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NCERT solution class 7 chapter 4 Simple equations 4.4 mathematics

EXERCISE 4.4


Page No 91:

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 I 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 of the number, the result is 23.

Answer:

(a) Let the number be x.

8 times of this number = 8x

8x + 4 = 60

8x = 60 − 4 (Transposing 4 to R.H.S.)

8x = 56

Dividing both sides by 8,

(b) Let the number be x.

One-fifth of this number = 

 (Transposing −4 to R.H.S.)

Multiplying both sides by 5,

(c) Let the number be x.

Three-fourth of this number = 

 (Transposing 3 to R.H.S.)

Multiplying both sides by 4, 

Dividing both sides by 3,

(d) Let the number be x.

Twice of this number = 2x

2x − 11 = 15

2x = 15 + 11 (Transposing −11 to R.H.S.)

2x = 26

Dividing both sides by 2,

x = 13

(e) Let the number of books be x.

Thrice the number of books = 3x

50 − 3x = 8

− 3x = 8 −50 (Transposing 50 to R.H.S.)

−3= −42

Dividing both sides by ­−3,

(f) Let the number be x.

Multiplying both sides by 5,

x + 19 = 40

x = 40 − 19 (Transposing 19 to R.H.S.)

= 21

(g) Let the number be x.

 of this number = 

Multiplying both sides by 2,

Dividing both sides by 5,

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 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 score be l.

2 × Lowest marks + 7 = Highest marks

2+ 7 = 87

2l = 87 − 7 (Transposing 7 to R.H.S.)

2l = 80

Dividing both sides by 2,

Therefore, the lowest score is 40.

(b) Let the base angles be equal to b.

The sum of all interior angles of a triangle is 180°.

b + b + 40° = 180°

2b + 40° = 180°

2b = 180º − 40º = 140º (Transposing 40º to R.H.S.)

Dividing both sides by 2,

Therefore, the base angles of the triangle are of 70º measure.

(c) Let Rahul’s score be x.

Therefore, Sachin’s score = 2x

Rahul’s score + Sachin’s score = 200 − 2

2x + x = 198

3x = 198

Dividing both sides by 3,

x = 66

Rahul’s score = 66

Sachin’s score = 2 × 66 = 132

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 year 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 was 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’s marbles equal x.

5 times the number of marbles Parmit has = 5x

5x + 7 = 37

5x = 37 − 7 = 30 (Transposing 7 to R.H.S.)

Dividing both sides by 5,

Therefore, Parmit has 6 marbles.

(ii) Let Laxmi’s age be x years.

3 × Laxmi’s age + 4 = Her father’s age

3x + 4 = 49

3x = 49 − 4 (Transposing 4 to R.H.S.)

3x = 45

Dividing both sides by 3,

x = 15

Therefore, Laxmi’s age is 15 years.

(iii) Let the number of fruit trees be x.

3 × Number of fruit trees + 2 = Number of non-fruit trees

3x + 2 = 77

3x = 77 − 2 (Transposing 2 to R.H.S.)

3x = 75

Dividing both sides of the equation by 3,

x = 25

Therefore, the number of fruit trees was 25.

Page No 92:

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 the number be x.

(7x + 50) + 40 = 300

7x + 90 = 300

7x = 300 − 90 (Transposing 90 to R.H.S.)

7x = 210

Dividing both sides by 7,

x = 30

Therefore, the number is 30.

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