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 the required number be x
Eight times a number = 8x
According to above statement we can write in the equation form as,
8x + 4 = 60
Transposing (4) from LHS to RHS it becomes (– 4)
8x = 60 – 4
8x = 56
Dividing both side by 8,
We get,
(8x/8) = 56/8
x = 7
(b) One-fifth of a number minus 4 gives 3.
Solution:
Let the required number be x
One-fifth of a number = (1/5) x
= x/5
According to the above statement, we can write in the equation form as,
(x/5) – 4 = 3
Transposing (– 4) from LHS to RHS it becomes (4)
x/5 = 3 + 4
x/5 = 7
Multiplying both sides by 5,
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 the required number to be x
Three-fourths of a number = (3/4) x
According 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
Multiplying both side by 4,
We get,
(3x/4) × 4 = 18 × 4
3x = 72
Dividing 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 the required number be x
Twice a number = 2x
According to above statement we can write in the equation form as,
2x –11 = 15
Transposing (-11) from LHS to RHS it becomes (11)
2x = 15 + 11
2x = 26
Dividing 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 the required number be x
Thrice the number = 3x
According to the above statement, we can write in the equation form as,
50 – 3x = 8
Transposing (50) from LHS to RHS it becomes (– 50)
– 3x = 8 – 50
-3x = – 42
Dividing 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 the required number to be x
According to above statement we can write in the equation form as,
(x + 19)/5 = 8
Multiplying both side by 5,
((x + 19)/5) × 5 = 8 × 5
x + 19 = 40
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 the required number be x
5/2 of the number = (5/2) x
According to above statement we can write in the equation form as,
(5/2) x – 7 = 23
Transposing -7 from LHS to RHS it becomes 7
(5/2) x = 23 + 7
(5/2) x = 30
Multiplying both side by 2,
((5/2) x) × 2 = 30 × 2
5x = 60
Dividing 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 the lowest score be x
According to the question, it is given,
The highest score is = 87
The 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
According to above statement we can write in the equation form as,
2x + 7 = Highest score
2x + 7 = 87
Transposing (7) from LHS to RHS it becomes (-7)
2x = 87 – 7
2x = 80
Dividing 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:
We know that the sum of angles of a triangle is 180°
According to the question it is given,
Vertex angle = 40°
The base angle is equal
Let the base angle be b
b + b + 40° = 180°
2b + 40°= 180°
Transposing (40) from LHS to RHS it becomes (-40)
2b = 180 – 40
2b = 140
Dividing both the side by 2
2b/2 = 140/2
b = 70o
Hence, the base angle of an isosceles triangle is 70o
(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 Rahul’s score be x
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
Dividing 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 the number of Parmit’s marbles = m
According to the question, it is given,
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
Transposing (7) from LHS to RHS it becomes (-7)
5m = 37 – 7
5m = 30
Dividing 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 be = ‘y’ years old
According to the question, it is given,
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
Transposing (4) from LHS to RHS it becomes (-4)
3y = 49 – 4
3y = 45
Dividing 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 trees be f.
According to the question it is given that,
3 × number of fruit trees + 2 = number of non-fruit trees
3f + 2 = 77
Transposing 2 from LHS to RHS it becomes -2
3f = 77 – 2
3f = 75
Dividing 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 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
Transposing (90) from LHS to RHS it becomes (-90)
7x = 300 – 90
7x = 210
Dividing both side by 7
7x/7 = 210/7
x = 30
Hence the required number is 30.
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