Exersice 3.4 Class 10 NCERT Solutions pdf with small and detailed answers

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Chapter 3-Pairs of linear equation in 1 variable Exercise Ex.3.4

here it is solution of question 1

Chapter 3 - Pairs of Linear Equations in Two Variables Exercise Ex. 3.4

Solution 1

 
        
(ii)      Elimination method:
          3x + 4y = 10      ...(1)
          2x - 2y = 2         ...(2)
          Multiplying equation (2) by 2, we obtain:
          4x - 4y = 4         ...(3)
          Adding equation (1) and (3), we obtain
          7x = 14
            x = 2
          Substituting the value of x in equation (1), we obtain:
          6 + 4y = 10
          4y = 4
            y = 1
          Hence, x = 2, y = 1

          Substitution method:
          From equation (2), we obtain:
          x = 1 + y            ...(4)
          Putting this value in equation (1), we obtain:
          3(1 + y) + 4y = 10.
                           7y = 7
                             y = 1
          Substituting the value of y in equation (4), we obtain:

         


(iv)      Elimination Method:
            
         

             Substitution method:
             From equation (2), we obtain:
             y = 3x - 9        ... (3)
             Putting this value in equation (1), we obtain:
             3x + 4(3x - 9) = -6
                            15x = 30
                                x = 2
             Substituting the value of x in equation (3), we obtain:
              y = 6 - 9 = -3
            
Concept insight: In order to solve the given pairs of equations by elimination method, it is required to eliminate one variable to get a linear equation in one variable. For elimination, we multiply one equation (or sometimes even both equations) by suitable constant to make the coefficient of one variable same in both the equations. The trick is to eliminate that variable which involves lesser computations.

In order to solve the given pair of equations by substitution method, we need to substitute the value of any one of the variable from any one of the equation. The trick here is to make sure you substitute the value of that variable which simplifies your calculations.
Note that the solution must be same in both the cases.

Chapter 3-Pairs of linear equation in 1 variable Exercise Ex.3.4

here it is solution of question 2

 

Solution 2

(i)    Let the fraction be

According to the question,


Subtracting equation (1) from equation (2), we obtain:
x = 3
Substituting this value of x in equation (1), we obtain:


Concept insight: This problem talks about a fraction. The numerator and denominator of the fraction are not known so we represent these as variables x and y respectively where variable y must be non zero. Then, a pair of linear equations can be formed from the given conditions. The pair of equations can then be solved by eliminating a suitable variable.

(ii)  Let present age of Nuri and Sonu be x and y respectively.

According to the question,


Subtracting equation (1) from equation (2), we obtain:
y = 20
Substituting the value of y in equation (1), we obtain:


Thus, the age of Nuri and Sonu are 50 years and 20 years respectively.

Concept insight: Here, Nuri's and Sonu's present age are not known. So, we will write both these in terms of variables. Then, using the given conditions, a pair of linear equations can be formed. The pair of equations can then be solved by eliminating a suitable variable.

(iii)  Let the units digit and tens digit of the number be x and y respectively.

Number = 10y + x
Number after reversing the digits = 10x + y

According to the question,
x + y = 9          ... (1)
9(10y + x) = 2(10x + y)
88y - 11x = 0
- x + 8y =0          ... (2)
Adding equations (1) and (2), we obtain:
9y = 9
y = 1
Substituting the value of y in equation (1), we obtain:
x = 8
Thus, the number is 10y + x = 10 x 1 + 8 = 18

Concept insight: This problem talks about a two digit number. Here, remember that a two digit number xy can be expanded as 10x + y. Then, using the two given conditions, a pair of linear equations can be formed which can be solved by eliminating one of the variables.

(iv)  Let the number of Rs 50 notes and Rs 100 notes be x and y respectively.
According to the question,

Multiplying equation (1) by 50, we obtain:
50x + 50 y = 1250           .... (3)
Subtracting equation (3) from equation (2), we obtain:

50y = 750
y = 15

Substituting the value of y in equation (1), we obtain:
x = 10
Hence, Meena received 10 notes of Rs 50 and 15 notes of Rs 100.

Concept insight: This problem talks about two types of notes, Rs 50 notes and Rs 100 notes. And the number of both these notes with Meena is not known. So, we denote the number of Rs 50 notes and Rs 100 notes by variables x and y respectively. Now two linear equations can be formed by the given conditions which can be solved by eliminating one of the variables.

(v)    Let the fixed charge for first three days and each day charge thereafter be Rs x and Rs y respectively.
According to the question,

Subtracting equation (2) from equation (1), we obtain:

2y = 6
y = 3
Substituting the value of y in equation (1), we obtain:

x + 12 = 27
x = 15

Hence, the fixed charge is Rs 15 and the charge per day is Rs 3.

Concept insight: Here, the fixed charges for the first three days and per day charges are not known so, they will be represented  using two different variables. The two equations can then be obtained by using the given conditions which can be solved by eliminating one of the variables.

 

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if you not read the previous blog please read it. it is the solution to exercise 1.1 AND EXERCISE 2.3

 
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