# NCERT Solutions for Class 10 Maths

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1.    Let the present age of Aftab and his daughter be x and y respectively.

Seven years ago, Age of Aftab = x – 7 and Age of his daughter = y – 7 According to the given condition,

(x – 7) = 7(y – 7)         ⇒         x – 7 = 7y – 49         ⇒          x – 7y = -42

Again, Three years hence, Age of Aftab = x + 3 and Age of his daughter = y + 3 According to the given condition,

(x + 3) =3 (y + 3)   ⇒     x + 3 = 3y + 9   ⇒    x – 3y = 6

Thus, the given conditions can be algebraically represented as:

x – 7y = -42     ⇒      x = -42 + 7y

Three solutions of this equation can be written in a table as follows:

And x – 3y = 6   ⇒   x = 6 + 3y

Three solutions of this equation can be written in a table as follows:

The graphical representation is as follows:

Concept insight: In order to represent the algebraic equations graphically the solution set of equations must be taken as whole numbers only for the accuracy. Graph of the two linear equations will be represented by a straight line.

2.    Let cost of 1 cricket bat = Rs x and let cost of 1 cricket ball= Rs y

According to given conditions, we have

3x + 6y = 3900     ⇒   x + 2y = 1300                … (1)

And   x + 3y = 1300            … (2)

To represent them graphically, we will find 3 sets of points which lie on the lines. For equation x + 2y = 1300, we have following points which lie on the line

x          0             1300

y          650            0

For equation x + 3y = 1300, we have following points which lie on the line

We plot the points for both of the equations and it is the graphical representation of the given situation.

It is clear that these lines intersect at B (1300, 0).

3.    Let cost of 1 kg of apples = Rs x and let cost of 1 kg of grapes= Rs y

According to given conditions, we have

2x +y = 160                            … (1)

4x + 2y = 300      ⇒          2x +y= 150                  … (2)

So, we have equations (1) and (2), 2x +y = 160 and 2x +y = 150 which represent

given situation algebraically.

For equation 2x + y = 160, we have following points which lie on the line

x   50   45

y   60   70

For equation 2x + y = 150, we have following points which lie on the line

x   50   40

y   50   70

We plot the points for both of the equations and it is the graphical representation of the given situation.

4.     (i) Let number of boys who took part in the quiz = x

Let number of girls who took part in the quiz = y

According to given conditions, we have

x + y = 10                      … (1)

And, y = x + 4       ⇒     x – y = -4          … (2)

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

x   0   10

y   10   0

For equation x – y = -4, we have 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, number of boys who took park in the quiz = 3 and, number of girls who took part in the quiz = 7.

(ii) Let cost of one pencil = Rs x and Let cost of one pen = Rs y

According to given conditions, we have

5x + 7y = 50                … (1)

7x + Sy = 46                 … (2)

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

x  10  3

y   0   5

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

x   8   3

y  -2  5

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

Therefore, cost of pencil = Rs 3 and, cost of pen = Rs 5

5.     (i) 5x – 4y + 8 = 0, 7x + 6y – 9 = 0

Comparing equation 5x – 4y + 8 = 0 with aix + bly + c1= 0 and 7x + 6y – 9 = 0 with a2x + b2y + c2 = 0,

We get, a1 = 5, b1 = -4, c1 = a2 = 7, b2 = 6, c2 = -9

Hence, lines have unique solution which means they intersect at one point.

(ii) 9x + 3y + 12 = 0, 18x + 6y + 24 = 0

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

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

Hence, lines are coincident.

(iii)  6x – 3y + 10 = 0, 2x -y + 9 = 0

Comparing equation 6x – 3y + 10 = 0 with a1x + b1y + c1 with a2x + b2y + c2 = 0,

We get, a1 = 6, b1 = -3, c1 = 10, a2 = 2, b2 = -1, c2 = 9

Hence, lines are parallel to each other.

6.    (i)  3x + 2y = 5, 2x – 3y = 7

Comparing equation 3x + 2y = 5 with a1x + b1y + c1= 0 and 2x – 3y = 7 with a2x + b2y + c2 = 0,

We get, a1 = 3, b1 = 2, c1 = 5, a2 = 2, b2 = -3, c2 = -7

Here  which means equations have unique solution.

Hence they are consistent.

(ii)  2x – 3y = 8, 4x – 6y = 9

Comparing equation 2x – 3y = 8 with a1x + b1y + c1 = 0 and 4x – 6y = 9 with a2x

+ b2y + c2 = 0,

We get, a1 = 2, b1 = -3, c1 = -8, a2 = 4, b2 = -6, c2 = -9

Therefore, equations have no solution because they are parallel.

Hence, they are inconsistent.

(iii) 9x – 10y = 14

Comparing equation  with a1x + b1y + c1 and 9x – 10y = 14 with a2x + b2y + c2 = 0

Therefore, equations have unique solution.

Hence, they are consistent.

(iv)  5x – 3y = 11, -10x + 6y = -22

Comparing equation 5x – 3y = 11 with a1x + b1y + c1 = 0 and -10x + 6y = -22 with a2x + b2y + c2 = 0,

We get, a1 = 5, b= -3, c1 = -11, a2 = -10, b2 = 6, c2 = 22

Therefore, the lines have infinite many solutions.

Hence, they are consistent.

7.    (i)  x +y = 5, 2x + 2y = 10

For equation x + y – 5 = 0, we have following points which lie on the line

x   0  5

y   5  0

For equation 2x + 2y – 10 = 0, we have following points which lie on the line

x   1   2

y   4   3

We can see that both of the lines coincide. Hence, there are infinite many solutions. Any point which lies on one line also lies on the other. Hence, by using equation (x + y – 5 = 0), we can say that x = 5 -y

We can assume any random values for y and can find the corresponding value of x using the above equation. All such points will lie on both lines and there will be

(ii) x -y = 8, 3x – 3y = 16

For x – y = 8, the coordinates are

x    0   8

y   -8   0

And for 3x – 3y = 16, the coordinates

Plotting these points on the graph, it is clear that both lines are parallel. So the two lines have no common point. Hence the given equations have no solution and lines are inconsistent.

(iii)  2x +y = 6, 4x – 2y = 4

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

x  0  3

y  6  0

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

x   0   1

y   -2  0

We can clearly see that lines are intersecting at (2, 2) which is the solution.

Hence x = 2 and y = 2 and lines are consistent.

(iv)  2x – 2y – 2 = 0, 4x – 4y – 5 = 0

For 2x – 2 – 2 = 0, the coordinates are:

x   2   0

y   0   -2

And for 4x – 4 – 5 = 0, the coordinates

Plotting these points on the graph, it is clear that both lines are parallel. So the two lines have no common point. Hence the given equations have no solution and lines are inconsistent.

8.    Let length of rectangular garden = x metres

Let width of rectangular garden = y metres

According to given conditions, perimeter = 36 m   ⇒     x + y = 36 …………. (i)

And x = y + 4          ⇒          x – y = 4 ………………………. (ii)

2x = 40      ⇒          x = 20 m

Subtracting eq. (ii) from eq. (i),

2y = 32      ⇒          y = 16 m

Hence, length = 20 m and width = 16 m

9.    (i) Let the second line be equal to a2x +b2y + c2 = 0

Comparing given line 2x + 3y – 8 = 0 with a1x + b1y + c1 = 0,

We get a1 = 2, b1= 3 and c1 = -8

Two lines intersect with each other if

So, second equation can be x + 2y = 3 because

(ii) Let the second line be equal to a2x +b2y + c2 = 0

Comparing given line 2x + 3y – 8 = 0 with a1x + b1y + c1 = 0, We get a1= 2, b1= 3 and c1 = -8

Two lines are parallel to each other if

So, second equation can be 2x + 3y – 2 = 0 because

(iii) Let the second line be equal to a2x +b2y + c2= 0

Comparing given line 2x + 3y – 8 = 0 with a1x + b1y + c1 = 0, We get a1 = 2, b1= 3 and c1 = -8

Two lines are coincident if

So, second equation can be 4x + 6y – 16 = 0 because

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

x  0  -1

y  1   0

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

x   4  0

y   0  6

We can see from the graphs that points of intersection of the lines with the x-axis are (-1, 0), (2, 3) and (4, 0).

11.    (i) x + y = 14        … (1)

x – y = 4          … (2)

x = 4 +y from equation (2)

Putting this in equation (1), we get

4 +y +y= 14        ⇒        2y = 10         ⇒        y = 5

Putting value of y in equation (1), we get

x + 5 = 14           ⇒         x = 14 – 5 = 9

Therefore, x = 9 and y = 5

(ii) s – t = 3                … (1)

Using equation (1), we can say that s = 3 + t

Putting this in equation (2), we get

Putting value oft in equation (1), we get

s – 6 = 3           ⇒         s = 3 + 6 = 9

Therefore, t = 6 and s = 9

(iii) 3x -y = 3 … (1)

9x – 3y = 9       … (2)

Comparing equation 3x -y = 3 with a1x +b1y +c1 = 0 and equation 9x – 3y = 9 with a2x + b2y + c2 = 0,

We get a1 = 3, b1 = -1, c1 =-3, a2=9,b2=-3 and c2 = -9

Here

Therefore, we have infinite many solutions for x and y

(iv) 0.2x + 0.3y = 1.3     …..(1)

0.4x + 0.5y = 2.3     …..(2)

Using equation (1), we can say that

(v)

Putting value of y in (1), we get x = 0

Therefore, x = 0 and y = 0

(vi)

Using equation (2), we can say that

Putting this in equation (1), we get

Putting value of y in equation (2), we get

Therefore, x = 2 and y = 3

12.   2x + 3y = 11 … (1)

2x – 4y = -24 … (2)

Using equation (2), we can say that

2x = -24 + 4y                 ⇒                             x = -12 + 2y

Putting this in equation (1), we get

2 (-12 + 2y) + 3y = 11             ⇒                       -24+4y+3y=11

⇒  7y = 35                                  ⇒                        y = 5

Putting value of y in equation (1), we get

2x + 3 (5) = 11                     ⇒                             2x + 15 = 11

⇒ 2x = 11 – 15 = -4              ⇒                               x = -2

Therefore, x = -2 and y = 5

Putting values of x and y my = mx + 3, we get

5 = m (-2) + 3                    ⇒                                5 = -2m + 3

⇒ -2m = 2            m = -1

13.   Let first number be x and second number be y.

According to given conditions, we have

x -y = 26          (assuming x > y)              (1)

x = 3y               (x > y)                         … (2)

Putting equation (2) in (1), we get

3y -y = 26                   ⇒         2y = 26             ⇒        y = 13

Putting value of y in equation (2), we get x = 3y = 3 × 13 = 39

Therefore, two numbers are 13 and 39.

(ii) Let smaller angle = x and let larger angle =y

According to given conditions, we have

y = x + 18 … (1)

Also, x +y = 1800 (Sum of supplementary angles)        … (2)

Putting (1) in equation (2), we get

x+x+ 18 = 180           ⇒         2x = 180 – 18 = 162            ⇒                 x = 81°

Putting value of x in equation (1), we get y = x + 18 = 81 + 18 = 99°

Therefore, two angles are 810 and 990.

(iii) Let cost of each bat = Rs x and let cost of each ball = Rs y

According to given conditions, we have

7x + 6y = 3800                                  (1)

And, 3x + 5y = 1750              … (2)

Using equation (1), we can say that

7x = 3800 – 6y  ⇒

Putting this in equation (2), we get 3

Putting value of y in (2), we get

3x + 250 = 1750

⇒ 3x = 1500        ⇒         x = 500

(iv) Therefore, cost of each bat = Rs 500 and cost of each ball = Rs 50

Let fixed charge = Rs x and let charge for every km = Rs y

According to given conditions, we have According to given conditions, we have

x+ 10y = 105           … (1)

x + 15y = 155           … (2)

Using equation (1), we can say that

x = 105 – 10y

Putting this in equation (2), we get

105 – 10y + 15y = 155         ⇒                5y = 50              ⇒          y = 10

Putting value of y in equation (1), we get

x + 10 (10) = 105                  ⇒               x = 105 – 100 = 5

Therefore, fixed charge = Rs 5 and charge per km = Rs 10

To travel distance of 25 Km, person will have to pay = Rs (x + 25y) = Rs (5 + 25 × 10) = Rs (5 + 250) = Rs 255

(v) Let numerator = x and let denominator = y

According to given conditions, we have

Using equation (1), we can say that

Putting value of x in equation (2), we get

(vi) Let present age of Jacob = x years

Let present age of Jacob’s son = y years

According to given conditions, we have

(x + 5) = 3 (y + 5)        … (1)

And, (x – 5) = 7 (y 5)       … (2)

From equation (1), we can say that

x + 5 = 3y + 15             ⇒         x = 10 + 3y

Putting value of x in equation (2) we get

10 + 3y – 5 = 7y – 35

⇒ -4y = -40               ⇒                   y = 10 years

Putting value of y in equation (1), we get

x + 5 = 3 (10 + 5) = 3 × 15 = 45

⇒ x = 45 – 5 = 40 years

Therefore, present age of Jacob = 40 years and, present age of Jacob’s son = 10 years

14.  (i) x + y = 5        … (1)

2x – 3y = 4          … (2)

Elimination method: Multiplying equation (1) by 2, we get equation (3)

2x + 2y = 10 … (3)

2x – 3y = 4 … (2)

Subtracting equation (2) from (3), we get

Putting value of y in (1), we get

Therefore,

Substitution method:

x +y = 5            … (1)

2x – 3y = 4       … (2)

From equation (1), we get, x = 5 – y

Putting this in equation (2), we get

2 (5 – y) – 3y = 4     ⇒     10 – 2y – 3y = 4

⇒ 5y = 6

Putting value of y in (1), we get

(ii) 3x + 4y = 10         … (1)

2x – 2y = 2                   … (2)

Elimination method:

Multiplying equation (2) by 2, we get (3)

4x – 4y = 4       … (3)

3x + 4y = 10 … (1)

Adding (3) and (1), we get

7x = 14                       ⇒                       x = 2

Putting value of x in (1), we get

3 (2) + 4y = 10          ⇒              4y = 10 – 6 = 4            ⇒        y = 1

Therefore, x = 2 and y = 1

Substitution method:

3x + 4y = 10                 … (1)

2x – 2y = 2                   … (2)

From equation (2), we get

2x = 2 + 2y          ⇒                  x = 1 +y             ….(3)

Putting this in equation (1), we get

3 (1 +y) + 4y = 10           ⇒                   3 + 3y + 4y = 10

⇒  7y = 7         ⇒          y = 1

Putting value of y in (3), we get x = 1 + 1 = 2

Therefore, x = 2 and y = 1

(iii) 3x – 5y – 4 = 0             … (1)

9x = 2y + 7                          … (2)

Elimination method:

Multiplying (1) by 3, we get (3)

9x – 15y – 12 = 0                      … (3)

9x – 2y – 7 = 0                          … (2)

Subtracting (2) from (3), we get

Substitution Method:

3x – 5y – 4 = 0             … (1)

9x = 2y + 7                   … (2)

From equation (1), we can say that

(iv)

Elimination method:

Multiplying equation (2) by 2, we get (3)

Therefore, x = 2 and y = -3

Substitution method:

From equation (2), we can say that x =

Putting this in equation (1), we get

Putting value of y in (1), we get

Therefore, x = 2 and y = -3

15.  (i) Let numerator = x and let denominator = y

According to given condition, we have

⇒   x+ 1=y – 1 and 2x =y + 1

⇒   x -y = -2           … (1) and 2x -y = 1                 … (2)

So, we have equations (1) and (2), multiplying equation (1) by 2 we get (3)

2x – 2y = -4                 ….(3)

2x -y = 1                      … (2)

Subtracting equation (2) from (3), we get – y = – 5    ⇒    y = 5

Putting value of y in (1), we get

x – 5 = -2               ⇒                   x = -2 + 5 = 3

Therefore, fraction =

(iii) Let present age of Nuri = x years and let present age of Sonu = y years

5 years ago, age of Nuri = (x – 5) years

5 years ago, age of Sonu = (y – 5) years

According to given condition, we have

(x – 5) = 3 (y – 5)       ⇒        x – 5 = 3y – 15         ⇒      x – 3y = -10 … (1)

10 years later from present, age of Nuri = (x + 10) years

10 years later from present, age of Sonu = (y + 10) years

According to given condition, we have

(x + 10) = 2 (y + 10)       ⇒          x + 10 = 2y + 20          ⇒             x – 2y = 10        … (2)

Subtracting equation (1) from (2), we get

y = 10 – (-10) = 20 years

Putting value of y in (1), we get

x – 3 (20) = -10         ⇒               x – 60 = -10        ⇒          x = 50 years

Therefore, present age of Nuri = 50 years and present age of Sonu = 20 years

(iii) Let digit at ten’s place = x and Let digit at one’s place = y

According to given condition, we have x+y= 9                       … (1)

And 9 (10x +y) = 2 (10y + x)           ⇒           90x + 9y = 20y + 2x

⇒  88x = 11y        ⇒           8x =y       ⇒      8x -y = 0         … (2)

Adding (1) and (2), we get

9x = 9                ⇒                   x = 1

Putting value of x in (1), we get

1 +y = 9         ⇒             y = 9 – 1 = 8

Therefore, number = 10x +y = 10 (1) + 8 = 10 + 8 = 18

(iv) Let number of Rs 100 notes = x and let number of Rs 50 notes = y

According to given conditions, we have x +y = 25                 … (1)

and 100x + 50y = 2000          ⇒          2x +y = 40      … (2)

Subtracting (2) from (1), we get

-x = -15                  ⇒                 x = 15

Putting value of x in (1), we get

15 +y = 25              ⇒                 y = 25 – 15 = 10

Therefore, number of Rs 100 notes = 15 and number of Rs 50 notes = 10

(v) Let fixed charge for 3 days = Rs x Let additional charge for each day thereafter = Rs y

According to given condition, we have

x + 4y = 27       … (1)

x + 2y = 21       … (2)

Subtracting (2) from (1), we get

2y = 6                  ⇒                     Y = 3

Putting value of y in (1), we get

x + 4 (3) = 27         ⇒             x = 27 – 12 = 15

Therefore, fixed charge for 3 days = Rs 15 and additional charge for each day after 3 days = Rs 3

16.   (i) x – 3y – 3 = 0

3x – 9y – 2 = 0

Comparing equation x – 3y – 3 = 0 with a1x +b1y + c1 = 0 and 3x – 9y – 2 = 0 with a2x + b2y + c2 = 0,

We get a1 = 1, b= -3, c1 = -3, a2 = 3, b2 = -9, c2 = -2

Here  this means that the two lines are parallel.

Therefore, there is no solution for the given equations i.e. it is inconsistent.

(ii)  2x +y = 5

3x + 2y = 8

Comparing equation 2x +y = 5 with a1x +b1y + c1= 0 and 3x + 2y = 8 with a2x + b2y + c2 = 0,

We get a1 = 2, b1 = 1, c= -5, a2 = 3, b2 = 2, c2 = -8

Here  L this means that there is unique solution for the given equations.

(iii) 3x – 5y = 20

6x – 10y = 40

Comparing equation 3x – 5y = 20 with a1x +b1y + c1 = 0 and 6x – 10y = 40 with a2x + b2y + c2 = 0,

We get a1 = 3, b1 = -5,  c= -20, a2 = 6, b2 = -10, c2 = -40

Here

It means lines coincide with each other.

Hence, there are infinite many solutions.

(iv) x – 3y – 7 = 0

3x – 3y – 15 = 0

Comparing equation x – 3y – 7 = 0 with a1x +b1y + c1 = 0 and 3x – 3y – 15 = 0 with a2x + b2y + c2 = 0,

We get a1 = 1, b1 = -3, c1 = -7, a2 = 3, b2 = -3, c2 = -15

Here   this means that we have unique solution for these equations.

⇒        x  = 4 and y = -1

17.   (i)  Comparing equation 2x + 3y – 7 = 0 with a1x + b1y + c1= 0 and (a – b) x + (a + b)

y – 3a – b + 2 = 0 with a2x + b2y + c2 = 0

We get a1 = 2, b1=3 and c1 = -7, a2 = (a – b), b2 = (a + b) and c2 = 2 – b – 3a

Linear equations have infinite many solutions if

⇒     2a + 2b = 3a – 3b          and        6 – 3b – 9a = -7a – 7 b

⇒     a = 5b              … (1)      and      -2a = -4b – 6           … (2)

Putting (1) in (2), we get

-2 (5b) = -4b – 6            ⇒                    -10b + 4b = -6

⇒ -6b = -6                     ⇒                         b = 1

Putting value of b in (1), we get

a = 5b = 5 (1) = 5

Therefore, a = 5 and b = 1

(ii) Comparing (3x +y – 1 = 0) with a1x + b1y + c1 = 0 and (2k – 1)x + (k – 1)y -2k –1 = 0) with a2x + b2 + c= 0,

We get a1 = 3, b1 = 1 and c1 = -1, a2 = (2k – 1), b2 = (k – 1) and c2 = -2k – 1

Linear equations have no solution if

18.   Substitution Method

8x + 5y = 9           … (1)

3x + 2y = 4        … (2)

From equation (1),

Putting this in equation (2), we get

Putting value of x in (1), we get

8 (-2) + 5y = 9      ⇒       5y = 9 + 16 = 25     ⇒       y = 5

Therefore, x = -2 andy = 5

Cross multiplication method

8x + 5y = 9           … (1)

3x + 2y = 4       … (2)

19.   (i) Let fixed monthly charge = Rs x and let charge of food for one day = Rs y

According to given conditions,

x + 20y = 1000           … (1),

and x + 26y = 1180 … (2)

Subtracting equation (1) from equation (2), we get

6y= 180                   ⇒                 y= 30

Putting value of y in (1), we get

x + 20 (30) = 1000        ⇒            x = 1000 – 600 = 400

Therefore, fixed monthly charges = Rs 400 and, charges of food for one day = Rs 30

(ii) Let numerator = x and let denominator =y

According to given conditions,

⇒   3x – 3 =y          …(1)    4x =y + 8        … (1)

⇒   3x -y = 3          … (1)   4x -y = 8          … (2)

Subtracting equation (1) from (2), we get

4x -y – (3x -y) = 8 – 3           ⇒            x = 5

Putting value of x in (1), we get

3(5) – y = 3        ⇒                      15 – y = 3         ⇒               y = 12

Therefore, numerator = 5 and, denominator = 12

It means fraction =

(iii) Let number of correct answers = x and let number of wrong answers =y According to given conditions,

3x -y = 40 … (1)

And, 4x – 2y = 50         … (2)

From equation (1),                   y = 3x – 40

Putting this in (2), we get

4x – 2 (3x – 40) = 50        ⇒         4x – 6x + 80 = 50

⇒   -2x = -30                     ⇒                        x = 15

Putting value of x in (1), we get

3 (15) -y = 40         ⇒                  45 -y = 40        ⇒           y = 45 – 40 = 5

Therefore, number of correct answers = x = 15 and number of wrong answers = y = 5

Total questions = x +y = 15 + 5 = 20

(iv) Let speed of car which starts from part A = x km/hr

Let speed of car which starts from part B = y km/hr

According to given conditions,

⇒        x + y =100         … (2)

Adding (1) and (2), we get

2x = 120                 ⇒          x = 60 km/hr

Putting value of x in (1), we get

60 -y = 20              ⇒           y = 60 – 20 = 40 km/hr

Therefore, speed of car starting from point A = 60 km/hr And, Speed of car starting from point B = 40 km/hr

(v)  Let length of rectangle = x units and Let breadth of rectangle = y units

Area = xy square units, According to given conditions,

xy – 9 = (x – 5) (y + 3)

xy – 9 = xy + 3x- 5y- 15         ⇒         3x- 5y= 6         … (1)

And, xy + 67 = (x + 3) (y + 2)

xy + 67 = xy + 2x + 3y + 6     ⇒        2x + 3y = 61 … (2)

From equation (1),

Putting this in (2), we get

⇒  19y = 171                            ⇒          y = 9 units

Putting value of y in (2), we get

2x + 3 (9) = 61         ⇒         2x = 61 – 27 = 34         ⇒          x = 17 units

Therefore, length = 17 units and, breadth = 9 units

20.  (i)

⇒  3p + 2q = 12 and 6 (2p + 3q) = 13 (6)

⇒  3p + 2q = 12 and 2p + 3q =13

⇒  3p + 2q – 12 = 0          … (3)     and        2p + 3q – 13 = 0          … (4)

(ii)

Putting this in (1) and (2), we get

2p + 3q = 2      … (3)

4p – 9q = -1 … (4)

Multiplying (3) by 2 and subtracting it from (4), we get

4p – 9q + 1 – 2 (2p + 3q – 2) = 0

⇒ 4p – 9q + 1 – 4p – 6q + 4 = 0

Putting value of q in (3), we get

2p + 1 = 2        2p = 1  p = 1/2

Putting values of p and q in  we get

(iii)

Putting (3) in (1) and (2), we get

4p + 3y = 14 … (4) 3p – 4y = 23 … (5)

Multiplying (4) by 3 and (5) by 4, we get

3 (4p + 3y – 14 = 0) and, 4 (3p – 4y – 23 = 0)

⇒ 12p + 9y – 42 = 0 … (6)      ⇒      12p – 16y – 92 = 0 … (7)

Subtracting (7) from (6), we get

9y – (-16y) – 42 – (-92) = 0

⇒ 25y + 50 =0                      ⇒                   y = 50 – 25 = -2

Putting value of y in (4), we get

4p + 3 (-2) = 14                ⇒                    4p – 6 = 14

⇒4p = 20                           ⇒            p = 5

Putting value of p in (3), we get

(iv)

Putting this in (1) and (2), we get

5p+q=2                     ⇒                          5p + q – 2 = 0 … (3)

And, 6p – 3q = 1            ⇒            6p – 3q – 1 = 0             … (4)

Multiplying (3) by 3 and adding it to (4), we get

3 (5p + q – 2) + 6p – 3q – 1 = 0

15p + 3q – 6 + 6p – 3q – 1 = 0

(v) 7x – 2y = 5xy             … (1)

8x + 7y = 15xy                (2)

Dividing both the equations by xy, we get

Putting these in (3) and (4), we get

7q – 2p = 5      … (5)

8q + 7p = 15 … (6)

From equation (5),

⇒  65q = 30 + 35 = 65             ⇒                 q = 1

Putting value of q in (5), we get

7 (1) – 2p = 5              ⇒              2p = 2                ⇒           p=1

Putting value of p and q in , we get x = 1 and y = 1

(vi) 6x+ 3y- 6xy = 0 … (1)

2x + 4y – 5xy = 0         … (2)

Dividing both the equations by xy, we get

Putting these in (3) and (4), we get

6q + 3p – 6 = 0             … (5)

2q+ 4p- 5 = 0             … (6)

From (5),

3p = 6 – 6q                    ⇒             p= 2 – 2q

Putting this in (6), we get

2q + 4 (2 – 2q) – 5 = 0                          2q + 8 – 8q – 5 = 0

⇒  -6q = -3                   ⇒            q = 1/2

Putting value of q in (p = 2 – 2q), we get

p= 2 – 2 (1/2) = 2 – 1 = 1

Putting values of p and q in , we get x = 1 and y = 2

(vii)

Putting this in (1) and (2), we get

10p + 2q = 4 … (3)

15p – Sq = -2 … (4)

From equation (3),

2q = 4 – 10p                    ⇒                     q= 2 – Sp         … (5)

Putting this in (4), we get

15p – 5 (2 – 5p) = -2       ⇒                 15p – 10 + 25p = -2

⇒     40p = 8

Putting value of p in (5), we get

Putting x = 3 in (7), we get

3 -y = 1                       ⇒                         y = 3 – 1 = 2

Therefore, x = 3 and y = 2

(viii)

⇒   4p + 4q = 3       … (3)     and        4p – 4q = -1 … (4)

Adding (3) and (4), we get

8p = 2                    ⇒                               p = 1/4

Putting value of p in (3), we get

4 (1/4) + 4q = 3                   ⇒               1 + 4q = 3

⇒  4q = 3 – 1 = 2                  ⇒                q=1/2

Adding (5) and (6), we get

6x = 6                         ⇒                                         x = 1

Putting x = 1 in (5) , we get

3 (1) +y 4                  ⇒             y = 4 – 3 = 1

Therefore, x = 1 and y = 1

21.    (i) Let speed of rowing in still water = x km/h

Let speed of current =y km/h

So, speed of rowing downstream = (x +y) km/h

And, speed of rowing upstream = (x -y) km/h

According to given conditions,

Adding (1) and (2), we get

2x = 12                   ⇒                            x = 6

Putting x = 6 in (1), we get

6 +y = 10               ⇒                           y=10 – 6=4

Therefore, speed of rowing in still water = 6 km/h

Speed of current = 4 km/h

(ii) Let time taken by 1 woman alone to finish the work = x days

Let time taken by 1 man alone to finish the work =y days

So, 1 woman’s 1 day work =  part of the work

And, 1 man’s 1 day work =  part of the work

So, 2 women’s 1 day work =  part of the work

And, 5 men’s 1 day work =  part of the work

Therefore, 2 women and 5 men’s 1 day work = part of the work… (1)

It is given that 2 women and 5 men complete work in = 4 days

It means that in 1 day, they will be completing  part of the work … (2)

Clearly, we can see that (1) = (2)

⇒ 8p + 20q = 1 … (5)    and      9p + 18q = 1 … (6)

Multiplying (5) by 9 and (6) by 8, we get

72p + 180q = 9            … (7)

72p + 144q = 8            … (8)

Subtracting (8) from (7), we get

36q = 1

Putting this in (6), we get

Therefore, 1 woman completes work in = 18 days

And, 1 man completes work in = 36 days

(iii) Let speed of train = x km/h and let speed of bus =y km/h

According to given conditions,

Putting this in the above equations, we get

60p + 240q = 4                       … (1)

And 100p + 200q =           …(2)

Therefore, speed of train = 60 km/h

And, speed of bus = 80 km/h

22. Let the age of Ani and Biju be x years and y years respectively.

Age of Dharam = 2x years and Age of Cathy =  years

According to question,                        x – y = 3       … (1)

And                       ⇒          4x – y = 60 … (2)

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

3x = 60 – 3 = 57             ⇒             x =

Age of Ani = 19 years

Age of Biju = 19 – 3 = 16 years

Again, According to question,                         y – x = 3      … (3)

And                           ⇒                    4x – y = 60 … (4)

Adding (3) and (4), we obtain:

3x = 63               ⇒                x = 21

Age of Ani = 21 years

Age of Biju = 21 + 3 = 24 years

23.  Let the money with the first person and second person be Rs x and Rs y respectively. According to the question,

x + 100 = 2(y – 100)   ⇒    x + 100 = 2y – 200    ⇒                                             x – 2y = – 300                          … (1)

Again, 6(x – 10) = (y + 10)     ⇒          6x – 60 = y + 10     ⇒       6x – y = 70            … (2)
Multiplying equation (2) by 2, we obtain:

12x – 2y = 140        … (3)

Subtracting equation (1) from equation (3), we obtain:

11x = 140 + 300             ⇒          11x = 440             ⇒         x = 40

Putting the value of x in equation (1), we obtain:

40 – 2y = -300         ⇒         40 + 300 = 2y       ⇒          2y = 340

⇒   y = 170

Thus, the two friends had Rs 40 and Rs 170 with them.

24.  Let the speed of the train be x km/h and the time taken by train to travel the given distance be t hours and the distance to travel be d km.

According to the question

⇒(x+10)(t -2) = d         ⇒          xt +10t -2x -20 = d

⇒-2x +10t = 20        ……(2)        [Using eq. (1)]

⇒  3x-10t =30 ……. (3)              [Using eq. (1)]

Adding equations (2) and (3), we obtain:

x = 50

Substituting the value of x in equation (2), we obtain:

(-2) x (50) + 10t = 20        ⇒             -100 + 10t = 20

⇒ 10t = 120                                   t = 12

From equation (1), we obtain:

d = xt = 50 x 12 = 600

Thus, the distance covered by the train is 600 km.

25.  Let the number of rows be x and number of students in a row be y.

Total number of students in the class = Number of rows x Number of students in a row = xy

According to the question,

Total number of students = (x – 1) (y + 3) xy = (x – 1) (y + 3)

⇒ xy = xy – y + 3x – 3

⇒ 3x – y – 3 = 0

⇒ 3x – y = 3         … (1)

Total number of students = (x + 2) (y – 3) xy = xy + 2y – 3x – 6

⇒ 3x – 2y = -6 … (2)

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

y= 9

Substituting the value of y in equation (1), we obtain:

⇒ 3x – 9 = 3

⇒ 3x = 9 + 3 = 12

⇒ x = 4

Number of rows = x = 4

Number of students in a row = y = 9

Hence, Total number of students in a class = xy = 4 x 9 = 36

26. ∠C=3 ∠B=2 (∠A+ ∠B)

Taking                   3 ∠B= 2(∠A+ ∠B)

⇒ ∠B= 2∠A

⇒ 2∠A – ∠B= 0 ……………. (1)

We know that the sum of the measures of all angles of a triangle is 180°.

∠A+ ∠B+ ∠C=180°

⇒ ∠A+ ∠B+ 3∠B= 180°

⇒ ∠A+ 4∠B= 180° ……………… (2)

Multiplying equation (1) by 4, we obtain:

8∠A- 4∠B= 0 ………………………………….. (3)

Adding equations (2) and (3), we get

9∠A= 180°     ⇒          ∠A= 20°

From eq. (2), we get,

20° + 4 ∠B = 180°    ⇒        ∠B = 40°

And ∠C = 3 × 40°= 120°

Hence the measures of ∠A, ∠B and ∠C are 20°, 40° and 120° respectively.

27.  5x – y = 5         y = 5x -5

Three solutions of this equation can be written in a table as follows:

3x – y =3          y = 3x-3

It can be observed that the required triangle is A ABC.

The coordinates of its vertices are A (1, 0), B (0, -3), C (0, -5).

28.  (i) px + qy = p-q   … (1)

qx- py = p+q      … (2)

Multiplying equation (1) by p and equation (2) by q, we obtain:

p2x+ pqy = – pq          … (3)

q2 x- pqy = pq + q2     … (4)

Adding equations (3) and (4), we obtain:

p2x+q2x=p2+q2         ⇒                 (p2 + q2)x = p2 + q2

Substituting the value of x in equation (1), we obtain:

p(1)+qy=p-q              ⇒           qy= q            ⇒        y = -1

Hence the required solution is x = 1 and y = -1.

(ii) ax + by = c          … (1)

bx+ay= 1+c           … (2)

Multiplying equation (1) by a and equation (2) by b, we obtain:

a2x+aby =ac              ….(3)

b2x + aby = b+bc      ….(4)

Subtracting equation (4) from equation (3),

Substituting the value of x in equation (1), we obtain:

(iii)

ax + by = a2+ b2                                       …. (2)

Multiplying equation (1) and (2) by b and a respectively, we obtain:

b2 x — aby = 0                  …. (3)

a2 x + aby = a3 + ab2      …. (4)

Adding equations (3) and (4), we obtain:

b2x + a2x = a3 + ab2       ⇒     x (b2 + a2)= a(a+b2)        ⇒       x = a

Substituting the value of x in equation (1), we obtain:

b(a) – ay = 0         ab — ay =0           ⇒      y = b

(iv) (a—b)x+(a+b)y=a2-2ab—b2 … (1)

(a +b)(x+ y)= a2 + b2    ⇒          (a+b)x+(a+b)y=a2 +b2              …(2)

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

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

⇒ (a – b – a – b) x = —2ab – 2b2               ⇒       – 2bx = -2b (a + b)

⇒ x = a + b

Substituting the value of x in equation (1), we obtain:

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

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

(v) 152x – 378y = -74     … (1)

378x + 152y = -604     … (2)

Adding the equations (1) and (2), we obtain:

226x – 226y = -678      ⇒      x + y = 3     … (3)

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

530x –  530y = 530            x – y = 1       … (4)

Adding equations (3) and (4), we obtain:

2x = 4                       x = 2

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

29.  We know that the sum of the measures of opposite angles in a cyclic quadrilateral is 180°.

∴ ∠A+ ∠C = 180°                        4y + 20-4x = 180°

—4x+ 4y =160°                       x— y = —40°                … (1)

Also ∠B+ ∠D= 180°                   3y-5-7x +5=180°

⇒  -7x + 3y =180°                                                     … (2)

Multiplying equation (1) by 3, we obtain: 3x – 3y = —120°      …(3)

Adding equations (2) and (3), we obtain:

4x = 60°                       x = -15°

Substituting the value of x in equation (1), we obtain:

15 – y = -40°           y = -15+ 40 = 25

∴ ∠A= 4y+20= 4×25+20 =120°

∠B=  3y —5 = 3 x 25 – 5 = 70°

∠C=  —4x = —4x(-15) = 60°

∠D = —7x+5=-7(-15)+5=110°

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