Class 6 - Mathematics
Chapter - Simple Equations : Exercise 4.1
Top Block 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 = 0 | |
| 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 | x = -5 | |
| 9. | m⁄3 = 2 | m = -6 | |
| 10. | m⁄3 = 2 | m = 0 | |
| 11. | m⁄3 = 2 | m = 6 |
Answer :
| S. No. | Equation | Value | Say, whether the Equation is satisfied. (Yes / No) |
|---|---|---|---|
| 1. | x + 3 = 0 | x = 3 | Yes |
| 2. | x + 3 = 0 | x = 0 | No |
| 3. | x + 3 = 0 | x = -3 | No |
| 4. | x – 7 = 1 | x = 7 | No |
| 5. | x – 7 = 1 | x = 8 | Yes |
| 6. | 5x = 25 | x = 0 | No |
| 7. | 5x = 25 | x = 5 | Yes |
| 8. | 5x = 25 | x = -5 | No |
| 9. | m⁄3 = 2 | m = -6 | No |
| 10. | m⁄3 = 2 | m = 0 | No |
| 11. | 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 ( n = 1)
Putting n=1 in L.H.S.,
1 + 5 = 6
∵ L.H.S. ≠ R.H.S.,
∴ n = 1 is not the solution of given equation.
(b) 7n + 5 = 19 (n = –2)
Putting n = –2 in L.H.S.,
7 (–2) + 5 = –14 + 5 = –9
∵ L.H.S. ≠ R.H.S.,
∴ n = –2 is not the solution of given equation.
(c) 7n + 5 = 19 (n = 2)
Putting n = 2 in L.H.S.,
7(2) + 5 = 14 + 5 = 19
∵ L.H.S. = R.H.S.,
∴ n = 2 is the solution of given equation.
(d) 4p – 3 = 13 (p = 1)
Putting p = 1 in L.H.S.,
4(1) – 3 = 4 – 3 = 1
∵ L.H.S. ≠ R.H.S.,
∴ p = 1 is not the solution of given equation.
(e) 4p – 3 = 13( p = –4)
Putting p = –4 in L.H.S.,
4(–4) – 3 = –16 – 3 = –19
∵ L.H.S. ≠ R.H.S.,
∴ p = –4 is not the solution of given equation.
(f) 4p – 3 = 13 ( p = 0)
Putting p = 0 in L.H.S.,
4(0) – 3 = 0 – 3 = –3
∵ L.H.S. ≠ R.H.S.,
∴ p = 0 is not the solution of given equation.
Question : 3.Solve the following equations by trial and error method:
(i) 5p + 2 = 17
(ii) 3m – 14 = 4
Answer :
(i) 5p + 2 = 17
Putting p = –3 in L.H.S. 5(–3) + 2 = –15 + 2 = –13
∵ –13 ≠ 17 Therefore, p = –3 is not the solution.
Putting p = –2 in L.H.S. 5(–2) + 2= –10 + 2 = –8
∵ –8 ≠ 17 Therefore, p = –2 is not the solution.
Putting p = –1 in L.H.S. 5(–1) + 2 = –5 + 2 = –3
∵ –3 ≠ 17 Therefore, p = –1 is not the solution.
Putting p = 0 in L.H.S. 5 (0) + 2 = 0 + 2 = 2
∵ 2 ≠ 17 Therefore, p=0 is not the solution.
Putting p = 1 in L.H.S. 5(1) + 2 = 5 + 2 = 7
∵ 7 ≠ 17 Therefore, p = 1 is not the solution.
Putting p = 2 in L.H.S. 5(2) + 2 = 10 + 2 = 12
∵ 12 ≠ 17 Therefore, p = 2 is not the solution.
Putting p = 3 in L.H.S. 5(3) + 2 = 15 + 2 = 17
∵ 17 = 17 Therefore, p = 3 is the solution.
(ii) 3m – 14 = 4
Putting m = –2 in L.H.S. 3(–2) – 14 = –6 – 14 = – 20
∵ –20 ≠ 4 Therefore, m = –2 is not the solution.
Putting m = –1 in L.H.S. 3(–1) – 14 = –3 – 14 = –17
∵ –17 ≠ 4 Therefore, m = –1 is not the solution.
Putting m = 0 in L.H.S. 3(0) – 14 = 0 – 14 = –14
∵ –14 ≠ 4 Therefore, m = 0 is not the solution.
Putting m = 1 in L.H.S. 3(1) – 14 = 3 – 14 = –11
∵ –11 ≠ 4 Therefore, m = 1 is not the solution.
Putting m = 2 in L.H.S. 3(2) – 14 = 6 – 14 = –8
∵ –8 ≠ 4 Therefore, m = 2 is not the solution.
Putting m = 3 in L.H.S. 3(3) – 14 = 9 – 14 = –5
∵ –5 ≠ 4 Therefore, m = 3 is not the solution.
Putting m = 4 in L.H.S. 3(4) – 14 = 12 – 14 = –2
∵ –2 ≠ 4 Therefore, m = 4 is not the solution.
Putting m = 5 in L.H.S. 3(5) – 14 = 15 – 14 = 1
∵ 1 ≠ 4 Therefore, m = 5 is not the solution.
Putting m = 6 in L.H.S. 3(6) – 14 = 18 – 14 = 4
∵ 4 = 4 Therefore, m = 6 is the solution.
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) 10a = 70
(iv) b5 = 6
(v) 34t = 15
(vi) 7m + 7 = 77
(vii) x4 – 4 = 4
(viii) 6y – 6 = 60
(ix) z3 + 3 = 30
Question : 5. Write the following equations in statement form:
(i) p + 4 = 15
(ii) m – 7 = 3
(iii) 2m = 7
(iv) m5 = 3
(v) 3m5 = 6
(vi) 3p + 4 = 25
(vii) 4p – 2 = 18
(viii) p2 + 2 = 8
Answer :
(i) The sum of numbers p and 4 is 15.
(ii) 7 subtracted from m is 3.
(iii) Two times m is 7.
(iv) The number m is divided by 5 gives 3.
(v) Three-fifth of the number m is 6.
(vi) Three times p plus 4 gets 25.
(vii) If you take away 2 from 4 times p, you get 18.
(viii) If you added 2 to half is p, you get 8.
Mddle block 1
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. (Tale 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 are 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 angle. (Let the base angle be b in degrees. Remember that the sum of angles of a triangle is 180o .)
Answer :
(i) Let m be the number of Parmit’s marbles.
∴ 5m + 7 = 37
(ii) Let the age of Laxmi be y years
.
∴ 3y + 4 = 49
(iii) Let the lowest score be l.
∴ 2l + 7 = 87
(iv) Let the base angle of the isosceles triangle be b, so vertex angle = 2b.
∴ 2b + b + b = 180o ⇒ 4b = 180o [Angle sum property of a ∆]