NCERT solutions for class 10 Maths
Chapter 3
Pair of Linear Equations in Two Variables
Exercise 3.7
Ex 3.7 Question 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 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.
3p = 60 – 3 = 57
Age of Biju = 19 – 3 = 16 years
Case II: Biju is 3 years older than Ani.
q – p = 3 … (iii)
2p – q/2 = 30
4p – q = 60 … (iv)
Adding (iii) and (iv),
3p = 63
p = 21
Age of Ani = 21 years
Age of Biju = 21 + 3 = 24 years
Ex 3.7 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 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.
Ex 3.7 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 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.
Ex 3.7 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 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
Ex 3.7 Question 5.
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 180O.]
∠ A +∠B+ ∠C = 180O
∠A + ∠B +3 ∠B = 180O
∠A + 4 ∠B = 180O……..(ii)
Multiplying 4 to equation (i) ,
8 ∠A – 4 ∠B = 0……..(iii)
Adding equations (iii) and (ii)
9 ∠A = 180O
∠A = 20O
From equation (ii),
20O+ 4∠B = 180O
∠B = 40O
3∠B =∠C
∠C = 3 x 40 = 120O
Therefore, ∠A = 20O
∠B=40O
∠C = 120O
Ex 3.7 Question 6.
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).
Ex 3.7 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) 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
Ex 3.7 Question 8.
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.