Q.3 Pick out the solution from the values given in the bracket next to each equation.
Show that the other values do not satisfy the equation.
(a) 5m = 60 (10, 5, 12, 15)
(b) n + 12 (12, 8, 20, 0)
(c) p – 5 = 5 (0, 10, 5 ,– 5)
(d) \(\frac{q}{2} \) = 7 (7, 2, 10, 14)
(e) r – 4 = 0 (4, -4, 8, 0)
(f) x + 4 = 2 (-2, 0, 2, 4)


Answer :

(a) 5m = 60 (10, 5 , 12, 15) For m = 10, LHS = 5 x 10 = 50, RHS = 60 Here, LHS \(\neq \)RHS
\(\therefore \)m = 10 is not the solution of the equation
For m = 5, LHS = 5×5 = 25, RHS = 60
Here, LHS \(\neq \) RHS
\(\therefore \) m = 5 is not the solution of the equation
For m = 12, LHS = 5 x 12 = 60, RHS = 60
Here, LHS = RHS
\(\therefore \) m = 12 is the solution of the equation
For m = 15 LHS = 5 x 15 = 75, RHS = 60
Here, LHS \(\neq \) RHS
\(\therefore \) m = 15 is not the solution of the equation
(b) n + 12 = 20 (12, 8, 20, 0)
For n = 12, LHS = 12 + 12 = 24, RHS = 20
Here, LHS \(\neq \) RHS
\(\therefore \) n = 12 is not the solution of the equation
For n = 8, LHS = 8 + 12 = 20, RHS = 20
Here, LHS = RHS
\(\therefore \) n = 8 is the solution of the equation
For n = 20, LHS = 20 + 12 = 32, RHS = 20
Here, LHS \(\neq \) RHS
\(\therefore \) n = 20 is not the solution of the equation
For n = 0, LHS = 0 + 12 – 12, RHS = 20
Here, LHS \(\neq \) RHS
\(\therefore \) n= 0 is not the solution of the equation
(c) p – 5 = 5 (0, 10, 5, -5)
For p = 0, LHS = 0 – 5 = -5, RHS = 5
Here, LHS \(\neq \) RHS
\(\therefore \) p = 0 is not the solution of the equation
For p = 10, LHS = 10 – 5 = 5, RHS = 5
Here, LHS = RHS
\(\therefore \) p = 10 is the solution of the equation
For p = 5, LHS = 5-5-0, RHS = 5
Here LHS \(\neq \) RHS
\(\therefore \) p = 5 is not the solution of the equation
For p = 5, LHS = 5 – 5 = 0, RHS = 5
Here, LHS \(\neq \) RHS
\(\therefore \) p = -5 is not the solution of the equation
(d) q/2 = 7 (7, 2, 10, 14)
For q = 7, LHS = 7/2 , RHS = 7
Here LHS \(\neq \) RHS
\(\therefore \) q = 7 is not the solution of the equation
For q = 2, LHS = 2/2 = 1, RHS = 7
Here, LHS \(\neq \) RHS
\(\therefore \) q = 2 is not the solution of the equation
For q = 10, LHS = 10/2 = 5, RHS = 7
Here, LHS \(\neq \) RHS
For q = 14, LHS = 14/2 = 7, RHS = 7
Here, LHS = RHS
\(\therefore \) q = 14 is the solution of the equation
(e) r – 4 = 0 (4, -4, 8, 0)
For r = 4, LHS = 4 – 4 = 0, RHS = 0
Here, LHS = RHS
\(\therefore \) r = 4 is the solution of the equation
For r = -4, LHS = -4 – 4 = -8, RHS = 0
Here, LHS \(\neq \) RHS
\(\therefore \) r = -4 is not the solution of the equation
For r = 8, LHS = 8 – 4 = 4, RHS = 0
Here, LHS \(\neq \) RHS
For r = 8 is not the solution of the equation
For r = 0, LHS = 0 – 4 = – 4, RHS = 0
Here, LHS \(\neq \) RHS
\(\therefore \) r = 0 is not the solution of the equation
(f) x + 4 = 2 (-2, 0, 2, 4)
For x = -2, LHS = -2 + 4 = 2, RHS = 2
Here, LHS – RHS
\(\therefore \) x = -2 is the solution of the equation
For x = 0, LHS = 0 + 4 – 4, RHS = 2
Here, LHS \(\neq \) RHS
\(\therefore \) x = 0 is not the solution of the equation
For x = – 2, LHS = -2 + 4 – 6, RHS = 2
Here, LHS \(\neq \) RHS
\(\therefore \) x = 2 is not the solution of the equation
For r = 4, LHS = 4 + 4 = 8, RHS = 2
Here, LHS \(\neq \) RHS
\(\therefore \)x = 4 is not the solution of the equation

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