Simple Equations NCERT Solution for Class 7 Maths Chapter 4 Exercise 4.3
Simple Equations Class 7 Ex. 4.3;
Simple Equations Class 7 Ex. 4.1
Simple Equations Class 7 Ex. 4.2
Exercise 4.3
Simple Equations Chapter 4 NCERT Math Ex.-4.3.
Question 1:- Set up the equation and solve them to find the unknown number in the following of 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 numbers, the result is 23.
Solution 1:-
(a) Let the required number be x.
So, the required equation will be
8x + 4 = 60
8x = 60 – 4
⇒ 8x = 56
⇒ (8x)/8=(56)/8 (Dividing both sides by 8)
⇒ x = 7
Thus, x = 7; required unknown number.
(b) Let the required number be x.
So, the required equation will be
(1/5)x – 4 = 3
⇒ x/5 =4 + 3 (Transposing 4 to RHS)
⇒ x/5= 7
⇒ (x/5)X5 =7X5 (Multiplying both sides by 5)
⇒ x = 35; required unknown number,
(c) Let the required number be x.
(3/4)x + 3 = 21 is the required equation.
Solving the equation, we have
(3/4)x = 21-3
(3/4)x = 18
x = 18(4/3)
x= 6×4
x = 24. (required unknown number).
(d) Let, the required unknown number be x.
2x -11 = 15 ( the required equations ).
Solving the equation, we have
2x – 11= 15
⇒ 2x = 15 + 11 (Transposing 11 to RHS)
⇒ 2x = 26
⇒ (2x)/2=26/2 (Dividing both sides by 2)
⇒ x = 13 is the required unknown number,
(e) Let the required number be x.
50 – 3x = 8 ( required equations ).
Solving the equation, we have
50 – 3x = 8
-(3)x = 8 – 50 (Transposing 50 to RHS)
⇒ -(3)x = -(42)
⇒ −(3x)/−3=−(42)/−(3) (Dividing both sides by -3)
⇒ x = 14 is the required unknown number.
(f) Let the required number be x.
(x+19)/5 = 8 is the required equation.
Solving the equation, we have
(x+19)/5 = 8
⇒ (x+19)/5 × 5 = 8 × 5(Multiplying both sides by 5)
⇒ x + 19 = 40
x = 40 – 19 (Transposing 19 to R.H.S)
So, x = 21 ( the required unknown number).
(g) Let the required number be x.
(5/2)x – 7 = 23 is the required equation.
Solving the equation, we have
(5/2)x = 23+7
(5/2)x =30
x= 30x(2/5)
x= 6×2
x = 12. ( the required unknown number ).
Question 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 angle 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?
Solution 2:-
(a) Let the lowest score be x.
The required equation ⇒ 2x + 7 = 87
2x = 87 – 7 [ Transposing 7 to R.H.S ]
⇒ 2x = 80
⇒ (2x)/2=80/2 [ Dividing both sides by 2 ]
The required lowest marks ⇒ x = 40.
(b) Let each base angle be x degrees.
a + b + c = 180°
a = 40°
b = c = x
Sum of all angles of the triangle = 180 degree
x + x + 40 = 180°
⇒ 2x + 40° = 180°
Solving the equation, we have
2x + 40° = 180°
2x = 180° – 40° [ Transposing 40° to R.H.S ]
2x = 140°
⇒ (2x)/2=140/2 [ Dividing both sides by 2 ]
⇒ x = 70°
So, the required each of the base angle = 70°
(c) Let the runs scored by Rahul = x
Runs scored by Sachin = 2x
x + 2x = 200 – 2
Solving the equation, we have
3x = 198
⇒ (3x)/3=198/3 (Dividing both sides by 3)
⇒ x = 66
So, the run that scored by Rahul is 66 and the runs scored by the Sachin = 2 × 66 = 132.
Question 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?
Solution 3:-
(i) Let be the number of marbles with Parmit be x.
Number of marbles that Irfan has
5x + 7 = 37
⇒ 5x = 37 – 7 (Transposing 7 to R.H.S)
⇒ 5x = 30
⇒ (5x)/5=30/5 (Dividing both sides by 5)
⇒ x = 6
Thus, the required number of marbles that Parmit have = 6.
(ii) Let Laxmi’s age be x years.
Laxmi’s Father’s age = 3x + 4
3x + 4 = 49
⇒ 3x = 49 – 4 [ Transposing to R.H.S ]
⇒ 3x = 45
⇒ (3x)/3=45/3 [ Dividing both sides by 3 ]
⇒ x = 15
Thus, the age of Laxmi should be = 15 years.
(iii) Let the number of planted fruit tree be x.
Number of non-fruit trees = 3x + 2 = 77
⇒ 3x = 77 – 2 [ Transposing 2 to R.H.S ]
⇒ 3x = 75
⇒ (3x)/3=75/3 [ Dividing both sides by 3 ]
⇒ x = 25
Thus, the required number of fruit tree planted = 25.
Question 4 :- 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!
Solution 4:-
Suppose, my identity number is x.
7x + 50 + 40 = 300
7x + 90 = 300
⇒ 7x = 300 – 90 (Transforming 90 to RHS)
⇒ 7x = 210
⇒ (7x)/7 = 210/7 (Dividing both sides by 7)
⇒ x = 30
Thus, my identity is 30.
Simple Equations Chapter 4 NCERT Math Ex.-4.3.
Simple Equations Chapter 4 NCERT Math Ex.-4.3.
Simple Equations Chapter 4 NCERT Math Ex.-4.3.