# HBSE 7th Class Maths Solutions Chapter 4 Simple Equations Ex 4.1

Haryana State Board HBSE 7th Class Maths Solutions Chapter 4 Simple Equations Ex 4.1 Textbook Exercise Questions and Answers.

## Haryana Board 7th Class Maths Solutions Chapter 4 Simple Equations Exercise 4.1

Question 1.
Complete the last column of the table.

 Equation Value Say, whether the equation is satisfied, (Yes/No) 1. x + 3 = 0 x = 3 2. x + 3 = 0 x = 0 3. x + 3 = 0 x = -3 4. x – 7 = 1 x = 7 5. x – 7 = 1 x = 8 6. 5x = 25 x = 0 7. 5x = 25 x = 5 8. 5x = 25 m x = — 5 9. $$\frac{m}{3}$$ = 2 m = – 6 10. $$\frac{m}{3}$$ = 2 m = 0 11. $$\frac{m}{3}$$ = 2 m = 6

Solution:
1. Put x = 3, x + 3 = 0++
3 + 3 = 0
⇒ 6 ≠ 0 → No.

2. Put x = 0, x + 3 = 0
0 + 3 = 0
⇒ 3 ≠ 0 → No.

3. Put x = – 3, x + 3 = 0
-3 + 3 = 0
⇒ 0 ≠ 0 → Yes.

4. Put x = 7, x – 7 = 1
7 – 7 = 1
⇒ 0 ≠ 1 → No.

5. Put x = 8, x – 7 = 1
8 – 7 = 1
⇒ 1 = 1 → Yes.

6. Put x = 0, 5x = 25
5×0 = 25
⇒ 0 ≠ 25 → No.

7. Put x = 5, 5x = 25
5 x 5 = 25
⇒ 25 = 25 → Yes.

8. Put x = – 5, 5x = 25
5 x (- 5) = 25
⇒ -25 ≠ 25 → No.

9. Put m = – 6, $$\frac{m}{3}$$ = 2
$$\frac{-6}{3}$$ = 2
⇒ -2 ≠ 2 → No.

10. Put m = 0, $$\frac{m}{3}$$ = 2
$$\frac{0}{3}$$ = 2
⇒ 0 ≠ 2→ No.

11. Put m = 6, $$\frac{m}{3}$$ = 2
$$\frac{6}{3}$$ = 2
⇒ 2 = 2→ 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)
4p – 3 = 13 (P = 0)
Solution:
(a) Put n = 1, n + 5 = 19
1 + 5 = 19
1 + 5 = 19
6 ≠ 19
So, the value given in the brackets is not a solution of equation.

(b) Put n = – 2, In+ 5 =19
7 x (- 2) + 5 = 19 -14 + 5 =19
-11 ≠ 19
So, the value given in the brackets is not a solution of equation.

(c) Put n = 2, 7n + 5 =19
7 x 2 + 5 =19
14 + 5 = 19
19 = 19
So, the value given in the brackets is a solution of equation.

(d) Put p = 1, 4p – 3 =13
4 x 1 – 3 = 13
4 – 3 =13
⇒ 1 ≠ 13
So, the value given in the brackets is a not solution of equation.

(e) Put p = – 4, 4p – 3 =13
4 x (- 4) – 3 =13
-16 – 3 =13
-19 ≠ 13
So, the value given in the brackets is not a solution of equation.

(f) Put p = 0, 4p – 3 =13
4 x 0 – 3 = 13
0 – 3 = 13
-3 ≠ 13
So, the value given in the brackets is not a solution of equation.

Question 3.
Solve the following equations by a trial and error method.
(i) 5p + 2 = 17 (ii) 3m -14 = 4
Solution:
(i) 5p + 2 = 17

 p L.H.S. R.H.S. 1 5 x 1+ 2 = 5 + 2 =7 17 2 5 x 2 + 2 = 10 + 2 = 12 17 3 5 x 3 + 2 = 15 + 2= 17 17

When p = 3, then L.H.S. = R.H.S.
∴ p = 3 is the solution of the equation.

(ii) 3m – 14 = 4

 m L.H.S. R.H.S. 1 3 x 1-14 = 3 – 14 = -11 4 2 3 x 2-14 = 6 – 14 = -8 4 3 3 x 3-14 = 9 – 14 = -5 4 4 3 x 4 -14 = 12 – 14 = -2 4 5 3 x 5 – 14 = 15 – 14 = 1 4 6 3 x 6 – 14 = 18 – 14 = 4 4

When m = 6, then L.H.S. = R.H.S.
∴ m = 6 is the solution of equation.

Question 4.
Write equations for the following statements :
(i) The sum of numbers x and 4 is 9.
(ii) The difference ofy and 2 is 8.
(Ui) Ten times a is 70.
(iv) The number b divided by 5 gives 6*.
(v) Three fourth oft is 15.
(vi) Seven times m plus 7 gets you 77.
(vii) One fourth of a number x minus 4 leaves 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.
Solution:
(i) x + 4 = 9
(ii) y – 2 = 8
(iii) 10a = 70
(iv) b/5 = 6
(v) $$\frac{3}{4}$$t = 15
(vi) 7m+ 7 = 77
(vii) $$\frac{1}{4}$$n – 4 = 4
(viii) 6y – 6 = 60
(ix) $$z + 3 Question 5. Write the following equations in statement forms : (i) p + 4 = 15 (ii) m-7 = 3 (iii) 2m = 7 (iv) [latex]\frac{m}{5}$$ = 3
(v) $$\frac{3m}{5}$$ = 6
(vi) 3p + 4= 25
(vii) 4p-2 = 18
(viii) $$\frac{p}{2}$$ + 2 = 8
Solution:
(i) The sum of numbers p and 4 is 15.
(ii) Taking away 7 from m gives 3.
(iii) Two times a number m is 7.
(iv) One fifths of m is 3.
(v) Three fifth of m is 6.
(ut) Add 4 to three times p to get 25.
(vii) Taking away 2 from four times ofp gives 18.
(viii) Half of a number p plus 2 is 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 Permit’s marbles).
(ii) Laxmi’s father is 49 years old. He is 4 years older than three times Laxmi’s age. (Take Laxmi’ 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 l).
(iv) In an isosceles triangle, the vertex angle is twice either base angel. (Let the base angle be b in degrees. Remember that the sum of angles of a triangles is 180 degrees).
Solution:
(i) Let us take m to be the number of parmit’s marbles.
Hence the required equation, 5m + 7 = 37
(ii) Let us take y to be the Laxmi’s age.
Hence the required equation, 3y + 4 = 49
(iii) Let us take the lowest score to be l.
Hence the required equation, 2l + 7 = 87
(iv) Let the base angle be b°,
Hence the required equation, 2 b° + b° + b°= 180 or, 4 b° =180°