EXERCISE 4.4
1) Set up equations and solve them to find the unknown numbers in the following equations:
a) Add 4 to eight times a number, you get 60.
Ans: Let the number = 8x
8x + 4 = 60
8x = 60 – 4
8x = 56
x = 56/4
x = 7
Therefore, the number is 7
b) One – fifth of a number minus 4 gives 3.
Ans: Let the number be x
One – fifth of a number = 1/5× x = x/5
x/5 – 4 = 3
x/5 = 3 + 4
x/5= 7
x = 7 × 5
x = 35
c) If I take three- fourths of a number and add 3 to it , I get 21.
Ans: Let the number be x
Three – fourth of the number = 3/4 × x = 3x/4
3x/4 + 3 = 21
3x/4 = 21 – 3
3x/4 = 18
3x = 18 × 4 → 3x = 72
x = 72/3 → x = 24
Therefore, the is 24
d) When I subtract 11 from twice a number, the result was
Ans: Let the number be x
11 subtracted from twice a number = 2x – 11
2x – 11 = 15
2x = 15 + 11 → 2x = 26 → x = 26/2
x = 13
Therefore, the number is 13.
e) Munna subtracts thrice the number of notebooks he has from 50, he finds the result to be 8.
Ans:
Let the number be x
Thrice the number = 3x
Subtracting it from 50 gives
50– 3x = 8
-3x = 8 – 50 → -3x = -42
x = (-42)/(-3) → x = 14
Therefore, the number is 14.
f) Ibenhal thinks of a number. If he adds 19 to it and divides the sum by 5, she will get 8.
Ans: Let the number be x
Adding 19 to the number = x + 19
(x+19)/5 = 8
x + 19 = 8 × 5
x + 19 = 40 → x = 40 – 19 → x = 21
Therefore, he number is 21.
g) Anwar thinks of a number. If he takes away 7 from 5/2 of the number, the result is 23.
Ans: Let the number be x
Taking away 7 from 5/2 of the number gives
5x/2 – 7 = 23
5x/2 = 23 + 7
5x/2= 30 → 5x = 30 × 2 → 5x = 60
x = 60/5 → x = 12
Therefore, the 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?
Ans: Let the lowest score be x
The highest score is 87
Twice the lowest marks plus 7 = highest marks
2x + 7 = 87
2x = 87 – 7
2x = 80 → x = 80/2 → x = 4
Hence, the lowest score is 40. – same as question 6-iii-in exercise 4.1
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 180ᵒ).
Ans: Let the base angle be x
The sum of the angles of a triangle is 1800.
Therefore, x + x + 40 = 180
2x + 40 = 180
2x = 180 – 40
2x = 140 → x = 140/2
x = 70
Therefore, each base angle of the 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?
Ans: Let the runs scored by Rahul be x
The runs scored by Sachin = 2x
Together their sum fell two short of a double century
Therefore, x + 2x = 200 – 2
3x = 198
x = 198/3 → x = 66
Therefore, Rahul’s score is 66
Sachin’s score is 2x = 2 × 66 = 132
3) Solve the following:
a) Irfan says that he has 7 marbles more than five times the marble Parmit has. Irfan has 37 marbles. How many marbles does Parmit have?
Ans: Let the marbles with Parmit be x
7 marbles more than five times Parmit’s marbles = Irfan’s marbles
Therefore, 5x + 7 = 37
5x = 37 – 7
5x = 30 → x = 30/5 → x = 6
Therefore, Parmit has 6 marbles.
b) Laxmi’s father is 49 years old. He is 4 years older than three times Laxmi’s age. What is Laxmi’sage?
Ans: Let Laxmi’s age be x
Her father is 4 years older than 3 times Laxmi’s age
3x + 4 = 49
3x = 49 – 4
3x = 45 → x = 45/3 → x = 15
Therefore, Laxmi’s age is 15 years.
c) 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?
Ans: Let the no. of fruits trees be f.
Two more than 3 times the number of fruit trees = Non-fruit trees
Therefore, 3f + 2 = 77
3f = 77 – 2 → 3f = 75 → f = 75/3
f = 25
Therefore, the no. of fruit trees are 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: Let the number be x
Seven times over = 7x
Adding a fifty
7x + 50 = 300 – 40
7x = 300 – 40 – 50
7x = 300 – 90
7x = 210 → x = 210/7 → x = 30
Therefore, the required number is 30.