NCERT Solution class 7 Mathematics EXERCISE- 4.4
CHAPTER – 4
EXERCISE – 4.4
Q1. Set up equation 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 statements can be written in the equation form as,
= 8x + 4 = 60
By transferring 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 statements can be written in the equation form as,
= (x/5) – 4 = 3
By transferring – 4 from LHS to RHS it becomes 4
= x/5 = 3 + 4
= x/5 = 7
Multiply both sides by 5,
Then we get,
= (x/5) x 5 = 7 x 5
= x = 35
(c) If 1 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 statements can be written by in the equation form as,
= (3/4)x + 3 = 21
By transferring 3 from LHS to RHS it becomes -3
= (3/4)x = 21 – 3
= (3/4) x =18
Multiply both sides by 4.
Then we get,
= (3x/4) x 4 = 18 x 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 statements 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 statements 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 statements can be written in the equation form as,
= ( x+19)/5 = 8
Multiply both sides by 5,
= (( x+190)/5) x 5 = 8 x 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 numbers , the result is 23.
SOLUTION:-
Let us assume the required number be x,
5/2 of the number = (5/2)x
The given above statements 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 sides by 2,
= ((5/2 x) x 2 = 30 x 2
= 5x = 60
= x = 12
Q2. Solve the following:
(i) 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 are 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 statements 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
= x = 80/2
= x = 40
Hence, the lowest score is 40
(B) In an isosceles triangle, the base angles are equal. The vertex angle is 400. What are the base angles of the triangle ?( Remember , the sum of three angles of a triangle is 180o).
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 + 400 = 1800
= 2b +400 = 1800
= 2b = 1800 – 400
= 2b = 1400
= b = 1400/2
= b = 700
Hence 700 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 score be x
Then,
Sachin scored twice as many runs as Rahul is 2x
Together, their runs fell two short a double century.
= Rahul score + Sachin score = 200 – 2
= x + 2x = 198
= 3x = 198
Divide both the side by 3,
= 3x/3 = 198/3
= x= 66
So, Rahul score is 66
And Sachin score is 2x = 2 x 66 = 132
Q3. Solve the following:
(i) Ifran says that he has 7 marbles more than five times the marbles Parmit has. Ifran has 37 marbles. How many marbles does Parmit have?
SOLUTION:-
Let us assume number of Parmit marbles = m
From the question it is given that,
Then,
Ifran has 7 marbles more than five times the marbles Parmit has
= 5 x Number of Parmit marbles + 7 = Total number of marbles Ifran having
= ( 5 x 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, Parmit has 6 marbles
(ii) Laxmi father is 49 years old. He is 4 years older than three times Laxmi age. What is Laxmi age?
SOLUTION:-
Let Laxmi age to be = y years old
From the question it is given that,
Lakshmi father is 4 years older than three times of her age
= 3 x Laxmi age + 4 = Age of Lakshmi father
= (3 x y) + 4 = 49
= 3y + 4 = 49
By transposing 4 from LHS to RHS it becomes -4
= 3y = 49 – 4
= 3y = 45
Divide both theside by 3
= 3y/3 = 45/3
= y = 15
So, Lakshmi age is 15 years.
(iii) People of sundargram planted trees in the village garden. Some of the trees were fruits trees. The number of non – fruits trees were two more than three times the number if fruit trees. What was the number of fruit trees planted if he number of non – fruits trees planted was 77 ?
SOLUTION:-
Let the number of fruit trees be f.
From the question it is given that,
3 x number of fruit trees + 2 = number of non – fruits trees
= 3f + 2 = 77
By transposing 2 from LHS to RHS it becomes -2
= 3f = 77 – 2
= 3f = 75
= f = 25
SO, NUMBER OF FRUIT TREE WAS 25.
Q4. 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