NCERT solution class 7 chapter 4 simple equations exercise 4.1 mathematics

Exercise 4.1


QUESTION 1

Complete the last column of the table

s.no Equation Value Say , whether the equation is satisfied (Yes/NO)
(i) x + 3 = 0 x = 3
(ii) x + 3 = 0 x = 0
(iii) x + 3 = 0 x = -3
(iv) x – 7 = 1 x = 7
(v) x – 7 = 1 x = 8
(vi) 5x = 25 x = 0
(vii) 5x = 25 x = 5
(viii) 5x = 25 x = -5
(ix) \dfrac{m}{3} = 2 m = -6
(x) \dfrac{m}{3} = 2 m = 0
(xi) \dfrac{m}{3} = 2 m = 6

 

Sol :

(i) x + 3 = 0

L.H.S. = x + 3

By putting x = 3,

L.H.S = 3 + 3 = 6 ≠ R.H.S

∴ No, the equation is not satisf‌ied

 

(ii) x + 3

L.H.S. = x + 3

By putting x = 0,

L.H.S. = 0 + 3 = 3 ≠ R.H.S.

∴ No, the equation is not satisf‌ied.

 

(iii) x + 3 = 0

L.H.S = x = -3

By putting x = -3,

L.H.S. = – 3 + 3 = 0 = R.H.S.

∴ Yes, the equation is satisf‌ied.

 

(iv) x – 7 = 1

L.H.S. = x – 7

By putting x = 7

L.H.S. = 7 – 7 = 0 ≠ R.H.S.

∴ No, the equation is not satisf‌ied.

 

(v) x – 7 = 1

L.H.S = x – 7

By putting x =  8,

L.H.S. = 8 – 7 =1 = R.H.S.

∴ Yes, the equation is satisf‌ied.

 

(vi) 5x = 25

L.H.S. = 5x

By putting x = 0 ,

L.H.S. = 5 \times 0 = 0 ≠ R.H.S.

∴ No, the equation is not satisf‌ied.

 

(vii) 5x = 25

L.H.S. = 5x

By putting x = 5,

L.H.S. = 5 \times 5 = 25 = R.H.S.

∴ Yes, the equation is satisf‌ied.

 

(viii) 5= 25

L.H.S. =5x

By putting x = -5,

L.H.S. = 5 \times (-5) = -25 \times R.H.S.

∴ No, the equation is not satisf‌ied.

 

(ix) \dfrac{m}{3}=2

L.H.S = \dfrac{m}{3}

By putting m = -6

L.H.S =\dfrac{-6}{3} = -2 ≠ R.H.S

∴ No, the equation is not satisf‌ied. I

 

(x) \dfrac{m}{3}=2

L.H.S = \dfrac{m}{3}

By putting = 0

L.H.S \dfrac{0}{3} = 0 ≠ R.H.S

∴ No , the equation is not satisfied

 

(xi) \dfrac{m}{3}=2

L.H.S =\dfrac{m}{3}

By putting m = 6

L.H.S =\dfrac{6}{3} = 2 = R.H.S

∴ Yes, the equation is satisf‌ied.

 


 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?

Sol :

Let the lowest score be l .

2 \times Lowest marks + 7 = Highest marks

2l+7=87

2l= 87 — 7 (Transposing 7 to R.H.S.)

2l = 80

Dividing both sides by 2

\dfrac{2l}{2}=\dfrac{80}{2}

l = 40

Therefore , the lowest score is 40

 

(b) In an isosceles triangle, the base angles are equal. The vertex angle is 40°. What are the base angle of the triangle ? (Remember , the sum of three angles of a triangle is 180°) .

Sol :

Let the base angles be equal to b

The sum of interior angles of a triangle is 180°

b + b + 40° = 180°

2b +40° = 180°

2b = 180° – 40° (transposing 40° to RHS)

2b = 140°

Dividing both sides by 2 ,

\dfrac{2b}{2}=\dfrac{140°}{2}

b = 70°

Therefore, the base angles of the triangle are of 70° measure

 

(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?

Sol :

Let Rahul’s score be x

Therefore, Sachin’s score = 2x

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

2x +  x = 198

3x = 198

\dfrac{3x}{3}=\dfrac{198}{3}( Dividing both sides by 3 )

x = 66

Rahul’s score = 66  , Sachin’s score = 2 \times = 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 ?

Sol :

Let Parmit’s marbles = x

5 times the number of marbles Parmit has = 5x

5x + 7 = 37

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

5= 30

\dfrac{5x}{5}=\dfrac{30}{5} ( Dividing both sides by 5 )

x = 6

Therefore, Parmit has 6 marbles.

 

(ii) Laxmi’s father is 49 year old. He is 4 years older than three times Laxmi’s age. What is Laxmi’s age ?

Sol :

Let Laxmi’s age be x years

3 \times Laxmi’s age + 4 = Her father’s age

3x + 4 = 49

3x = 49 – 4

3x = 45

( Dividing both sides by 3)

\dfrac{3x}{3}=\dfrac{45}{3}

x = 15

Therefore , Laxmi’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 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?

Sol :

Let the number of fruit trees be x

3 \times number of fruit trees + 2 = Number of non-fruit trees

3+ 2 = 77

3x = 77 – 2 ( Transposing 2 to R.H.S )

3x = 75

Dividing both sides of the equation by 3

\dfrac{3x}{3}=\dfrac{75}{3}

x = 25

Therefore , the number of fruit trees was 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!

Sol :

Let the number be X.

(7x + 50) + 40 = 300

7x +  90 = 300

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

7x= 210

Diyiding both sides by 7,

\dfrac{7x}{7}=\dfrac{210}{7}

x = 30

Therefore , the number is 30

 


QUESTION 5

Write the following equations in statement forms: 

(i) p + 4 = 15

Sol :

The sum of p and 4 is 1 5.

 

(ii) m — 7 = 3

Sol :

7 subtracted from m is 3.

 

(iii) 2m = 7

Sol :

Twice of a number m is 7.

 

(iv) \dfrac{m}{5}=3

Sol :

One-f‌ifth of m is 3.

 

(v) \dfrac{3m}{5}=6

Sol :

Three-f‌ifth of m is 6.

 

(vi) 3p + 4 = 25

Sol :

Three times of a number p, when added to 4, gives 25.

 

(vii) 4p — 2 = 18

Sol :

When 2 is subtracted from four times of a number p, it gives 18.

 

(viii) \dfrac{P}{2}+2=8

Sol :

When 2 is added to half of a number p it gives 8.

 


QUESTION 6

Set up an equation in the following cases:

(i) Irfan says that he has 7 marbles more than five times the marbles Parmit has. Irfan has 3 marbles. (Take m to be the number of Parmit’s marbles.)

Sol :

Let Parmit has m marbles.

5×number of marbles Parmit has + 7 = Number of marbles irfan has

5×m + 7 = 37

5m + 7 = 37

 

(ii) Laxmi’s father is 49 years old. He is 4 years older than three times Laxmi’s age. (Take Laxmi’s age to be y years.)

Sol :

Let Laxmi be y years old.

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

3× y + 4 = 49

3y + 4 = 49

 

(iii) 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. (Take the lowest score to be I.)

Sol :

Let the lowest marks be l

2× Lowest marks + 7 = Highest marks

l + 7 = 87

2l + 7 = 87

 

(iv) In an isosceles triangle, the vertex angle is twice either base angle. (Let the base angle be it in degrees. Remember that the sum of angles of a triangle is 180 degrees.)

Sol :

An isosceles triangle has two of its angles of equal measure.

Let base angle be b

Vertex angle = 2×Base angle = 2b

Sum of all interior angles of a Δ = 180°

b + b + 2b = 180°

4b = 180°


 

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