EXERCISE 4.1

EXERCISE 4.1

1). Complete the last column of the table.

Ans:
I) X+3=0
L.H.S X+3
Substituting the value of X=3
LHS, 3+3=6
Therefore, LHS ≠ 0
Therefore, LHS ≠ RHS
II) X+3=0
Substituting the value of X=0
LHS, X+3
0+3= 3
Therefore, LHS≠ 0
Therefore, LHS ≠RHS
III) X+3=0
Substituting the value of X = -3
LHS, X+3
-3 +3 =0
Therefore LHS =0
Therefore LHS = RHS
IV)X-7=1
Substituting the value of X=7
LHS, X – 7
7 – 7=0
Therefore LHS≠1
Therefore LHS≠RHS
V)X – 7 = 1
Substituting the value of X = 8
LHS, X – 7
8 – 7 = 1
Therefore, LHS = 1
Therefore, LHS = RHS
VI)5X = 25
Substituting the value of X = 0
LHS, 5X
5 x 0 = 0
Therefore LHS ≠ 25
Therefore, LHS ≠ RHS
VII)5X = 25
Substituting the value of X = 5
LHS , 5X
5 x 5 = 25
Therefore , LHS = 25
Therefore, LHS = RHS
VIII)5X = 25
Substituting the value of X = -5
LHS, 5X
5 x (-5) = -25
Therefore, LHS ≠ 25
Therefore, LHS ≠ RHS
IX) m/3= 2
Substituting the value of m= -6
LHS, m/3= (-6)/3= -2
Therefore, LHS ≠2
Therefore, LHS ≠RHS
X) m/3= 2
Substituting the value of m=0
LHS, m/3= 0/3=0
Therefore, LHS ≠2
Therefore, LHS ≠RHS
XI)m/3= 2
Substituting the value of m=6
LHS, m/3= 6/3= 2
Therefore, LHS = 2
Therefore, LHS = RHS

2) Check whether the value given in the brackets is a solution to the given equations or not:
a) n+ 5 = 19 ( n=1)
Ans: LHS = n+5
Substituting the value of n=1
LHS = n+5 = 1+5 = 6
Therefore, LHS ≠ RHS
Hence, the value of n =1 is not a solution to the given equation.
b) 7n+5= 19 ( n-2)
Ans: LHS = 7n+5
Substituting the value of n= -2
LHS = 7n+5 = 7x (-2) +5 = -14 + 5 = -9
Therefore LHS ≠ RHS
Hence, the value of n = -2 is not a solution to the given equation.
c) 7n+5=19 (n=2)
Ans: LHS = 7n+5
Substituting the value of n=2
LHS = 7n+5 = 7x (2) + 5 = 14+ 5 = 19
Therefore, LHS = RHS
Hence, the value of n=2 is a solution to the given equation.
d) 4p-3 = 13 ( p = 1 )
Ans: LHS = 4p – 3
Substituting the value of p = 1
LHS = 4p – 3= 4x 1 – 3 = 4 – 3 = 1
Therefore, LHS ≠ RHS
Hence, the value of p=1 is not a solution to the given equation.
e) 4p – 3 = 13 ( p = -4 )
Ans: LHS = 4p – 3
Substituting the value of p = -4
LHS = 4p – 3 = 4x (-4) – 3 = -16 – 3= -19
Therefore, LHS ≠ RHS
Hence, the value of p = -4 is not a solution to the given equation.
f) 4p – 3= 13 ( p = 0)
Ans: substituting the value of p=0
LHS = 4p – 3 = 4×0 – 3= 0 – 3 = – 3
Therefore LHS ≠RHS
Hence, the value of p=0 is not the solution to the given equation.

3) Solve the following by trial and error method.
a) 5p + 2 = 17
Ans: Taking the value of p = 0
LHS, = 5p + 2
= (5 x 0) + 2 = 0 + 2 = 2
Therefore, LHS ≠RHS
Hence, p =0 is not a solution to the equation.
Taking the value of p = 1
LHS = 5p + 2
= (5 x 1) + 2 = 5 + 2 = 7
Therefore, LHS ≠RHS
Hence p=1 is not a solution to the equation.
Taking the value of p = 2
LHS = 5p + 2
= (5 x 2) + 2 = 10 + 2 = 12
Therefore LHS ≠RHS
Hence, p=2 is not a solution to the equation.
Taking the value of p = 3
LHS = 5p + 2
= (5 x 3) + 2 = 15 + 2 = 17
Therefore LHS = RHS
Hence, p=3 is a solution to the equation.
b) 3m – 14 = 4
Ans: Taking the value of m = 0
LHS = 3m – 14
= (3 x 0) – 14 = 0 – 14 = -14
Therefore, LHS ≠ RHS
Hence, m=0 is not a solution to the equation.
Taking m = 1
LHS = 3m – 14
= (3 x 1 ) – 14 = 3 – 14 = -11
Therefore, LHS ≠RHS
Hence, m=1 is not a solution to the equation.
Taking m=2
LHS = 3m – 14
= (3 x 2) – 14 = 6 – 14 = – 8
Therefore, LHS ≠RHS
Hence m=2 is not a solution to the equation.
Taking m=3
LHS = 3m – 14
= (3 x 3) – 14 = 9 – 14 = -5
Therefore, LHS ≠ RHS
Hence, m=3 is not a solution to the equation.
Taking m=4
LHS = 3m – 14
= (3 x 4) – 14 = 12 – 14 = -2
Therefore, LHS≠RHS
Hence, m=4 is not a solution to the equation.
Taking m=5
LHS = 3m – 14
= (3 x 5) – 14 = 15 – 14 = 1
Therefore, LHS≠RHS
Hence, m=5 is not a solution to the equation.
Taking m = 6
LHS = 3m – 14
= (3 x 6) – 14 = 18 – 14 = 4
Therefore, LHS = RHS
Hence, m=6 is a solution to the equation.

4) Write equations for the following statements:
a) The sum of numbers x and 4 is 9.
Ans: x + 4 = 9
b) 2 subtracted from y is 8.
Ans: y – 2 = 8
c) Ten times a is 70.
Ans: 10a = 70
d) The number b divided by 5 gives 6.
Ans: b/5= 6
e) Three – fourth of t is 15.
Ans: 3t/4= 15
f) Seven times m plus 7 gets you 77
Ans: 7m + 7 = 77
f) One – fourth of a number x minus 4 gives 4.
Ans: x/4– 4 = 4
g) If you take away 6 from 6 times 6, you get 60.
Ans: 6y – 6 = 60
h) If you add 3 to one- third of z, you get 30.
Ans: z/3+ 3 = 30

5) Write the following equations in statement forms.
i) p + 4 = 15
Ans: The sum of p and 4 is 15
ii) m – 7 = 3
Ans: 7 subtracted from m = 3
iii) 2m = 7
Ans: 2 times m is 7
iv) m/5= 3
Ans: One- fifth of m is 3
v) 3m/5= 6
Ans: Three-fifth of m is 6
vi) 3p + 4 = 25
Ans: Three times p plus 4 gives 25
vii) 4p – 2 = 18
Ans: 2 subtracted from 4 times p is 18
viii) p/2+ 2 = 8
Ans: 2 added to half of p gives 8

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 n to be the number of Parmit’s marbles.)
Ans: Let the number of Parmit’s marbles be m
Irfan has 7 marbles more than 5 times Parmit’s marbles
5m + 7
Total no. of marbles with Irfan is 37
Therefore, 5m + 7 = 37
ii) Laxmi’s father is 49 years old. He is 4 years older than three Laxmi’s age. (Take Laxmi’s age to be y years.)
Ans: Let Laxmi’s age be ‘y’ years old
Laxmi’s father is 4 years older than 3 times Laxmi’s age
3y + 4 = Age of Laxmi’s father
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.)
Ans: Let the lowest score be l
Twice the lowest marks plus 7 = highest marks
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).
Ans: Let the base angle be b
Vertex angle = 2 x b
= 2b
We know the sum of angles of a triangle is 180’.
b + b + 2b = 180
4b = 180