# Simple Equations NCERT Solutions Class 7 Maths

## Exercise 4.1

Question 1
Complete the last column of the table.

Question 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)
(a) LHS 1+ 5 =6 &ne RHS
So it is not the solution
(b) LHS 7 * (-2)+5=--9 &ne RHS
So it is not the solution
(c) LHS 7 * (2) + 5=19 = RHS
So it is the solution
(d) LHS 4* 1 -3 =1 &ne RHS
So it is not the solution
(e) LHS 4* (-4) -3 =-19 &ne RHS
So it is not the solution
(f) LHS 4* 0 -3 =-3 &ne RHS
So it is not the solution

Question 23
Solve the following equations by trial and error method:
(i) 5p + 2 = 17
(ii) 3m – 14 = 4
(i)

(ii)

Question 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) 3/4 t=15
(vi) 7m + 7 =77
(vii) x/4 -4=4
(viii) 6y - 6=60
(ix) z/3 + 3 =30

Question 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 4 is 15.
(ii) 7 subtracted from m is 3.
(iii) Two times m is 7.
(iv) The number m is divided by 5 gives 3.
(v) Three-fifth of the number m is 6.
(vi) Three times p plus 4 gets 25.
(vii) If you take away 2 from 4 times p, you get 18.
(viii) If you added 2 to half is p, you get 8

Question 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) 5m + 7 =37
(ii) 3y + 4=49
(iii) 2l + 7=87
(iv) Base angle =b
Vertex angle = 2b
Now sum of angles of a triangle is 180 degrees
b + b + 2b=180
3b=180

## Exercise 4.2

Question 1
Give first the step you will use to separate the variable and then solve the equations:
(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
Subtracting 1 both sides
x+1-1=0-1
x=-1

(c) x-1=5
x-1+1=5+1
x=6

(d) x+6=2
Subtracting 6 both sides
x+6-6=2-6
x=-4

(e) y-4=-7
y-4+4=-7+4
y=-3

(f) y-4=4
y-4+4=4+4
y=8

(g) y+4=4
Subtracting 4 both sides
y+4-4=4-4
y=0

(h) y+4=-4
Subtracting 4 both sides
y+4-4=-4-4
y=-8

Question 2
Give first the step you will use to separate the variable and then solve the equations
(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) 3l=42
Dividing both sides by 3
3l/3=42/3
l=14

(b) b/2=6
Multiplying both sides by 2
(b/2)*2=6*2
b=12

(c) p/7=4
Multiplying both sides by 7
(p/7)*7=4*7
p=28

(d) 4x=25
Dividing both sides by 4
x=25/4

(e) 8y=36
Dividing both sides by 8
8y/8=36/8
y=9/2

(f) z/3=5/4
Multiplying both sides by 3
(z/3)*3=(5/4)*3
z=15/4

(g) a/5=7/15
Multiplying both sides by 5
(a/5)*5=(7/15)*5
a=73

(h) 20t=-10
Dividing both sides by 20
20t/20=-10/20
t=-1/2

Question 3
Give first the step you will use to separate the variable and then solve the equations
(a) 3n-2=46
(b) 5m+7=17
(c) 20p/3=40
(d) 3p/10=6
(a) 3n-2=46
3n-2+2=46+2=48
Dividing both sides by 3
3n/3=48/3
n=16

(b) 5m+7=17
Subtracting 7 both sides
5m+7-7=17-7=17-7
5m=10
Dividing both sides by 5
5m/5=10/5
m=2

(c) 20p/3=40
Multiplying both sides by 3
(20p/3)*3 = 40*3
20p = 120
Dividing both sides by 20
20p/20=120/20
p=6

(d) 3p/10=6
Multiplying both sides by 10
(3p/10)*10 = 6*10
3p=60
Dividing both sides by 3
3p/3 = 60/3
p=20

Question 4
Solve the following equation:
(a) 10 p = 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
Dividing both sides by 10
10p/10=100/10
p=10

(b) 10p+10=100
Subtracting both sides 10
10p+10-10 = 100-10
10p=90
Dividing both sides by 10
10p/10=90/10
p=9

(c) p/4=5
Multiplying both sides by 4
p/4 * 4 = 5*4
p = 20

(d) -p/3=5
Multiplying both sides by – 3
-p/3*(-3) = 5*(-3)
p = -15

(e) 3p/4 =6
Multiplying both sides by 4
(3p/4) * 4 = 6*4
3p = 24
Dividing both sides by 3
3p/3 = 24/3
p=8

(f) 3s = -9
Dividing both sides by 3
3s/3 = -9/3
s = -3

(g) 3s+12=0
Subtracting both sides 10
3s+12-12=0-12
3s=-12
Dividing both sides by 3
3s/3 = -12/3
s=-4

(h) 3s=0
Dividing both sides by 3
3s/3=0/3
s=0

(i) 2q = 6
Dividing both sides by 2
2q/2 = 6/2
q = 3

(j) 2q-6 = 0
2q-6+6 = 0+6
2q = 6
Dividing both sides by 2
2q/2 = 6/2
q=3

(k) 2q+6 = 0
Subtracting both sides 6
2q+6-6 = 0-6
2q = -6
Dividing both sides by 2
2q/2 = -6/2
q = -3

(l) 2q+6 = 12
Subtracting both sides 6
2q+6-6 = 12-6
2q = 6
Dividing both sides by 2
2q/2 = 6/2
q = 3

## Exercise 4.3

Question 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-fourth of a number and add 3 to it, I get 21.
(d) When I 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 divides the sum by 5, she will get 8.
(g) Anwer thinks of a number. If he takes away 7 from 5/2 of the number, the result is 11/2.
(a) Let the number be x.
According to the question
8x+4=60
Transposing 4 to other side
8x = 60-4
8x = 56
Dividing both sides by 8
x = 56/8
x = 7

(b) Let the number be y.
According to the question
y/5-4=3
Transposing 4 to other side
y/5 = 3+4
y/5 = 7
Multiplying both sides by 5
y = 7*5
y = 35

(c) Let the number be z.
According to the question
$\frac {3}{4} z+3=21$
Transposing 3 to other side
$\frac {3}{4} z=21 -3$
$\frac {3}{4} z=18$
Multiplying both sides by 4
3z = 18*4
3z = 72
Dividing both sides by 3
z = 72/3
z = 24

(d) Let the number be x
According to the question
2x-11=15
Transposing 11 to other side
2x = 15+11
2x = 26
Dividing both sides by 2
x = 26/2
x = 13

(e) Let the number be z.
According to the question
50-3z=8
Transposing 50 to other side
-3z = 8-50
-3z = -42
Dividing both sides by -3
z = -42/-3
z = 14

(f) Let the number be x.
According to the question,
$\frac {x+19}{5}=8$
Multiplying both sides by 5
x+19 = 8*5
x+19 = 40
x = 40-19
x = 21

(g) Let the number be x.
According to the question
$\frac {5}{2} x-7=\frac {11}{2}$
$\frac {5}{2}x = \frac {11}{2} + 7$
$\frac {5}{2} x = \frac {25}{2}$
Multiplying both sides by 2
5x = 25
5x = 25
Dividing both sides by 5
x = 25/5
x = 5

Question 2
Solve the following:
(a) The teacher tells the class that the highest marks obtained by a student in her class are twice the lowest marks plus 7. The highest score is 87. What is the lowest score?
(b) In an isosceles triangle, the base angles 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 marks be m.
According to the question
$2m+7=87$
$2m = 87-7$
$2m = 80$
Dividing both sides by 2
$m = 40$
Hence, the lowest score is 40.

(b) Let the base angle of the triangle be x.
Given, a=40° , b =x ,c=x
Angle sum property of a triangle
a+b+c=180
40+x+x=180
40 +2x = 180
2x= 180 - 40
2x = 140
Dividing both sides by 2
x = 140/2
x = 70
Therefore , the base angles of the isosceles triangle are 70° each.

(c) Let the score of Rahul be x runs then Sachin’s score will be 2x.
According to the question
x+2x =198
3x = 198
Dividing both sides by 3
x = 198/3
x = 66
Thus, Rahul’s score = 66 runs
And Sachin’s score = 2 * 66 = 132 runs.

Question 3
Solve the following:
(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 Sundergram planted a total of 102 trees in the 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?
(i) Let the number of marbles Parmit has be m.
According to the question
5m+7=37
Transposing 7 to other side
5m=37-7
5m=30
Dividing both sides by 5
m=30/5
m=6
Hence, Parmit has 6 marbles.

(ii) Let the age of Laxmi be y years.
Then her father’s age = (3y+4) years
According to question
3y+4=49
Transposing 4 to other side
3y=49-4
3y=45
Dividing both sides by 3
y=45/3
y=15
Therefore, the age of Laxmi is 15 years
.
(iii) Let the number of fruit trees be x.
Then the number of non-fruits tree = 3x+2
According to the question $x+3x+2 = 102$
4x+2 =102
Transposing 2 to other side
4x = 102-2
4x = 100
Dividing both sides by 4
x = 100/4
x = 25
Thus, the number of fruit trees are 25.

Question 4
Solve the following riddle:
I am a number, Tell my identity!
Take me seven times over, And add a fifty!
To reach a triple century, You still need forty!
Let the number be n.
According to the question
7n+50+40 = 300
7n+90 = 300
Transposing 90 to other side
7n = 300-90
7n = 210
Dividing both sides by 7
n = 210/7
n = 30
Thus, the required number is 30.

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