# NCERT Solution for Class 10 Mathematics Chapter 3 Pair of Linear Equations in Two Variables

 Chapter Name NCERT Solution for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Topics Covered Short Revision for the ChapterNCERT Exercise Solution Related Study NCERT Solution for Class 10 MathsNCERT Revision Notes for Class 10 MathsImportant Questions for Class 10 MathsMCQ for Class 10 MathsNCERT Exemplar Questions For Class 10 Maths

## Short Revision for Pair of Linear Equations in Two Variables

1. A pair of two linear equations is either consistent or inconsistent.  consistency provides either a unique solution or infinitely many solutions.
2. The two lines represented by a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
(i) intersect, if a1/a2 ≠ b1/b2
In this case, the pair of equations is consistent with unique solution.
(ii) coincident, if a1/a2 = b1/b2 = c1/c2 .
In this case, the pair of equations is consistent with infinitely many solutions.
(iii) are parallel, if a1/a2 = b1/b2 ≠ c1 = c2 .
In this case, the pair of equations is inconsistent and has no solution.
3.  A pair of linear equations in two variables can be solved by graphical method or algebraic method.
4. Algebraic method provides the better results than graphical method.

### NCERT Exercises Solution

#### Exercise 3.1

1. 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.

To represent the given situation graphically, we plot the points A(0,6), B(-42,0) and C(42, 12) to get the line AB and the points C(42, 12), D(6,0) and E(0, -2) to get the line CD as shown :

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

3. 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

1. 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.
(ii) 5 pencils and 7 pens together cost 50, whereas 7 pencils and 5 pens together cost 46. Find the cost of one pencil and that of one pen.

Solution

(i) Let number of boys = x and number of girls = y
We have, x + y = 10 ....(i)
and y = x + 4 ....(ii)
Let us plot the two equations on the graph.
Some points on line (i) are :

The two lines meet at A (3, 7), i.e., x = 3, y = 7
Hence, number of boys = 3, number of girls = 7.
(ii) Let cost of one pencil  = ₹ x
and cost of one pen  = ₹ y
According to the given condition, we have
5x + 7y = 50 ..... (i)
and 7x + 5y = 46 .... (ii)

2. On comparing the ratios a1/a2 , b1/b2 , c1/c2 find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident:
(i) 5x – 4y + 8 = 0
7x + 6y – 9 = 0
(ii) 9x + 3y + 12 = 0
18x + 6y + 24 = 0
(iii) 6x – 3y + 10 = 0
2x – y + 9 = 0

3. On comparing the ratio, (a1/a2) , (b1/b2) , (c1/c2) find out whether the following pair of linear equations are consistent, or inconsistent.
(i) 3x + 2y = 5 ; 2x – 3y = 7
(ii) 2x – 3y = 8 ; 4x – 6y = 9
(iii)(3/2)x+(5/3)y = 7; 9x – 10y = 14
(iv) 5x – 3y = 11 ; – 10x + 6y = –22
(v)(4/3)x+2y = 8 ; 2x + 3y = 12

4. Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically:
(i) x + y = 5, 2x + 2y = 10
(ii) x – y = 8, 3x – 3y = 16
(iii) 2x + y – 6 = 0, 4x – 2y – 4 = 0
(iv) 2x – 2y – 2 = 0, 4x – 4y – 5 = 0

5. 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 length of the garden = x m
and breadth = y m; half the perimeter = 36 m
We have x + y = 36 ...(i)
and x = y + 4 ...(ii)
Solving (i) and (ii), we get x = 20, y = 16
Therefore, length and breadth of the garden are 20 m and 16 m respectively.

6. 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

7. 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.

On plotting these points on a graph paper, we get the coordinates of the vertices of a triangle ABC are A(2, 3), B(-1, 0), C(4, 0).

#### Exercise 3.3

1. Solve the following pair of linear equations by the substitution method
(i) x + y = 14
x – y = 4
(ii) s – t = 3
(s/3) + (t/2) = 6
(iii) 3x – y = 3
9x – 3y = 9
(iv) 0.2x + 0.3y = 1.3
0.4x + 0.5y = 2.3
(v) √2 x+√3 y = 0
√3 x-√8 y = 0
(vi) (3x/2) – (5y/3) = -2
(x/3) + (y/2) = (13/6)

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

3. 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.
(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.
(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.
(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?
(v) A fraction becomes 9/11 , if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes 5/6. Find the fraction.
(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?

#### Exercise 3.4

1. 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
(iv) x/2+ 2y/3 = -1 and x-y/3 = 3

2. Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method:
(i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes if we only add 1 to the denominator. What is the fraction?
(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?
(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.
(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.
(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.

Solution

(i) Let fraction be x/y
According to given conditions,
(x+1)/(y-1) = 1 ⇒ x + 1 = y - 1
⇒ x - y = -2 ...(i)
and x/(y + 1) = 1/2 ⇒ 2x = y + 1
⇒ 2x - y = 1 ...(ii)
Solving (i) and (ii), we get
x = 3, y = 5
Hence, required fraction is 3/5.
(ii) Let present age of Nuri = x years
and present age of Sonu = y years
Five years ago,
Nuri's age = (x - 5) years and Sonu's age = (y - 5) years
∴ x - 5 = 3(y - 5) ⇒ x - 3y = -10 ....(i)
Ten years later,
Nuri's age = (x + 10) years and Sonu's age = (y + 10)years
∴ x + 10 = 2(y + 10) ⇒ x -2y = 10 ....(iii)
Solving (i) and (ii), we get
x = 50, y = 20
Hence, Nuri's age = 50 years and Sonu's age  = 20 years

#### Exercise 3.5

1. Which of the following pairs of linear equations has unique solution, no solution, or infinitely many solutions. In case there is a unique solution, find it by using cross multiplication method.
(i) x – 3y – 3 = 0 and 3x – 9y – 2 = 0

(ii) 2x + y = 5 and 3x + 2y = 8
(iii) 3x – 5y = 20 and 6x – 10y = 40

(iv) x – 3y – 7 = 0 and 3x – 3y – 15 = 0

So, the pair of equations is consistent with infinitely many solutions.We can consider one equation say 3x - 5y = 20.
Let y = k

So, the pair of linear equations is consistent and has a unique solution.
Solve for x and y by using cross - multiplication method to get x = 4 and y = -1.

2. (i) For which values of a and b does the following pair of linear equations have an infinite number of solutions?
2x + 3y = 7
(a – b) x + (a + b) y = 3a + b – 2
(ii) For which value of k will the following pair of linear equations have no solution?
3x + y = 1
(2k – 1) x + (k – 1) y = 2k + 1

3. Solve the following pair of linear equations by the substitution and cross-multiplication methods:
8x + 5y = 9
3x + 2y = 4

4. Form the pair of linear equations in the following problems and find their solutions (if they exist) by any algebraic method:
(i) A part of monthly hostel charges is fixed and the remaining depends on the number of days one has taken food in the mess. When a student A takes food for 20 days she has to pay Rs.1000 as hostel charges whereas a student B, who takes food for 26 days, pays Rs.1180 as hostel charges. Find the fixed charges and the cost of food per day.
(ii) A fraction becomes 1/3 when 1 is subtracted from the numerator and it becomes 1/4 when 8 is added to its denominator. Find the fraction.
(iii) Yash scored 40 marks in a test, getting 3 marks for each right answer and losing 1 mark for each wrong answer. Had 4 marks been awarded for each correct answer and 2 marks been deducted for each incorrect answer, then Yash would have scored 50 marks. How many questions were there in the test?
(iv) Places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the two cars?
(v) The area of a rectangle gets reduced by 9 square units, if its length is reduced by 5 units and breadth is increased by 3 units. If we increase the length by 3 units and the breadth by 2 units, the area increases by 67 square units. Find the dimensions of the rectangle.

(iii) Let number of Yash's right answer = x
and that of wrong answers = y
Total number of questions = x + y
We have 3x - y = 40 ...(i)
and 4x - 2y = 50, i.e., 2x - y = 25 ...(ii)
Solving (i) and (ii), we get x = 15 , y = 5
∴ Total number of questions = 20.
(iv) Let speed of car starting from A  = x km/hr and speed of car starting from B = y km/hr.
We have 5x - 5y = 100 ⇒ x - y = 20 ...(i)
and x + y = 100 ....(ii)
Solving (i) and (ii), we get x = 60, y = 40.
Hence, speed of car starting from place A = 60 km/hr
and speed of car starting from place B = 40 km/hr.
(v) Let length of the rectangle = x units
and breadth of the rectangle = y units.
According to given conditions, we get,

#### Exercise 3.6

1. Solve the following pairs of equations by reducing them to a pair of linear equations:
(i) 1/2x + 1/3y = 2
1/3x + 1/2y = 13/6

(ii) 2/√x + 3/√y = 2
4/√x + 9/√y = -1

(iii) 4/x + 3y = 14
3/x -4y = 23

(iv) 5/(x-1) + 1/(y-2) = 2
6/(x-1) – 3/(y-2) = 1

(v) (7x-2y)/ xy = 5
(8x + 7y)/xy = 15

(vi) 6x + 3y = 6xy
2x + 4y = 5xy

(vii) 10/(x+y) + 2/(x-y) = 4
15/(x+y) – 5/(x-y) = -2

(viii) 1/(3x+y) + 1/(3x-y) = 3/4
1/2(3x+y) – 1/2(3x-y) = -1/8

2. Formulate the following problems as a pair of equations, and hence find their solutions:
(i) Ritu can row downstream 20 km in 2 hours, and upstream 4 km in 2 hours. Find her speed of rowing in still water and the speed of the current.
(ii) 2 women and 5 men can together finish an embroidery work in 4 days, while 3 women and 6 men can finish it in 3 days. Find the time taken by 1 woman alone to finish the work, and also that taken by 1 man alone.
(iii) Roohi travels 300 km to her home partly by train and partly by bus. She takes 4 hours if she travels 60 km by train and the remaining by bus. If she travels 100 km by train and the remaining by bus, she takes 10 minutes longer. Find the speed of the train and the bus separately.

Solution

(i) Let speed of rowing in still water = x km/hr
and speed of current = y km/hr
∴ Speed along downstream = (x+ y)km/hr
and speed along upstream = (x - y) km/hr

Solving (i) and (ii), we get x = 6, y = 4.
Therefore, speed of rowing in still water = 6 km/hr
and speed of stream (current) = 4 km/hr
(ii) Let 1 man finishes the work in x days
and 1 woman finishes the work in y days .
So, 1 man's 1 day's work = 1/x
∴ 5 men's 1 day's work = 5/x
And 1 woman's one day's work = 1/y
∴ 2 women's one day's work = 2/y
∴ 2 women's and 5 men's 1 day's work = 2/y + 5/x
Since 2 women and 5 men finish the work in 4 days.

Solving for x and y, we get x = 36, y = 18
Hence, 1 man can finish the work in 36 days and 1 woman can finish the work in 18 days.
(iii) Let speed of the train = x km/hr
and speed of the bus = y km/hr
According to given conditions,

#### Exercise 3.7

1. The ages of two friends Ani and Biju differ by 3 years. Ani’s father Dharam is twice as old as Ani and Biju is twice as old as his sister Cathy. The ages of Cathy and Dharam differ by 30 years. Find the ages of Ani and Biju.

2. One says, “Give me a hundred, friend! I shall then become twice as rich as you”. The other replies, “If you give me ten, I shall be six times as rich as you”. Tell me what is the amount of their (respective) capital? [From the Bijaganita of Bhaskara II] [Hint : x + 100 = 2(y – 100), y + 10 = 6(x – 10)].

3. A train covered a certain distance at a uniform speed. If the train would have been 10 km/h faster, it would have taken 2 hours less than the scheduled time. And, if the train were slower by 10 km/h; it would have taken 3 hours more than the scheduled time. Find the distance covered by the train.

4. The students of a class are made to stand in rows. If 3 students are extra in a row, there would be 1 row less. If 3 students are less in a row, there would be 2 rows more. Find the number of students in the class.

5. In a ∆ABC, ∠ C = 3 ∠ B = 2 (∠A + ∠ B). Find the three angles.

6. Draw the graphs of the equations 5x – y = 5 and 3x – y = 3. Determine the co-ordinates of the vertices of the triangle formed by these lines and the y axis.

7. Solve the following pair of linear equations:
(i) px + qy =p – q
qx – py = p + q

(ii) ax + by = c
bx + ay = 1 + c

(iii) x/a – y/b = 0
ax + by = a2 + b2

(iv) (a – b)x + (a + b) y = a2 – 2ab – b2
(a + b)(x + y) = a2 + b2

(v) 152x – 378y = – 74
–378x + 152y = – 604

8. ABCD is a cyclic quadrilateral (see Fig. 3.7). Find the angles of the cyclic quadrilateral.