# NCERT solutions for class 7 maths chapter 4 Simple Equations

NCERT Solutions for Class 7 Maths Chapter 4 - Simple Equations are meticulously crafted to assist students in their exam preparations and assignments. These solutions provide comprehensive guidance for students aiming to excel in Mathematics and secure impressive grades.

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### Exercise 4.1

Question 1 :

Complete the last column of the table.

 S. No. Equation Value Say whether the equation is satisfied (Yes/No) (i) x + 3 = 0 x = 3 (ii) x + 3 = 0 x = 0 (iii) x + 3 = 0 x = -3 (iv) x – 7 = 1 x = 7 (v) x – 7 = 1 x = 8 (vi) 5x = 25 x = 0 (vii) 5x = 25 x = 5 (viii) 5x = 25 x = -5 (ix) (m/3) = 2 m = – 6 (x) (m/3) = 2 m = 0 (xi) (m/3) = 2 m = 6

(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 ≠ RHS

∴ 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 ≠ RHS

∴ 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

∴ 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 ≠ RHS

∴ 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

∴ 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 ≠ RHS

∴ 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

∴ 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 ≠ RHS

∴ 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 ≠ RHS

∴ 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 ≠ RHS

∴ 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

∴ Yes, the equation is satisfied.

 S. No. Equation Value Say whether the equation is satisfied (Yes/No) (i) x + 3 = 0 x = 3 No (ii) x + 3 = 0 x = 0 No (iii) x + 3 = 0 x = -3 Yes (iv) x – 7 = 1 x = 7 No (v) x – 7 = 1 x = 8 Yes (vi) 5x = 25 x = 0 No (vii) 5x = 25 x = 5 Yes (viii) 5x = 25 x = -5 No (ix) (m/3) = 2 m = – 6 No (x) (m/3) = 2 m = 0 No (xi) (m/3) = 2 m = 6 Yes

Question 2 :

Solve the following equations by trial and error method.

(i) 5p + 2 = 17

(ii) 3m – 14 = 4

(i) LHS = 5p + 2

By substituting the value of p = 0,

Then,

LHS = 5p + 2

= (5 × 0) + 2

= 0 + 2

= 2

By comparing LHS and RHS,

2 ≠ 17

LHS ≠ RHS

Hence, the value of p = 0 is not a solution to the given equation.

Let, p = 1

LHS = 5p + 2

= (5 × 1) + 2

= 5 + 2

= 7

By comparing LHS and RHS,

7 ≠ 17

LHS ≠ RHS

Hence, the value of p = 1 is not a solution to the given equation.

Let, p = 2

LHS = 5p + 2

= (5 × 2) + 2

= 10 + 2

= 12

By comparing LHS and RHS,

12 ≠ 17

LHS ≠ RHS

Hence, the value of p = 2 is not a solution to the given equation.

Let, p = 3

LHS = 5p + 2

= (5 × 3) + 2

= 15 + 2

= 17

By comparing LHS and RHS,

17 = 17

LHS = RHS

Hence, the value of p = 3 is a solution to the given equation.

(ii) LHS = 3m – 14

By substituting the value of m = 3,

Then,

LHS = 3m – 14

= (3 × 3) – 14

= 9 – 14

= – 5

By comparing LHS and RHS,

-5 ≠ 4

LHS ≠ RHS

Hence, the value of m = 3 is not a solution to the given equation.

Let, m = 4

LHS = 3m – 14

= (3 × 4) – 14

= 12 – 14

= – 2

By comparing LHS and RHS,

-2 ≠ 4

LHS ≠ RHS

Hence, the value of m = 4 is not a solution to the given equation.

Let, m = 5

LHS = 3m – 14

= (3 × 5) – 14

= 15 – 14

= 1

By comparing LHS and RHS,

1 ≠ 4

LHS ≠ RHS

Hence, the value of m = 5 is not a solution to the given equation.

Let, m = 6

LHS = 3m – 14

= (3 × 6) – 14

= 18 – 14

= 4

By comparing LHS and RHS,

4 = 4

LHS = RHS

Hence, the value of m = 6 is a solution to the given equation.

Question 3 :

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 are 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) From the question, it is given that,

Number of Parmit’s marbles = m

Then,

Irfan has 7 marbles, more than five times the marbles Parmit has.

= 5 × Number of Parmit’s marbles + 7 = Total number of marbles Irfan having

= (5 × m) + 7 = 37

= 5m + 7 = 37

(ii) From the question, it is given that,

Let Laxmi’s age be = y years old

Then,

Lakshmi’s father is 4 years older than three times her age.

= 3 × Laxmi’s age + 4 = Age of Lakshmi’s father

= (3 × y) + 4 = 49

= 3y + 4 = 49

(iii) From the question, it is given that,

Highest score in the class = 87

Let the lowest score be l.

= 2 × Lowest score + 7 = Highest score in the class

= (2 × l) + 7 = 87

= 2l + 7 = 87

(iv) From the question, it is given that,

We know that the 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 4 :

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 = n + 5

By substituting the value of n = 1,

Then,

LHS = n + 5

= 1 + 5

= 6

By comparing LHS and RHS,

6 ≠ 19

LHS ≠ RHS

Hence, the value of n = 1 is not a solution to the given equation n + 5 = 19.

(b) LHS = 7n + 5

By substituting the value of n = -2,

Then,

LHS = 7n + 5

= (7 × (-2)) + 5

= – 14 + 5

= – 9

By comparing LHS and RHS,

-9 ≠ 19

LHS ≠ RHS

Hence, the value of n = -2 is not a solution to the given equation 7n + 5 = 19.

(c) LHS = 7n + 5

By substituting the value of n = 2,

Then,

LHS = 7n + 5

= (7 × (2)) + 5

= 14 + 5

= 19

By comparing LHS and RHS,

19 = 19

LHS = RHS

Hence, the value of n = 2 is a solution to the given equation 7n + 5 = 19.

(d) LHS = 4p – 3

By substituting the value of p = 1,

Then,

LHS = 4p – 3

= (4 × 1) – 3

= 4 – 3

= 1

By comparing LHS and RHS,

1 ≠ 13

LHS ≠ RHS

Hence, the value of p = 1 is not a solution to the given equation 4p – 3 = 13.

(e) LHS = 4p – 3

By substituting the value of p = – 4,

Then,

LHS = 4p – 3

= (4 × (-4)) – 3

= -16 – 3

= -19

By comparing LHS and RHS,

-19 ≠ 13

LHS ≠ RHS

Hence, the value of p = -4 is not a solution to the given equation 4p – 3 = 13.

(f) LHS = 4p – 3

By substituting the value of p = 0,

Then,

LHS = 4p – 3

= (4 × 0) – 3

= 0 – 3

= -3

By comparing LHS and RHS,

– 3 ≠ 13

LHS ≠ RHS

Hence, the value of p = 0 is not a solution to the given equation 4p – 3 = 13.

Question 5 :

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) The above statement can be written in the equation form as,

= x + 4 = 9

(ii) The above statement can be written in the equation form as,

= y – 2 = 8

(iii) The above statement can be written in the equation form as,

= 10a = 70

(iv) The above statement can be written in the equation form as,

= (b/5) = 6

(v) The above statement can be written in the equation form as,

= ¾t = 15

(vi) The above statement can be written in the equation form as,

Seven times m is 7m.

= 7m + 7 = 77

(vii) The above statement can be written in the equation form as,

One-fourth of a number x is x/4.

= x/4 – 4 = 4

(viii) The above statement can be written in the equation form as,

6 times y is 6y.

= 6y – 6 = 60

(ix) The above statement can be written in the equation form as,

One-third of z is z/3.

= 3 + z/3 = 30

Question 6 :

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) Twice of number m is 7.

(iv) The number m divided by 5 gives 3.

(v) Three-fifth of m is 6.

(vi) Three times p plus 4 gives you 25.

(vii) Four times p minus 2 gives you 18.

(viii) If you add half of a number p to 2, you get 8.

### Exercise 4.2

Question 1 :

Give first the step you will use to separate the variable and then solve the 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) Now, we have to divide both sides of the equation by 3.

Then, we get

= 3l/3 = 42/3

= l = 14

(b) Now, we have to multiply both sides of the equation by 2.

Then, we get

= b/2 × 2= 6 × 2

= b = 12

(c) Now, we have to multiply both sides of the equation by 7.

Then, we get

= p/7 × 7= 4 × 7

= p = 28

(d) Now, we have to divide both sides of the equation by 4.

Then, we get

= 4x/4 = 25/4

= x = 25/4

(e) Now, we have to divide both sides of the equation by 8.

Then, we get

= 8y/8 = 36/8

= x = 9/2

(f) Now, we have to multiply both sides of the equation by 3.

Then, we get

= (z/3) × 3 = (5/4) × 3

= x = 15/4

(g) Now, we have to multiply both sides of the equation by 5.

Then, we get

= (a/5) × 5 = (7/15) × 5

= a = 7/3

(h) Now, we have to divide both sides of the equation by 20.

Then, we get

= 20t/20 = -10/20

= x = – ½

Question 2 :

Give the steps you will use to separate the variable and then solve the equation.

(a) 3n – 2 = 46

(b) 5m + 7 = 17

(c) 20p/3 = 40

(d) 3p/10 = 6

(a) First, we have to add 2 to both sides of the equation.

Then, we get

= 3n – 2 + 2 = 46 + 2

= 3n = 48

Now,

We have to divide both sides of the equation by 3.

Then, we get

= 3n/3 = 48/3

= n = 16

(b) First, we have to subtract 7 from both sides of the equation.

Then, we get

= 5m + 7 – 7 = 17 – 7

= 5m = 10

Now,

We have to divide both sides of the equation by 5.

Then, we get

= 5m/5 = 10/5

= m = 2

(c) First, we have to multiply both sides of the equation by 3.

Then, we get

= (20p/3) × 3 = 40 × 3

= 20p = 120

Now,

We have to divide both sides of the equation by 20.

Then, we get

= 20p/20 = 120/20

= p = 6

(d) First, we have to multiply both sides of the equation by 10.

Then, we get

= (3p/10) × 10 = 6 × 10

= 3p = 60

Now,

We have to divide both sides of the equation by 3.

Then, we get

= 3p/3 = 60/3

= p = 20

Question 3 :

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) Now,

We have to divide both sides of the equation by 10.

Then, we get

= 10p/10 = 100/10

= p = 10

(b) First, we have to subtract 10 from both sides of the equation.

Then, we get

= 10p + 10 – 10 = 100 – 10

= 10p = 90

Now,

We have to divide both sides of the equation by 10.

Then, we get

= 10p/10 = 90/10

= p = 9

(c) Now,

We have to multiply both sides of the equation by 4.

Then, we get

= p/4 × 4 = 5 × 4

= p = 20

(d) Now,

We have to multiply both sides of the equation by – 3.

Then, we get

= – p/3 × (- 3) = 5 × (- 3)

= p = – 15

(e) First, we have to multiply both sides of the equation by 4.

Then, we get

= (3p/4) × (4) = 6 × 4

= 3p = 24

Now,

We have to divide both sides of the equation by 3.

Then, we get

= 3p/3 = 24/3

= p = 8

(f) Now,

We have to divide both sides of the equation by 3.

Then, we get

= 3s/3 = -9/3

= s = -3

(g) First, we have to subtract 12 from both sides of the equation.

Then, we get

= 3s + 12 – 12 = 0 – 12

= 3s = -12

Now,

We have to divide both sides of the equation by 3.

Then, we get

= 3s/3 = -12/3

= s = – 4

(h) Now,

We have to divide both sides of the equation by 3.

Then, we get

= 3s/3 = 0/3

= s = 0

(i) Now,

We have to divide both sides of the equation by 2.

Then, we get

= 2q/2 = 6/2

= q = 3

(j) First, we have to add 6 to both sides of the equation.

Then, we get

= 2q – 6 + 6 = 0 + 6

= 2q = 6

Now,

We have to divide both sides of the equation by 2.

Then, we get

= 2q/2 = 6/2

= q = 3

(k) First, we have to subtract 6 from both sides of the equation.

Then, we get

= 2q + 6 – 6 = 0 – 6

= 2q = – 6

Now,

We have to divide both sides of the equation by 2.

Then, we get

= 2q/2 = – 6/2

= q = – 3

(l) First, we have to subtract 6 from both sides of the equation.

Then, we get

= 2q + 6 – 6 = 12 – 6

= 2q = 6

Now,

We have to divide both sides of the equation by 2.

Then, we get

= 2q/2 = 6/2

= q = 3

Question 4 :

Give 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) We have to add 1 to both sides of the given equation.

Then, we get

= x – 1 + 1 = 0 + 1

= x = 1

(b) We have to subtract 1 from both sides of the given equation.

Then, we get

= x + 1 – 1 = 0 – 1

= x = – 1

(c) We have to add 1 to both sides of the given equation.

Then, we get

= x – 1 + 1 = 5 + 1

= x = 6

(d) We have to subtract 6 from both sides of the given equation.

Then, we get

= x + 6 – 6 = 2 – 6

= x = – 4

(e) We have to add 4 to both sides of the given equation.

Then, we get

= y – 4 + 4 = – 7 + 4

= y = – 3

(f) We have to add 4 to both sides of the given equation.

Then, we get

= y – 4 + 4 = 4 + 4

= y = 8

(g) We have to subtract 4 from both sides of the given equation.

Then, we get

= y + 4 – 4 = 4 – 4

= y = 0

(h) We have to subtract 4 from both sides of the given equation.

Then, we get

= y + 4 – 4 = – 4 – 4

= y = – 8

### Exercise 4.3

Question 1 :

Solve the following equations:

(a) 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) Let us divide both the sides by 2,

= (2(x + 4))/2 = 12/2

= x + 4 = 6

By transposing 4 from LHS to RHS it becomes -4

= x = 6 – 4

= x = 2

(b) Let us divide both the sides by 3,

= (3(n – 5))/3 = 21/3

= n – 5 = 7

By transposing -5 from LHS to RHS it becomes 5

= n = 7 + 5

= n = 12

(c) Let us divide both the sides by 3,

= (3(n – 5))/3 = – 21/3

= n – 5 = -7

By transposing -5 from LHS to RHS it becomes 5

= n = – 7 + 5

= n = – 2

(d) Let us divide both the sides by -4,

= (-4(2 + x))/ (-4) = 8/ (-4)

= 2 + x = -2

By transposing 2 from LHS to RHS it becomes – 2

= x = -2 – 2

= x = – 4

(e) Let us divide both the sides by 4,

= (4(2 – x))/ 4 = 8/ 4

= 2 – x = 2

By transposing 2 from LHS to RHS it becomes – 2

= – x = 2 – 2

= – x = 0

= x = 0

Question 2 :

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) Let us divide both the sides by 5,

= 4/5 = (5(p – 2))/5

= 4/5 = p -2

By transposing – 2 from RHS to LHS it becomes 2

= (4/5) + 2 = p

= (4 + 10)/ 5 = p

= p = 14/5

(b) Let us divide both the sides by 5,

= – 4/5 = (5(p – 2))/5

= – 4/5 = p -2

By transposing – 2 from RHS to LHS it becomes 2

= – (4/5) + 2 = p

= (- 4 + 10)/ 5 = p

= p = 6/5

(c) By transposing 4 from RHS to LHS it becomes – 4

= 16 – 4 = 3(t + 2)

= 12 = 3(t + 2)

Let us divide both the sides by 3,

= 12/3 = (3(t + 2))/ 3

= 4 = t + 2

By transposing 2 from RHS to LHS it becomes – 2

= 4 – 2 = t

= t = 2

(d) By transposing 4 from LHS to RHS it becomes – 4

= 5(p – 1) = 34 – 4

= 5(p – 1) = 30

Let us divide both the sides by 5,

= (5(p – 1))/ 5 = 30/5

= p – 1 = 6

By transposing – 1 from RHS to LHS it becomes 1

= p = 6 + 1

= p = 7

(e) By transposing 16 from RHS to LHS it becomes – 16

= 0 – 16 = 4(m – 6)

= – 16 = 4(m – 6)

Let us divide both the sides by 4,

= – 16/4 = (4(m – 6))/ 4

= – 4 = m – 6

By transposing – 6 from RHS to LHS it becomes 6

= – 4 + 6 = m

= m = 2

Question 3 :

(a) Construct 3 equations starting with x = 2

(b) Construct 3 equations starting with x = – 2

(a) First equation is,

Multiply both sides by 6

= 6x = 12 … [equation 1]

Second equation is,

Subtracting 4 from both sides,

= 6x – 4 = 12 -4

= 6x – 4 = 8 … [equation 2]

Third equation is,

Divide both sides by 6

= (6x/6) – (4/6) = (8/6)

= x – (4/6) = (8/6) … [equation 3]

(b) First equation is,

Multiply both sides by 5

= 5x = -10 … [equation 1]

Second equation is,

Subtracting 3 from both sides,

= 5x – 3 = – 10 – 3

= 5x – 3 = – 13 … [equation 2]

Third equation is,

Dividing both sides by 2

= (5x/2) – (3/2) = (-13/2) … [equation 3]

Question 4 :

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) By transposing (5/2) from LHS to RHS it becomes -5/2

Then,

= 2y = (37/2) – (5/2)

= 2y = (37-5)/2

= 2y = 32/2

Now,

Divide both sides by 2,

= 2y/2 = (32/2)/2

= y = (32/2) × (1/2)

= y = 32/4

= y = 8

(b) By transposing 28 from LHS to RHS it becomes -28

Then,

= 5t = 10 – 28

= 5t = – 18

Now,

Divide both sides by 5,

= 5t/5= -18/5

= t = -18/5

(c) By transposing 3 from LHS to RHS it becomes -3

Then,

= a/5 = 2 – 3

= a/5 = – 1

Now,

Multiply both sides by 5,

= (a/5) × 5= -1 × 5

= a = -5

(d) By transposing 7 from LHS to RHS it becomes -7

Then,

= q/4 = 5 – 7

= q/4 = – 2

Now,

Multiply both sides by 4,

= (q/4) × 4= -2 × 4

= a = -8

(e) First we have to multiply both the sides by 2,

= (5x/2) × 2 = – 5 × 2

= 5x = – 10

Now,

We have to divide both the sides by 5,

Then we get,

= 5x/5 = -10/5

= x = -2

(f) First we have to multiply both the sides by 2,

= (5x/2) × 2 = (25/4) × 2

= 5x = (25/2)

Now,

We have to divide both the sides by 5,

Then we get,

= 5x/5 = (25/2)/5

= x = (25/2) × (1/5)

= x = (5/2)

(g) By transposing (19/2) from LHS to RHS it becomes -19/2

Then,

= 7m = 13 – (19/2)

= 7m = (26 – 19)/2

= 7m = 7/2

Now,

Divide both sides by 7,

= 7m/7 = (7/2)/7

= m = (7/2) × (1/7)

= m = ½

(h) By transposing 10 from LHS to RHS it becomes – 10

Then,

= 6z = -2 – 10

= 6z = – 12

Now,

Divide both sides by 6,

= 6z/6 = -12/6

= m = – 2

(i) First we have to multiply both the sides by 2,

= (3l/2) × 2 = (2/3) × 2

= 3l = (4/3)

Now,

We have to divide both the sides by 3,

Then we get,

= 3l/3 = (4/3)/3

= l = (4/3) × (1/3)

= x = (4/9)

(j) By transposing -5 from LHS to RHS it becomes 5

Then,

= 2b/3 = 3 + 5

= 2b/3 = 8

Now,

Multiply both sides by 3,

= (2b/3) × 3= 8 × 3

= 2b = 24

And,

Divide both sides by 2,

= 2b/2 = 24/2

= b = 12

### Exercise 4.4

Question 1 :

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 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 us assume the lowest score is x

From the question it is given that,

The highest score is = 87

Highest marks obtained by a student in her class is twice the lowest marks plus 7= 2x + 7

5/2 of the number = (5/2) x

The above given statement can be written in the equation form as,

Then,

= 2x + 7 = Highest score

= 2x + 7 = 87

By transposing 7 from LHS to RHS it becomes -7

= 2x = 87 – 7

= 2x = 80

Now,

Divide both the sides by 2

= 2x/2 = 80/2

= x = 40

Hence, the lowest score is 40

(b) From the question it is given that,

We know that, the sum of angles of a triangle is 180o

Let base angle be b

Then,

= b + b + 40o = 180o

= 2b + 40 = 180o

By transposing 40 from LHS to RHS it becomes -40

= 2b = 180 – 40

= 2b = 140

Now,

Divide both the sides by 2

= 2b/2 = 140/2

= b = 70o

Hence, 70o is the base angle of an isosceles triangle.

(c) Let us assume Rahul’s score is x

Then,

Sachin scored twice as many runs as Rahul is 2x

Together, their runs fell two short of a double century,

= Rahul’s score + Sachin’s score = 200 – 2

= x + 2x = 198

= 3x = 198

Divide both the sides by 3,

= 3x/3 = 198/3

= x = 66

So, Rahul’s score is 66

And Sachin’s score is 2x = 2 × 66 = 132

Question 2 :

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 Sundargram planted 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 if the number of non-fruit trees planted was 77?

(i) Let us assume number of Parmit’s marbles = m

From the question it is given that,

Then,

Irfan has 7 marbles more than five times the marbles Parmit has

= 5 × Number of Parmit’s marbles + 7 = Total number of marbles Irfan has

= (5 × m) + 7 = 37

= 5m + 7 = 37

By transposing 7 from LHS to RHS it becomes -7

= 5m = 37 – 7

= 5m = 30

Divide both the sides by 5

= 5m/5 = 30/5

= m = 6

So, Permit has 6 marbles

(ii) Let Laxmi’s age be = y years old

From the question it is given that,

Lakshmi’s father is 4 years older than three times of her age

= 3 × Laxmi’s age + 4 = Age of Lakshmi’s father

= (3 × y) + 4 = 49

= 3y + 4 = 49

By transposing 4 from LHS to RHS it becomes -4

= 3y = 49 – 4

= 3y = 45

Divide both the sides by 3

= 3y/3 = 45/3

= y = 15

So, Lakshmi’s age is 15 years.

(iii) Let the number of fruit tress be f.

From the question it is given that,

3 × number of fruit trees + 2 = number of non-fruit trees

= 3f + 2 = 77

By transposing 2 from LHS to RHS it becomes -2

=3f = 77 – 2

= 3f = 75

Divide both the sides by 3

= 3f/3 = 75/3

= f = 25

So, number of fruit tree was 25.

Question 3 :

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!

Let us assume the number is x.

Take me seven times over and add a fifty = 7x + 50

To reach a triple century you still need forty = (7x + 50) + 40 = 300

= 7x + 50 + 40 = 300

= 7x + 90 = 300

By transposing 90 from LHS to RHS it becomes -90

= 7x = 300 – 90

= 7x = 210

Divide both sides by 7

= 7x/7 = 210/7

= x = 30

Hence, the number is 30.

Question 4 :

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 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 and divides the sum by 5, she will get 8.

(g) Anwar thinks of a number. If he takes away 7 from 5/2 of the number, the result is 23.

(a) Let us assume the required number be x

Eight times a number = 8x

The given above statement can be written in the equation form as,

= 8x + 4 = 60

By transposing 4 from LHS to RHS it becomes – 4

= 8x = 60 – 4

= 8x = 56

Divide both side by 8,

Then we get,

= (8x/8) = 56/8

= x = 7

(b) Let us assume the required number be x

One-fifth of a number = (1/5) x = x/5

The given above statement can be written in the equation form as,

= (x/5) – 4 = 3

By transposing – 4 from LHS to RHS it becomes 4

= x/5 = 3 + 4

= x/5 = 7

Multiply both side by 5,

Then we get,

= (x/5) × 5 = 7 × 5

= x = 35

(c) Let us assume the required number is x

Three-fourths of a number = (3/4) x

The given above statement can be written in the equation form as,

= (3/4) x + 3 = 21

By transposing 3 from LHS to RHS it becomes – 3

= (3/4) x = 21 – 3

= (3/4) x = 18

Multiply both sides by 4,

Then we get,

= (3x/4) × 4 = 18 × 4

= 3x = 72

Then,

Divide both sides by 3,

= (3x/3) = 72/3

= x = 24

(d) Let us assume the required number is x

Twice a number = 2x

The given above statement can be written in the equation form as,

= 2x –11 = 15

By transposing -11 from LHS to RHS it becomes 11

= 2x = 15 + 11

= 2x = 26

Then,

Divide both sides by 2,

= (2x/2) = 26/2

= x = 13

(e) Let us assume the required number is x

Thrice the number = 3x

The given above statement can be written in the equation form as,

= 50 – 3x = 8

By transposing 50 from LHS to RHS it becomes – 50

= – 3x = 8 – 50

= -3x = – 42

Then,

Divide both sides by -3,

= (-3x/-3) = – 42/-3

= x = 14

(f) Let us assume the required number is x

The given above statement can be written in the equation form as,

= (x + 19)/5 = 8

Multiply both sides by 5,

= ((x + 19)/5) × 5 = 8 × 5

= x + 19 = 40

Then,

By transposing 19 from LHS to RHS it becomes – 19

= x = 40 – 19

= x = 21

(g) Let us assume the required number is x

5/2 of the number = (5/2) x

The given above statement can be written in the equation form as,

= (5/2) x – 7 = 23

By transposing -7 from LHS to RHS it becomes 7

= (5/2) x = 23 + 7

= (5/2) x = 30

Multiply both sides by 2,

= ((5/2) x) × 2 = 30 × 2

= 5x = 60

Then,

Divide both the sides by 5

= 5x/5 = 60/5

= x = 12