Linear equations in two variables- Exercise 3.7-Class 10

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 differs by 30 years. Find the ages of Ani and Biju.

Solution:

The difference between the ages of Biju and Ani is 3 years.

Either Biju is 3 years older than Ani or Ani is 3 years older than Biju.

However, it is obvious that in both cases, Ani’s father’s age will be 30 years more than that of Cathy’s age.

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

Therefore,

age of Ani’s father, Dharam = 2 × x = 2x years and the age of Biju’s sister Cathy y/2 years

By using the information given in the question,

Case (I):

When Ani is older than Biju by 3 years, x − y = 3 ………………….(1)

2x – ^y/₂ = 30

4x – y = 60 ————(2)

Subtracting (1) from (2)

3x = 60 – 3 = 57

x = ⁵⁷/₃ = 19

Therefore, age of Ani = 19 years

And age of Biju = 19 − 3 = 16 years

Case (II):

When Biju is older than Ani, y − x = 3 ……………….(i)

2x – ^y/₂ = 30

4x –y = 60 ———-(2)

Adding (1) and (2)

3x = 63

x = 21

Therefore, age of Ani = 21 years and

age of Bijju = 21 + 3 = 24 years

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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 those friends were having Rs x and y with them.

 

Using the information given in the question,

 

x + 100 = 2(y − 100) x + 100 = 2y – 200

or x − 2y = −300 ……………….(1)

6(x − 10) = (y + 10)

Or 6x − 60 = y + 10

Or 6x − y = 70 ………………………….(2)

 

Multiplying equation (2) by 2,

12x − 2y = 140———— (3)

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

11x = 140 + 300

11x = 440

x = 40

Using this in equation (1),

40 − 2y = −300

40 + 300 = 2y

2y = 340

y = 170

Therefore, those friends had Rs 40 and Rs 170 with them respectively.

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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 be x km/h and the time taken by train to travel the given distance be t hours and the distance to travel was d km. We know that,

Speed = ^{Distance travelled}/_{time taken to travel that distance}

x = ^d/_t

Or, d = xt ………………………………… (1)

Using the information given in the question,

(x+10) = ^d/_(t-2)

(x+10)(t-2) = d

xt – 2x + 10t – 20 = d

By using equation (1), we get

-2x + 10t = 20 ———-(2)

(x-10) = ^d/_(t+3)

(x-10)(t+3) = d

xt – 10t + 3x – 30 = d

By using equation (1), we obtain

3x − 10t = 30 ————(3)

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

x = 50

Using equation (2), we obtain

(−2) × (50) + 10t = 20

−100 + 10t = 20 10t = 120 t = 12 hours

From equation (1), we obtain

Distance to travel = d = xt

= 50 × 12

= 600 km

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

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

Total students of the class = Number of rows × Number of students in a row

= xy

Data given:

Condition 1:

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

 xy = (x − 1) (y + 3) = xy − y + 3x − 3

3x − y − 3 = 0——–(1)

Condition 2:

Total number of students = (x + 2) (y − 3)

xy = xy + 2y − 3x − 6

3x − 2y = −6 ……………………………….(2)

Subtracting equation (2) from (1),

(3x − y) − (3x − 2y) = 3 − (−6)

− y + 2y = 3 + 6 y = 9

By using equation (i), we obtain

3x − 9 = 3

3x = 9 + 3 = 12

x = 4

Number of rows = x = 4

Number of students in a row = y = 9

Number of total students in a class = xy = 4 × 9 = 36

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

Solution:

Given that,

∠C = 3∠B = 2(∠A + ∠B)

3∠B = 2(∠A + ∠B)

3∠B = 2∠A + 2∠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°. Therefore,

∠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 obtain

9 ∠A = 180°

 

∠A = 20°

From equation (2), we obtain

20° + 4 ∠B = 180°

4 ∠B = 160°

∠B = 40°

∠C = 3 ∠B

= 3 × 40° = 120°

Therefore, ∠A, ∠B, ∠C are 20°, 40°, and 120° respectively.

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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:

5x − y = 5

Or, y = 5x − 5

The solution table will be as follows.

x	0	1	2
y	-5	0	5

3x − y = 3

Or, y = 3x − 3

The solution table will be as follows.

x	0	1	2
y	-3	0	3

The graphical representation is,

It can be observed that the required triangle is ΔABC formed by these lines and y-axis.

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

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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

Solution:

(i) px + qy = p − q ……………………..… (1)

qx − py = p + q ……………………..… (2)

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

p²x + pqy = p² − pq …………………………… (3)

q²x − pqy = pq + q² ………………………..… (4)

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

p²x + q²x = p²+ q²

(p² + q²) x = p² + q²

x = (p²+ q²)/( p²+ q²) = 1

From equation (1), we obtain

p (1) + qy = p − q

qy = − q

y = − 1

(ii)ax + by = c ……………… (1)

bx + ay = 1 + c ……….. (2)

Multiplying equation (1) by a and equation (2) by b,

a²x + aby = ac ………………… (3)

b²x + aby = b + bc …………… (4)

Subtracting equation (4) from equation (3),

(a² − b²) x = ac − bc − b

x = [c(a-b)-b]/(a²-b²)

From equation (1),

ax + by = c

a{[c(a-b)-b]/(a²-b²)}+by = c

[ac(a-b)-b]/(a²-b²)]+by = c

by = c – [ac(a-b)-b]/(a²-b²)]

by = (a²c-b²c-a²c+abc+ab)/(a²-b²)

by = [abc-b²c+ab]/(a²-b²)

y= [c(a-b)+a]/(a²-b²)

(iii) ^x/_a – ^y/_b = 0

Or, bx − ay = 0 ……………. (1)

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 + a² x = a³ + ab²

x (b² + a²) = a (a² + b²) x = a

By using (1), we obtain b (a) − ay = 0

ab − ay = 0

ay = ab

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

Using equation (1), we obtain

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

a² – b² + (a + b) y = a²− 2ab – b²

(a + b) y = − 2ab

y = ^-2ab/_(a+b)

(v) 152x – 378y = -74

76x – 189y = -37

x = ^(189y-37)/₇₆————(1)

-378x + 152y = -604

-189x + 76y = -302 ————(2)

Substituting the value of x in equation (2),

-189[^(189y-37)/₇₆] + 76y = -302

− (189)² y + 189 × 37 + (76)² y = − 302 × 76

189 × 37 + 302 × 76 = (189)² y − (76)² y

6993 + 22952 = (189 − 76) (189 + 76) y

29945 = (113) (265) y

y = 1

From equation (1),

x = ^189(1)-37­/₇₆

x = ^189-37/₇₆ = ¹⁵²/₇₆

x = 2

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8. ABCD is a cyclic quadrilateral finds the angles of the cyclic quadrilateral.

Solution:

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

Therefore, ∠A + ∠C = 180

4y + 20 − 4x = 180

− 4x + 4y = 160

x − y = − 40 ………………………(i)

Also, ∠B + ∠D = 180

3y − 5 − 7x + 5 = 180

− 7x + 3y = 180 …………………..(ii)

Multiplying equation (i) by 3,

3x − 3y = − 120 ……………………(iii)

Adding equations (ii) and (iii),

− 7x + 3x = 180 − 120

− 4x = 60

x = −15

By using equation (i),

x − y = − 40

−15 − y = − 40

y = −15 + 40 = 25

∠A = 4y + 20 = 4(25) + 20 = 120°

∠B = 3y − 5 = 3(25) − 5 = 70°

∠C = − 4x = − 4(− 15) = 60°

∠D = − 7x + 5 = − 7(−15) + 5 = 110° 

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