NCERT Solutions Maths Class 10 Chapter 3

# NCERT Solutions Maths Class 10 Chapter 3

## NCERT Solutions Maths Class 10 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 Rs 50, whereas 7 pencils and 5 pens together cost Rs. 46. Find the cost of one pencil and that of one pen.

Solution:  (i) Let the number of boys who took part in the quiz be x and the number of girls who took part in the quiz be y.

According to the first condition,

= 10    …….. (1)

According to the second condition,

+ 4

– = −4    …….. (2)

For equation x + y = 10, we have the following points which lie on the line.

 x 0 10 y 10 0

For equation x – y = –4, we have the following points which lie on the line.

 x 0 –4 y 4 0

We plot the points for both of the equations to find the solution.

We can clearly see that the intersection point of two lines is (3, 7).

Therefore, the number of boys who took part in the Mathematics quiz is 3 and the number of girls who took part in the Mathematics quiz is 7.

(ii) Let the cost of one pencil be Rs x and the cost of one pen be Rs y.

According to the first condition,

5+ 7= 50   ……… (1)

For equation 5x + 7y = 50, we have the following points which lie on the line.

 x 10 3 y 0 5

According to the second condition,

7+ 5= 46   ……… (2)

For equation 7x + 5y = 46, we have the following points which lie on the line.

 x 8 3 y –2 5

We plot the points for both of the equations to find the solution.

We can clearly see that the intersection point of two lines is (3, 5).

Therefore, the cost of one pencil is Rs 3 and the cost of one pen is Rs 5.

2. On comparing the ratios a1/a2, b1/b2 and c1/c2, find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident:

(i) 5− 4+ 8 = 0

7+ 6– 9 = 0

(ii) 9+ 3+ 12 = 0

18+ 6+ 24 = 0

(iii) 6− 3+ 10 = 0

2– + 9 = 0

Solution:  (i) 5− 4+ 8 = 0  …… (1)

7+ 6– 9 = 0  …… (2)

Comparing equation 5− 4+ 8 = 0 with a1x + b1y + c1 = 0 and 7+ 6– 9 = 0 with  a2x + b2y + c2 = 0, we get

a1 = 5,  b1 = –4,  c1 = 8,  a2 = 7,  b2 = 6 and c2 = –9

We have, a1/a2 ≠ b1/b2 because 5/7 ≠ 4/6.

Hence, the equations have unique solution which means the lines representing them intersect at a point.

(ii) 9+ 3+ 12 = 0  …… (1)

18+ 6+ 24 = 0  …… (2)

Comparing equation 9+ 3+ 12 = 0 with a1x + b1y + c1 = 0 and 18+ 6+ 24 = 0 with a2x + b2y + c2 = 0, we get

a1 = 9,  b1 = 3,  c1 = 12,  a2 = 18,  b2 = 6 and c2 = 24

We have, a1/a2 = b1/b2 = c1/c2 because 9/18 = 3/6 = 12/24.

Hence, the lines are coincident.

(iii) 6− 3+ 10 = 0  …… (1)

2– + 9 = 0  …… (2)

Comparing equation 6− 3+ 10 = 0 with a1x + b1y + c1 = 0 and 2– + 9 = 0 with a2x + b2y + c2 = 0, we get

a1 = 6, b1 = –3, c1 = 10, a2 = 2, b2 = –1 and c2 = 9

We have, a1/a2 = b1/b2 ≠ c1/c2 because 6/2 = –3/–1 10/9.

Hence, the lines are parallel to each other.

3. On comparing the ratios a1/a2, b1/b2 and c1/c2, find out whether the following pair of linear equations are consistent, or inconsistent.

(i) 3+ 2= 5; 2− 3= 7

(ii) 2− 3= 8; 4− 6= 9

(iii) (3/2)x + (5/3)y = 7; 9− 10= 14

(iv) 5− 3= 11; −10+ 6= −22

(v) (4/3)x + 2y = 8; 2x + 3y = 12

Solution: (i) 3+ 2= 5  …… (1)

2− 3= 7  …… (2)

Comparing equation 3+ 2= 5 with a1x + b1y + c1 = 0 and 2− 3= 7 with a2x + b2y + c2 = 0, we get

a1 = 3,  b1 = 2,  c1 = −5,  a2 = 2,  b2 =  −3 and c2 =  −7

We have, a1/a2 = 3/2 and b1/b2 = 2/−3

Here, a1/a2 ≠ b1/b2 which means equations have a unique solution.

Hence, the given pair of linear equations are consistent.

(ii) 2− 3= 8  …… (1)

4− 6= 9  …… (2)

Comparing equation 2− 3= 8 with a1x + b1y + c1 = 0 and 4− 6= 9 with

a2x + b2y + c2 = 0, we get

a1 = 2,  b1 = −3,  c1 = −8,  a2 = 4,  b2 = −6 and c2 = −9

Here, a1/a2 = b1/b2 ≠ c1/c2 because 1/2 = 1/2 ≠ −8/−9

Therefore, the equations have no solution because they are parallel.

Hence, the given pair of linear equations are inconsistent.

(iii) (3/2)x + (5/3)y = 7  …… (1)

9− 10= 14  …… (2)

Comparing equation (3/2)x + (5/3)y = 7 with a1x + b1y + c1 = 0 and 9− 10= 14 with a2x + b2y + c2 = 0, we get

a1 = 3/2, b1 = 5/3, c1 = −7,  a2 = 9,  b2 = −10 and c2 = −14

We have, a1/a2 = 1/6 and b1/b2 = −1/6

Here, a1/a2 ≠ b1/b2 which means equations have a unique solution.

Hence, the given pair of linear equations are consistent.

(iv) 5− 3= 11  …… (1)

−10+ 6= −22  …… (2)

Comparing equation 5− 3= 11 with a1x + b1y + c1 = 0 and −10+ 6= −22 with a2x + b2y + c2 = 0, we get

a1 = 5,  b1 = −3,  c1 = −11, a2 = −10,  b2 = 6 and c2 = 22

We have, a1/a2 = −1/2, b1/b2 = −1/2 and c1/c2 = −1/2

Here, a1/a2 = b1/b2 = c1/c2

Therefore, the lines are coincident and they have infinitely many solutions.

Hence, the given pair of linear equations are consistent.

(v) (4/3)x + 2y = 8  …… (1)

2x + 3y = 12  …… (2)

Comparing equation (4/3)x + 2y = 8 with a1x + b1y + c1 = 0 and 2x + 3y = 12 with a2x + b2y + c2 = 0, we get

a1 = 4/3,  b1 = 2,  c1 = −8, a2 = 2,  b2 = 3 and c2 = −12

We have, a1/a2 = 2/3, b1/b2 = 2/3 and c1/c2 = 2/3

Here, a1/a2 = b1/b2 = c1/c2

Therefore, the lines are coincident and they have infinitely many solutions.

Hence, the given pair of linear equations are consistent.

4. Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically:

(i) = 5, 2+ 2= 10

(ii) – = 8, 3− 3= 16

(iii) 2– 6 = 0, 4− 2– 4 = 0

(iv) 2− 2– 2 = 0, 4− 4– 5 = 0

Solution: (i) = 5   …… (1)

2+ 2= 10   …… (2)

Comparing equation x + y = 5 with a1x + b1y + c1 = 0 and 2x + 2y = 10 with a2x + b2y + c2 = 0, we get

a1 = 1,  b1 = 1,  c1 = −5, a2 = 2,  b2 = 2 and c2 = −10

We have, a1/a2 = 1/2, b1/b2 = 1/2 and c1/c2 = 1/2

Here, a1/a2 = b1/b2 = c1/c2

Therefore, the lines are coincident and they have infinitely many solutions.

Hence, the given pair of linear equations are consistent.

For equation x + y = 5, we have following points which lie on the line.

 x 0 5 y 5 0

For equation 2x + 2y = 10, we have following points which lie on the line.

 x 1 2 y 4 3

We plot the points for both of the equations to find the solution.

Any point which lies on one line also lies on the other.

We can take any random value for y and find the corresponding value of x using the given equation. All such points will lie on both lines and there will be infinite number of solutions for these equations.

(ii) – = 8   …….. (1)

3− 3= 16   …….. (2)

Comparing equation x − y = 8 with a1x + b1y + c1 = 0 and 3x − 3y = 16 with a2x + b2y + c2 = 0, we get

a1 = 1,  b1 = −1,  c1 = −8, a2 = 3,  b2 = −3 and c2 = −16

We have, a1/a2 = 1/3, b1/b2 = 1/3 and c1/c2 = 1/2

Here, a1/a2 = b1/b2 ≠ c1/c2

Therefore, the lines are parallel and they have no solution.

Hence, the given pair of linear equations are inconsistent.

(iii) 2– 6 = 0     ….. (1)

4− 2– 4 = 0     ….. (2)

Comparing equation 2x + y – 6 = 0 with a1x + b1y + c1 = 0 and 4x − 2y – 4 = 0 with a2x + b2y + c2 = 0, we get

a1 = 2,  b1 = 1,  c1 = −6, a2 = 4,  b2 = −2 and c2 = −4

We have, a1/a2 = 1/2 and b1/b2 = −1/2

Here, a1/a2 ≠ b1/b2

Therefore, the lines intersect at a point and they have a unique solution.

Hence, the given pair of linear equations are consistent.

For equation 2x + y – 6 = 0, we have the following points which lie on the line.

 x 0 2 3 y 6 2 0

For equation 4x – 2y – 4 = 0, we have the following points which lie on the line.

 x 0 2 1 y –2 2 0

We plot the points for both of the equations to find the solution.

We can clearly see that lines are intersecting at (2, 2).

Hence, x = 2 and y = 2 are the solution of the given linear equations.

(iv) 2− 2– 2 = 0     ….. (1)

4− 4– 5 = 0     ….. (2)

Comparing equation 2x − 2y – 8 = 0 with a1x + b1y + c1 = 0 and 4x − 4y – 5 = 0 with a2x + b2y + c2 = 0, we get

a1 = 2,  b1 = −2,  c1 = −2, a2 = 4,  b2 = −4 and c2 = −5

We have, a1/a2 = 1/2, b1/b2 = 1/2 and c1/c2 = 2/5

Here, a1/a2 = b1/b2 ≠ c1/c2

Therefore, the lines are parallel and they have no solution.

Hence, the given pair of linear equations are inconsistent.

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 the length of the rectangular garden be x metres and the width of the rectangular garden be y metres.

According to the first condition, half perimeter = 36 m.

x + y = 36  ……(i)

According to the second condition, x = y + 4

– = 4  ……..(ii)

Adding equations (i) and (ii), we get

2x = 40

x = 20 m

Putting the value of x in equation (i), we get

20 + y = 36

y = 16 m

Hence, the length of the rectangular garden is 20 m and its width is 16 m.

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

Solution: (i) Let the second linear equation be a2x + b2y + c2 = 0.

Comparing the given linear equation 2x + 3y – 8 = 0 with a1x + b1y + c1 = 0, we get

a1 = 2, b1 = 3 and c1 = –8

We know that, two lines intersect with each other if a1/a2 ≠ b1/b2

So, the second equation can be 3x + 2y = 5 because a1/a2 ≠ b1/b2

(ii) Let the second linear equation be a2x + b2y + c2 = 0.

Comparing the given linear equation 2x + 3y – 8 = 0 with a1x + b1y + c1 = 0, we get

a1 = 2, b1 = 3 and c1 = –8

We know that, two lines are parallel with each other if a1/a2 = b1/b2 ≠ c1/c2

So, the second equation can be 4x + 6y = 3 because a1/a2 = b1/b2 ≠ c1/c2

(iii) Let the second linear equation be a2x + b2y + c2 = 0.

Comparing the given linear equation 2x + 3y – 8 = 0 with a1x + b1y + c1 = 0, we get

a1 = 2, b1 = 3 and c1 = –8

We know that, two lines coincide with each other if a1/a2 = b1/b2 = c1/c2

So, the second equation can be 6x + 9y = 24 because a1/a2 = b1/b2 = c1/c2

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.

Solution: For equation x – y + 1 = 0, we have the following points which lie on the line.

 x 0 –1 2 y 1 0 3

For equation 3x + 2y – 12 = 0, we have the following points which lie on the line.

 x 4 0 2 y 0 6 3

Now, let us plot these points to find the two lines.

We can observe from the above graphs that points of intersection of the lines with the x-axis are (–1, 0), (2, 3) and (4, 0). Triangle PBC is shaded in the above graph.