RBSE Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.4

Rajasthan Board RBSE Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.4 Textbook Exercise Questions and Answers.

RBSE Class 7 Maths Solutions Chapter 4 Simple Equations Ex 4.4

Question 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.
(b) One-fifth of a number minus 4 gives 3.
(c) If take three-fourths of a number and add 3 to it, I get 21.
(d) When I subtracted 11 from twice a number, the result was 15.
(e) Munna subtracts thrice the number of notebooks he has from 50, he finds the result to be 8.
(f) Ibenhal thinks of a number. If she adds 19 to it and divides the sum by 5, she will get 8.
(g) Anwar thinks of a number. If he takes away 7 from \(\frac{5}{2}\) of the number, the result is 23.
Answer:
(a) Let the number be x.
According to the question,
8x + 4 = 60
⇒ 8x = 60 - 4 ⇒ 8x = 56
⇒ x = \(\frac{56}{8}\) ⇒ x = 7
∴ The required number is 7.

(b) Let the number be x.
According to the question,
\(\frac{1}{5}\) × x - 4 = 3 ⇒ \(\frac{x}{5}\) - 4 = 3
⇒ \(\frac{x}{5}\) = 4 + 3
⇒ \(\frac{x}{5}\) = 7 ⇒ x = 7 × 5 = 35

(c) Let the number be x.
According to the question,
\(\frac{3}{4}\)x + 3 = 21
\(\frac{3x}{4}\) + 3 = 21
⇒ \(\frac{3x}{4}\) = 21 - 3 = 18
3x = 18 × 4 ⇒ x = \(\frac{18 \times 4}{3}\) = 2 4
∴ The required number is 24.

(d) Let the number be x.
According to question, we get
2x - 11 = 15
⇒ 2x = 15 + 11 = 26
x = \(\frac{26}{2}\) = 13
∴ The required number is 13.

(e) Let the number be x.
According to the question, we get
50 - 3x = 8
⇒ - 3x = 8 - 50
⇒ - 3x = - 42
Multiply - 1 to both sides,
(- 1) × (- 3x) = (- 1) × (- 42)
⇒ 3x = 42
⇒ x = \(\frac{42}{3}\) = 14
∴ The number is 14.

(f) Let the number be x.
According to the question, we get
\(\frac{x+19}{5}\) = 8 ⇒ x + 19 = 40
x = 40 - 19 = 21
∴ The number is 21.

(g) Let the number be x.
Then, according to the question, we get
\(\frac{5}{2}\)x - 7 = 23
⇒ \(\frac{5}{2}\)x = 23 + 7 = 30
⇒ 5x = 30 × 2 = 60
⇒ x = \(\frac{60}{5}\) = 12
∴ The number is 12.

Question 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?
(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 the three angles of a triangle is 180°.)
(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?
Answer:
(a) Let the lowest marks be x.
Twice the lowest marks is 2x.
Highest score obtained by a student is twice the lowest marks plus 7 = 2x + 7
But the highest marks is 87,
∴ 2x + 7 = 87
⇒ 2x = 87 - 7 ⇒ 2x = 80
⇒ x = \(\frac{80}{2}\) = 40.
∴ The lowest score is 40 marks.

(b) Let the base angle be x, vertex angle is 40° (sum of angles of a triangle is 180°).
∴ 2x + 40° = 180°
⇒ 2x = 180° - 40°
⇒ 2x = 140a
⇒ x = 70°
∴ The base angles are 70°.

(c) Let 'x' runs be the score of Rahul.
Then '2x' runs be the score of Sachin.
According to the question,
x + 2x = 200 - 2 ⇒ 3x = 198
⇒ x = \(\frac{198}{3}\) = 66
Therefore, Rahul scores 66 runs.
Sachin scores 2 × 66 = 132 runs.

Question 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?
(ii) Laxmi’s father is 49 years old. He is 4 years older than three times Laxmi’s age. What is Laxmi’s age?
(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?
Answer:
(i) Let Parmit has m marbles.
Then, five times the marble Parmit has = 5 m.
Irfan has seven marbles more than Parmit is 5m + 7.
Since, it is given that the total number of marbles Irfan has is 37.
∴ 5m + 7 = 37 ⇒ 5m = 37 - 7
⇒ 5m = 30
⇒ m = \(\frac{30}{5}\) ⇒ m = 6.

(ii) Let the age of Laxmi be y years.
Three times Laxmi’s age is 3y years. Laxmi’s father is 4 years older than three times Laxmi’s age then father's age is 3y + 4 years.
∴ The age of Laxmi’s father is 49 years.
∴ 3y + 4 = 49 ⇒ 3y = 49 - 4
⇒ 3y = 45
⇒ y = \(\frac{45}{3}\)
⇒ y = 15
Laxmi’s age is 15 years.

(iii) Let the numbers of fruit trees planted be x.
∴ Non fruit trees are 3x + 2.
But, the number of non-fruit trees be 77.
⇒ 3x + 2 = 77
⇒ 3x = 77 - 2 = 75
⇒ x = \(\frac{75}{3}\) = 25
∴ Number of fruit trees are 25.

Question 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!
Answer:
Let ‘x’ be the number.
According to the question,
7x +50 = 300 - 40
⇒ 7x + 50 = 260
⇒ 7x = 260 - 50
⇒ 7x = 210
⇒ x = \(\frac{210}{3}\) = 30
So, the number is 30.