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Home»Class 7»NCERT Solutions for Class 7 Maths Exercise 4.3 Simple Equations
Class 7

NCERT Solutions for Class 7 Maths Exercise 4.3 Simple Equations

7 Mins Read

Transposition is a critical technique in algebra for 7th graders. It involves moving a term from one side of an equation to the other, while changing its sign. This helps in isolating the variable (like x, y, etc.) for easy solving.

How Transposition Works:

When transposing, if you move a positive term to the other side, it becomes negative, and vice versa. This simplifies the equation.

Example 1: Solve 4m – 7 = 9

Transpose -7 to the RHS: Change -7 to +7 when moving.
Original LHS: 4m – 7.
Transpose -7: 4m = 9 + 7.
Simplified Equation: 4m = 16.

Solve for m: Divide both sides by 4.
m = 16 / 4.
m = 4.

Example 2: Solve 6y + 8 = 20

Transpose +8 to the RHS: Change +8 to -8 when moving.
Original LHS: 6y + 8.
Transpose +8: 6y = 20 – 8.
Simplified Equation: 6y = 12.

Solve for y: Divide both sides by 6.
y = 12 / 6.
y = 2.

Checking Your Solution
Verify your solution by substituting it back into the original equation.

For y = 2:

Original equation: 6y + 8 = 20.
Substitute y: 6 x 2 + 8 = 20.
Simplify: 12 + 8 = 20.
As 20 = 20, our solution y = 2 is correct!
Transposition makes solving equations more straightforward. Practice with various equations to master this skill!

NCERT Solutions for Class 7 Maths Exercise 4.3 Chapter 4 Simple Equations

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-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.

Solutions

(a) Eight times a number plus 4 equals 60

Equation: 8x + 4 = 60
Transpose +4 to RHS: 8x = 60 – 4
Simplify: 8x = 56
Solve for x: x = 56 / 8
x = 7

(b) One-fifth of a number minus 4 equals 3

Equation: (1/5)y – 4 = 3
Transpose -4 to RHS: (1/5)y = 3 + 4
Simplify: (1/5)y = 7
Solve for y: y = 7 * 5
y = 35

(c) Three-fourths of a number plus 3 equals 21

Equation: (3/4)z + 3 = 21
Transpose +3 to RHS: (3/4)z = 21 – 3
Simplify: (3/4)z = 18
Solve for z: z = 18 / (3/4)
z = 24

(d) Twice a number minus 11 equals 15

Equation: 2w – 11 = 15
Transpose -11 to RHS: 2w = 15 + 11
Simplify: 2w = 26
Solve for w: w = 26 / 2
w = 13

(e) Munna subtracts thrice the number of notebooks from 50, result is 8

Equation: 50 – 3n = 8
Transpose -3n to RHS: 50 = 8 + 3n
Simplify: 50 = 3n + 8
Transpose +8 to LHS: 3n = 50 – 8
Simplify: 3n = 42
Solve for n: n = 42 / 3
n = 14

(f) Ibenhal thinks of a number, adds 19, divides by 5, gets 8

Equation: (u + 19) / 5 = 8
Multiply both sides by 5: u + 19 = 8 * 5
Simplify: u + 19 = 40
Transpose +19 to RHS: u = 40 – 19
u = 21

(g) Anwar thinks of a number, takes away 7 from 5/2 of it, result is 23

Equation: (5/2)v – 7 = 23
Transpose -7 to RHS: (5/2)v = 23 + 7
Simplify: (5/2)v = 30
Solve for v: v = 30 / (5/2)
v = 12

2. 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?

Solutions

(a) Finding the Lowest Score in a Class

Equation: Highest Score = 2 * Lowest Score + 7
Given Highest Score = 87
Set up the equation: 87 = 2 * Lowest Score + 7
Transpose +7 to RHS: 2 * Lowest Score = 87 – 7
Simplify: 2 * Lowest Score = 80
Solve for Lowest Score: Lowest Score = 80 / 2
Lowest Score = 40

(b) Calculating Base Angles of an Isosceles Triangle

In an isosceles triangle, the base angles are equal.
Given Vertex Angle = 40°
Sum of angles in a triangle = 180°
Let each base angle be x
Equation: Vertex Angle + 2 * Base Angle = 180°
Set up the equation: 40° + 2x = 180°
Simplify: 2x = 180° – 40°
2x = 140°
Solve for x: x = 140° / 2
x = 70°
Each base angle = 70°

(c) Runs Scored by Sachin and Rahul

Sachin scored twice as many runs as Rahul.
Together, they scored just two less than 200 runs.
Let Rahul’s runs = y, then Sachin’s runs = 2y
Equation: Rahul’s Runs + Sachin’s Runs = 198
Set up the equation: y + 2y = 198
Simplify: 3y = 198
Solve for y: y = 198 / 3
y = 66
Rahul scored 66 runs, Sachin scored 132 runs (2 * 66).

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 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?

Solutions

(i) Finding the Number of Marbles Parmit Has

Irfan’s marbles = 5 times Parmit’s marbles + 7.
Irfan has 37 marbles.
Let Parmit’s marbles = x.
Equation: Irfan’s Marbles = 5x + 7.
Set up the equation: 37 = 5x + 7.
Transpose +7 to RHS: 5x = 37 – 7.
Simplify: 5x = 30.
Solve for x: x = 30 / 5.
x = 6.
Parmit has 6 marbles.

(ii) Calculating Laxmi’s Age

Laxmi’s father’s age = 49 years.
Father’s age = 3 times Laxmi’s age + 4.
Let Laxmi’s age = y.
Equation: Father’s Age = 3y + 4.
Set up the equation: 49 = 3y + 4.
Transpose +4 to RHS: 3y = 49 – 4.
Simplify: 3y = 45.
Solve for y: y = 45 / 3.
y = 15.
Laxmi’s age = 15 years.

(iii) Number of Fruit Trees Planted in Sundargram

Non-fruit trees = 3 times fruit trees + 2.
Number of non-fruit trees = 77.
Let fruit trees = z.
Equation: Non-Fruit Trees = 3z + 2.
Set up the equation: 77 = 3z + 2.
Transpose +2 to RHS: 3z = 77 – 2.
Simplify: 3z = 75.
Solve for z: z = 75 / 3.
z = 25.
Number of fruit trees = 25.

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Amit

Amit, a BE in Mechanical Engineering, is a math enthusiast dedicated to making math fun and accessible for kids in classes 1 to 10. With a knack for simplifying complex concepts, Amit offers easy-to-understand solutions, fostering a love for math in young minds across India.

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