Class 7, Maths, Chapter 4, Exercise 4.1 Solutions
Q.1. Complete the last column of the table.
Q.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)
Ans:
(a) n + 5 = 19 (n = 1)
Putting n = 1 in LHS = n + 5
Then, = 1 + 5
= 6 ≠ 19 (RHS)
Since LHS ≠ RHS
Hence, n = 1 is not the solution of the given equation.
(b) 7n + 5 = 19; (n = -2)
Putting n = – 2 in LHS = 7n + 5
Then, = 7 × (-2) + 5
= -14 + 5
= -9 ≠ 19 (RHS)
Since, LHS ≠ RHS
Hence, n = -2 is not the solution of the given equation.
(c) 7n+ 5 = 19; (n = 2)
Putting n = 2 in LHS = 7n + 5
Then, = (7 × 2) + 5
= 14 + 5
= 19 = 19 (RHS)
Since, LHS = RHS
Hence, n = 2 is the solution of the given equation.
(d) 4p – 3 = 13; (p = 1)
Putting p = 1 in LHS = 4p -3
Then, = (4 ×1) – 3
= 4 – 3
= 1 ≠ 13 (RHS)
Since, LHS ≠ RHS
Hence, p = 1 is not the solution of the given equation.
(e) 4p – 3 = 13; (p = -4)
Putting p = -4 in LHS = 4p -3
Then, = 4 × (-4) – 3
= -16 – 3
= -19 ≠ 13 (RHS)
Since, LHS ≠ RHS
Hence, p = – 4 is not the solution of the given equation.
(f) 4p – 3 = 13; (p = 0)
Putting p = 0 in LHS = 4p – 3
Then, = 4 × (0) – 3
= 0 – 3
= – 3 ≠ 13 (RHS)
Since, LHS ≠ RHS
Hence, p = 0 is not the solution of the given equation.
Q.3. Solve the following equations by trial and error method:
(i) 5p + 2 = 17
(ii) 3m – 14 = 4
Ans:
(i) 5p + 2 = 17
Putting p = 1 in LHS = 5p +2
Then, = (5 ×1) + 2
= 5+2 = 7 ≠ 17 (RHS)
Since, LHS ≠ RHS
Therefore, p =1 is not the solution.
Putting p = 2 in LHS = 5p +2
Then, = (5 × 2) + 2
= 10 + 2 = 12 ≠ 17 (RHS)
Since, LHS ≠ RHS
Therefore, p =2 is not the solution.
Putting p = 3, LHS = 5p + 2
Then, = (5 × 3) + 2
= 15 + 2 = 17 = 17 (RHS)
Since, LHS = RHS
Therefore, the given equation is satisfied for p = 3. Thus, p = 3 is the required solution.
(ii) 3m – 14 = 4
Putting m = 1, LHS = 3m – 14
Then, = (3 ×1) – 14
= 3 – 14 = -11 ≠ 4 (RHS)
Since, LHS ≠ RHS
Therefore, m = 1 is not the solution.
Putting m = 2, LHS = 3m – 14
Then, = (3 × 2) – 14 = 6 – 14
= -8 ≠ 4 (RHS)
Since, LHS ≠ RHS
Therefore, m = 2 is not the solution.
Putting m = 3, LHS = 3m – 14
Then, = (3 × 3) – 14
= 9 – 14
= -5 ≠ 4 (RHS)
Since, LHS ≠ RHS
Therefore, m = 3 is not the solution.
Putting m = 4, LHS = 3m – 14
Then, = (3 × 4) – 14
= 12 – 14
= -2 ≠ 4 (RHS)
Since, LHS ≠ RHS
Therefore, m = 4 is not the solution.
Putting m = 5, LHS = 3m – 14
Then, = (3 × 5) – 14
= 15 – 14
= 1 ≠ 4 (RHS)
Since, LHS ≠ RHS
Therefore, m = 5 is not the solution.
Putting m = 6, LHS = 3m – 14
Then, = (3 × 6) – 14
= 18 – 14
= 4 = 4 (RHS)
Since, LHS = RHS
Therefore, m = 6 satisfied the equation. So, m = 6 is the solution.
Q.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
Ans:
(i) The sum of numbers x and 4 is 9.
x + 4 = 9
(ii) 2 subtracted from y is 8.
y – 2 = 8
(iii) Ten times a is 70.
10a = 70
(iv) The number b divided by 5 gives 6.
$\frac{\mathsf{b}}{\mathsf{5}}\mathsf{=6}$
(v) Three-fourth of t is 15.
$\frac{\mathsf{3}}{\mathsf{4}}\mathsf{t=15}$
(vi) Seven times m plus 7 gets you 77.
7m + 7 = 77
(vii) One-fourth of a number x minus 4 gives 4.
$\frac{1}{\mathsf{4}}\mathsf{x-4=4}$
(viii) If you take away 6 from 6 times y, you get 60.
6y – 6 = 60
(ix) If you add 3 to one-third of z, you get 30
$\frac{\mathsf{z}}{\mathsf{3}}\mathsf{+3=30}$
Q.5. Write the following equations in statement forms:
(i) p + 4 = 15
(ii) m – 7 = 3
(iii) 2m = 7
(iv) $\frac{m}{5}=3$
(v) $\frac{3m}{5}=6$
(vi) 3p+4=25
(vii) 4p-2=18
(viii) $\frac{p}{2}+2=8$
Ans:
(i) p + 4 = 15
The sum of p and 4 is 15.
(ii) m – 7 = 3
7 subtracted from m is 3
(iii) 2m = 7
Two times of m is 7.
(iv) $\frac{m}{5}=3$
The number m divided by 5 gives 3.
(v) $\frac{3m}{5}=6$
Three-fifth of the number m gives 6.
(vi) 3p+4=25
Three times p plus 4 gives 25.
(vii) 4p-2=18
Take away 2 from Four times of p gives 18.
(viii) $\frac{\mathsf{p}}{\mathsf{2}}\mathsf{+2=8}$
Two added to half of number p gives 8.
Q.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).
Ans:
(i) Let m be the number Parmit’s marbles.
Irfan has 7 marbles more than five times than Parmit has.
∴ Number of marble Irfan has = 5m + 7
Total number of marble Irfan has = 37.
Therefore, the required equation is 5m + 7 = 37
(ii) Let the age of Laxmi be y years.
Laxmi’s father is 4 years older than three times Laxmi’s age
∴ Laxmi’s father’s age = 3y + 4
And age of the Laxmi’s father = 49
Therefore, the required equation is 3y + 4 = 49
(iii) Let the lowest score be l.
Highest marks obtained by a student is twice the lowest marks plus 7.
∴ The highest score = 2l + 1
But, the highest score obtained = 87.
Therefore, the required equation is 2l + 1 = 87
(iv) Let each base angle be ‘b’ degrees.
The vertex angle is twice either base angle.
∴ Vertex angle of the triangle = 2b
Sum of the angles of a triangle = 180°
2b + b +b = 1800
4b = 1800
∴ Required equation is 4b = 180°