Class 7 Maths Chapter 4 Question Answers - Simple Equations

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)
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)
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)
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)
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)
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 +
= (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
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 ≠ 4LHS ≠ 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.

Question 3. 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.
The above statement can also be written in the equation form as,
= y – 2 = 8
(iii) Ten times a is 70.
The above statement can also be written in the equation form,
= 10a = 70
(iv) The number b, when divided by 5, gives 6.
The above statement can also be written in the equation form as,
= (b/5) = 6
(v) Three-fourths of t is 15.
The above statement can also be written in the equation form,
= ¾t = 15
(vi) Seven times m plus seven will get you 77.
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.
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.
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.
The above statement can also be written in the equation form as,
One-third of z is z/3.
= 3 + z/3 = 30

Question 4. 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.)
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.)
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).
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

Question 5. 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
Seven subtracted from m is 3.
(iii) 2m = 7
Twice of number m is 7.
(iv) m/5 = 3
The number m divided by the number 5 gives 3.
(v) (3m)/5 = 6
Three-fifth of m is 6.
(vi) 3p + 4 = 25
Three times p plus four will give you 25.
(vii) 4p – 2 = 18
Four times p minus 2 gives you the number 18.
(viii) p/2 + 2 = 8
If you add half of the number p to 2, you get 8.