**Simple Equations**

*(NCERT Exercise 4.4)**:*

**Important Points**❶ An equation which has only one variable whose power is one is called a

*SIMPLE EQUATION*.

❷ In a simple equation keeping only the variable in the LHS of the equation and moving the constant to RHS (right hand side) is called the

*method of TRANSPOSITION*.

❸ Rules for method of Transposition are:

- A positive constant in the LHS becomes a negative constant when it is taken to the RHS and vice versa.
- A negative constant in the LHS becomes a positive constant when it is taken to the RHS and vice versa.
- The coefficient of the variable in tbe LHS becomes t.he multiplicative inverse when it is taken to the RHS :and vice versa.

**Q1: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.**

Answer: Let number be = y

Eight times a number = 8y

Add 4 to to above result = 8y + 4

Required equation is:

**8y + 4 = 60**

⇒ 8y = 60 - 4 (Transposing 4 to RHS)

⇒ 8y = 56

⇒ y = 56/8

⇒

**y = 7**

*Verification*: LHS = 8y + 4 = 8 × 7 + 4 = 56 + 4 = 60 = RHS

**(b) One fifth of a number minus 4 gives 3.**

Answer: Let the number be = p

One fifth of number = p/5

∴ Required equation is:

**p/5 - 4 = 3**

Multiply by 5 to both sides,

⇒ (p/5 - 4) × 5 = 3 × 5

⇒ 5 × p/5 - 4 × 5 = 15

⇒ p - 20 = 15

⇒ p = 15 + 20 (transposing -20 to RHS)

⇒

**p = 35**

**(c) If I take three fourths of a number and count up 3 more, I get 21.**

Answer:Let the number be = t

Three fourth of number

*t*= 3t/4

∴ Required equation is:

**3t/4 + 3 = 21**

Multiply by 4 to both sides

⇒ 4 × (3t/4 + 3 ) = 21 × 4

⇒

⇒ 3t + 12 = 84

Transpose 12 to RHS

⇒ 3t = 84 - 12 = 72

⇒ t = 72/3

⇒

**t = 24**

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

Answer: Let a number be y.

The desired equation is:

**2y - 11 = 15**

⇒ 2y = 15 + 11 (Transposing -11 to RHS)

⇒ 2y = 26

⇒ y = 26/2

⇒

**y = 13**

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

Answer:Let the number be = z

Thrice the number = 3z

Desired equation is:

**50 - 3z = 8**

Transposing 50 to RHS

⇒ -3z = 8 - 50

⇒ -3z = -42

⇒ 3z = 42

⇒ z = 42/3

⇒

**z = 14**

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

Answer: Let the number is = m

Desired equation is:

**(m + 19)/5 = 8**

Multiply by 5 to both sides

⇒ 5 × (m+19)/5 = 5 × 8

⇒ m + 19 = 40

⇒ m = 40 - 19 (Transposing 19 to RHS)

⇒

**m = 21**

**(g) Anwar thinks of a number. If he takes away 7 from 5/2 of the number, the result is 11/2 .**

Answer: Let the number be = t

5/2 of the number = 5t/2

The desired equation is:

**Q2: 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?**

Answer: Let the lowest score = y

Required equation is:

2 × lowest marks + 7 = 87

⇒ 2 × y + 7 = 87

⇒ 2y = 87 -7 (transposing 7 to RHS)

⇒ 2y = 80

Dividing both sides by 2,

⇒ 2y/2 = 80/2

∴ y = 40

Thus 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°).**

Answer: Let the base angle = b°

Since, the sum of three angles of a triangle is 180°

Required equation:

2 × base angles + 40° = 180°

⇒ 2b + 40° = 180°

Transposing 40° to LHS

⇒ 2b = 180° - 40°

⇒ 2b = 140°

Dividing both sides by 2,

⇒ 2b/2 = 140°/2

⇒ b = 70°

This the base angle is 70°

**(c) Smita’s mother is 34 years old. Two years from now mother’s age will be 4 times Smita’s present age. What is Smita’s present age?**

Answer: Let Smita's present age = m years

Present age of Smita's mother = 34 years

After 2 yeats, Smita's mother will be = 34+ 2 = 36 years

Required equation is

4 × Smita's Present age = 36 years

4 × m = 36

⇒ 4m = 36

Dividing by 4 to both sides

⇒ 4m/4 = 36/4

⇒ m = 9

Thus Smita's present age = 9

**(d) 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: Rahul scored the less number of runs.

Let Rahul's score = r runs

Sachin's score = 2 times of Rahul score = 2×r = 2r

Required equation is:

Rahul score + Sachin score + 2 = Double Century

⇒ r + 2r + 2 = 200

⇒ 3r + 2 = 200

Transposing 2 to RHS

⇒ 3r = 200 - 2

⇒ 3r = 198

Divide both sides by 3

⇒ 3r/3 = 198/3

⇒ r = 66

Thus Rahul score = 66 ... (answer)

and Sachin score = 2r = 2× 66 = 132 ... (answer)

**Q3: 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?**

Answer: Let the number of marbles Parmit has = n

Number of marbles Irfan has = 37

Required Equation is:

5 times of Parmit Marbles + 7 = Irfan's Marbles

⇒ 5 × n + 7 = 37

⇒ 5n + 7 = 37

Transposing 7 to RHS

⇒ 5n = 37 - 7

⇒ 5n = 30

Dividing by 5 to both sides

⇒ 5n/5 = 30/5

⇒ n = 6

Thus Parmit 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?**

Answer: Let Laxmi's age = m years

Age of Laxmi's father = 49

Required Equation is:

Three times Laxmi Age + 4 = Laxmi's Father Age

⇒ 3 × m + 4 = 49

⇒ 3m + 4 = 49

Transposing 4 to RHS

⇒ 3m = 49 - 4

⇒ 3m = 45

Divide by 3 to both sides,

⇒ 3m/3 = 45/3

⇒ m = 15

⇒

Thus Laxmi's age = 15 years.

**(iii) Maya, Madhura and Mohsina are friends studying in the same class. In a class test in geography, Maya got 16 out of 25. Madhura got 20. Their average score was 19. How much did Mohsina score?**

Answer: Let Mohsina's score = t

Total score = 25

Maya score = 16

Madhura score = 20

Average score = 19

Required equation is:

(Maya Score + Madhura score + Mohsina Score) ÷ 3 = Average score

⇒ (16 + 20 + t )/3 = 19

Multiple both sides by 3,

⇒

⇒ 16 + 20 + t = 57

⇒ t + 36 = 57

Transposing 36 to RHS

⇒ t = 57 - 36

⇒ t = 21

Thus, Mohsina scored 21 marks.

**(iv) People of Sundargram 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?**

Answer: Let number of fruit trees = u.

Total number of trees = 102

Non-fruit trees = 3 × fruit trees + 2 = 3u + 2

Required equation is:

Fruit Trees + Non-Fruit Trees = 102

⇒ u + (3u + 2) = 102

⇒ u + 3u + 2 = 102

⇒ 4u + 2 = 102

Transposing 2 to RHS

4u = 102 -2

4u = 100

Divide by 4 to both sides

4u/4 = 100/4

⇒ u = 25

Thus number of fruit trees planted = 25

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