Important Questions: Simple Equations | Mathematics (Maths) Class 7 PDF Download

Q1: Check whether the given value in the brackets is the correct solution to the below-given equation or not:
(a) n + 5 = 19 (n = 1)
Ans:
LHS = n + 5
By substituting the value of n = 1
Then we get,
LHS = n + 5
= 1 + 5
= 6
By comparing the LHS and RHS
6 ≠ 19
LHS ≠ RHS
Hence we get that the value of n = 1 is not the solution to the given equation above n + 5 = 19.

(b) 7n + 5 = 19 (n = – 2)
Ans:
LHS = 7n + 5
Now, by substituting the value of n = -2
Then we get,
LHS = 7n + 5
= (7 × (-2)) + 5
= – 14 + 5
= – 9
By comparing the LHS and RHS
-9 ≠ 19
LHS ≠ RHS
Hence we know that the value of n = -2 is not the solution to the above-given equation 7n + 5 = 19.

(c) 7n + 5 = 19 (n = 2)
Ans:
LHS = 7n + 5
By substituting the value of n as = 2
Then,
LHS = 7n + 5
= (7 × (2)) + 5
= 14 + 5
= 19
By comparing both the LHS and the RHS
19 = 19
LHS = RHS
Hence, the value of n = 2 is the solution to the given equation is 7n + 5 = 19.

(d) 4p – 3 = 13 (p = 1)
Ans:
LHS = 4p – 3
By substituting the value of p = 1
Then,
LHS = 4p – 3
= (4 × 1) – 3
= 4 – 3
= 1
By comparing both the LHS and the RHS
1 ≠ 13
LHS ≠ RHS
Hence then, the value of p = 1 is not the solution to the above-given equation 4p – 3 = 13.

(e) 4p – 3 = 13 (p = – 4)
Ans:
LHS = 4p – 3
By substituting the value of p as = – 4
Then,
LHS = 4p – 3
= (4 × (-4)) – 3
= -16 – 3
= -19
By comparing both the LHS and the RHS
-19 ≠ 13
LHS ≠ RHS
Hence then, the value of p = -4 is not the solution to the above-given equation 4p – 3 = 13.

(f) 4p – 3 = 13 (p = 0)
Ans:
LHS = 4p – 3
By substituting the value of p = 0
Then,
LHS = 4p – 3
= (4 × 0) – 3
= 0 – 3
= -3
By comparing the LHS and RHS
– 3 ≠ 13
LHS ≠ RHS
Hence then, the value of p = 0 is not the solution to the above-given equation 4p – 3 = 13.

Q2: Solve the following equations by the trial and error method:
(i) 5p + 2 = 17
Ans:
LHS = 5p + 2
By substituting the value of p as = 0
Then,
LHS = 5p + 2
= (5 × 0) + 2
= 0 + 2
= 2
By comparing both the LHS and RHS
2 ≠ 17
LHS ≠ RHS
Hence then, the value of p = 0 is not the solution to the above-given equation.
Let, p = 1
LHS = 5p + 2
= (5 × 1) + 2
= 5 + 2
= 7
By comparing the LHS and the RHS
7 ≠ 17
LHS ≠ RHS
Hence then, the value of p = 1 is not the solution to the above-given equation.
Let, p = 2
LHS = 5p + 2
= (5 × 2) + 2
= 10 + 2
= 12
By comparing both the LHS and the RHS
12 ≠ 17
LHS ≠ RHS
Hence then, the value of p = 2 is not the solution to the above-given equation.
Let, p = 3
LHS = 5p + 2
= (5 × 3) + 2
= 15 + 2
= 17
By comparing both the LHS and the RHS
17 = 17
LHS = RHS
Hence, the value of p = 3 is the solution given to the given equation.

(ii) 3m – 14 = 4
Ans:
LHS = 3m – 14
By substituting the value of m as = 3
Then,
LHS = 3m – 14
= (3 × 3) – 14
= 9 – 14
= – 5
By comparing both the LHS and the RHS
-5 ≠ 4
LHS ≠ RHS
Hence then, the value of m = 3 is not the solution to the above-given equation.
Let, m = 4
LHS = 3m – 14
= (3 × 4) – 14
= 12 – 14
= – 2
By comparing both the LHS and the RHS
-2 ≠ 4
LHS ≠ RHS
Hence then, the value of m = 4 is not the solution to the above-given equation.
Let, m = 5
LHS = 3m – 14
= (3 × 5) – 14
= 15 – 14
= 1
By comparing both the LHS and the RHS
1 ≠ 4
LHS ≠ RHS
Hence then, the value of m = 5 is not the solution to the above-given equation.
Let, m = 6
LHS = 3m – 14
= (3 × 6) – 14
= 18 – 14
= 4
By comparing both the LHS and RHS
4 = 4
LHS = RHS
Hence, the value of m = 6 is the solution to the above-given equation.

Q3: Write the equations for the following statements:
(i) The sum of the numbers x and 4 is 9.
Ans:
The above statement can also be written in the equation form as,
= x + 4 = 9

(ii) 2 subtracted from y is 8.
Ans:
The above statement can also be written in the equation form as,
= y – 2 = 8

(iii) Ten times a is 70.
Ans:
The above statement can also be written in the equation form,
= 10a = 70

(iv) The number b, when divided by 5, gives 6.
Ans:
The above statement can also be written in the equation form as,
= (b/5) = 6

(v) Three-fourths of t is 15.
Ans:
The above statement can also be written in the equation form,
= ¾t = 15

(vi) Seven times m plus seven will get you 77.
Ans:
The above statement can also be written in the equation form,
Seven times m will be 7m
= 7m + 7 = 77

(vii) One-fourth of the number x minus 4 gives 4.
Ans:
The above statement can also be written in the equation form,
One-fourth of the number x is x/4.
= x/4 – 4 = 4

(viii) If you take away six from 6 times y, you get 60.
Ans:
The above statement can also be written in the equation form as,
Six times y is 6y.
= 6y – 6 = 60

(ix) If you add 3 to the one-third of z, you get 30.
Ans:
The above statement can also be written in the equation form as,
One-third of z is z/3.
= 3 + z/3 = 30

Q4: Set up an equation form in the following cases given below:
(i) Irfan says that he has seven marbles, more than five times the marbles Permit has. Irfan has 37 marbles. (Take m to be the number of Permit’s marbles.)
Ans:
From the question, it is given that,
Number of Permit’s marbles = m
Then,
Irfan has seven marbles, more than five times the marbles Permit has
= 5 × Number of Permit’s marbles + 7 = Total number of the marbles Irfan having
= (5 × m) + 7 = 37
= 5m + 7 = 37

(ii) Laxmi’s father is 49 years old. He is four years older than Laxmi three times. (Take Laxmi’s age to be y years.)
Ans:
From the question, it is given that,
Let Laxmi’s age be = y years old.
Then,
Lakshmi’s father is four years older than three times her age.
= 3 × Laxmi’s age + 4 = Age of her father
= (3 × y) + 4 = 49
= 3y + 4 = 49

(iii) A teacher tells her class that the highest marks obtained by a student in the class are twice the lowest marks obtained plus 7. The highest score is 87. (Take the lowest score to be denoted as l.)
Ans:
From the question, it is given that,
Highest score in the class = 87
Let the lowest score be equal to the l
= 2 × Lowest score + 7 = Highest score in class
= (2 × l) + 7 = 87
= 2l + 7 = 87

(iv) In the isosceles triangle, the vertex angle is twice the base angle. (Let the base angle be denoted as b in degrees. Remember that the sum of the angles of the triangle is 180 degrees).
Ans:
From the above question, it is given that,
We know that the total sum of angles of a triangle is 180o.
Let the base angle be b
Then,
Vertex angle = 2 × base angle = 2b
= b + b + 2b = 180o
= 4b = 180o

Q5: Write the following equations in statement forms:
(i) p + 4 = 15
Ans: The sum of the numbers p and 4 is 15.

(ii) m – 7 = 3
Ans: Seven subtracted from m is 3.

(iii) 2m = 7
Ans: Twice of number m is 7.

(iv) m/5 = 3
Ans: The number m divided by the number 5 gives 3.

(v) (3m)/5 = 6
Ans: Three-fifth of m is 6.

(vi) 3p + 4 = 25
Ans: Three times p plus four will give you 25.

(vii) 4p – 2 = 18
Ans: Four times p minus 2 gives you the number 18.

(viii) p/2 + 2 = 8
Ans: If you add half of the number p to 2, you get 8.

Q6: Give the steps which are used to separate the variable and then solve the equation:
(a) 3n – 2 = 46
Ans:
First, we have to add two to both the side of the equation,
Then, we receive,
= 3n – 2 + 2 = 46 + 2
= 3n = 48
Now,
We have to divide both of these sides of the equation by 3,
Then, we get,
= 3n/3 = 48/3
= n = 16

(b) 5m + 7 = 17
Ans:
First, we have to subtract seven from the both sides of the equation,
Then, we receive,
= 5m + 7 – 7 = 17 – 7
= 5m = 10
Now,
We have to divide both these sides of the equation by 5,
Then, we receive,
= 5m/5 = 10/5
= m = 2

(c) 20p/3 = 40
Ans:
First, we have to multiply both of these sides of the equation by 3,
Then, we get,
= (20p/3) × 3 = 40 × 3
= 20p = 120
Now,
We have to divide both of these sides of the equation by 20,
Then, we receive,
= 20p/20 = 120/20
= p = 6

(d) 3p/10 = 6
Ans:
First, we have to multiply both of these sides of the equation by 10,
Then, we receive,
= (3p/10) × 10 = 6 × 10
= 3p = 60
Now,
We have to divide both of these sides of the equation by 3,
Then, we get,
= 3p/3 = 60/3
= p = 20

Q7: Solve the following equations:
(a) 10p = 100
Ans:
Now,
We have to divide both of these sides of the equation by 10,
Then, we get,
= 10p/10 = 100/10
= p = 10

(b) 10p + 10 = 100
Ans:
Firstly we have to subtract 10 from both sides of the following equation,
Then, we get,
= 10p + 10 – 10 = 100 – 10
= 10p = 90
Now,
We have to divide both of these sides of the equation by 10,
Then, we get,
= 10p/10 = 90/10
= p = 9

(c) p/4 = 5
Ans:
Now,
We have to multiply both of these sides of the equation by 4,
Then, we get,
= p/4 × 4 = 5 × 4
= p = 20

(d) – p/3 = 5
Ans:
Now,
We have to multiply both of these sides of the equation by – 3,
Then, we receive,
= – p/3 × (- 3) = 5 × (- 3)
= p = – 15

(e) 3p/4 = 6
Ans:
First, we have to multiply both of these sides of the equation by 4,
Then, we get,
= (3p/4) × (4) = 6 × 4
= 3p = 24
Now,
We have to divide both of these sides of the equation by 3,
Then, we get,
= 3p/3 = 24/3
= p = 8

(f) 3s = – 9
Ans:
Now,
We have to divide both of these sides of the equation by 3,
Then, we get,
= 3s/3 = -9/3
= s = -3

(g) 3s + 12 = 0
Ans:
First, we have to subtract the number 12 from the both sides of the equation,
Then, we get,
= 3s + 12 – 12 = 0 – 12
= 3s = -12
Now,
We have to divide both of these sides of the equation by 3,
Then, we get,
= 3s/3 = -12/3
= s = – 4

(h) 3s = 0
Ans:
Now,
We have to divide both of these sides of the equation by 3,
Then, we get,
= 3s/3 = 0/3
= s = 0

(i) 2q = 6
Ans:
Now,
We have to divide both of these sides of the equation by 2,
Then, we get,
= 2q/2 = 6/2
= q = 3

(j) 2q – 6 = 0
Ans:
Firstly we have to add 6 to both the sides of the following equation,
Then, we get,
= 2q – 6 + 6 = 0 + 6
= 2q = 6
Now,
We have to divide both of these sides of the equation by 2,
Then, we get,
= 2q/2 = 6/2
= q = 3

(k) 2q + 6 = 0
Ans:
First, we have to subtract the number 6 from the both sides of the following equation,
Then, we get,
= 2q + 6 – 6 = 0 – 6
= 2q = – 6
Now,
We have to divide both of these sides of the equation by the number 2,
Then, we get,
= 2q/2 = – 6/2
= q = – 3

(l) 2q + 6 = 12
Ans:
First, we have to subtract the number 6 to the both sides of the following equation,
Then, we get,
= 2q + 6 – 6 = 12 – 6
= 2q = 6
Now,
We have to divide both of these sides of the equation by 2,
Then, we get,
= 2q/2 = 6/2
= q = 3

Q8: Solve the following equations:
(a) 2y + (5/2) = (37/2)
Ans: By transposing the fraction (5/2) from LHS to RHS, so it becomes -5/2
Then,
= 2y = (37/2) – (5/2)
= 2y = (37-5)/2
= 2y = 32/2
Now,
Divide both the side by the number 2,
= 2y/2 = (32/2)/2
= y = (32/2) × (1/2)
= y = 32/4
= y = 8

(b) 5t + 28 = 10
Ans: By transposing 28 from the LHS to RHS, it becomes -28
Then,
= 5t = 10 – 28
= 5t = – 18
Now,
Dividing both these sides by 5,
= 5t/5= -18/5
= t = -18/5

(c) (a/5) + 3 = 2
Ans: By transposing three from the LHS to RHS, it becomes -3
Then,
= a/5 = 2 – 3
= a/5 = – 1
Now,
Multiply both these sides by 5,
= (a/5) × 5= -1 × 5
= a = -5

(d) (q/4) + 7 = 5
Ans: By transposing the number 7 from the LHS to RHS, it becomes -7
Then,
= q/4 = 5 – 7
= q/4 = – 2
Now,
Multiply both these sides by 4,
= (q/4) × 4= -2 × 4
= a = -8

(e) (5/2) x = -5
Ans:
First, we have to multiply both these sides by 2,
= (5x/2) × 2 = – 5 × 2
= 5x = – 10
Now,
We have to divide both these sides by 5,
Then we get,
= 5x/5 = -10/5
= x = -2

(f) (5/2) x = 25/4
Ans:
First, we have to multiply both these sides by 2,
= (5x/2) × 2 = (25/4) × 2
= 5x = (25/2)
Now,
We have to divide both these sides by 5,
Then we get,
= 5x/5 = (25/2)/5
= x = (25/2) × (1/5)
= x = (5/2)

Q9: Solve the following equations:
(a) 2(x + 4) = 12
Ans:
Let us divide both these sides by 2,
= (2(x + 4))/2 = 12/2
= x + 4 = 6
By transposing four from the LHS to RHS, it becomes -4
= x = 6 – 4
= x = 2

(b) 3(n – 5) = 21
Ans:
Let us divide both these sides by 3,
= (3(n – 5))/3 = 21/3
= n – 5 = 7
By transposing -5 from the LHS to the RHS, it becomes 5
= n = 7 + 5
= n = 12

(c) 3(n – 5) = – 21
Ans:
Let us divide both these sides by 3,
= (3(n – 5))/3 = – 21/3
= n – 5 = -7
By transposing -5 from the LHS to the RHS, it becomes 5
= n = – 7 + 5
= n = – 2

(d) – 4(2 + x) = 8
Ans:
Let us divide both these sides by -4,
= (-4(2 + x))/ (-4) = 8/ (-4)
= 2 + x = -2
By transposing two from the LHS to the RHS, it becomes – 2
= x = -2 – 2
= x = – 4

(e) 4(2 – x) = 8
Ans:
Let us divide both these sides by 4,
= (4(2 – x))/ 4 = 8/ 4
= 2 – x = 2
By transposing the number 2 from the LHS to the RHS, it becomes – 2
= – x = 2 – 2
= – x = 0
= x = 0

Q10: Solve the following equations:
(a) 4 = 5(p – 2)
Ans:
Let us divide both these sides by 5,
= 4/5 = (5(p – 2))/5
= 4/5 = p -2
By transposing – 2 from the RHS to the LHS, it becomes 2
= (4/5) + 2 = p
= (4 + 10)/ 5 = p
= p = 14/5

(b) – 4 = 5(p – 2)
Ans:
Let us divide both these sides by 5,
= – 4/5 = (5(p – 2))/5
= – 4/5 = p -2
Now by transposing – 2 from the RHS to the LHS, it becomes 2
= – (4/5) + 2 = p
= (- 4 + 10)/ 5 = p
= p = 6/5

(c) 16 = 4 + 3(t + 2)
Ans: By transposing four from the RHS to the LHS, it becomes – 4
= 16 – 4 = 3(t + 2)
= 12 = 3(t + 2)
Let us divide both these side by 3,
= 12/3 = (3(t + 2))/ 3
= 4 = t + 2
By transposing 2 from the RHS to the LHS it becomes – 2
= 4 – 2 = t
= t = 2

(d) 4 + 5(p – 1) =34
Ans:
By transposing 4 from the LHS to the RHS it becomes – 4
= 5(p – 1) = 34 – 4
= 5(p – 1) = 30
Let us divide both these side by 5,
= (5(p – 1))/ 5 = 30/5
= p – 1 = 6
By transposing – 1 from the RHS to the LHS it becomes 1
= p = 6 + 1
= p = 7

(e) 0 = 16 + 4(m – 6)
Ans: By transposing 16 from the RHS to the LHS it becomes – 16
= 0 – 16 = 4(m – 6)
= – 16 = 4(m – 6)
Let us divide both these side by the number 4,
= – 16/4 = (4(m – 6))/ 4
= – 4 = m – 6
Now by transposing – 6 from the RHS to the LHS it becomes 6
= – 4 + 6 = m
= m = 2