NCERT Solutions: Pair of Linear Equations in Two Variables (Exercise 3.7)

# Pair of Linear Equations in Two Variables (Exercise 3.7) NCERT Solutions - Mathematics (Maths) Class 10

Q.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:
The age difference between Ani and Biju is 3 yrs.
Either Biju is 3 years older than that of Ani or Ani is 3 years older than Biju. From both the cases we find out that Ani’s father’s age is 30 yrs more than that of Cathy’s age.
Let the ages of Ani and Biju be A and B respectively.
Therefore, the age of Dharam = 2 x A = 2A yrs.
And the age of Biju sister Ann B / 2 yrs
By using the information that is given,
Case I:
When Ani is older than that of Biju by 3 yrs then A – B = 3.............(1)
2A − B / 2 = 30
4A – B = 60...........(2)
By subtracting the equations (1) and (2) we get,
3A = 60 – 3 = 57
A = 57 / 3 = 19
Therefore, the age of Ani = 19 yrs
And the age of Biju is 19 – 3 = 16 yrs.
Case II:
When Biju is older than Ani, B – A = 3 ...........(1)
2A − B / 2 = 30
4A – B = 60..............(2)
Adding the equation (1) and (2) we get,
3A = 63
A = 21
Therefore, the age of Ani is 21 yrs
And the age of Biju is 21 + 3 = 24 yrs.

Q.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 Sangam have Rs A with him and Reuben have Rs B with him.
Using the information that is given we get,
A + 100 = 2(B – 100) ⇒ A + 100 = 2B – 200
Or A – 2B = -300.............(1)
And
6(A – 10) = ( B + 10 )
Or 6A – 60 = B + 10
Or 6A – B = 70............(2)
When equation (2) is multiplied by 2 we get,
12A – 2B = 140...............(3)
When equation (1) is subtracted from equation (3) we get,
11A = 140 + 300
11A = 440
⇒ A = 440 / 11 = 40
Using A = 40 in equation (1) we get,
40 – 2B = -300
40 + 300 = 2B
2B = 340 B = 170
Therefore, Sangam had Rs 40 and Reuben had Rs 170 with them.

Q.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 A km/hr and the time taken by the train to travel a distance be N hours and the distance to travel be X hours.
Speed of the train = Distance travelled by train / Time taken to travel that distance
A = N (distance) / X (time)
Or, N = AX.............(1)
Using the information that is given, we get:
(A + 10) = X / (N - 2)
(A + 10) (N – 2) = X
AN + 10N – 2A – 20 = X
By using the equation (1) we get,
– 2A + 10N = 20 .............(2)
(A - 10) = X / (N + 3)
(A - 10) (N + 3) = X
AN - 10N + 3A – 30 = X
By using the equation (1) we get,
3A – 10N = 30...........(3)
Adding equation (2) and equation (3) we get,
A = 50
Using the equation (2) we get,
(-2) x (50) + 10N = 20
-100 + 10N = 20
=> 10N = 120
N = 12hours
From the equation (1) we get, Distance travelled by the train, X = AN
= 50 x 12
= 600 km
Hence, the distance covered by the train is 600km.

Q.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 A and the number of students in a row be B.
Total number of students = Number of rows x Number of students in a row
= AB
Using the information, that is given,
First Condition:
Total number of students = (A – 1) ( B + 3)
Or AB = ( A – 1 )(B + 3) = AB – B + 3A – 3
Or 3A – B – 3 = 0
Or 3A – Y = 3..................(1)
Second condition:
Total Number of students = (A + 2 ) ( B – 3 )
Or AB = AB + 2B – 3A – 6
Or 3A – 2B = -6.................(2)
When equation (2) is subtracted from (1)
(3A – B) – (3A – 2B) = 3 – (-6)
-B + 2B = 3 + 6B = 9
By using the equation (1) we get,
3A – 9 =3
3A = 9 + 3 = 12
A = 4
Number of rows, A = 4
Number of students in a row, B = 9
Number of total students in a class = AB = 4 x 9 = 36

Q.5. In a ∆ABC, ∠ C = 3 ∠ B = 2 (∠A + ∠ B). Find the three angles.
Solution:
Given,
∠C = 3 ∠B = 2(∠B + ∠A)
∠B = 2 ∠A + 2∠B
∠B = 2 ∠A
∠A – ∠B = 0................(i)
We know, the sum of all the interior angles of a triangle is 180º.
Thus, ∠ A +∠B+ ∠C = 180º
∠A + ∠B + 3∠B = 180º
∠A + 4∠B = 180º.............(ii)
Multiplying 4 to  equation (i), we get
8∠A – 4∠B = 0..............(iii)
Adding equations (iii) and (ii) we get
9∠A = 180º
∠A = 20º
Using this in equation (ii), we get
20º + 4∠B = 180º
∠B = 40º
3∠B =∠C
∠C = 3 x 40 = 120º
Therefore, ∠A = 20º
∠B = 40º
∠C = 120º

Q.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.
Solution:
Given,
5x – y = 5
=> y = 5x – 5
Its solution table will be.

Also given,3x – y = 3
y = 3x – 3

The graphical representation of these lines will be as follows:
From the above graph we can see that the triangle formed is ∆ABC by the lines and the y axis. Also the coordinates of the vertices are A(1, 0), C(0, -5) and B(0, -3).

Q.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 p to equation (1) and q to equation (2), we get
p2x + pqy = p2 − pq ………… (iii)
q2x − pqy = pq + q2 ………… (iv)
Adding equation (iii) and equation (iv), we get
p2x + q2x = p2  + q2
(p2 + q2) x = p2 + q2
x = (p2 + q2) / p2 + q2 = 1
From equation (i), we get
p(1) + qy = p – q
qy = p - q - p
qy = -q
y = -1
(ii) ax + by = c…………………(i)
bx + ay = 1 + c………… ..(ii)
Multiplying a to equation (i) and  b to equation (ii), we obtain
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) / (a– b2)
x = c(a - b) – b / (a2 + b2)
From equation (i), we obtain
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 – b2c + 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) respectively, we get
b2x − aby = 0 …………… (iii)
a2x + aby = a3 + ab3 …….. (iv)
Adding equations (iii) and (iv), we get
b2x + a2x = a3 + ab2
x (b2 + a2) = a (a2 + b2) x = a
Using equation (i), we get
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)  = a2 + b2
(a + b) y + (a + b) x  = a2 + b2 ………………… (ii)
Subtracting equation (ii) from equation (i), we get
(a − b) x − (a + b) x = (a2 − 2ab − b2) − (a2 + b2)
x(a − b − a − b) = − 2ab − 2b2
− 2bx = − 2b (b + a) x = b + a
Substituting this value in equation (i), we get
(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
76x − 189y = − 37
x = (189y - 137) / 76……………..…(i)
− 378x + 152y = − 604
− 189x + 76y = − 302 ………….. (ii)
Using the value of x in equation (ii), we get
−189(189y − 37/76) + 76y = − 302
− (189)2y + 189 × 37 + (76)2y = − 302 × 76
189 × 37 + 302 × 76 = (189)2y − (76)2y
6993 + 22952 = (189 − 76) (189 + 76) y
29945 = (113) (265) y
y = 1
Using equation (i), we get
x = (189 - 37) / 76
x = 152 / 76 = 2

Q.8. ABCD is a cyclic quadrilateral (see Fig. 3.7). Find the angles of the cyclic quadrilateral.
Solution:
It is known that the sum of the opposite angles of a cyclic quadrilateral is 180o
Thus, we have
∠C +∠A = 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 3 to equation (1), we get
3x − 3y = − 120 ………(3)
Adding equation (2) to equation (3), we get
− 7x + 3x = 180 – 120
− 4x = 60
x = −15
Substituting this value in equation (i), we get
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.

The document Pair of Linear Equations in Two Variables (Exercise 3.7) NCERT Solutions | Mathematics (Maths) Class 10 is a part of the Class 10 Course Mathematics (Maths) Class 10.
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## Mathematics (Maths) Class 10

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## FAQs on Pair of Linear Equations in Two Variables (Exercise 3.7) NCERT Solutions - Mathematics (Maths) Class 10

 1. What are linear equations in two variables?
Ans. Linear equations in two variables are equations that involve two variables and have a degree of 1. They can be represented in the form of ax + by = c, where a, b, and c are constants and x and y are the variables. The solution to a pair of linear equations is the common values of x and y that satisfy both equations simultaneously.
 2. How can we solve a pair of linear equations in two variables?
Ans. There are several methods to solve a pair of linear equations in two variables, such as the substitution method, elimination method, and graphical method. In the substitution method, we solve one equation for one variable and substitute it into the other equation. In the elimination method, we eliminate one variable by multiplying the equations with suitable constants and then subtracting them. The graphical method involves plotting the equations on a graph and finding the point of intersection.
 3. Can a pair of linear equations have no solution?
Ans. Yes, a pair of linear equations can have no solution. This happens when the lines represented by the equations are parallel and never intersect. In this case, the equations are inconsistent, and there is no common solution for the variables. Graphically, the lines will be parallel and never cross each other.
 4. What is the significance of the point of intersection in a pair of linear equations?
Ans. The point of intersection of a pair of linear equations represents the common solution for the variables in both equations. It is the point where the lines represented by the equations meet or cross each other. This point satisfies both equations simultaneously. By finding the point of intersection, we can determine the values of x and y that satisfy both equations.
 5. Can a pair of linear equations have infinitely many solutions?
Ans. Yes, a pair of linear equations can have infinitely many solutions. This happens when the lines represented by the equations coincide or overlap each other. In this case, the equations are dependent, and any point on the line represents a valid solution for the variables. Graphically, the lines will be superimposed on each other, and every point on the line will satisfy both equations.

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