# Pair of Linear Equations in Two Variables Chapter 3 NCERT Solutions For Class 10 CBSE Mathematics.

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## Solution

Let the present age of Aftab be x  years and his daughter be y years then represent their ages 7years later and 3 years ago in term of x and y.

Therefore, 7 years ago, age of Aftab =(x−7)years and his daughter =(y−7)years.

x−7=7(y−7)

x−7=7y−49

x−7y−7+49=0

x−7y+42=0…………………………………..(1)

After 3 years from now, age of Aftab =(x+3) years and his daughter =(y+3) years.

x+3=3(y+3)

x+3=3y+9

x−3y+3−9=0

x−3y−6=0………………………………………(2)

Now by representing the equations in their intercept forms, we get

## Solution

Let the cost of one bat as ₹ x and the cost of one ball as ₹ y.

The cost of 3 bats and 6 balls is ₹ 3900.

3x+6y=3900

3(x+2y)=3900

x+2y=1300…………………………………(1)

Also, the cost of 1 bat and 3 balls is ₹ 1300.

x+3y=1300………………………………….(2)

Now by representing the equations in their intercept forms, we get

## Solutions:

Let the cost of 1 kg of apples be ‘Rs. x’ And, cost of 1 kg of grapes be ‘Rs. y’.

According to the question, the algebraic representation is

2x+y=160……………………………..(1)

And

4x+2y=300……………………………..(2)

## Solutions

(i) Let there be x number of girls and y number of boys. As per the given question, the algebraic expression can be represented as follows.

x + y = 10…………………………..(1)

x – y = 4………………………………(2)

Now, for x + y = 10 or x=10−y, the solutions are;

For x – y = 4 or x = 4 + y, the solutions are;

It can be seen that the given lines cross each other at point (7, 3). Therefore, there are 7 girls and 3 boys in the class.

(ii) Let 1 pencil costs Rs.x and 1 pen costs Rs.y. According to the question, the algebraic expression can be represented as;

5x + 7y = 50 ………………………………(1)

7x + 5y = 46………………………………..(2)

5x + 7y = 50 or x = (50 -7y)/5, the solutions are;

7x + 5y = 46 or x =(46 – 5y)/7, the solutions are;

it is can be seen that the given lines cross each other at point (3, 5). So, the cost of a eraser is 3/- and cost of a chocolate is 5/-

## Solutions:

Comparing these equations with 𝑎1x + 𝑏1y + 𝑐1=0 And

𝑎2x + 𝑏2y + 𝑐2 = 0

(i)

𝑎1 = 5, 𝑏1 = −4, 𝑐1= 8

𝑎2 = 7, 𝑏2 = 6, 𝑐2 = −9

𝑎1/ 𝑎2 = 5/ 7 , 𝑏1/ 𝑏2 = − 4/ 6 = − 2/ 3 , 𝑐1/ 𝑐2 = 8/ −9

Since, 𝑎1 /𝑎2 ≠ 𝑏1/ 𝑏2 So, the pairs of equations given in the question have a unique solution and the lines cross each other at exactly one point.

(ii) 𝑎1 = 9, 𝑏1 = 3, 𝑐1= 12

𝑎2 = 18, 𝑏2 = 6, 𝑐2 = 24

𝑎1 /𝑎2 = 9 /18 = 1/ 2 , 𝑏1/ 𝑏2 = 3 /6 = 1 /2 , 𝑐1/ 𝑐2 = 12 /24 = 1/ 2

Since, 𝑎1 /𝑎2 = 𝑏1 /𝑏2 = 𝑐1/ 𝑐2 So, the pairs of equations given in the question have infinite possible solutions and the lines are coincident.

(iii) 𝑎1 = 6, 𝑏1 = −3, 𝑐1= 10

𝑎2 = 2, 𝑏2 = −1, 𝑐2 = 9

𝑎1 /𝑎2 = 6/ 2 = 3/ 1 , 𝑏1/ 𝑏2 = −3/ −1 = 3/ 1 , 𝑐1 /𝑐2 = 10/ 9 Since, 𝑎1/ 𝑎2 = 𝑏1/ 𝑏2 ≠ 𝑐1/ 𝑐2 So, the pairs of equations given in the question are parallel to each other and the lines never intersect each other at any point and there is no possible solution for the given pair of equations.

## Solutions

(I)

𝑎1 = 3, 𝑏1 = 2, 𝑐1= -5

𝑎2 = 2, 𝑏2 = −3, 𝑐2 = −7

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

Since, 𝑎1/ 𝑎2 ≠ 𝑏1/ 𝑏2

So, the given equations intersect each other at one point and they have only one possible solution. The equations are consistent.

(ii)

𝑎1 = 2, 𝑏1 = −3, 𝑐1= -8

𝑎2 = 4, 𝑏2 = −6, 𝑐2 = −9

𝑎1/ 𝑎2 = 2/ 4 = 1/ 2 , 𝑏1/ 𝑏2 = 3/ 6 = 1/ 2 , 𝑐1/ 𝑐2 = 8 /9 Since, 𝑎1/ 𝑎2 = 𝑏1/ 𝑏2 ≠ 𝑐1 /𝑐2

So, the equations are parallel to each other and they have no possible solution.

Hence, the equations are inconsistent.

(iii)

𝑎1 = 3/ 2 , 𝑏1 = 5/ 3 , 𝑐1= -7

𝑎2 = 9, 𝑏2 = −10, 𝑐2 = 14

𝑎1/ 𝑎2 = 32/ 9  , 𝑏1/ 𝑏2 = 53/(−10) = − 1/ 2 , 𝑐1 /𝑐2 = −7/ 14 = − 1/ 2

Since, 𝑎1/ 𝑎2 ≠ 𝑏1/ 𝑏2

So, the equations are intersecting each other at one point and they have only one possible solution. Hence, the equations are consistent.

### YOU ARE READING: NCERT Solutions For Class 10 CBSE Mathematics Chapter 3 Pair of Linear Equations in Two Variables.

(iv)

𝑎1 = 5, 𝑏1 = −3, 𝑐1= 11

𝑎2 = −10, 𝑏2 = 6, 𝑐2 = −22

𝑎1/ 𝑎2 = 5 /−10 = − 1/ 2 , 𝑏1/ 𝑏2 = −3/ 6 = − 1/ 2 , 𝑐1/ 𝑐2 = 11/ −22 = − 1/ 2

Since, 𝑎1/ 𝑎2 = 𝑏1/ 𝑏2 = 𝑐1/ 𝑐2

These linear equations are coincident lines and have infinite number of possible solutions. Hence, the equations are consistent.

(v)

𝑎1 = 4/3 , 𝑏1 = 2, 𝑐1= 8

𝑎2 = 2, 𝑏2 = 3, 𝑐2 = 12

𝑎1/ 𝑎2 = 4 /3X2 = 2/ 3 , 𝑏1/ 𝑏2 = 2/ 3 , 𝑐1 /𝑐2 = 8/ 12 = 2/ 3

Since, 𝑎1 /𝑎2 = 𝑏1/ 𝑏2 = 𝑐1/ 𝑐2.

These linear equations are coincident lines and have infinite number of possible solutions. Hence, the equations are consistent.

## Solutions

(I)

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

𝑆𝑖𝑛𝑐𝑒, 𝑎1/ 𝑎2 = 𝑏1/ 𝑏2 = 𝑐1/ 𝑐2

∴ The equations are coincident and they have infinite number of possible solutions.

So, the equations are consistent.

For, x + y = 5 or x = 5 – y

For 2x + 2y = 10

From the figure, we can see, that the lines are overlapping each other.

Therefore, the equations have infinite possible solutions.

(ii)

𝑎1/ 𝑎2 = 1/ 3 , 𝑏1/ 𝑏2 = −1/ −3 = 1/ 3 , 𝑐1 /𝑐2 = 8 /16 = 1/ 2

𝑆𝑖𝑛𝑐𝑒, 𝑎1 /𝑎2 = 𝑏1/ 𝑏2 ≠ 𝑐1 /𝑐2

The equations are parallel to each other and have no solutions.

Hence, the pair of linear equations is inconsistent.

### YOU ARE READING: NCERT Solutions For Class 10 CBSE Mathematics Chapter 3 Pair of Linear Equations in Two Variables.

(iii)

a1/𝑎2 = 2/ 4 = 1/ 2 , 𝑏1/ 𝑏2 = 1/ −2 , 𝑐1 /𝑐2 = −6/ −4 = 3/ 2

𝑆𝑖𝑛𝑐𝑒, 𝑎1 /𝑎2 ≠ 𝑏1/ 𝑏2

The given linear equations are intersecting each other at one point and have only one solution.

Hence, the pair of linear equations is consistent.

Now, for 2x + y – 6 = 0 or y = 6 – 2x

And for 4x – 2y – 4 = 0

It can be seen that these lines are intersecting each other at only one point,(2,2).

(iv)

𝑎1/ 𝑎2 = 2/ 4 = 1/ 2 , 𝑏1 /𝑏2 = −2/ −4 = 1/ 2 , 𝑐1/ 𝑐2 = 2/ 5

𝑆𝑖𝑛𝑐𝑒, 𝑎1 /𝑎2 = 𝑏1 /𝑏2 ≠ 𝑐1/ 𝑐2

Thus, these linear equations have parallel and have no possible solutions.

Hence, the pair of linear equations are inconsistent.

## Solutions:

Let us consider. The width of the garden is x and length is y.

Now, according to the question,

we can express the given condition as;

y – x = 4 and y + x = 36

Now, taking y – x = 4 or y = x + 4

For y + x = 36, y = 36 – x

the lines intersect each other at a point(16, 20).

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

## Solutions:

(I)

To find another linear equation in two variables such that the geometrical representation of the pair so formed is intersecting lines, it should satisfy below condition;

𝑎1 /𝑎2 ≠ 𝑏1/ 𝑏2

Thus, another equation could be 2x – 7y + 9 = 0,

such that; 𝑎1/ 𝑎2 = 2 /2 = 1 𝑎𝑛𝑑 𝑏1 /𝑏2 = 3 /−7

Clearly, you can see another equation satisfies the condition.

(ii)

To find another linear equation in two variables such that the geometrical representation of the pair so formed is parallel lines, it should satisfy below condition;

𝑎1/ 𝑎2 = 𝑏1/ 𝑏2 ≠ 𝑐1 /𝑐2

Thus, another equation could be 6x + 9y + 9 = 0,

such that; 𝑎1 /𝑎2 = 2 /6 = 1/ 3 , 𝑏1/ 𝑏2 = 3 /9 = 1/ 3 , 𝑐1/ 𝑐2 = − 8/ 9

Clearly, you can see another equation satisfies the condition.

### YOU ARE READING: NCERT Solutions For Class 10 CBSE Mathematics Chapter 3 Pair of Linear Equations in Two Variables.

(iii)

To find another linear equation in two variables such that the geometrical representation of the pair so formed is coincident lines, it should satisfy below condition;

𝑎1/ 𝑎2 = 𝑏1/ 𝑏2 = 𝑐1/ 𝑐2

Thus, another equation could be 4x + 6y – 16 = 0,

such that; 𝑎1 /𝑎2 = 2/ 4 = 1/ 2 , 𝑏1/ 𝑏2 = 3/ 6 = 1/ 2 , 𝑐1 /𝑐2 = −8/ −16 = 1/ 2

Clearly, you can see another equation satisfies the condition.

## Solution:

Given, the equations for graphs are x – y + 1 = 0 and 3x + 2y – 12 = 0.

For, x – y + 1 = 0 or x = 1+y

For, 3x + 2y – 12 = 0

it can be seen that these lines are intersecting each other at point (2, 3) and x-axis at (−1, 0) and (4, 0).

Therefore, the vertices of the triangle are (2, 3), (−1, 0), and (4, 0).

## Solutions:

(i)

From 1st equation, we get, x = 14 – y

Now, substitute the value of x in second equation to get,

(14 – y) – y = 4

14 – 2y = 4

2y = 10 Or y = 5

By the value of y, we can now find the exact value of x;

∵ x = 14 – y

∴ x = 14 – 5 Or x = 9

Hence, x = 9 and y = 5.

(ii)

From 1st equation, we get, s = 3 + t …………………….(1)

Now, substitute the value of s in second equation to get t ,

(3+t)/3 + (t/2) = 6

⇒ (2(3+t) + 3t )/6 = 6

⇒ (6+2t+3t)/6 = 6

⇒ (6+5t) = 36

⇒5t = 30

⇒t = 6

Now, substitute the value of t in equation (1) s = 3 + 6 = 9

Therefore, s = 9 and t = 6.

### YOU ARE READING: NCERT Solutions For Class 10 CBSE Mathematics Chapter 3 Pair of Linear Equations in Two Variables.

(iii)

From 1st equation, we get, x = (3+y)/3

Now, substitute the value of x in the given second equation to get,

9(3+y)/3 – 3y = 9

⇒9 +3y -3y = 9

⇒ 9 = 9

Therefore, y has infinite values and since, x = (3+y) /3, so x also has infinite values.

(iv)

From 1st equation, we get,

x = (1.3- 0.3y)/0.2 ………………….(1)

Now, substitute the value of x in the given second equation to get,

0.4(1.3-0.3y)/0.2 + 0.5y = 2.3

⇒2(1.3 – 0.3y) + 0.5y = 2.3

⇒ 2.6 – 0.6y + 0.5y = 2.3

⇒ 2.6 – 0.1 y = 2.3

⇒ 0.1 y = 0.3

⇒ y = 3

Now, substitute the value of y in equation (1), we get,

x = (1.3-0.3(3))/0.2 = (1.3-0.9)/0.2 = 0.4/0.2 = 2

Therefore, x = 2 and y = 3.

(v)

From 1st equation, we get, x = – (√3/√2)y ………………………..(1)

Putting the value of x in the given second equation to get,

√3(-√3/√2)y – √8y = 0

⇒ (-3/√2)y- √8 y = 0

⇒ y = 0

Now, substitute the value of y in equation (1), we get, x = 0.

## YOU ARE READING: NCERT Solutions For Class 10 CBSE Mathematics Chapter 3 Pair of Linear Equations in Two Variables.

(vi)

From 1st equation, we get,

(3/2)x = -2 + (5y/3)

⇒ x = 2(-6+5y)/9 = (-12+10y)/9 ………………………(1)

Putting the value of x in the given second equation to get,

((-12+10y)/9)/3 + y/2 = 13/6

⇒y/2 = 13/6 –( (-12+10y)/27 ) + y/2 = 13/6

substituting the value of y in equation (1), we get, (3x/2) – 5(3)/3 = -2 ⇒ (3x/2) – 5 = -2 ⇒ x = 2 Therefore, x = 2 and y = 3.

## Solution:

2x + 3y = 11…………………………..(I)

2x – 4y = -24………………………… (II)

From equation (II),

we get x = (11-3y)/2 ………………….(III)

Substituting the value of x in equation (II),we get

2(11-3y)/2 – 4y = 24

11 – 3y – 4y = -24 -7y = -35 y = 5……………………………………..(IV)

Putting the value of y in equation (III), we get

x = (11-3×5)/2 = -4/2 = -2

Hence, x = -2, y = 5

Also, y = mx + 3 5 = -2m +3 -2m = 2 m = -1

Therefore the value of m is -1.

## Solution:

Let the two numbers be x and y respectively, such that y > x.

According to the question,

y = 3x ……………… (1)

y – x = 26 …………..(2)

Substituting the value of (1) into (2), we get

3x – x = 26 x = 13 ……………. (3)

Substituting (3) in (1), we get

y = 39 Hence, the numbers are 13 and 39.

## Solution:

Let the larger angle by x ° and smaller angle be y ° .

We know that the sum of two supplementary pair of angles is always 180° .

According to the question,

x + y = 180°……………. (1)

x – y = 18°……………..(2)

From (1), we get

x = 180° – y …………. (3)

Substituting (3) in (2), we get

180° – y – y =18°

162° = 2y

y = 81o ………….. (4)

Using the value of y in (3), we get

x = 180° – 81° = 99°

Hence, the angles are 99° and 81° .

## Solution:

Let the cost a bat be x and cost of a ball be y.

According to the question,

7x + 6y = 3800 ………………. (I)

3x + 5y = 1750 ………………. (II)

From (I), we get

y = (3800-7x)/6………………..(III)

Substituting (III) in (II). we get,

3x+5(3800-7x)/6 =1750

⇒3x+ 9500/3 – 35x/6 = 1750

⇒3x- 35x/6 = 1750 – 9500/3

⇒(18x-35x)/6 = (5250 – 9500)/3

⇒-17x/6 = -4250/3

⇒-17x = -8500 x = 500 ……………………….. (IV)

Substituting the value of x in (III), we get

y = (3800-7 ×500)/6 = 300/6 = 50

Hence, the cost of a bat is Rs 500 and cost of a ball is Rs 50.

## Solution:

Let the fixed charge be Rs x and per km charge be Rs y.

According to the question,

x + 10y = 105 …………….. (1)

x + 15y = 155 …………….. (2)

From (1), we get

x = 105 – 10y ………………. (3)

Substituting the value of x in (2), we get

105 – 10y + 15y = 155

5y = 50 y = 10 …………….. (4)

Putting the value of y in (3), we get

x = 105 – 10 × 10 = 5

Hence, fixed charge is Rs 5 and per km charge = Rs 10

Charge for 25 km = x + 25y = 5 + 250 = Rs 255

## Solution:

Let the fraction be x/y.

According to the question,

(x+2) /(y+2) = 9/11

11x + 22 = 9y + 18

11x – 9y = -4 …………….. (1)

(x+3) /(y+3) = 5/6

6x + 18 = 5y +15

6x – 5y = -3 ………………. (2)

From (1), we get

x = (-4+9y)/11 …………….. (3)

### YOU ARE READING: NCERT Solutions For Class 10 CBSE Mathematics Chapter 3 Pair of Linear Equations in Two Variables.

Substituting the value of x in (2), we get

6(-4+9y)/11 -5y = -3

-24 + 54y – 55y = -33

-y = -9

y = 9 ………………… (4)

Substituting the value of y in (3), we get

x = (-4+9×9 )/11 = 7

Hence the fraction is 7/9.

## Solutions:

Let the age of Jacob and his son be x and y respectively.

According to the question,

(x + 5) = 3(y + 5) x – 3y = 10 …………………………………….. (1)

(x – 5) = 7(y – 5) x – 7y = -30 ………………………………………. (2)

From (1), we get x = 3y + 10 ……………………. (3)

Substituting the value of x in (2), we get

3y + 10 – 7y = -30 -4y = -40 y = 10 ………………… (4)

Substituting the value of y in (3), we get

x = 3 x 10 + 10 = 40

Hence, the present age of Jacob’s and his son is 40 years and 10 years respectively.

## Solutions:

(i)

By the method of elimination.

x + y = 5 ……………………………….. (i)

2x – 3y = 4 ……………………………..(ii)

When the equation (i) is multiplied by 2, we get

2x + 2y = 10 ……………………………(iii)

When the equation (ii) is subtracted from (iii) we get,

5y = 6 y = 6/5 ………………………………………(iv)

Substituting the value of y in eq. (i) we get,

x=5−6/5 = 19/5

∴x = 19/5 , y = 6/5

By the method of substitution.

From the equation (i), we get:

x = 5 – y………………………………….. (v)

When the value is put in equation (ii) we get,

2(5 – y) – 3y = 4

-5y = -6

y = 6/5

### YOU ARE READING: NCERT Solutions For Class 10 CBSE Mathematics Chapter 3 Pair of Linear Equations in Two Variables.

When the values are substituted in equation (v),

we get: x =5− 6/5 = 19/5

∴x = 19/5 ,y = 6/5.

(ii)

By the method of elimination. 3x + 4y = 10……………………….(i)

2x – 2y = 2 ………………………. (ii)

When the equation (i) and (ii) is multiplied by 2, we get:

4x – 4y = 4 ………………………..(iii)

When the Equation (i) and (iii) are added, we get:

7x = 14

x = 2 ……………………………….(iv)

Substituting equation (iv) in (i) we get,

6 + 4y = 10

4y = 4

y = 1

Hence, x = 2 and y = 1

By the method of Substitution From equation (ii) we get,

x = 1 + y……………………………… (v)

Substituting equation (v) in equation (i) we get,

3(1 + y) + 4y = 10

7y = 7

y = 1

When y = 1 is substituted in equation (v) we get,

x= 1 + 1 = 2

Therefore, x = 2 and y = 1

## YOU ARE READING: NCERT Solutions For Class 10 CBSE Mathematics Chapter 3 Pair of Linear Equations in Two Variables.

(iii)

By the method of elimination:

3x – 5y – 4 = 0 ………………………………… (i)

9x = 2y + 7

9x – 2y – 7 = 0 …………………………………(ii)

When the equation (i) and (iii) is multiplied we get,

9x – 15y – 12 = 0 ………………………………(iii)

When the equation (iii) is subtracted from equation (ii) we get,

13y = -5

y = -5/13 ………………………………………….(iv)

When equation (iv) is substituted in equation (i) we get,

3x +25/13 −4=0

3x = 27/13

x =9/13

∴x = 9/13 and y = -5/13

By the method of Substitution:

From the equation (i) we get,

x = (5y+4)/3 …………………………………………… (v)

Putting the value (v) in equation (ii) we get,

9(5y+4)/3 −2y −7=0

13y = -5

y = -5/13

Substituting this value in equation (v) we get,

x = (5(-5/13)+4)/3

x = 9/13

∴x = 9/13, y = -5/13

(iv)

By the method of Elimination.

3x + 4y = -6 …………………………. (i)

x-y/3 = 3

3x – y = 9 ……………………………. (ii)

## YOU ARE READING: NCERT Solutions For Class 10 CBSE Mathematics Chapter 3 Pair of Linear Equations in Two Variables.

When the equation (ii) is subtracted from equation (i) we get,

-5y = -15

y = 3 ………………………………….(iii)

When the equation (iii) is substituted in (i) we get,

3x – 12 = -6

3x = 6

x = 2

Hence, x = 2 , y = -3

By the method of Substitution:

From the equation (ii) we get,

x = (y+9)/3…………………………………(v)

Putting the value obtained from equation (v) in equation (i) we get,

3(y+9)/3 +4y =−6

5y = -15

y = -3

When y = -3 is substituted in equation (v) we get,

x = (-3+9)/3 = 2

Therefore, x = 2 and y = -3 2.

## Solution:

Let the fraction be a/b

According to the given information,

(a+1)/(b-1) = 1 => a – b = -2 ………………………………..(i)

a/(b+1) = 1/2 => 2a-b = 1…………………………………(ii)

When equation (i) is subtracted from equation (ii) we get,

a = 3 …………………………………………………..(iii)

When a = 3 is substituted in equation (i) we get,

3 – b = -2

-b = -5

b = 5

Hence, the fraction is 3/5.

## Solution:

Let us assume, present age of Nuri is x And present age of Sonu is y.

According to the given condition, we can write as;

x – 5 = 3(y – 5)

x – 3y = -10…………………………………..(1)

Now, x + 10 = 2(y +10)

x – 2y = 10…………………………………….(2)

Subtract eq. 1 from 2, to get, y = 20 ………………………………………….(3)

Substituting the value of y in eq.1, we get,

=> x – 3X20 = -10

x – 60 = -10

x = 50

Therefore, Age of Nuri is 50 years Age of Sonu is 20 years.

## Solution:

Let the unit digit and tens digit of a number be x and y respectively.

Then, Number (n) = 10B + A N after reversing order of the digits = 10A + B

According to the given information,

A + B = 9…………………….(i)

9(10B + A) = 2(10A + B)

88 B – 11 A = 0

-A + 8B = 0 ………………………………………………………….. (ii)

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

9B = 9

B = 1……………………………………………………………………….(3)

Substituting this value of B, in the equation (i) we get A= 8

Hence the number (N) is 10B + A = 10 x 1 +8 = 18

## Solution:

Let the number of Rs.50 notes be A and the number of Rs.100 notes be B

According to the given information,

A + B = 25 ……………………………………………………………………….. (i)

50A + 100B = 2000 ………………………………………………………………(ii)

When equation (i) is multiplied with (ii) we get,

50A + 50B = 1250 …………………………………………………………………..(iii)

Subtracting the equation (iii) from the equation (ii) we get,

50B = 750

B = 15

Substituting in the equation (i) we get,

A = 10

Hence, Manna has 10 notes of Rs.50 and 15 notes of Rs.100.

## Solution:

Let the fixed charge for the first three days be Rs.A and the charge for each day extra be Rs.B.

According to the information given,

A + 4B = 27 …………………………………….…………………………. (i)

A + 2B = 21 ……………………………………………………………….. (ii)

When equation (ii) is subtracted from equation (i) we get,

2B = 6

B = 3 …………………………………………………………………………(iii)

Substituting B = 3 in equation (i) we get,

A + 12 = 27

A = 15

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

## Solutions:

(i)

a1/a2=1/3 , b1/b2= -3/-9 =1/3, c1/c2=-3/-2 = 3/2

(a1/a2) = (b1/b2) ≠ (c1/c2)

Since, the given set of lines are parallel to each other they will not intersect each other and therefore there is no solution for these equations.

(ii)

a1/a2 = 2/3 , b1/b2 = 1/2 , c1/c2 = -5/-8

(a1/a2) ≠ (b1/b2)

Since they intersect at a unique point these equations will have a unique solution by cross multiplication method: x/(b1c2-c1b2) = y/(c1a2 – c2a=) = 1/(a1b2-a2b1)

x/(-8-(-10)) = y/(15+16) = 1/(4-3)

x/2 = y/1 = 1

∴ x = 2 and y =1

(iii)

(a1/a2) = 3/6 = 1/2 (b1/b2) = -5/-10 = 1/2 (c1/c2) = 20/40 = 1/2

a1/a2 = b1/b2 = c1/c2

Since the given sets of lines are overlapping each other there will be infinite number of solutions for this pair of equation.

### YOU ARE READING: NCERT Solutions For Class 10 CBSE Mathematics Chapter 3 Pair of Linear Equations in Two Variables.

(iv)

(a1/a2) = 1/3 (b1/b2) = -3/-3 = 1 (c1/c2) = -7/-15

a1/a2 ≠ b1/b2

Since this pair of lines are intersecting each other at a unique point, there will be a unique solution.

By cross multiplication,

x/(45-21) = y/(-21+15) = 1/(-3+9)

x/24 = y/ -6 = 1/6

x/24 = 1/6 and y/-6 = 1/6

∴ x = 4 and y = 1.

## Solution:

(i) 3y + 2x -7 =0

(a + b)y + (a-b)y – (3a + b -2) = 0

a1/a2 = 2/(a-b) , b1/b2 = 3/(a+b) , c1/c2 = -7/-(3a + b -2)

For infinitely many solutions,

a1/a2 = b1/b2 = c1/c2

Thus 2/(a-b) = 7/(3a+b– 2)

6a + 2b – 4 = 7a – 7b

a – 9b = -4 ……………………………….(i)

2/(a-b) = 3/(a+b)

2a + 2b = 3a – 3b

a – 5b = 0 ……………………………….….(ii)

Subtracting (i) from (ii), we get

4b = 4

b =1

Substituting this eq. in (ii), we get

a -5 x 1= 0

a = 5

Thus at a = 5 and b = 1

the given equations will have infinite solutions.

## YOU ARE READING: NCERT Solutions For Class 10 CBSE Mathematics Chapter 3 Pair of Linear Equations in Two Variables.

(ii)

3x + y -1 = 0

(2k -1)x + (k-1)y – 2k -1 = 0

a1/a2 = 3/(2k -1) , b1/b2 = 1/(k-1), c1/c2 = -1/(-2k -1) = 1/( 2k +1)

For no solutions a1/a2 = b1/b2 ≠ c1/c2

3/(2k-1) = 1/(k -1) ≠ 1/(2k +1)

3/(2k –1) = 1/(k -1)

3k -3 = 2k –1

k =2

Therefore, for k = 2 the given pair of linear equations will have no solution.

## Solution:

8x + 5y = 9 …………………..(1)

3x + 2y = 4 ……………….….(2)

From equation (2) we get x = (4 – 2y )/ 3 ……………………. (3)

Using this value in equation 1, we get

8(4-2y)/3 + 5y = 9

32 – 16y +15y = 27

-y = -5

y = 5 ……………………………….(4)

Using this value in equation (2), we get

3x + 10 = 4

x = -2

Thus, x = -2 and y = 5.

Now, Using Cross Multiplication method:

8x +5y – 9 = 0

3x + 2y – 4 = 0

x/(-20+18) = y/(-27 + 32 ) = 1/(16-15)

-x/2 = y/5 =1/1

∴ x = -2 and y =5.

## Solutions:

(i) Let x be the fixed charge and y be the charge of food per day.

According to the question,

x + 20y = 1000……………….. (i)

x + 26y = 1180………………..(ii)

Subtracting (i) from (ii) we get

6y = 180

y = Rs.30

Using this value in equation (ii) we get

x = 1180 -26 * 30

x= Rs.400.

Therefore, fixed charges is Rs.400 and charge per day is Rs.30.

(ii) Let the fraction be x/y.

So, as per the question given,

(x -1)/y = 1/3

=> 3x – y = 3…………………(1)

x/(y + 8) = 1/4

=> 4x –y =8 ………………..(2)

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

x = 5 ………………………………………….(3)

Using this value in equation (2), we get,

(4×5)– y = 8

y= 12

Therefore, the fraction is 5/12.

(iii) Let the number of right answers is x and number of wrong answers be y

According to the given question;

3x−y=40……..(1)

4x−2y=50

⇒2x−y=25…….(2)

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

x = 15 ….….(3)

Putting this in equation (2), we obtain;

30 – y = 25 Or y = 5

Therefore, number of right answers = 15 and number of wrong answers = 5

Hence, total number of questions = 20

## YOU ARE READING: NCERT Solutions For Class 10 CBSE Mathematics Chapter 3 Pair of Linear Equations in Two Variables.

(iv) Let the speed of 1st car and 2nd car be u km/h and v km/h respectively.

According to the given information,

When the cars travel in the same direction at different speeds, they meet in 5 hours.

Therefore, distance travelled by 1st car =5ukm

and distance travelled by 2nd car =5vkm

5u−5v=100

5(u−v)=100

u−v=20…(1)

When the cars travel towards each other at different speeds, they meet in 1 hour

therefore, distance travelled by 1st car =ukm

and distance travelled by 2nd car =vkm

u+v=100…(2)

Adding both the equations, we obtain

2u=120

u=60

Substituting this value in equation (2), we obtain

60+v=100

v=40

Equations are u−v=20 and u+v=100 where the speed of 1st car and 2nd car be u km/h and v km/h respectively.

Hence, speed of the 1st car =60km/h and speed of the 2nd car =40km/h

## YOU ARE READING: NCERT Solutions For Class 10 CBSE Mathematics Chapter 3 Pair of Linear Equations in Two Variables.

(v) Let, The length of rectangle = x unit And breadth of the rectangle = y unit

Now, as per the question given,

(x – 5) (y + 3) = xy -9

3x – 5y – 6 = 0……………………………(1)

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

2x + 3y – 61 = 0…………………………..(2)

Using cross multiplication method, we get,

x/(305 +18) = y/(-12+183) = 1/(9+10)

x/323 = y/171 = 1/19

Therefore, x = 17 and y = 9.

Hence, the length of rectangle = 17 units And breadth of the rectangle = 9 units.

## Solution:

Let us assume 1/x = m and 1/y = n ,

then the equation will change as follows.

m/2 + n/3 = 2

⇒ 3m+2n-12 = 0…………………….(1)

m/3 + n/2 = 13/6

⇒ 2m+3n-13 = 0……………………….(2)

Now, using cross-multiplication method, we get,

m/(-26-(-36) ) = n/(-24-(-39)) = 1/(9-4)

m/10 = n/15 = 1/5

m/10 = 1/5 and n/15 = 1/5

So, m = 2 and n = 3

1/x = 2 and 1/y = 3

x = 1/2 and y = 1/3

## Solution:

Substituting 1/√x = m and 1/√y = n in the given equations, we get

2m + 3n = 2 ………………………..(i)

4m – 9n = -1 ………………………(ii)

Multiplying equation (i) by 3, we get

6m + 9n = 6 ………………….…..(iii)

Adding equation (ii) and (iii), we get

10m = 5

m = 1/2…………………………….…(iv)

Now by putting the value of ‘m’ in equation (i), we get

2×1/2 + 3n = 2

3n = 1

n = 1/3

m =1/√x

½ = 1/√x

x = 4

n = 1/√y

1/3 = 1/√y

y = 9

Hence, x = 4 and y = 9

## Solution:

Putting 1/𝑥 = 𝑚 in the given equation we get,

So, 4m + 3y = 14

=> 4m + 3y – 14 = 0 ……………..…..(1)

3m – 4y = 23

=> 3m – 4y – 23 = 0 ……………………….(2)

By cross-multiplication, we get,

m/(-69-56) = y/(-42-(-92)) = 1/(-16-9)

-m/125 = y/50 = -1/ 25

-m/125 = -1/25 and y/50 = -1/25

m = 5 and b = -2 m = 1/x = 5

So , x = 1/5 y = -2

## Solution:

Substituting 1/(x-1) = m and 1/(y-2) = n in the given equations, we get,

5m + n = 2 …………………………(i)

6m – 3n = 1 ……………………….(ii)

Multiplying equation (i) by 3, we get

15m + 3n = 6 …………………….(iii)

Adding (ii) and (iii) we get

21m = 7

m = 1/3

### YOU ARE READING: NCERT Solutions For Class 10 CBSE Mathematics Chapter 3 Pair of Linear Equations in Two Variables.

Putting this value in equation (i), we get

5×1/3 + n = 2

n = 2- 5/3 = 1/3

m = 1/ (x-1)

⇒ 1/3 = 1/(x-1)

⇒ x = 4

n = 1/(y-2)

⇒ 1/3 = 1/(y-2)

⇒ y = 5

Hence, x = 4 and y = 5

## Solution:

(7x-2y)/ xy = 5

7/y – 2/x = 5…………………………..(i)

(8x + 7y)/xy = 15

8/y + 7/x = 15…………………………(ii)

Substituting 1/x =m in the given equation we get,

– 2m + 7n = 5

=> -2 + 7n – 5 = 0 ……..(iii)

7m + 8n = 15

=> 7m + 8n – 15 = 0 ……(iv)

By cross-multiplication method, we get,

m/(-105-(-40)) = n/(-35-30) = 1/(-16-49)

m/(-65) = n/(-65) = 1/(-65)

m/-65 = 1/-65

m = 1

n/(-65) = 1/(-65)

n = 1

m = 1 and n = 1

m = 1/x = 1

n = 1/x = 1

Therefore, x = 1 and y = 1

## Solution:

6x + 3y = 6xy

6/y + 3/x = 6

Let 1/x = m and 1/y = n

=> 6n +3m = 6

=>3m + 6n-6 = 0…………………….(i)

2x + 4y = 5xy

=> 2/y + 4/x = 5

=> 2n +4m = 5

or, 4m+2n-5 = 0……………………..(ii)

3m + 6n – 6 = 0

4m + 2n – 5 = 0

By cross-multiplication method, we get

m/(-30 –(-12)) = n/(-24-(-15)) = 1/(6-24)

m/-18 = n/-9 = 1/-18

m/-18 = 1/-18

m = 1

n/-9 = 1/-18

n = 1/2

m = 1 and n = ½

m = 1/x = 1 and n = 1/y = ½

x = 1 and y = 2

Hence, x = 1 and y = 2

## Solution:

Substituting 1/x+y = m and 1/x-y = n in the given equations, we get,

10m + 2n = 4

=> 10m + 2n – 4 = 0 ………………..…..(i)

15m – 5n = -2

=> 15m – 5n + 2 = 0 ……………………..(ii)

Using cross-multiplication method, we get,

m/(4-20) = n/(-60-(20)) = 1/(-50 -30)

m/-16 = n/-80 = 1/-80

m/-16 = 1/-80 and n/-80 = 1/-80

m = 1/5 and n = 1

m = 1/(x+y) = 1/5

x+y = 5 …………………………………………(iii)

n = 1/(x-y) = 1

x-y = 1……………………………………………(iv)

Adding equation (iii) and (iv), we get

2x = 6

=> x = 3 …….(v)

Putting the value of x = 3 in equation (3), we get

y = 2

Hence, x = 3 and y = 2

## Solution:

Substituting 1/(3x+y) = m and 1/(3x-y) = n in the given equations, we get,

m + n = 3/4 …………………………….…… (1)

m/2 – n/2 = -1/8

m – n = -1/4 …………………………..…(2)

Adding (1) and (2), we get

2m = 3/4 – ¼

2m = 1/2

Putting in (2), we get 1/4 – n = -¼

n = 1/4 + 1/4 = ½

m = 1/(3x+y) = 1/4

3x + y = 4 …………………………………(3)

n = 1/( 3x-y) = ½

3x – y = 2 ………………………………(4)

## YOU ARE READING: NCERT Solutions For Class 10 CBSE Mathematics Chapter 3 Pair of Linear Equations in Two Variables.

Adding equations (3) and (4), we get

6x = 6 x = 1 ……………………………….(5)

Putting in (3), we get

3(1) + y = 4

y = 1

Hence, x = 1 and y = 1

## Solutions:

(i) Let us consider, Speed of Ritu in still water = x km/hr Speed of Stream = y km/hr

Now, speed of Ritu during,

Downstream = x + y km/h

Upstream = x – y km/h

As per the question given,

2(x+y) = 20 Or x + y = 10……………………….(1)

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

Adding both the eq.1 and 2, we get

, 2x = 12

x = 6

Putting the value of x in eq.1, we get,

y = 4

Therefore, Speed of Ritu rowing in still water = 6 km/hr

Speed of Stream = 4 km/hr

(ii)

Let us consider,

Number of days taken by women to finish the work = x

Number of days taken by men to finish the work = y

Work done by women in one day = 1/x

Work done by women in one day = 1/y

As per the question given, 4(2/x + 5/y) = 1 (2/x + 5/y) = 1/4

And, 3(3/x + 6/y) = 1 (3/x + 6/y) = 1/3

Now, put 1/x=m and 1/y=n, we get,

2m + 5n = ¼

=> 8m + 20n = 1…………………(1)

3m + 6n =1/3

=> 9m + 18n = 1………………….(2)

## YOU ARE READING: NCERT Solutions For Class 10 CBSE Mathematics Chapter 3 Pair of Linear Equations in Two Variables.

Now, by cross multiplication method, we get here

, m/(20-18) = n/(9-8) = 1/ (180-144)

m/2 = n/1 = 1/36

m/2 = 1/36

m = 1/18

m = 1/x = 1/18

or x = 18

n = 1/y = 1/36

y = 36

Therefore, Number of days taken by women to finish the work = 18

Number of days taken by men to finish the work = 36.

(iii)

Let us consider,

Speed of the train = x km/h

Speed of the bus = y km/h

According to the given question,

60/x + 240/y = 4 …………………(1)

100/x + 200/y = 25/6 …………….(2)

Put 1/x=m and 1/y=n, in the above two equations;

60m + 240n = 4……………………..(3)

100m + 200n = 25/6

600m + 1200n = 25 ………………….(4)

## YOU ARE READING: NCERT Solutions For Class 10 CBSE Mathematics Chapter 3 Pair of Linear Equations in Two Variables.

Multiply eq.3 by 10, to get,

600m + 2400n = 40 ……………………(5)

Now, subtract eq.4 from 5, to get,

1200n = 15

n = 15/1200 = 1/80

Substitute the value of n in eq. 3, to get,

60m + 3 = 4

m = 1/60

m = 1/x = 1/60

x = 60 And y = 1/n

y = 80

Therefore, Speed of the train = 60 km/h

Speed of the bus = 80 km/h