Simple Equations
Exercise 4.4
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.
Let the given number be x
By the given condition of the problem
8x + 4 = 60
Transposing 4 to RHS
8x = 60 – 4
8x = 56
Transposing 8 to RHS
x = 56/8
x = 7
Ans: the given number is 7.
(b) One-fifth of a number minus 4 gives 3.
Let the given number be x
By the given condition of the problem
1/5 x – 4 = 3
Transposing 4 to RHS
1/5 x = 3 + 4
1/5 x = 7
Transposing 5 to RHS
x = 7 X 5
x = 35
Ans: the given number is 35.
(c) If I take three-fourths of a number and add 3 to it, I get 21.
Let the number be x
By the given condition of the problem
3/4 x + 3 = 21
Transposing 3 to RHS
3/4 x = 21 – 3
3/4 x = 18
Transposing ¾ to RHS
x = 18 X 4/3
x = 72/3
x = 24
Ans: the given number is 24.
(d) When I subtracted 11 from twice a number, the result was 15.
Let the number be x
By the given condition of the problem
2x – 11 = 15
Transposing 11 to RHS
2x = 15 + 11
2x = 26
Transposing 2 to RHS
x = 26/2
x = 13
Ans: the given number is 13.
(e) Munna subtracts thrice the number of notebooks he has from 50, he finds the result to be 8.
Let the number of notebooks be x
By the given condition of the problem
50 – 3x = 8
Transposing 50 to RHS
-3x = 8 – 50
-3x = -42
Transposing -3 to RHS
x = -42/-3
x = 14
Ans: the number of notebooks with Munna are 14.
(f) Ibenhal thinks of a number. If she adds 19 to it and divides the sum by 5, she will get 8.
Let the number be x
By the given condition of the problem
(x + 19)/5 = 8
Transposing 5 to RHS
(x + 19) = 8 X 5
x +19 = 40
Transposing 19 to RHS
x = 40 – 19
x = 21
Ans: the given number is 21.
(g) Anwar thinks of a number. If he takes away 7 from 5/2 of the number, the result is 23.
Let the number be x
By the given condition of the problem
5/2 x – 7 = 23
Transposing 7 to RHS
5/2 x = 23 + 7
5/2 x = 30
Transposing 5/2 to RHS
x = 30 X 2/5
x = 60/5
x = 12
Ans: the given number is 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?
Let the lowest score be x
Highest score = 87
By the given condition of the problem
2 x + 7 = 87
Transposing 7 to RHS
2 x = 87 – 7
2 x = 80
Transposing 2 to RHS
x = 80/2
x = 40
Ans: 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°).
Let the base angle of an isosceles triangle be x0
The vertex angle = 400
The Sum of the three angles of a triangle is 1800 and base angles of isosceles triangles are equal.
By the given condition of the problem
x + x + 40 = 180
2x + 40 = 180
Transposing 40 to RHS
2x = 180 – 40
2x = 140
Transposing 2 to RHS
x = 140/2
x = 70
Ans: The base angles are 700 each.
(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 the runs scored by Rahul be x
Runs scored by Sachin = 2x
By the given condition of the problem
x + 2x = 200 – 2
x + 2x = 180
3x = 180
Transposing 3 to RHS
x = 180/3
x = 60
Transposing 2 to RHS
x = 140/2
x = 70
Ans: The runs scored by Rahul = 60.
The runs scored by Sachin = 2x = 2 X 60 = 120.
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?
Let the marbles with Parmit be x
The marbles with Irfan = 37
By the given condition of the problem
5x + 7 = 37
Transposing 7 to RHS
5x = 37 – 7
5x = 30
Transposing 5 to RHS
x = 30/5
x = 6
Ans: The marbles with Parmit are = 6.
(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 the age of Laxmi be x years
Age of Laxmi’s father = 49 years
By the given condition of the problem
3x + 4 = 49
Transposing 4 to RHS
3x = 49 – 4
3x = 45
Transposing 3 to RHS
x = 45/3
x = 15
Ans: The age of Laxmi 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 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?
Let the number of fruit trees be x
the number of non-fruit trees = 77
By the given condition of the problem
3x + 2 = 77
Transposing 2 to RHS
3x = 77 – 2
3x = 75
Transposing 3 to RHS
x = 75/3
x = 25
Ans: The number of fruit trees = 25.
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!
Ans: 30
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