NCERT Class 10 Maths Solutions Chapter - 3 Pairs of Linear Equations in Two Variables, Ex-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 differs by 30 years. Find the ages of Ani and Biju.

Solution

The difference between the ages of Ani and Biju is given as 3 years. So, either Biju is 3 years older than Ani or Ani is 3 years older than Biju. 

Let the age of Ani and Biju be x years and y years respectively.
Age of Dharam = 2 * x = 2x years

Case I: Ani is older than Biju by 3 years
x - y = 3        ... (1)

4x - y    = 60    ... (2)

Subtracting (1) from (2), we obtain:
3x = 60 - 3 = 57

Age of Ani = 19 years
Age of Biju = 19 - 3 = 16 years

Case II: Biju is older than Ani by 3 years
y - x = 3        ... (3)

4x - y = 60        ... (4)

Adding (3) and (4), we obtain:
3x = 63
x = 21

Age of Ani = 21 years
Age of Biju = 21 + 3 = 24 years

Concept Insight: In this problem, ages of Ani and Biju are the unknown quantities. So, we represent them by variables x and y. Now, note that here it is given that the ages of Ani and Biju differ by 3 years. So, it is not mentioned that which one is older. So, the most important point in this question is to consider both cases  Ani is older than Biju and  Biju is older than Ani. For second condition the relation  on the ages of Dharam and Cathy can be implemented . Pair of linear equations can be solved using a suitable algebraic method.

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?

Solution

Let the money with the first person and second person be Rs x and Rs y respectively.

According to the question,
x + 100 = 2(y - 100)
x + 100 = 2y - 200
x - 2y = - 300        ... (1)

6(x - 10) = (y + 10)
6x - 60 = y + 10
6x - y = 70            ... (2)

Multiplying equation (2) by 2, we obtain:
12x - 2y = 140        ... (3)
Subtracting equation (1) from equation (3), we obtain:
11x = 140 + 300
11x = 440
x = 40
Putting the value of x in equation (1), we obtain:
40 - 2y = -300
40 + 300 = 2y
2y = 340
y = 170

Thus, the two friends had Rs 40 and Rs 170 with them.

Concept insight: This problem talks about the amount of capital with two friends. So, we will represent them by variables x and y respectively. Now, using the given conditions, a pair of linear equations can be formed which can then be solved easily using elimination method.

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/h and the time taken by train to travel the given distance be t hours and the distance to travel be d km.

Or, d = xt        ... (1)

According to the question,

By using equation (1), we obtain:
3x - 10t = 30        ... (3)

Adding equations (2) and (3), we obtain:
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 (1), we obtain:
d = xt = 50 x 12 = 600 

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

Concept insight: To solve this problem, it is very important to remember the relation, . 

Now, all these three quantities are unknown. So, we will represent these by three different variables. By using the given conditions, a pair of equations will be obtained. Mind one thing that the equations obtained will not be linear. But they can be reduced to linear form by using the fact that, . 

Then two linear equations can be formed which can be solved easily by elimination method.

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 number of students in a row be y.
Total number of students in the class = Number of rows x Number of students in a row
                                                     = xy

According to the question,
Total number of students = (x - 1) (y + 3)
                                xy = (x - 1) (y + 3)

                                    = xy - y + 3x - 3
                      3x - y - 3 = 0
                           3x - y = 3        ... (1)

Total number of students = (x + 2) (y - 3)
                                xy = xy + 2y - 3x - 6
                          3x - 2y = -6    ... (2)

Subtracting equation (2) from (1), we obtain:
y = 9

Substituting the value of y in equation (1), we obtain:
3x - 9 = 3
3x = 9 + 3 = 12
x = 4

Number of rows = x = 4
Number of students in a row = y = 9 

Total number of students in a class = xy = 4 x 9 = 36

Concept insight: This problem talks about some number of students who are made to stand in a row. So, in order to know the total number of students, it is must  to know the number of students standing in each row and the total number of rows. So, these two quantities are represented by variables x and y. Use the conditions given in the question to obtain a pair of equations. The pair of equations can then be solved easily by eliminating a suitable variable.

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

Solution

C = 3B = 2(A + B) 

3B = 2(A + B) 
B = 2A 
2A - B = 0 ... (1) 

We know that the sum of the measures of all angles of a triangle is 180°.

A + B + C = 180° 
A + B + 3B = 180° 
A + 4B = 180° ... (2) 

Multiplying equation (1) by 4, we obtain: 
8A - 4B = 0 ... (3) 

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

9A = 180° A = 20° 

From equation (2), we obtain: 
20° + 4B = 180° 4B = 160° 
B = 40° 
C = 3B = 3 x 40° = 120° 
Thus, the measure of A, B and C are 20°, 40°, and 120° respectively. 

Concept insight: Here the problem is about three angles of a triangle. By simplifying the given relation, a linear equation is obtained. Using the angle sum property of triangles, i.e., the sum of all the three angles of a triangle is 180° another relation is obtained. So, from these two relations, a pair of linear equations will be obtained which can be easily solved by elimination method.

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.

Solution

Three solutions of this equation can be written in a table as follows: 

x	0	1	2
y	-5	0	5

x	0	1	2
y	-3	0	3

The graphical representation of the two lines will be as follows:

It can be observed that the required triangle is ABC.
The coordinates of its vertices are A (1, 0), B (0, -3), C (0, -5).

Concept insight: In order to find the coordinates of the vertices of the triangle so formed, find the points where the two lines intersects the y-axis and also where the two lines intersect each other. Here, note that the coordinates of the intersection of lines with y-axis is taken and not with x-axis, this is because the question says to find the triangle formed by the two lines and the y-axis.

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

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, we obtain:
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

Substituting the value of x in 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, we obtain:
a²x + aby = ac         ... (3)
b²x + aby = b + bc    ... (4)
Subtracting equation (4) from equation (3),
(a² - b²)x = ac - bc - b

Substituting the value of x in equation (1), we obtain:

bx - ay  = 0                         ... (1)    
ax + by = a² + b²                 ... (2)    
Multiplying equation (1) and (2) by b and a respectively, we obtain:
b²x - aby = 0                       ... (3)
a²x + aby = a³ + ab²            ... (4)
Adding equations (3) and (4), we obtain:
b²x + a²x = a³ + ab²
x(b² + a²) = a(a² + b²)
x = a
Substituting the value of x in equation (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

^{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
Substituting the value of x in 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}

Concept insight: In all the above parts, the given pair of equations can be easily solved by eliminating a suitable variable. Although one can use any of the other methods but the elimination method is the best suited over here.

 

(v)

 152x - 378y = -74                    ... (1)
-378x + 152y = -604                 ... (2)
Adding the equations (1) and (2), we obtain:
-226x - 226y = -678

Subtracting the equation (2) from equation (1), we obtain:
530x - 530y = 530

Adding equations (3) and (4), we obtain:
2x = 4

x = 2
Substituting the value of x in equation (3), we obtain:
y = 1

Concept insight: Here, since the coefficients of x and y in the two equations are interchanging this types of problems can be easily  solved by once adding the given equations and then  subtracting the given equations. By doing this  an equation of the form x - y = a and x + y = b, where a and b are constants will be obtained. And these equations can now be solved easily by elimination method.

Question 8.  ABCD is a cyclic quadrilateral. Find the angles of the cyclic quadrilateral.

Solution

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

A + C = 180 
4y + 20 - 4x = 180 
-4x + 4y = 160
x - y = -40            ... (1)
Also, B + D = 180
3y - 5 - 7x + 5 = 180
-7x + 3y = 180        ... (2)
Multiplying equation (1) by 3, we obtain: 

3x - 3y = -120        ... (3)
Adding equations (2) and (3), we obtain:
-4x = 60
x = -15
Substituting the value of x in equation (1), we obtain:
-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° 

Concept insight: The most important idea to solve this problem is by using the fact that the sum of the measures of opposite angles in a cyclic quadrilateral is 180°. By using this relation, two linear equations can be obtained which can be solved easily by eliminating a suitable variable.