Rajasthan Board RBSE Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7 Textbook Exercise Questions and Answers.

## RBSE Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.7

Question 1.

The ages of two friends Ani and Biju differ by 3 years. Ani’s father Dharim istwice 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.

Solution:

Let the age of Ani = x years

and the age of Biju = y years

Age of Dharam = 2x years

Age of Cathy = \(\frac{1}{2}\)y years

According to the first condition, either

(Age of Ani) – (Age of Biju) = 3

⇒ x – y = 3 ….(1)

or y – x = 3 or -x + y = 3 …(2)

According to the second condition either,

(Age of Dharam) – (Age of Cathy) = 30

⇒ 2x – \(\frac{y}{2}\) = 30

⇒ \(\frac{4x-y}{2}\) = 30

⇒ 4x – y = 60

Subtracting (1) from equation (3)

Substituting this value of x in equation (1)

19 – y = 3

or – y = 3 – 19

or – y = -16

or y = 16

So, x = 19, y = 16 is the required solutions.

On adding equation (2) and (3)

-x + y = 3

4x – y = 60

3x = 63

or x = \(\frac{63}{3}\) = 21

On substituting the value of x in (2), we get

– 21 + y= 3

or y = 3 + 21 = 24

So, another required solution is x = 21, y= 24

Hence, Present age of Ani is 19 years and Present age of Biju is 16 years or Present age of Ani is 21 years and present age of Biju is 24 years.

Question 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)].

Solution:

Let the capital of one friend = Rs. x

and the capital of other friend = Rs. y

According to the first condition,

x + 100= 2 (y – 100)

or x + 100 = 2y – 200

or x – 2y = – 200 – 100

or x – 2y = – 300 … ..(1)

According to the second condition,

y + 10 = 6 (x – 10)

or y + 10 = 6x – 60

or 6x – y = 10 + 60

or 6x – y = 70 …(2)

Multiplying equation (1) by 6

6x – 12y = – 1800 ….(3)

Subtracting (2) from equation (3)

Substituting the value of y in equation (2)

6x – 170= 70

or 6x= 70 + 170

or 6x= 240

or x = \(\frac{240}{6}\) = 60

Hence, their capitals are respectively Rs. 40 and Rs. 170.

Question 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.

Solution:

Let the speed of the train = x km/hr

and time taken by the train = y hours

Distance covered by the train = (Speed) × (Time)

= (xy) km

According to the first condition,

(x + 10) (y – 2) = xy

or xy – 2x + 10y – 20 = xy

or -2x + 10y – 20 = 0

or x – 5y + 10 = 0 ….(1)

According to the second condition,

(x – 10) (y + 3) = xy

or xy + 3x – 10y – 30 = xy

or 3x – 10y – 30 = 0 ….(2)

Multiplying equation (1) by 3

3x – 15y + 30 = 0 ….(3)

Subtracting (2) from equation (3)

Substituting this value of y in equation (1)

x – 5 × 12 + 10 = 0

or x – 60 + 10 = 0

or x – 50 = 0

or x = 50

∴ Speed of the train =50 km/hr

Time taken by the train

= 12 hours

Hence, the distance covered by the train

= (50 × 12) km = 600 km

Question 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.

Solution:

Let the number of students in each row = x

and the number of rows = y

∴ Number of students in the class = xy

According to the first condition,

(x + 3) (y – 1) = xy

or xy – x + 3y – 3 = xy

or -x + 3y – 3 = 0

or x – 3y + 3 = 0 ….(1)

According to the second condition,

(x – 3) (y + 2) = xy

or xy + 2x – 3y – 6 = xy

or 2x – 3y – 6 = 0 ….(2)

Subtracting (1) from equation (2)

Substituting x in equation (I)

9 – 3y + 3 = 0

or – 3y + 12 = 0

or – 3y = – 12

or y = \(\frac{12}{3}\) = 4

∴ Number of students in each row = 9

and number of rows = 4

Hence, number of students in the class = 9 × 4 = 36

Question 5.

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

Solution:

In △ABC,

∠C = 3∠B = 2 (∠A + ∠B)

I II III

From II and III

3∠B = 2 (∠A + ∠B)

or 3∠B = 2∠A + 2∠B

or 3∠B – 2∠B = 2∠A

or ∠B = 2∠A …. (1)

Again from I and II

∠C = 3∠B

or ∠C = 3 (2∠A) [Using (1)]

or ∠C = 6∠A …. (2)

The sum of the three angles of a triangle is 180°

or ∠A + ∠B + ∠C = 180°

or ∠A + 2∠A + 6∠A = 180°

or 9∠A = 180°

∴ ∠A = \(\frac{180^{\circ}}{9}\) = 20°

∠A = 20°; ∠B = 2 × 20° = 40°

∠C = 6 × 20° = 120°

Question 6.

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

Solution:

The given pair of linear equations is

5x – y = 5

and 3x – y = 3

or 5x – y = 5

or x = \(\frac{5+y}{5}\) …(1)

Putting y = 0 in equation (1),

x = \(\frac{5+0}{5}\) = \(\frac{5}{5}\) = 1

Putting y = -5 in equation (1),

x = \(\frac{5-5}{5}\) = \(\frac{0}{5}\) = 0

Putting y = 5 in equation (1),

x = \(\frac{5+5}{5}\) = \(\frac{10}{5}\) = 2

x |
1 | 0 | 2 |

y |
0 | – 5 | 5 |

Plotting the points A (1, 0), B (0, – 5); C (2, 5) on the graph paper, we obtain the line corresponding to the equation 5x – y = 5

and 3x – y = 3

or 3x = 3 + y

x = \(\frac{3+y}{3}\) …(2)

Putting y = 0 in equation (2),

x = \(\frac{3+0}{3}\) = \(\frac{3}{3}\) = 1

Putting y = – 3 in equation (2),

x = \(\frac{3-3}{3}\) = \(\frac{0}{3}\) = 0

Putting y = 3 in equation (2),

x = \(\frac{3+3}{3}\) = \(\frac{6}{3}\) = 2

x |
1 | 0 | 2 |

y |
0 | – 3 | 3 |

Plotting the points A (1, 0), D (0, – 3), E (2, 3) on the graph paper, we obtain the line corresponding to the equation 3x – y = 3

From the graph it is clear that the given lines intersect at A (1, 0), △ABD formed by these lines and y-axis has been shaded. The vertices of △ABD are : A (1, 0), B (0, – 5) and D (0, – 3)

Area of △ABD = \(\frac{1}{2}\) × BD × OA

= \(\frac{1}{2}\) × 2 × 1

= 1 square unit

Question 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) \(\frac{x}{a}\) – \(\frac{y}{b}\) = 0

ax + by = a^{2} + b^{2}

(iv) (a – b) x + (a + b) y = a^{2} – 2ab – b^{2}

(a + b) (x + y) = a^{2} + b^{2}

(v) 152x – 378y = – 74

– 378+ + 152y = – 604

Solution:

(i) The given pair of linear equations is

px + qy = p – q …(1)

and qx – py = p + q …(2)

Multiplying (1) by q and (2) by p

Substituting this value of y in equations (1)

px + q (-1) = p – q

or px – q = p – q

or px = p – q + q

or px = p

or x = 1

Hence, x – 1 and y = – 1

(ii) The given pair of linear equation is

ax + by = c

and bx + ay = 1 + c

or ax + by – c = 0

and bx + ay – (1 + c) = 0

Hence x = a and y = b

(iv) The given pair of linear equations is

(a – b)x + (a + b)y = a^{2} – 2ab – b^{2}

or ax – bx + ay + by = a^{2} – 2ab – b^{2} …(1)

and (a + b)(x + y) = a^{2} + b^{2}

or ax + bx + ay + by = a^{2} + b^{2} …(2)

Subtracting (2) from equation (1)

Substituting this value of x in equation (1)

(a – b)(a + b) + (a + b)y = a^{2} – 2ab – b^{2}

or a^{2} – b^{2} + (a + b)y = a^{2} – 2ab – b^{2}

or (a + b)y = a^{2} – 2ab – b^{2} – a^{2} + b^{2}

or (a + b) y = -2ab

or

Hence, x = a + b and y = \(\frac{-2 a b}{a+b}\)

(v) The given pair of linear equations is

152x – 378y = -74 ….(1)

-378x + 152y = -604 …(2)

Adding equation (1) and equation (2)

-226x – 226v = -678

⇒ -226(x + y) = -678

⇒ x + y = \(\frac{-678}{226}\)

⇒ x + y = 3 …(3)

Subtracting equation (2) from equation (1)

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

⇒ 530x – 530y = -74 + 604 = 530

⇒ 530(x – y) = 530

∴ x – y = 1 …(4)

Adding equation (4) to equation (3)

x + y + x – y = 3 + 1

or 2x = 4 ∴ x = 2

Putting x = 2 ¡n equation (3)

x + y = 3

⇒ 2 + y = 3

⇒ y = 3 – 2 = 1

∴ y = 1

Hence the required solution of the pair of equation is

x = 2 and y = 1

Question 8.

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

Solution:

The given figure ABCD is a cyclic quadrilateral.

We know that the sum of the opposite angles of a cyclic quadrilateral is 18

∴ ∠A + ∠C = 180°

and ∠B + ∠D = 1800 …(2)

Putting the values in equation (1)

4y + 20° – 4x = 180°

⇒ -4x + 4y = 160°

or -x + y = 40° ….(3)

Similarly,

∠B + ∠D = 180°

⇒ 3y – 5° – 7x + 5°= 180°

⇒ -7x + 3y = 180° …(4)

Multiplying equation (3) by 3 and subtracting from equation (4)

Putting the value of x in equation (3)

-(-15°) + y = 40°

⇒ 15° + y = 40°

⇒ y = 40° – 150 = 25°

Then the values of the angles of the cyclic quadrilateral will be

∠A = 4y + 20

= 4 × 25 + 20

= 1oo + 20 = 120°

∠B = 3y – 5

= 3 × 25 – 5 = 75 – 5 = 70°

∠C = -4x = -4 × (-15°) = 60°

∠D = -7x + 5

= -7 × -15 + 5

= 105 + 5 = 110°

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