Chapter 4: Simple Equations

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) One-fifth of a number minus 4 gives 3.

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) If I take three-fourths of a number and add 3 to it, I get 21.

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) When I subtracted 11 from twice a number, the result was 15.

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) Munna subtracts thrice the number of notebooks he has from 50, and he finds the result to be 8.

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) Ibenhal thinks of a number. If she adds 19 to it and divides the sum by 5, she will get 8.

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

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

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) 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°.)

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) 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?

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 sides by 3.

= 3x/3 = 198/3

= x = 66

So, Rahul’s score is 66.

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

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) Laxmi’s father is 49 years old. He is 4 years older than three times Laxmi’s age.

What is Laxmi’s age?

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) 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?

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.

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.

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Dropped Topics – 4.6 From solution to equation