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Exercise 3.1 

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Q1. Aftab tells his daughter, “Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be.” (Isn’t this interesting?) Represent this situation algebraically and graphically.

Solution:

Let Aftab present age be x years

Her daughter present age be y years

Situation – I

7 years ago Aftab age = x – 7 years

And her daughter age = y – 7 years

Q2. The coach of a cricket team buys 3 bats and 6 balls for Rs 3900. Later, she buys another bat and 2 more balls of the same kind for Rs 1300. Represent this situation algebraically and geometrically.

Q3. The cost of 2 kg of apples and 1kg of grapes on a day was found to be Rs 160. After a month, the cost of 4 kg of apples and 2 kg of grapes is Rs 300. Represent the situation algebraically and geometrically.

 

Exercise 3.2

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Q1. Form the pair of linear equations in the following problems, and find their solutions graphically.

(i) 10 students of Class X took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz.

Solution:

Let the numbers of girls be x.  

And the number of girls is y.

According to question

The total numbers of boys and girls is 10. 

∴ x + y = 10 ……………….. (i)  

There are 4 more girls than boys.

∴  x – y = 4  ……………….. (ii)

Table for equation (i)  

Graph for these tables will be as follow - 

(ii) 5 pencils and 7 pens together cost Rs 50, whereas 7 pencils and 5 pens together cost Rs 46. Find the cost of one pencil and that of one pen.

Solution:

Q5. Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m.  Find the dimensions of the garden.

Solution: Let the length of rectangular garden be x m.  

And breadth = y m

Semi-perimeter = 36 m  

Q6. Given the linear equation 2x + 3y – 8 = 0, write another linear equation in two variables such that the geometrical representation of the pair so formed is:

(i) Intersecting lines

(ii) parallel lines

(iii) Coincident lines

Q7. Draw the graphs of the equations x – y + 1 = 0 and 3x + 2y – 12 = 0. Determine the coordinates of the vertices of the triangle formed by these lines and the x-axis, and shade the triangular region.

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Exercise 3.3

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Q1. Solve the following pair of linear equations by the substitution method.

Solution Q1:

(i) x + y = 14  ............ (i)

       x – y = 4    ............ (ii)

Using substitution method–

From Equation (ii)

x – y = 4      

x = 4 + y

Now putting value of x as 4 + y in equation (i)  

x + y = 14

Or  (4 + y) + y = 14

Or  4 + 2y = 14

Or  2y = 14 - 4

Or  2y = 10

Now putting value of y in equation (ii)

   x = 4 + y

Or  x = 4 + 5 = 9

Therefore solution of given pair of linear equation is –

So x = 9, and y = 5 Answer

Solution: (iii) 3x – y = 3....... (i)  

9x – 3y = 9....... (ii)

Using substitution method–

From Equation (i)

3x – y = 3

Or 3x – 3 = y

Or y = 3x – 3

Now putting the value of y in equation (ii)

9x – 3y = 9

Or 9x – 3(3x – 3) = 9

Or 9x – 9x + 9 = 9

Or 9 = 9

Or x = 0 and y = 3x – 3 Answer

Therefore solution of given pair of linear equation is –

x = 0 and y = 3x – 3

Q2. Solve 2x + 3y = 11 and 2x – 4y = – 24 and hence find the value of ‘m’ for which y = mx + 3.

Solution:

2x + 3y = 11 ............. (i)

2x – 4y = –24 ........... (ii)

From equation (i)

2x + 3y = 11

⇒ 2x = 11 – 3y

Now, getting the value of m we put the value of x and y in y = mx + 3

Q3. Form the pair of linear equations for the following problems and find their solution by substitution method.

(i) The difference between two numbers is 26 and one number is three times the other. Find them.

Solution: Let the first number be x and the second be y.|

According to question,

Situation-I

x – y = 26 ............. (i)

Situation-II

x = 3y     ............. (ii)

Now, putting x = 3y in equation (i)  

   x – y = 26

⇒ 3y – y = 26

⇒ 2y = 26

⇒ y = 13

Now, y = 13 putting in equation (ii)

   x = 3y

⇒ x = 3 × 13

      = 39

Therefore, the first number is 39 and the second number is 13.

(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.

Solution:

Let the larger angle of two supplementary angles be x.

And the smaller is y.

Therefore, Situation-I

x – y = 18°  ............... (i)

Situation-II

x + y = 180°  ........... (ii)  (The sum of two supplementary angles is 180° )

Now From equation (i)

    x – y = 18°

(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.

Solution:

Let the larger angle of two supplementary angles be x.

And the smaller is y.

Therefore, Situation-I

x – y = 18°  ............... (i)

Situation-II

x + y = 180°  ........... (ii)  (The sum of two supplementary angles is 180° )

Now From equation (i)

    x – y = 18°

(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.

Solution:

Let the larger angle of two supplementary angles be x.

And the smaller is y.

Therefore, Situation-I

x – y = 18°  ............... (i)

Situation-II

x + y = 180°  ........... (ii)  (The sum of two supplementary angles is 180° )

Now From equation (i)

    x – y = 18°

⇒ x = 18° + y

Now putting the value of x in equation (ii)

   x + y = 180°

⇒ 18° + y + y = 180°

⇒ 18° + 2y = 180°

⇒ 2y = 180° – 18°

x = 18° + y

Now putting the value of x in equation (ii)

   x + y = 180°

⇒ 18° + y + y = 180°

⇒ 18° + 2y = 180°

⇒ 2y = 180° – 18°

x = 18° + y

Now putting the value of x in equation (ii)

   x + y = 180°

⇒ 18° + y + y = 180°

⇒ 18° + 2y = 180°

⇒ 2y = 180° – 18°

⇒ 2y = 162°

⇒ y = 81°

Putting the value of y in equation (i)

⇒ x = 18° + y

⇒ x = 18° + 81°

⇒ x = 99°

Therefore, the larger angle is 99° and the smaller is 81° 

(iii) The coach of a cricket team buys 7 bats and 6 balls for Rs 3800. Later, she buys 3 bats and 5 balls for Rs 1750. Find the cost of each bat and each ball.

Solution:

Let the cost of a bat be x.  

And the cost of a ball be y.

Situation I

    7 bats + 6 balls = 3800

⇒  7x + 6y = 3800 ......... (i)

Situation II

3 bats +5 balls = 1750

⇒  3x + 5y = 1750 ......... (ii)

From equation (ii)

    3x + 5y = 1750

⇒ 3x = 1750 – 5y 

(iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is Rs 105 and for a journey of 15 km, the charge paid is Rs 155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km?

Situation I

x + 10y = 105 ........... (i)

Situation II

x + 15y = 155 ............ (ii)

From equation (i)

x + 10y = 105

⇒  x = 105 – 10y

Now putting the value of x in equation (ii)

x + 15y = 155

⇒(105 – 10y) + 15y = 155

⇒  105 + 5y = 155

⇒  5y = 155 –105

⇒  5y = 50

Now, putting y = 10 in equation (i)  

⇒  x = 105 – 10y

⇒  x =105 –10(10)

⇒  x = 105 –100 = 5

​The charge for 25 km = x + 25y

                  = 5 + 25(10)

                  = 5 + 250

                  = 255 Rupees

(vi) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?

Solution: Let the present age of Jacob be x years.

And his son present age be y years.

Situation I

Five years hence, Jacob’s age = x + 5 years

And his son age will be = y + 5 years

Therefore, x + 5 = 3(y + 5)

⇒  x + 5 = 3y + 15

⇒  x – 3y = 15– 5

⇒  x – 3y = 10  ........... (i)

Situation II

Five years ago, Jacob’s age = x – 5 years

And his son’s age  = y – 5 years   

Then,  x – 5 = 7(y – 5)

⇒  x – 5 = 7y –35

⇒  x – 7y = 5 –35

⇒  x – 7y = – 30  ........... (ii)

From equation (ii)

x – 7y = – 30

Þx = 7y –30

Now, putting the value of x  in equation (i)

x – 3y = 10 

⇒ (7y – 30) – 3y = 10

⇒ 4y = 10 + 30

⇒ 4y = 40

⇒ y = 10

Putting y = 10 in equation (ii)

⇒ x = 7(10) – 30

⇒ x = 70– 30 = 40

Therefore, Jacob’s present age 40 years and his son’s age 10 years. 

EXERCISE 3.4

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Q1. Solve the following pair of linear equations by the elimination method and the substitution method :

(i) x + y = 5 and 2x - 3y = 4      

(ii) 3x + 4y = 10 and 2x - 2y = 2

(iii) 3x - 5y - 4 = 0 and 9x = 2y + 7

Solution:

Q1. (i) x + y = 5 .......... (i)

        2x - 3y = 4 ......... (ii)

Equation (i) × 3 = 3x + 3y = 15  ...(iii)

Equation (ii) ×1 = 2x - 3y = 4  ... (iv)

[Note: Here the coefficient of y has been equal and the signs are unlike, therefore these will be added]

Now adding Equation (iii) and (iv) 

Solution: (ii) 3x + 4y = 10 ........ (i)

2x - 2y = 2   ........ (ii)

Equation (i) × 1 ⇒ 3x + 4y = 10 ........ (iii)

Equation (ii) ×2 ⇒ 4x -4y = 4........ (iv)

Adding equation (i) and (ii)   

Now putting the value of x = 2 in equation (i)

3x + 4y = 10

⇒ 3(2) + 4y = 10

⇒ 6 + 4y = 10

⇒4y = 10- 6

⇒ 4y = 4

⇒ y = 1

Therefore, solution of given pair of linear equation are x = 2 and y = 1

Solution:(iii) 3x - 5y - 4 = 0

Or       3x – 5y = 4 ......... (i)

         9x = 2y + 7

Or       9x - 2y = 7 ......... (ii)

Equation (i) × 3  ⇒ 9x - 15y = 12 ..... (iii)

Equation (ii) × 1 ⇒ 9x - 2y = 7     .....(iv)

Subtracting equation (iv) from equation (iii)  

Or      3x + 4y = - 6 .......... (i)

Or      3x - y = 9  ............. (ii)

Subtracting equation (ii) from equation (i)

Now, putting y = - 3 in equation (i)  

        3x + 4y = - 6 

⇒   3x + 4(-3) = - 6 

⇒   3x - 12 = - 6 

⇒   3x = 12 - 6 

⇒   3x = 6 

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⇒   x = 2

Therefore, solution of given pair of linear equation are x = 2 and y = - 3

Q2. Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method :

Solution: Let the numerator be x and denominator of fraction be y.

(Note: Here in equation (i) and (ii) has the equal coefficients of y, therefore it need not to equalize them.)

Now, subtracting equation (ii) f rom equation (i) 

∴  x = 3

Now, putting the value of x = 3 in equation (i)

     x - y = - 2

⇒  3 - y = - 2 

⇒   y = 3 + 2 

⇒   y = 5 

(ii) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?

Solution: Let the Nuri’s age be x years  

And Sonu’s age be y years

Situation I  

Five years ago,

Nuri’s age = x - 5 years

Sonu’s age = y - 5 years

According to question,

⇒ x - 5 = 5(y - 5)

⇒ x - 5 = 5y - 25

⇒ x - 5y = 5 - 25

⇒ x - 5y = - 20 ............ (i)

Situation II  

Ten years later,

Nuri’s age = x + 10 years

Sonu’s age = y + 10 years

According to question,

⇒ x + 10 = 2(y + 10)

⇒ x + 10 = 2y + 20

⇒ x - 2y = 20 - 10

⇒ x - 2y = 10  ............ (ii)

(Since the coefficients of x are automatically equal, the coefficients will not equal.)

Now, subtracting equation (ii) from equation (i)

Putting y = 10 in equation (i)

     x -  5y = -  20

Or   x -  5(10) = -  20

Or   x -  50 = -  20

Or   x = 50 -  20

Or x = 30

Therefore Nuri’s age is 30 years and Sonu’s age is 10 years.

(iii) The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.

Solution: Let the unit digit of the required number be x.

And ten’s digit be y.

Then the real number = 10y + x,

And reversed number = 10x + y 

Situation I

            x + y = 9 ........... (i)

Situation II

            9(number) = 2(reversed number)

⇒      9(10y + x) = 2(10x + y)

⇒       90y + 9x = 20x + 2y

⇒       20x - 9x + 2y - 90y = 0

⇒       11x - 88y = 0

⇒       x - 8y = 0

⇒       x = 8y ........... (ii)

Putting x = 8y in equation (i)  

      x + y = 9

Or   8y + y = 9

Or    9y = 9

Or     y =  = 1

Putting y = 1 in equation (ii)  

x = 8y = 8 × 1 = 8

Therefore, required number = 10y + x

                 = 10 × 1 + 8

                 = 18 Answer

(iv) Meena went to a bank to withdraw Rs 2000. She asked the cashier to give her Rs 50 and Rs 100 notes only. Meena got 25 notes in all. Find how many notes of Rs 50 and Rs 100 she received.

Solution: Let the numbers of 50 notes = x

And number of 100 rupees notes = y

Situation I

Total numbers of notes  = 25

∴ x + y = 25  ........... (i)

Situation II

     x notes of 50 + y notes of 100 = 2000  

Therefore,  50x + 100y = 2000

Or    x + 2y = 40   ........... (ii) (simplifying)

Subtracting equation(ii) from equation (i)  

∴  y = 15

Now putting y = 15 in equation (i)

     x + y = 25

Or   x + 15 = 25

Or   x = 25 - 10

Or   x = 10

∴ Number of 50 notes is 10 and 100 notes is 15.

(v) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid Rs 27 for a book kept for seven days, while Susy paid Rs 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.

Putting y = 3 in equation (i)

            x + 7y = 27

Or        x + 7(3) = 27

Or    x + 21 = 27

Or    x = 27 - 21

Or    x = 6

Therefore, fixed charge = ₹ 6 and additional charge = ₹ 3

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