NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations

Question

NCERT Solutions Class 7 Maths Chapter 4 Simple Equations here are made by experts in the field. Our experts have prepared these solutions in very simple language to make it easy for students to learn and comprehend. These solutions are precise and it is according to the latest syllabus provided by the CBSE. Our NCERT solutions are the best way to assist the students in their CBSE exam preparation and as well as for competitive exams like JEE Mains, JEE Advance or any other similar exams. These solutions are explained in the step-wise method.

Our NCERT Solutions Class 7 is all you need for your overall preparation for all the topics needed. Important topics mentioned here are: 

• Variable
• An Equation
• Solving an Equation
• Forming an Equation
• Application of Simple Equations to Practical Situations

These topics are very important according to exam perspective. Experts at Sarthak advise learners to go through all the topics of NCERT Solutions Class 7 Maths to gain complete clarity of this chapter. Every year number of questions are asked from this chapter which carries a significant weightage in the CBSE board examination. One must refer to our solutions for better understanding, solving questions, revision, completing assignments, doing homework with ease. Students can also find solutions for NCERT intext questions, exercises, and back-of-chapter questions. These solutions give not only provide the required solutions but also give you a deep understanding of all the related concepts.

Answer 1

NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations

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
(vi)	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

Answer:

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
(vi)	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

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)

Answer:

(a) n + 5 = 19 (n = 1)

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) 7n + 5 = 19 (n = – 2)

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) 7n + 5 = 19 (n = 2)

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) 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 LHS and RHS,

1 ≠ 13

LHS ≠ RHS

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

Answer 2

(e) 4p – 3 = 13 (p = – 4)

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) 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 LHS and RHS,

– 3 ≠ 13

LHS ≠ RHS

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

3. Solve the following equations by trial and error method.

(i) 5p + 2 = 17

(ii) 3m – 14 = 4

Answer:

(i) 5p + 2 = 17

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) 3m – 14 = 4

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.

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-fourths 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.

Answer:

(i) The sum of numbers x and 4 is 9.

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

= x + 4 = 9

(ii) 2 subtracted from y is 8.

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

= y – 2 = 8

(iii) Ten times a is 70.

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

= 10a = 70

(iv) The number b divided by 5 gives 6.

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

= (b/5) = 6

(v) Three-fourths of t is 15.

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

= 3/4 t = 15

(vi) Seven times m plus 7 gets you 77.

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

Seven times m is 7m.

= 7m + 7 = 77

(vii) One-fourth of a number x minus 4 gives 4.

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) If you take away 6 from 6 times y, you get 60.

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

6 times y is 6y.

= 6y – 6 = 60

(ix) If you add 3 to one-third of z, you get 30.

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

One-third of z is z/3.

= 3 + z/3 = 30

Answer 3

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

Answer:

(i) p + 4 = 15

The sum of numbers p and 4 is 15.

(ii) m – 7 = 3

7 subtracted from m is 3.

(iii) 2m = 7

Twice of number m is 7.

(iv) m/5 = 3

The number m divided by 5 gives 3.

(v) (3m)/5 = 6

Three-fifth of m is 6.

(vi) 3p + 4 = 25

Three times p plus 4 gives you 25.

(vii) 4p – 2 = 18

Four times p minus 2 gives you 18.

(viii) p/2 + 2 = 8

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

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 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).

Answer:

(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 180°

Let the base angle be b.

Then,

Vertex angle = 2 × base angle = 2b

= b + b + 2b = 180°

= 4b = 180°

7. 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

Answer:

(a) x – 1 = 0

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

Then, we get

= x – 1 + 1 = 0 + 1

= x = 1

(b) x + 1 = 0

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

Then, we get

= x + 1 – 1 = 0 – 1

= x = – 1

(c) x – 1 = 5

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

Then, we get

= x – 1 + 1 = 5 + 1

= x = 6

(d) x + 6 = 2

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

Then, we get

= x + 6 – 6 = 2 – 6

= x = – 4

(e) y – 4 = – 7

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

Then, we get

= y – 4 + 4 = – 7 + 4

= y = – 3

(f) y – 4 = 4

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

Then, we get

= y – 4 + 4 = 4 + 4

= y = 8

(g) y + 4 = 4

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

Then, we get

= y + 4 – 4 = 4 – 4

= y = 0

(h) y + 4 = – 4

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

Then, we get

= y + 4 – 4 = – 4 – 4

= y = – 8

Answer 4

8. 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

Answer:

(a) 3l = 42

Now, we have to divide both sides of the equation by 3.

Then, we get

= 3l/3 = 42/3

= l = 14

(b) b/2 = 6

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

Then, we get

= b/2 × 2= 6 × 2

= b = 12

(c) p/7 = 4

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

Then, we get

= p/7 × 7= 4 × 7

= p = 28

(d) 4x = 25

Now, we have to divide both sides of the equation by 4

Then, we get

= 4x/4 = 25/4

= x = 25/4

(e) 8y = 36

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

Then, we get

= 8y/8 = 36/8

= x = 9/2

(f) (z/3) = (5/4)

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) (a/5) = (7/15)

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) 20t = – 10

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

Then, we get

= 20t/20 = -10/20

= x = – 1/2

9. 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

Answer:

(a) 3n – 2 = 46

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) 5m + 7 = 17

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) 20p/3 = 40

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) 3p/10 = 6

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

Answer 5

10. 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

Answer:

(a) 10p = 100

Now,

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

Then, we get

= 10p/10 = 100/10

= p = 10

(b) 10p + 10 = 100

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) p/4 = 5

Now,

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

Then, we get

= p/4 × 4 = 5 × 4

= p = 20

(d) – p/3 = 5

Now,

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

Then, we get

= – p/3 × (- 3) = 5 × (- 3)

= p = – 15

(e) 3p/4 = 6

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) 3s = – 9

Now,

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

Then, we get

= 3s/3 = -9/3

= s = -3

(g) 3s + 12 = 0

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) 3s = 0

Now,

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

Then, we get

= 3s/3 = 0/3

= s = 0

(i) 2q = 6

Now,

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

Then, we get

= 2q/2 = 6/2

= q = 3

(j) 2q – 6 = 0

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) 2q + 6 = 0

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) 2q + 6 = 12

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

Answer 6

11. 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

Answer:

(a) 2y + (5/2) = (37/2)

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) 5t + 28 = 10

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) (a/5) + 3 = 2

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) (q/4) + 7 = 5

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) (5/2) x = -5

First, we have to multiply both sides by 2.

= (5x/2) × 2 = – 5 × 2

= 5x = – 10

Now,

We have to divide both sides by 5.

Then, we get

= 5x/5 = -10/5

= x = -2

(f) (5/2) x = 25/4

First, we have to multiply both sides by 2.

= (5x/2) × 2 = (25/4) × 2

= 5x = (25/2)

Now,

We have to divide both sides by 5.

Then, we get

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

= x = (25/2) × (1/5)

= x = (5/2)

(g) 7m + (19/2) = 13

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) 6z + 10 = – 2

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) (3/2) l = 2/3

First, we have to multiply both sides by 2.

= (3l/2) × 2 = (2/3) × 2

= 3l = (4/3)

Now,

We have to divide both sides by 3.

Then, we get

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

= l = (4/3) × (1/3)

= x = (4/9)

(j) (2b/3) – 5 = 3

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

12. 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

Answer:

(a) 2(x + 4) = 12

Let us divide both 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) 3(n – 5) = 21

Let us divide both 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) 3(n – 5) = – 21

Let us divide both 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) – 4(2 + x) = 8

Let us divide both 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) 4(2 – x) = 8

Let us divide both 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

Answer 7

13. 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)

Answer:

(a) 4 = 5(p – 2)

Let us divide both 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) – 4 = 5(p – 2)

Let us divide both 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) 16 = 4 + 3(t + 2)

By transposing 4 from RHS to LHS, it becomes – 4

= 16 – 4 = 3(t + 2)

= 12 = 3(t + 2)

Let us divide both 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) 4 + 5(p – 1) = 34

By transposing 4 from LHS to RHS, it becomes – 4

= 5(p – 1) = 34 – 4

= 5(p – 1) = 30

Let us divide both 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) 0 = 16 + 4(m – 6)

By transposing 16 from RHS to LHS, it becomes – 16

= 0 – 16 = 4(m – 6)

= – 16 = 4(m – 6)

Let us divide both 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

14. (a) Construct 3 equations starting with x = 2

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

Answer:

(a) Construct 3 equations starting with x = 2

The first equation is,

Multiply both sides by 6.

= 6x = 12 … [equation 1]

The second equation is,

Subtracting 4 from both sides,

= 6x – 4 = 12 -4

= 6x – 4 = 8 … [equation 2]

The third equation is,

Divide both sides by 6.

= (6x/6) – (4/6) = (8/6)

= x – (4/6) = (8/6) … [equation 3]

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

The first equation is,

Multiply both sides by 5.

= 5x = -10 … [equation 1]

The second equation is,

Subtracting 3 from both sides,

= 5x – 3 = – 10 – 3

= 5x – 3 = – 13 … [equation 2]

The third equation is,

Dividing both sides by 2.

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

15. 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, and 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.

Answer:

(a) Let us assume the required number is 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 sides by 8.

Then, we get

= (8x/8) = 56/8

= x = 7

(b) Let us assume the required number is 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 sides 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

Answer 8

(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 sides by 5.

= 5x/5 = 60/5

= x = 12

16. 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?

Answer:

(a) Let us assume the lowest score is x.

From the question, it is given that

The highest score is = 87

The highest marks obtained by a student in her class are twice the lowest marks plus 7= 2x + 7

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

The given above 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 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 180^o

Let the base angle be b.

Then,

= b + b + 40^o = 180^o

= 2b + 40 = 180^o

By transposing 40 from LHS to RHS, it becomes -40

= 2b = 180 – 40

= 2b = 140

Now,

Divide both sides by 2.

= 2b/2 = 140/2

= b = 70^o

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

(c) Let us assume Rahul’s score is x.

Then,

Sachin’s 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 sides by 3.

= 3x/3 = 198/3

= x = 66

So, Rahul’s score is 66.

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

17. 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 was 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?

Answer:

(i) Let us assume the 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 having

= (5 × m) + 7 = 37

= 5m + 7 = 37

By transposing 7 from LHS to RHS, it becomes -7

= 5m = 37 – 7

= 5m = 30

Divide both 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 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 sides by 3.

= 3y/3 = 45/3

= y = 15

So, Lakshmi’s age is 15 years.

(iii) Let the number of fruit trees 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 sides by 3.

= 3f/3 = 75/3

= f = 25

So, the number of fruit trees was 25.

18. 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!

Answer:

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.