NCERT solutions for class 10 Maths chapter 3 Pair of Linear Equations in Two Variables (Exercise 3.7)

NCERT solutions for class 10 Maths

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Chapter 3

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Pair of Linear Equations in Two Variables

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Exercise 3.7

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Ex 3.7 Question 1.

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

Solution:

Let the ages of Ani and Biju be p and q respectively.

∴ Dharam’s age = 2 x p = 2p years

Cathy’s age = q/2 years

Case I – Ani is 3 years older than Biju.

4p – q = 60        … (iv)

Adding (iii) and (iv),

3p = 63

p = 21

Age of Ani = 21 years

Age of Biju = 21 + 3 = 24 years

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Ex 3.7 Question 2.

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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 money with the first person = ₹ x and the second person = ₹ y respectively.

x + 100 = 2(y – 100) ⇒ x + 100 = 2B – 200

Or x – 2y = -300 ……..(i)

6(x – 10) = ( y + 10 )

Or 6x – 60 = y + 10

Or 6x – y = 70 ……..(ii)

Multiplying equation (ii) by 2,

12A – 2B = 140 ………(iii)

Subtracting equation (i) from equation (iii),

11x = 140 + 300

11x = 440

⇒ x = 440/11 = 40

Putting the value of x in equation (i)

40 – 2y = -300

40 + 300 = 2y

2y = 340

y = 170

Therefore, the first person had Rs 40 and the second had Rs 170 with them.

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Ex 3.7 Question 3.

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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/hr and the time taken by the train to travel a distance be t hours and the distance to travel be d hours.

Speed of the train = Distance travelled by train / Time taken to travel that distance

x = d/t

Or, d = xt ………(i)

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

(x + 10) (t – 2) = d

xt + 10N – 2x – 20 = d

By using the equation (i)

– 2x + 10t = 20 ……..(i)

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

(x – 10) (t + 3) = d

xt – 10t + 3x – 30 = d

By using the equation (i),

3x – 10t = 30 ……..(iii)

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

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 (i),

d = xt = 50 x 12 = 600

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

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Ex 3.7 Question 4.

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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 the number of students in a row be y.

Total number of students = Number of rows x Number of students in a row

=xy

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

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

3x – y – 3 = 0

3x – y = 3 ……..(i)

Total Number of students = (x + 2 ) ( y – 3 )

xy = xy + 2y – 3x – 6

3x – 2y = -6 …….(ii)

Subtracting equation (ii) from (i),

(3x – y) – (3x – 2y) = 3 – (-6)

-y + 2y = 3 + 6y = 9

Substituting the value of y in equation (i),

3x – 9 =3

3x = 9+3 = 12

x = 4

Number of rows,x = 4

Number of students in a row, yB = 9

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

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Ex 3.7 Question 5.

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

Solution:

∠C = 3 ∠B = 2(∠B + ∠A)

∠B = 2 ∠A+2 ∠B

∠B=2 ∠A

∠A – ∠B= 0 ……..(i)

[The sum of all the interior angles of a triangle is 180^O.]

∠ A +∠B+ ∠C = 180^O

∠A + ∠B +3 ∠B = 180^O

∠A + 4 ∠B = 180^O……..(ii)

Multiplying 4 to  equation (i) ,

8 ∠A – 4 ∠B = 0……..(iii)

Adding equations (iii) and (ii)

9 ∠A = 180^O

∠A = 20^O

From equation (ii),

20^O+ 4∠B = 180^O

∠B = 40^O

3∠B =∠C

∠C = 3 x 40 = 120^O

Therefore, ∠A = 20^O

∠B=40^O

∠C = 120^O

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Ex 3.7 Question 6.

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

Solutions:

5x – y = 5 ⇒ y = 5x – 5

3x – y = 3 ⇒ y = 3x – 3

Graphical representation-

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

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Ex 3.7 Question 7.

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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……………(i)

qx – py = p + q……………….(ii)

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

p²x + pqy = p² − pq ………… (iii)

q²x − pqy = pq + q² ………… (iv)

Adding equation (iii) and equation (iv),

p²x + q² x = p²  + q²

(p² + q² ) x = p² + q²

x = (p² + q²)/ p² + q² = 1

Substituting the value of x in equation (i),

p(1) + qy = p – q

qy = p-q-p

qy = -q

y = -1

(ii) ax + by= c………(i)

bx + ay = 1+ c…………..(ii)

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

a²x + aby = ac ………………… (iii)

b²x + aby = b + bc…………… (iv)

Subtracting equation (iv) from equation (iii),

(a² – b²) x = ac − bc– b

x = (ac − bc– b)/ (a² – b²)

x = c(a-b) –b / (a²+b²)

Putting the value of x in equation (1)

ax +by = c

a{c(a−b)−b)/ (a² – b²)} +by=c

ac(a−b)−ab/ (a² – b²)+by=c

by=c–ac(a−b)−ab/(a² – b²)

by=abc – b² c+ab/a²-b²

^{y = c(a-b)+a/}a²-b²

(iii) x/a – y/b = 0

ax + by = a² + b²

x/a – y/b = 0

⇒ bx − ay = 0 ……. (i)

ax + by = a² + b² …….. (ii)

Multiplying a and b to equation (i) and (ii)

b²x − aby = 0 …………… (iii)

a²x + aby = a ³ + ab³ …… (iv)

Adding equation (iii) and (iv),

b²x + a²x = a ³ + ab²

x (b² + a²) = a (a² + b²) x = a

Putting the value of x in equation (i),

b(a) − ay = 0

ab − ay = 0

ay = ab,

y = b

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

(a + b)(x + y) = a² + b²

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

(x + y)(a + b)  = a ² + b²

(a + b) y + (a + b) x  = a² + b² ………………… (ii)

Subtracting equation (ii) from equation (i),

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

x(a − b − a − b) = − 2ab − 2b²

− 2bx = − 2b (b + a)

x = b + a

Substituting this value in equation (i),

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

a² − b² + y(a + b)  = a²− 2ab – b²

(a + b) y = − 2ab

y = -2ab/(a+b)

(v) 152x − 378y = − 74

− 378x + 152y = − 604

152x − 378y = − 74 ⇒ 76x − 189y = − 37

x =(189y-137)/76……………..…(i)

− 378x + 152y = − 604

− 189x + 76y = − 302 ………….. (ii)

Putting the value of x in equation (ii),

−189(189y−37/76)+76y=−302

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

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

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

29945 = (113) (265) y

y = 1

Substituting the value of y in equation (i),

x = (189-37)/76

x = 152/76 = 2

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Ex 3.7 Question 8.

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ABCD is a cyclic quadrilateral (see Fig. 3.7). Find the angles of the cyclic quadrilateral.

Solution:

∠A +∠C = 180

4y + 20− 4x = 180

− 4x + 4y = 160

x − y = − 40 ……………(1)

And, ∠B + ∠D = 180

3y − 5 − 7x + 5 = 180

− 7x + 3y = 180 ………..(2)

Multiplying  equation (I) by 3,

3x − 3y = − 120 ………(3)

Adding equation (ii) and equation (iii),

− 7x + 3x = 180 – 120

− 4x = 60

x = −15

Substituting this value of x in equation (i),

x − y = − 40

-y−15 = − 40

y = 40-15

= 25

∠A = 4y + 20 = 20+4(25) = 120°

∠B = 3y − 5 = − 5+3(25) = 70°

∠C = − 4x = − 4(− 15) = 60°

∠D = 5-7x

∠D= 5− 7(−15) = 110°

Hence, all the angles are measured.

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