Chapter 4 Simple Equations Ex 4.1 Solutions
Question - 1 : - Complete the last column of the table. | S. No. | Equation | Value | Say, whether the equation is satisfied. (Yes/No) |
| (i) | x + 3 = 0 | x = 3 | |
| (ii) | x + 3 = 0 | x = 0 | |
| (iii) | x + 3 = 0 | x = -3 | |
| (iv) | x – 7 = 1 | x = 7 | |
| (v) | x – 7 = 1 | x = 8 | |
| (vi) | 5x = 25 | x = 0 | |
| (vii) | 5x = 25 | x = 5 | |
| (viii) | 5x = 25 | x = -5 | |
| (ix) | (m/3) = 2 | m = – 6 | |
| (x) | (m/3) = 2 | m = 0 | |
| (xi) | (m/3) = 2 | m = 6 | |
Answer - 1 : -
(i) x + 3 = 0
LHS = x + 3
By substituting thevalue of x = 3
Then,
LHS = 3 + 3 = 6
By comparing LHS andRHS
LHS ≠ RHS
∴No, the equation isnot satisfied.
(ii) x + 3 = 0
LHS = x + 3
By substituting thevalue of x = 0
Then,
LHS = 0 + 3 = 3
By comparing LHS andRHS
LHS ≠ RHS
∴No, the equation isnot satisfied.
(iii) x + 3 = 0
LHS = x + 3
By substituting thevalue of x = – 3
Then,
LHS = – 3 + 3 = 0
By comparing LHS andRHS
LHS = RHS
∴Yes, the equation issatisfied
(iv) x – 7 = 1
LHS = x – 7
By substituting thevalue of x = 7
Then,
LHS = 7 – 7 = 0
By comparing LHS andRHS
LHS ≠ RHS
∴No, the equation isnot satisfied
(v) x – 7 = 1
LHS = x – 7
By substituting thevalue of x = 8
Then,
LHS = 8 – 7 = 1
By comparing LHS andRHS
LHS = RHS
∴Yes, the equation issatisfied.
(vi) 5x = 25
LHS = 5x
By substituting thevalue of x = 0
Then,
LHS = 5 × 0 = 0
By comparing LHS andRHS
LHS ≠ RHS
∴No, the equation isnot satisfied.
(vii) 5x = 25
LHS = 5x
By substituting thevalue of x = 5
Then,
LHS = 5 × 5 = 25
By comparing LHS andRHS
LHS = RHS
∴Yes, the equation issatisfied.
(viii) 5x = 25
LHS = 5x
By substituting thevalue of x = -5
Then,
LHS = 5 × (-5) = – 25
By comparing LHS andRHS
LHS ≠ RHS
∴No, the equation isnot satisfied.
(ix) m/3 = 2
LHS = m/3
By substituting thevalue of m = – 6
Then,
LHS = -6/3 = – 2
By comparing LHS andRHS
LHS ≠ RHS
∴No, the equation isnot satisfied.
(x) m/3 = 2
LHS = m/3
By substituting thevalue of m = 0
Then,
LHS = 0/3 = 0
By comparing LHS andRHS
LHS ≠ RHS
∴No, the equation isnot satisfied.
(xi) m/3 = 2
LHS = m/3
By substituting thevalue of m = 6
Then,
LHS = 6/3 = 2
By comparing LHS andRHS
LHS = RHS
∴Yes, the equation issatisfied.
| S. No. | 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 | Yes |
| (viii) | 5x = 25 | x = -5 | No |
| (ix) | (m/3) = 2 | m = – 6 | No |
| (x) | (m/3) = 2 | m = 0 | No |
| (xi) | (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 - 2 : -
(a)
LHS = n + 5
By substituting the value of n = 1
Then,
LHS = n + 5
= 1 + 5
= 6
By comparing LHS and RHS
6 ≠ 19
LHS ≠ RHS
Hence, the value of n = 1 is not a solution to the given equation n + 5 = 19.
(b)
LHS = 7n + 5
By substituting the value of n = -2
Then,
LHS = 7n + 5
= (7 × (-2)) + 5
= – 14 + 5
= – 9
By comparing LHS and RHS
-9 ≠ 19
LHS ≠ RHS
Hence, the value of n = -2 is not a solution to the given equation 7n + 5 = 19.
(c)
LHS = 7n + 5
By substituting the value of n = 2
Then,
LHS = 7n + 5
= (7 × (2)) + 5
= 14 + 5
= 19
By comparing LHS and RHS
19 = 19
LHS = RHS
Hence, the value of n = 2 is a solution to the given equation 7n + 5 = 19.
(d)
LHS = 4p – 3
By substituting the value of p = 1
Then,
LHS = 4p – 3
= (4 × 1) – 3
= 4 – 3
= 1
By comparing LHS and RHS
1 ≠ 13
LHS ≠ RHS
Hence, the value of p = 1 is not a solution to the given equation 4p – 3 = 13.
(e)
LHS = 4p – 3
By substituting the value of p = – 4
Then,
LHS = 4p – 3
= (4 × (-4)) – 3
= -16 – 3
= -19
By comparing LHS and RHS
-19 ≠ 13
LHS ≠ RHS
Hence, the value of p = -4 is not a solution to the given equation 4p – 3 = 13.
(f)
LHS = 4p – 3
By substituting the value of p = 0
Then,
LHS = 4p – 3
= (4 × 0) – 3
= 0 – 3
= -3
By comparing LHS and RHS
– 3 ≠ 13
LHS ≠ RHS
Hence, the value of p = 0 is not a solution to the given equation 4p – 3 = 13.
Question - 3 : - Solve the following equations by trial and error method:
(i) 5p + 2 = 17
(ii) 3m – 14 = 4
Answer - 3 : -
(i)
LHS = 5p + 2
By substituting thevalue of p = 0
Then,
LHS = 5p + 2
= (5 × 0) + 2
= 0 + 2
= 2
By comparing LHS andRHS
2 ≠ 17
LHS ≠ RHS
Hence, the value of p= 0 is not a solution to the given equation.
Let, p = 1
LHS = 5p + 2
= (5 × 1) + 2
= 5 + 2
= 7
By comparing LHS andRHS
7 ≠ 17
LHS ≠ RHS
Hence, the value of p= 1 is not a solution to the given equation.
Let, p = 2
LHS = 5p + 2
= (5 × 2) + 2
= 10 + 2
= 12
By comparing LHS andRHS
12 ≠ 17
LHS ≠ RHS
Hence, the value of p= 2 is not a solution to the given equation.
Let, p = 3
LHS = 5p + 2
= (5 × 3) + 2
= 15 + 2
= 17
By comparing LHS andRHS
17 = 17
LHS = RHS
Hence, the value of p= 3 is a solution to the given equation.
(ii)
LHS = 3m – 14
By substituting thevalue of m = 3
Then,
LHS = 3m – 14
= (3 × 3) – 14
= 9 – 14
= – 5
By comparing LHS andRHS
-5 ≠ 4
LHS ≠ RHS
Hence, the value of m= 3 is not a solution to the given equation.
Let, m = 4
LHS = 3m – 14
= (3 × 4) – 14
= 12 – 14
= – 2
By comparing LHS andRHS
-2 ≠ 4
LHS ≠ RHS
Hence, the value of m= 4 is not a solution to the given equation.
Let, m = 5
LHS = 3m – 14
= (3 × 5) – 14
= 15 – 14
= 1
By comparing LHS andRHS
1 ≠ 4
LHS ≠ RHS
Hence, the value of m= 5 is not a solution to the given equation.
Let, m = 6
LHS = 3m – 14
= (3 × 6) – 14
= 18 – 14
= 4
By comparing LHS andRHS
4 = 4
LHS = RHS
Hence, the value of m= 6 is a solution to the given equation.
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 - 4 : -
(i)
The above statement can be written in the equation form as,
= x + 4 = 9
(ii)
The above statement can be written in the equation form as,
= y – 2 = 8
(iii)
The above statement can be written in the equation form as,
= 10a = 70
(iv)
The above statement can be written in the equation form as,
= (b/5) = 6
(v)
The above statement can be written in the equation form as,
= ¾t = 15
(vi)
The above statement can be written in the equation form as,
Seven times m is 7m
= 7m + 7 = 77
(vii)
The above statement can be written in the equation form as,
One-fourth of a number x is x/4
= x/4 – 4 = 4
(viii)
The above statement can be written in the equation form as,
6 times of y is 6y
= 6y – 6 = 60
(ix)
The above statement can be written in the equation form as,
One-third of z is z/3
= 3 + z/3 = 30
Question - 5 : - Write the following equations in statement forms:
(i) p + 4 = 15
(ii) m – 7 = 3
(iii) 2m = 7
(iv) m/5 = 3
(v) (3m)/5 = 6
(vi) 3p + 4 = 25
(vii) 4p – 2 = 18
(viii) p/2 + 2 = 8
Answer - 5 : -
(i)
The sum of numbers p and 4 is 15.
(ii)
7 subtracted from m is 3.
(iii)
Twice of number m is 7.
(iv)
The number m divided by 5 gives 3.
(v)
Three-fifth of m is 6.
(vi)
Three times p plus 4 gives you 25.
(vii)
Four times p minus 2 gives you 18.
(viii)
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.)
(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 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 180 degrees).
Answer - 6 : -
(i)
From the question it isgiven that,
Number of Parmit’smarbles = m
Then,
Irfan has 7 marblesmore than five times the marbles Parmit has
= 5 × Number ofParmit’s marbles + 7 = Total number of marbles Irfan having
= (5 × m) + 7 = 37
= 5m + 7 = 37
(ii)
From the question itis given that,
Let Laxmi’s age to be= y years old
Then,
Lakshmi’s father is 4years older than three times of her age
= 3 × Laxmi’s age + 4= Age of Lakshmi’s father
= (3 × y) + 4 = 49
= 3y + 4 = 49
(iii)
From the question itis given that,
Highest score in theclass = 87
Let lowest score be l
= 2 × Lowest score + 7= Highest score in the class
= (2 × l) + 7 = 87
= 2l + 7 = 87
(iv)
From the question itis given that,
We know that, the sumof angles of a triangle is 180o
Let base angle be b
Then,
Vertex angle = 2 ×base angle = 2b
= b + b + 2b = 180o
= 4b = 180o