NCERT solutions for class 7 Maths Chapter 4 Simple Equations

Solution for Exercise 4.2

Q.1 Given first the step you will use to separate the variable and then solve the equation:
(a) x – 1 = 0
(b) x + 1 = 0
(c) x – 1 = 5
(d) x + 6 = 2
(e) y – 4 = -7
(f) y -4 = 4
(g) y + 4 = 4
(h) y + 4 = -4

(a) x-1=0
= x – 1 + 1 = 0 + 1

= x = 1
(b) x + 1 = 0
Or, x=0-1 = -1
(c) x-1=5
Or, x= 5+1 = 6
(d) x+6= 2
Or, x=2-6= -4
(e) y-4 =-7
Or, y=-7+4=-3
(f) y-4=4
Or, y=4+4=8
(g) y+4=4
Or, y=4-4=0
(h) y+4=-4
Or, y=-4-4=-8

Q.2 Give first the step you will use to separate the variable and then solve the following equation:
(a) 3l = 42
(b) b/2= 6
(c) p/7=4
(d) 4x =25
(e) 8y=36
(f) z/3=5/4
(g) a/5=7/15
(h) 20t=-10

(a) 3I =42
Or, I=42/3 = 14
(b) b/2 = 6
Or, b=6×2=12
(c) p/7 = 4
Or, p= 4×7 = 28
(d) 4x=25
Or, x=25/4
(e) 8y=36
Or, y= 36 /8 = 36 ÷4/8÷4 = 9/2
(f) z/3= 5/4
Or, z= (5/4) ×3 =15/4
(g) a/5=7/15
Or, a= (7/15)× 5 =7/3
(h) 20t = -10
Or, t = -10 /20 = -1/2

Q.3 Give the steps you will use to separate the variables and then solve the equation:
(a) 3n – 2 = 46
(b) 5m + 7 = 17
(c) 20p/3=40
(d) 3p/10=6

(a) 3n-2 =46
Or, 3n = 46+2
Or, n = 48/3 = 16
(b) 5m +7 = 17
Or, 5m = 17 -7
Or, m = 10/5 = 2
(c) 20p/3 = 40
Or, 20p= 40×3
Or, p= 120 /20 = 6
(d) 3p/10 = 6
Or, 3p = 6×10
Or, p = 60 /3 = 20

Q.4 Solve the following equations:
(a) 10p = 100
(b) 10p + 10 = 100
(c) p/4=5
(d) -p/3=5
(e) 3p/4=6
(f) 3s = -9
(g) 3s + 12 = 0
(h) 3s = 0
(i) 2q = 6
(j) 2q – 6 = 0
(k) 2q + 6 = 0
(l) 2q + 6 = 12

(a) 10p=100
Or, p=100/10 = 10
(b) 10p+ 10 = 100
Or, 10p=100-10
Or, 10p=90
Or, p = 90/10 = 9
(c) p/4 = 5
Or, p= 5×4 = 20
(d) -p/3 = 5
Or, -p= 5×3
Or, -p= 15
Or, p=-15
(e) 3p/4 = 6
Or, 3p= 6×4
Or, p= 24/3
Or, p= 8
(f) 3s= -9
Or, s = -9/3 = -3
(g) 3s+12 =0
Or, 3s = -12
Or, s = -12 /3
Or, s = -4
(h) 3s = 0
Or, s =0/3
Or, s= 0
(i) 2q =6
Or, q =6/2= 3
(j) 2q-6= 0
Or, 2q= 6
Or, q= 6/2
Or, q=3
(k) 2q+6= 0
Or, 2q = 0-6
Or, q = -6/2 = -3
(l) 2q+6 = 12
Or, 2q= 12-6
Or, 2q= 6
Or, q=6/2 =3

Solution for Exercise 4.1

Q.1 Complete the last column of the table.

(i) x + 3 = 0
LHS = x + 3
By substituting the value of x = 3
Then, LHS = 3 + 3 = 6
By comparing LHS and RHS
LHS $$\neq$$RHS
Therefore No,the equation is not satisfied
(ii) x + 3 = 0
LHS = x + 3
By substituting the value of x = 0
Then, LHS = 0 + 3 = 3
By comparing LHS and RHS
LHS $$\neq$$RHS
Therefore No, the equation is not satisfied.
(iii) x + 3 = 0
LHS = x + 3
By substituting the value of x = – 3 Then, LHS = – 3 + 3 = 0
By comparing LHS and RHS
LHS = RHS
Hence,
Yes, the equation is satisfied
(iv) x – 7 = 1
LHS = x – 7
By substituting the value of x = 7
Then, LHS = 7 – 7 = 0
By comparing LHS and RHS
LHS $$\neq$$ RHS
Hence
No, the equation is not satisfied
(v) x – 7 = 1
LHS = x – 7
By substituting the value of x = 8
Then, LHS = 8 – 7 = 1
By comparing LHS and RHS
LHS = RHS
Hence,
Yes, the equation is satisfied.
(vi) 5x = 25
LHS = 5x
By substituting the value of x = 0
Then,LHS = 5 × 0 = 0
By comparing LHS and RHS
LHS $$\neq$$RHS
Hence,
No, the equation is not satisfied.
(vii) 5x = 25
LHS = 5x
By substituting the value of x = 5
Then,LHS = 5 × 5 = 25
By comparing LHS and RHS
LHS = RHS
Hence,
Yes, the equation is satisfied.
(viii) 5x = 25
LHS = 5x
By substituting the value of x = -5
Then, LHS = 5 × (-5) = – 25
By comparing LHS and RHS
LHS $$\neq$$ RHS
Hence,
No, the equation is not satisfied.
(ix) m/3 = 2
LHS = m/3
By substituting the value of m = – 6
Then,LHS = -6/3 = – 2
By comparing LHS and RHS
LHS $$\neq$$RHS
Hence,
No, the equation is not satisfied.
(x) m/3 = 2
LHS = m/3
By substituting the value of m = 0
Then, LHS = 0/3 = 0
By comparing LHS and RHS
LHS $$\neq$$RHS
Hence,
No, the equation is not satisfied.
(xi) m/3 = 2
LHS = m/3
By substituting the value of m = 6
Then, LHS = 6/3 = 2
By comparing LHS and RHS
LHS = RHS
Hence,
Yes, the equation is satisfied.

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; in – -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)

(a) n + 5 = 19 (n = 1) Put n = 1 in LHS
1 + 5 = 6
RHS =19
Since LHS ? RHS
Thus n = 1 is not the solution of the given equation. "n+5=19"
(b) 7n + 5 = 19 (n = -2) Put n = -2 in LHS
-2×7 + 5 = -9
RHS =19
Since LHS $$\neq$$RHS
Thus n = -2 is not the solution of the given equation. "7n+5=19"
(c) 7n + 5 = 19 (n = 2) Put n = 2 in LHS
7×2 + 5 = 19
RHS =19
Since LHS = RHS
Thus n = 2 is the solution of the given equation. "7n+5=19"
(d) 4p-3 = 13 (p= 1) Put p = 1 in LHS
4×1-3 = 1
RHS =13
Since LHS $$\neq$$RHS
Thus p = 1 is not the solution of the given equation. "4p-3=13"
(e) 4p - 3= 13 (p = -4) Put p = -4 in LHS
4×(-4) - 3 = -19
RHS =13
Since LHS $$\neq$$RHS
Thus p = -4 is not the solution of the given equation. "4p-3=13"
(f) 4p-3=13 (p= 0) Put p = 0 in LHS
0×4-3 = -3
RHS =13
Since LHS $$\neq$$ RHS
Thus p = 0 is not the solution of the given equation. "4p-3=13"

Q.3 Solve the following equations by trial and error method:
(i) 5p + 2 = 17
(ii) 3m – 14 = 4

(i) 5p + 2 = 17
For p = 1, LHS
= 5 × 1 + 2 = 5 + 2 = 7 $$\neq$$RHS=17
For p = 2, LHS = 5 × 2 + 2
= 10 + 2 = 12 $$\neq$$ RHS=17
For p = 3, LHS = 5 × 3 + 2
= 15 + 2 = 17 = RHS
Since the given equation is satisfied for p = 3 Thus, p = 3 is the required solution.
(ii) 3m – 14 = 4
For m = 1, LHS = 3 × 1 – 14
= 3 – 14 = -11 $$\neq$$ RHS = 4
For m = 2, LHS = 3 × 2 – 14 = 6 – 14 = -8 $$\neq$$ RHS= 4
For m = 3, LHS = 3 × 3 – 14 = 9 – 14 = -5 $$\neq$$ RHS =4
Form m = 4, LHS = 3 × 4 – 14 = 12 – 14 = -2 $$\neq$$ RHS= 4
For m = 5, LHS = 3 × 5 – 14 = 15 – 14 = -1 $$\neq$$ RHS =4
For m = 6, LHS = 3 × 6 – 14 = 18 – 14 = 4 =RHS
Since, the given equation is satisfied for m = 6. Thus, m = 6 is the required 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.

(i) x+4=9
(ii) y-2 =8
(iii) 10a=70
(iv) b/5=6
(v) $$\frac{3}{4}$$ t=15
(vi) 7m+7=77
(vii) x/4 – 4 = 4
(viii) 6y - 6 =60
(ix) 3+z/3=30

Q.5 Write the following equations in statement forms:
(i) p + 4 = 15
(ii) m-7=3
(Iii) 2m=7
(iv) m/5= 3
(v) 3m/5=6
(vi) 3p+4=25
(vii) 4p -2 =18
(viii) p/2+2=8

(i) The sum of numbers p and four is fifteen.
(ii) seven subtract from m gives three.
(iii) two times of a number is seven.
(iv) The number m divided by five gives three.
(v) Three-fifth of m is six.
(vi) Three times p plus four gives you twenty five.
(vii) four time of p minus two gives eighteen.
(viii) half of p plus two is equals eight

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).

(i) Let m be the Parmit’s marbles.
Therefore, Irfan’s marble = 5m + 7
Total number of Irfan’s marble is given by 37.
Thus, the required equation is 5m + 7 = 37

(ii) Let Laxmi’s age bey years.
Therefore, Laxmi’s father’s age = 3y + 4
But the Laxmi’s father age is given by 49
Thus the required equation is 3y + 4 = 49

(iii) Let the lowest score be l. A
Therefore, The highest score = 2l + 7
But the highest score is given by 87.
Thus, the required equation is 2l + 7 = 87

(iv) Let each base angle be ‘b’ degrees. Therefore Vertex angle of the triangle = 2b Sum of the angles of a triangle = 180° Therefore Required equation is b + b + 2b = 180° or 4b = 180°

Solution for Exercise 4.3

Q.1 Solve the following equations:
(a) 2y + (5/2) = (37/2)
(b) 5t + 28 = 10
(c) (a/5) + 3 = 2
(d) (q/4) + 7 = 5
(e) (5/2) x = -5
(f) (5/2) x = 25/4
(g) 7m + (19/2) = 13
(h) 6z + 10 = – 2
(i) (3/2) l = 2/3
(j) (2b/3) – 5 = 3

(a) 2y + (5/2) = (37/2)
= 2y = (37/2) – (5/2)
= 2y = (37-5)/2
= 2y = 32/2
= y = 16/2 = 8
(b) 5t+28= 10
= 5t = 10 -28
= t= -18/5
(c) a/5+3 =2
= a/5= 2-3
= a=-1×5=-5
(d) q/4+7 = 5
= q/4=5-7
=a= -2 ×4= -8
(e) (5/2)x = -5
= 5x=-5×2
= x= -10/5=-2
(f) 5x/2=25
= 5x = 25×2
= x = 50/5 =10
(g) 7m+19/2 =13
= 7m = 13 -19/2 = 7m = (26-19)/2
= 7m= 7/2
= m = 7/2×1/7 =1/2
(h) 6z + 10 = -2
= 6z = -2 -10
= z = -12/ 6 = -2
(i) (3/2)I = 2/3
= I = 2/3 × 2/3 =4/9
(j) (2b/3 ) - 5 = 3
= 2b/3 = 3+5
= 2b/3 = 8
= b = (8×3)/2 = 12

Q.2 Solve the following equations:
(?) 2(x + 4) = 12
(b) 3(n – 5) = 21
(c) 3(n – 5) = -21
(d) -4(2 + x) = 8
(e) 4(2 – x) = 8

(a) 2(x+4) = 12
= x+4 = 12/2
= x = 6-4 = 2
(b) 3(n-5 ) = 21
= n-5 = 21 /3
= n =7+5= 12
(c) 3(n-5) = -21
= n-5 = -21 /3
= n= -7 +5 = = -2
(d) -4 (2+x) = 8
= 2+x = 8/-4
= x = -2 -2 = -4
(e) 4(2- x ) = 8
= 2- x = 8/4
= -x = 2-2
= x = 0

Q.3 Solve the following equations:
(a) 4 = 5(p- 2)
(b) -4 = 5(p – 2)
(c) 16 = 4 + 3 (t + 2)
(d) 4 + 5(p – 1) = 34
(e) 0 = 16 + 4(m – 6)

(a) 4 = 5 ( p-2 )
= 5 (p-2) = 4
= p-2 = 4/5
= p = (4/5)+2
= p = 14 /5
(b) -4 = 5(p – 2)
= 5 (p-2) = -4
= p-2 = -4/5
= p = (-4/5) + 2
= p = 6/ 5
(c) 16 = 4 +3 (t +2)
= 4+3 (t+2) = 16
= 3(t+2) = 16-4
= t+2 = 12/3
= t = 4-2= 2
(d) 4 + 5 (p-1) = 34
= 5(p-1) = 34 -4
= p-1 = 30 /5
= p = 6 +1 = 7
(e) 0 = 16 +4( m- 6)
= 16 + 4(m-6) = 0
= 4(m-6 ) = -16
= m-6 = -16/4
= m -4+6 = 2

Q.5 (a) Construct 3 equations starting with x = 2
(b) Construct 3 equations starting with x = -2.

(a) Possible equations are: 5x + 2 = 12;
x/7=2/7 ;
8x –7 = 9
(b) Possible equations are:
8x= -16;
3x + 9 = 3;
5x + 19 = 9

Solution for Exercise 4.4

Q.1 Set up equations and solve them to find the unknown numbers in the following cases:
(a) Add 4 to eight times a number; you get 60.
(b) One-fifth of a number minus 4 gives 3.
(c) If I take three-fourths of a number and add 3 to it, I get 21.
(d) When subtracted 11 from twice a number, the result was 15.
(e) Munna subtracts thrice the number of notebooks he has from 50, he finds the result to be 8.
(f) Ibenhal thinks of a number. If she adds 19 to it and divides the sum by 5, she will get 8.
(g) Anwar thinks of a number. If he takes away 7 from 52 of the numbers, the result is 23.

(a) Let the required number be x.
According to question,
8x +4 = 60
= 8x = 60 -4
= x = 56 /8
= x = 7
(b) Let the required number be x.
According to question,
(1/5)x - 4 = 3
= (1/5)x = 3+4
= x = 7 ×5 = 35
= x = 35
(c) Let the required number be x.
According to question,
(3/4)x +3 = 21
= (3/4)x = 21-3
= x = (18 ×4)/3 = 24
= x = 24
(d) Let the required unknown number be x.
According to question,
2x-11= 15
= 2x = 15+11
= x = 26 /2 = 13
= x =13
(e) Let the required number be x. According to question,
50 - 3x = 8
= -3x = 8 - 50
= x = -42 / -3 = 14
= x = 42
(f) Let the required number be x.
According to question,
( x+19)/5=8
= x +19 = 8×5
= x = 40 -19
= x= 21
(g) Let the required number be x.
According to question,
(5/2)x =23 +7
= (5/2)x = 30
= x = (30×2)/5
= x = 12

Q.2 Solve the following:
(a) 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. What is the lowest score?
(b) In an isosceles triangle, the base angle are equal. The vertex angle is 40°. What are the base angles of the triangle? (Remember, the sum of three angles of a triangle is 180°?)
(c) Sachin scored twice as many runs as Rahul. Together, their runs fell two short of a double century. How many runs did each one score?

(a) Let the lowest score be x.
Therefore Highest marks obtained = 2x + 7
According to question,
2x+7 = 87
= 2x = 87 - 7
= x = 80 /2
= x= 40
The lowest score = x= 40
(b) Let each base angle be x degrees.
Therefore Sum of all angles of the triangle (x + x + 40°) degrees.
We know that sum of all angle of a triangle is 180°
So, (x+x+40°) = 180°
= 2x +40° = 180°
= 2x = 180° - 40°
= x = 140° / 2 = 70°
Hence the base angle = x = 70 °
(c)Let the runs scored by Rahul = x
Runs scored by Sachin = 2x
According to question,
x+2x + 2 = 200
= 3x = 200 -2
= x = 198 / 3
= x = 66
runs scored by Rahul = x = 66
Runs scored by Sachin = 2x = 2×66 = 132

Q.3 (i) Irfan says that he has 7 marbles more than five times the marbles Parmit has. Irfan has 37 marbles. How many marbles does Parmit have?
(ii) Laxmi’s father is 49 years old. He is 4 years older than three times Laxmi’s age. What is Laxmi’s age?
(iii) People of Sundargram planted trees in a village garden. Some of the trees were fruit trees. The number of non-fruit trees were two more than three times the number of fruit trees. What was the number of fruit trees planted if the number of non-fruit trees planted was 77?

(i) Let the number of marbles with Parmit be x
So, Number of marbles that Irfan has = 5x + 7
According to question,
5x + 7 = 37
= 5x = 37 -7
= x = 30 / 5 = 6
number of marbles with Parmit =x = 6
(ii) Let Laxmi’s age be x years.
So, Father’s age = 3x + 4
3x+4 = 49
= 3x = 49 -4
= x= 45/ 3 = 15
Thus, the age of Laxmi = 15 years
(iii) Let the number of planted fruit tree be x.
So, Number of non-fruit trees = 3x + 2
3x +2 = 77
= 3x = 77-2
= 3x = 75
= x = 75/3= 25
Thus, the required number of fruit tree planted = 25

Q.4 Solve the following riddle:
I am a number,
Tell my identity!
Take me seven times over
To reach a triple century
You still need forty!