≠ 19
Therefore, the value n = 1 is not a solution to n + 5 = 19.

(b) 7n + 5 = 19 (n = – 2)
Answer:

LHS

RHS

7 x (-2) + 5
= -14 + 5

19

-9

19


-9 ≠ 19
Therefore, the value n = -2 is not a solution to 7n + 5 = 19.

(c) 7n + 5 = 19 (n = 2)
Answer:


LHS

RHS

7 x 2 + 5
= 14 + 5

19

9

19


9 = 19
Therefore, the value n = 2 is a solution to 7n + 5 = 19.

(d) 4p – 3 = 13 (p = 1)
Answer:


LHS

RHS

4 x 1 - 3

13

1

13


1 ≠ 13
Therefore, the value p = 1 is not a solution to 4p – 3 = 13.

(e) 4p – 3 = 13 (p = – 4)
Answer:

LHS

RHS

4 x (-4) - 3

13

= -16 - 3

13


-19 ≠ 13
Therefore, the value p = -4 is not a solution to 4p – 3 = 13.

(f) 4p – 3 = 13 (p = 0)
Answer:


LHS

RHS

4 x 0 - 3

13

= 0 - 3

13


-3 ≠ 13
Therefore, the value p = 0 is not a solution to 4p – 3 = 13.

Question 3: Solve the following equations by trial and error method:
(i) 5p + 2 = 17
Answer:

p

5p +2

= 17

0

= 5 x 0 + 2
= 0 + 2
= 2

No

1

= 5 x 1 + 2
= 5 + 2
= 7

No

2

= 5 x 2 + 2
= 10 + 2
= 12

No

3

= 5 x 3 + 2
= 15 + 2
= 17

Yes


(ii) 3m – 14 = 4
Answer:


 m

3m - 14

= 4

0

= 3 x 0 - 14
= 0 - 14
= -14

No

1

= 3 x 1 - 14
= 3 - 14
= -11

No

2

= 3 x 2 - 14
= 6 - 14
= -8

No

3

= 3 x 3 - 14
= 9 - 14
= -5

No

4

= 3 x 4 - 14
= 12 - 14
= -2

No

5

= 3 x 5 - 14
= 15 - 14
= 1

No

6

= 3 x 6 - 14
= 18 - 14
= 4

Yes


Question 4: Write equations for the following statements:
(i) The sum of numbers x and 4 is 9.
Answer:
x + 4 = 9

(ii) 2 subtracted from y is 8.
Answer:
y - 2 = 8

(iii) Ten times a is 70.
Answer:
10a = 70

(iv) The number b divided by 5 gives 6.
Answer:
b/5 = 6

(v) Three-fourth of t is 15.
Answer:
3/4 t = 15

(vi) Seven times m plus 7 gets you 77.
Answer:
7m + 7 =  77

(vii) One-fourth of a number x minus 4 gives 4.
Answer:
1/4 x - 4 = 4

(viii) If you take away 6 from 6 times y, you get 60.
Answer:
6y - 6 = 60

(ix) If you add 3 to one-third of z, you get 30.
Answer:
1/3 z + 3 = 30

Question 5: Write the following equations in statement forms:
(i) p + 4 = 15
Answer:
The sum of numbers p and 4 is 15.

(ii) m – 7 = 3
Answer:
7 subtracted from m is 3.

(iii) 2m = 7
Answer:
Twice of number m is 7.

(iv) m/5 = 3
Answer:
The number m divided by 5 gives 3.

(v) (3m)/5 = 6
Answer:
Three-fifth of m is 6.

(vi) 3p + 4 = 25
Answer:
Three times p plus 4 gives you 25.

(vii) 4p – 2 = 18
Answer:
Four times p minus 2 gives you 18.

(viii) p/2 + 2 = 8
Answer:
If you add half of a number p to 2, you get 8.

Question 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.)
Answer:

Number of marbles Parmit has = m
Irfan has 7 marbles more than five times the marbles Parmit has
= 5 x number of Parmit’s marbles + 7 = Total number of marbles Irfan having
= (5 x m) + 7 = 37
= 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.)
Answer:

Let Laxmi’s age be = y years old
Laxmi’s father is 4 years older than three times of her age
= 3 x Laxmi’s age + 4 = Age of Laxmi’s father
= (3 x y) + 4 = 49
= 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.)
Answer:

Highest score in the class = 87
Let the lowest score be l.
= 2 x lowest score + 7 = highest marks in the class
= (2 x l) + 7 = 87
= 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).
Answer:

Let the base angle be b.
Vertex angle = 2 x base angle = 2b
= b + b + 2b = 180°
= 4b = 180°

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