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
Statement can be written as,
8x + 4 = 60
8x = 60 – 4
8x = 56
\(x = \dfrac{56}{8}\)
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 =\( \dfrac{1}{5} x = \dfrac{x}{5}\)
Statement can be written as,
\(\dfrac{x}{5} – 4 = 3\)
\(\dfrac{x}{5} = 3 + 4\)
\(\dfrac{x}{5} = 7\)
\(x = 7 × 5\)
x = 35
(c) If I take three-fourths of a number and add 3 to it, I get 21.
Solution:-
Let us assume the required number is x
Three-fourths of a number = \(\dfrac{3}{4} x\)
The Statement can be written in the equation form as,
\(\dfrac{3}{4} x + 3 = 21\)
\(\dfrac{3}{4}) x = 21 – 3\)
\(\dfrac{3}{4}) x = 18\)
\(x = \dfrac{18 × 4 }{ 3}\)
\(x = 24\)
(d) When I subtracted 11 from twice a number, the result was 15.
Solution:-
Let us assume the required number is x
Twice a number = 2x
The Statement can be written as,
\(2x –11 = 15\)
2x = 15 + 11
2x = 26
\(x = \dfrac{26}{2}\)
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 is x
Thrice the number = 3x
The Statement can be written as,
50 – 3x = 8
– 3x = 8 – 50
-3x = – 42
\(x = \dfrac{– 42}{ -3}\)
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 is x
The given above statement can be written in the equation form as,
\( \dfrac{x + 19}{5} = 8\)
(x + 19) = 8 × 5
x + 19 = 40
x = 40 – 19
x = 21
(g) Anwar thinks of a number. If he takes away 7 from \(5\over2\) of the number, the result is 23.
Solution:-
Let us assume the required number is x
\(\dfrac{5}{2}\) of the number = \(\dfrac{5}{2}\) x
The given above statement can be written in the equation form as,
\(\dfrac{5}{2}\) x – 7 = 23
\(\dfrac{5}{2}\) x = 23 + 7
\(\dfrac{5}{2}\) x = 30
5x = 30 × 2
5x = 60
\(x = \dfrac{60}{5}\)
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 is x
The highest score is = 87
Highest marks obtained by a student in her class is twice the lowest marks plus 7
= 2x + 7
2x + 7 = Highest score
2x + 7 = 87
2x = 87 – 7
2x = 80
\(x = \dfrac{80}{2}\)
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:-
Let base angle be \(b\)
The sum of angles of a triangle is \(180^\circ\)
\(b + b + 40o = 180^\circ\)
\(2b + 40 = 180^\circ\)
2b = 180 – 40
2b = 140
\(b = \dfrac{140}{2}\)
\(b=70^\circ\)
Hence, \(70^\circ\) 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 is 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
\(x = \dfrac{198}{3}\)
x = 66
So, Rahul’s score is 66
And Sachin’s score is 2x = 2 × 66 = 132
(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
Irfan has 7 marbles more than five times the marbles Parmit has
= 5 × Number of Parmit’s marbles + 7 = Total number of marbles Irfan has
(5 × m) + 7 = 37
5m + 7 = 37
5m = 37 – 7
5m = 30
\(m =\dfrac{30}{5}\)
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
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
3y = 49 – 4
3y = 45
\(y = \dfrac{45}{3}\)
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:-
3 × number of fruit trees + 2 = number of non-fruit trees
\(3f + 2 = 77\)
3f = 77 – 2
3f = 75
\(f= \dfrac{75}{3}\)
f = 25
So, number of fruit tree was 25.
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:
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
7x = 300 – 90
7x = 210
\(x = \dfrac{210}{7}\)
x = 30
Hence the number is 30.
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