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.
Solution:-
Let us assume the required number be 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 side by 8,
Then we get,
= (8x/8) = 56/8
= x = 7
(b) One-fifth of a number minus 4 gives 3.
Solution:-
Let us assume the required number be 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 side 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.
Solution:-
Let us assume the required number be 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 side by 4,
Then we get,
= (3x/4) × 4 = 18 × 4
= 3x = 72
Then,
Divide both side by 3,
= (3x/3) = 72/3
= x = 24
(d) When I subtracted 11 from twice a number, the result was 15.
Solution:-
Let us assume the required number be 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 side by 2,
= (2x/2) = 26/2
= x = 13
(e) Munna subtracts thrice the number of notebooks he has from 50, he finds the result to be 8.
Solution:-
Let us assume the required number be 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 side 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.
Solution:-
Let us assume the required number be x
The given above statement can be written in the equation form as,
= (x + 19)/5 = 8
Multiply both side 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.
Solution:-
Let us assume the required number be 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 side by 2,
= ((5/2) x) × 2 = 30 × 2
= 5x = 60
Then,
Divide both the side by 5
= 5x/5 = 60/5
= x = 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?
Solution:-
Let us assume the lowest score be x
From the question it is given that,
The highest score is = 87
Highest marks obtained by a student in her class is 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 the side 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°).
Solution:-
From the question it is given that,
We know that, the sum of angles of a triangle is 180o
Let base angle be b
Then,
= b + b + 40o = 180o
= 2b + 40 = 180o
By transposing 40 from LHS to RHS it becomes -40
= 2b = 180 – 40
= 2b = 140
Now,
Divide both the side by 2
= 2b/2 = 140/2
= b = 70o
Hence, 70o 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?
Solution:-
Let us assume Rahul’s score be 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 the side by 3,
= 3x/3 = 198/3
= x = 66
So, Rahul’s score is 66
And Sachin’s score is 2x = 2 × 66 = 132
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?
Solution:-
Let us assume 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 the side 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?
Solution:-
Let Laxmi’s age to be = y years old
From the question it is given that,
Lakshmi’s father is 4 years older than three times of 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 the side 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 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:-
Let the number of fruit tress 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 the side by 3
= 3f/3 = 75/3
= f = 25
So, number of fruit tree was 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!
Solution:-
Let us assume the number be 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 side by 7
= 7x/7 = 210/7
= x = 30
Hence the number is 30.
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