Exercise 3.1
Multiple Choice Questions (MCQs)
Quesntion1. Graphically, the pair of equations
6x –3y + 10 = 0
2x – y + 9 = 0
represents two lines which are
(a)    intersecting at exactly one point
(b)   intersecting exactly two points
(c)    coincident
(d)   parallel
Solution: (d)  
Explanation:
On comparing with a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 we have a₁= 6, b₁= ‒3, c₁ = 10 and a₂ = 2, b₂y = ‒1, c₂ = 9.

So, the given system of equations have no solution, this will be possible if two lines never intersect each other.
In other words both the lines are parallel to each other.
Rough figure is shown below for reference,

Quesntion2.The pair of equations x + 2y + 5 = 0 and – 3x- 6y + 1 = 0 has
(a) a unique solution
(b) exactly two solutions
(c) infinitely many solutions
(d) no solution
Solution: (d)
Given, equations are x + 2y + 5 = 0 and – 3x –6y + 1 = 0
On comparing with a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, we have

Therefore, the pair of equations has no solution.
Quesntion3. If a pair of linear equations is consistent, then the lines will be
(a) parallel
(b) always coincident
(c) intersecting or coincident
(d) always intersecting
Solution: (c)
If two lines are parallel then, they have no solution pair of linear equations is inconsistent;
If two lines are coincident then, they have infinite solution and pair of linear equations is consistent;
If two lines are intersecting then, they have unique solution and pair of linear equations is consistent.
Quesntion4.The pair of equations y = 0 and y = ‒7 has
(a) one solution
(b) two solutions
(c) infinitely many solutions  
(d) no solution
Solution: (d)
If we draw a rough graph of both the lines (as shown below) we will find that both lines are parallel. So, they have no solution.

Quesntion5. The pair of equations x = a and y = b graphically represents lines which are
(a) parallel
(b) intersecting at (b, a)
(c) coincident 
(d) intersecting at (a, b)
Solution: (d)
x = a is the equation of a straight line which is parallel to the y–axis lying at a distance ‘a’ from it. Also, y = b is the equation of a straight line parallel to the x–axis lying at a distance ‘b’ from it. Thus the two lines come to intersect at a point (a, b).
Quesntion6. For what value of k, do the equations 3x –y + 8 = 0 and 6x –k y = -16 represent coincident lines?

Quesntion7. If the lines given by 3x+ 2ky = 2 and 2x + 5y = 1 are parallel, then the value of k is

Quesntion8. The value of c for which the pair of equations cx – y = 2 and 6x –2y = 3 will have infinitely many solutions is
(a) 3
(b) -3
(c) -12
(d) no value
Solution: (d)
Condition for two lines a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 to have infinitely many solutions is:

As, c has different values, therefore, there will be no value of c for which the pair of equations will have infinitely many solutions.
Quesntion9. One equation of a pair of dependent linear equations is – 5x + 7y – 2 = 0. The second equation can be
(a) 10x + 14y + 4 = 0 
(b) 10x -14y + 4 = 0
(c) 10x + 14y + 4=0
(d) 10x -14y + 4=0
Solution: (d)
Condition for two linear equations a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 to be dependent is:

Quesntion10. A pair of linear equations which has a unique solutionx = 2 and y = – 3 is
(a) x + y = 1 and 2x + 3y = – 5
(b) 2x + y = – 11 and 4x + 10y = – 22
(c) 2x – y = 1 and 3x – 2y = 0
(d) x –4y – 14 = 0 and 5x – y –13 = 0
Solution: (b)
If x = 2, y = ‒3 is a unique solution of a pair of equations, then these values will satisfy both pair of equations.
So, we will check by putting x = 2, y = ‒3 in all options and check which one of the four options is correct.
In option (b),
Putting x = 2, y = ‒3 in LHS of equation, 2x + y = – 11
⇒ 2(2) + 5(-3) = 4 – 15 = 11 = RHS
Putting x = 2, y = ‒3 in LHS of equation 4x+ 10y = ‒22
⇒4 (2) + 10(-3) = 8 – 30 = -22 = RHS
Quesntion11. If x = a and y = b is the solution of the equations x – y = 2and x + y = 4, then the values of a and b are, respectively
(a) 3 and 5
(b) 5 and 3
(c) 3 and 1
(d) 1 and -3
Solution: (c)   
As, x = a and y = b being the solution of the equations x – y = 2 and x + y = 4, must satisfy both the equations.
Thus, on putting x = a and y = b in both the equations, we get:
            a – b = 2 …(i)
And      a + b = 4 …(ii)
On solving equations (i) and (ii), we get 2a = 6 or a = 3.
⟹ a = 3 and b = 1.
Quesntion12. Aruna has only Rs. 1 and Rs. 2 coins with her. If the total number of coins that she has is 50 and the amount of money with her is Rs. 75, then the numbers of Rs. 1 and Rs.2 coins are, respectively
(a) 35 and 15
(b) 35 and 20
(c) 15 and 35
(d) 25 and 25
Solution: (d)
Let total number of Rs. 1 coins = x
And total number of Rs. 2 coins = y
Now, according to the conditions given in question x + y = 50 …(i)
Also,    x × 1+ y×2=75
Þ        x + 2y = 75      …(ii)
Solving (i) and (ii)
(x + 2y) – (x + y) = 75 – 50
Þ        y = 25
As, x + y =50
Therefore, for y = 25, we get x = 25.
Quesntion13. The father’s age is six times his son’s age. Four years hence, the age of the father will be four times his son’s age. The present ages (in year) of the son and the father are, respectively
(a) 4 and 24
(b) 5 and 30
(c) 6 and 36    
(d) 3 and 24
Solution: (c)
Let present age of father = x
And present age of son = y  
It is given in question that, x = 6y      …(i)
According to the question,
(x +4) = 4 (y + 4)
⇒ x – 4y = 12  …(ii)
Using equations (i) and (ii), we get:
        6y – 4y = 12
Þ    2y = 12
Þ    y = 6
If y = 6, then x = 36
Therefore, present age of father is 36 year and present age of son is 6 year.
Exercise 3.2
Very Short Answer Type Questions
Quesntion1. Do the following pair of linear equations have no solution? Justify your answer.


Hence, the given pair of linear equations is coincident and having infinitely many solutions.
Quesntion2. Do the following equations represent a pair of coincident lines? Justify your answer.

Therefore, the given pair of linear equations has unique solution.
(ii) Given, –2x – 3y -1 = 0 and 6y + 4x + 2 = 0 on comparing with a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, we have:

So, the given pair of linear equations is coincident.

Therefore, the given pair of linear equations has no solution.
Quesntion3. Are the following pair of linear equations consistent? Justify your answer.
(i) -3x -4y = 12 and 4y + 3x = 12
(ii) and
(iii) and
(iv) x + 3y = 11 and2(2x + 6)= 22
Solution:
To remember:

Therefore, the pair of linear equations has no solution or inconsistent.
(ii) Yes,
Given,

Therefore, the given pair of linear equations has infinitely many solutions orconsistent.
(iv) No

So, the pair of linear equation have no solution or inconsistent.
Exercise 3.3
Short Answer Type Questions (Q. No. 1 to 11)
Quesntion1. For which value(s) of l, do the pair of linear equations lx + y = l²and
x+ly = 1 have
(i) no solution?           
(ii) infinitely many solutions?
(iii) a unique solution?
Solution:
Given, lx + y = l² and x + ly –1                  ….(i)
Here, a₁ = l, b₁, = 1, c, = –l² and a₂ = 1, b₂ = l, c₂ = -1

(iii) Condition for a unique solution is:

Therefore, all real values of l except  ±1 equation will have unique solution.
Quesntion2. For which value (s) of k will the pair of equations
kx + 3y = k -3,
12x+ ky = k
have no solution?
Solution:
Given,
kx + 3y – (k –3) = 0and 12x + ky – k = 0
On comparing with a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 we have
Here, a₁ = k, b₁=3 and c₁ = -(k -3)
And a₂ = 12, b₂ = k and c₂ = -k         …..(i)
Condition for a pair of linear equations to have no solution is:

Thus, from (iii) and (iv), it’s clear that at k = – 6 the given pair of linear equations will have no solution.
Quesntion3. For which values of a and b will the following pair of linear equations has infinitely many solutions?
x + 2y = 1
(a – b) x + (a + b) y = a + b –2
Solution:
Given equation are:
x + 2y −1 = 0 and (a – b) x + (a + b) y – (a + b – 2) = 0 ….(i)
On comparing with a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 we have:

Quesntion4. Find the values of p in (i) to (iv) and p and q in (v) for the following pair of equations
(i) 3x –y – 5 = 0 and 6x – 2y – p = 0, if the Lines represented by these
equations are parallel.
(ii) –x + py = 1 and px- y = 1,if the pair of equations has no solution.
(iii) -3x + 5y = 7 and 2px – 3y =1
if the lines represented by these equations are intersecting at a unique point.
(iv) 2x + 3y -5 = 0 and px – 6y – 8 = 0,
if the pair of equations has a unique solution.
(v) 2x + 3y = 7 and 2px + py = 28 –qy,
if the pair of equations has infinitely many solutions.
Solution:
(i) Given, 3x – y -5 = 0 and 6x-2y – p = 0
On comparing with a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 we have:
c₁ = 5 and a₂ = 6, b₂ = -2, c₂ = –p
As, the lines represented by these equations are parallel, therefore

Therefore, the given pair of linear equations are parallel for all real values of p except 10.
(ii) Given, –x + py -1 = 0 and px-y-1 = 0      …..(i)
We have, a₁ = -1, b₁ = p, c₁= -1and a₂ = p, b₂ = -1 and c₂ = -1
As, the pair of linear equations has no solution i.e., both lines are parallel to each other.

Therefore, the given pair of linear equations has no solution for p = 1.
(iii) Given, -3x + 5y – 7 = 0 and 2px-3y-1 = 0
Here, a₁ = -3, b₁ = 5, c₁ = -7 and a₂ = 2p, b₂ = -3, c₂ = -1
As, the lines are intersecting at a unique point i.e., it has a unique solution.

Therefore, the lines represented by these equations are intersecting at a unique point for all great values of p except 9/10.
(iv) Given, 2x + 3y-5 = 0 and px –6y -8 = 0
Here,c₁ = -5 and a₂ = p, b₂ = -6, c₂ = −8
As, the pair of linear equations has a unique solution.

Given, the pair of linear equations has a unique solution for all values of p except -4.
(v)   Given, 2x + 3y =7and 2px + py = 28 -qy
Here, a₁ = 2, b₂, = 3, c_1 = -7 and a₂ = 2p, b₂ = (p + q), c₂ = -28
Since, the pair of equations has infinitely many solutions i.e., both lines are coincident.

⟹ 4 + q =12
Therefore, q = 8
Therefore, the pair of equations has infinitely many solutions for the values of p = 4and q = 8
Quesntion5. Two straight paths are represented by the equations x –3y = 2 and -2x + 6y = 5. Check whether the paths cross each other or not.
Solution:

Hence, two straight paths represented by the given equations never cross each other because they are parallel to each other.
Quesntion6. Write a pair of linear equations which has the unique solution
x = –1 and y 3. How many such pairs can you write?
Solution:

Hence, infinitely many pairs of linear equations are possible.
Quesntion7. If 2x + y = 23 and 4x – y = 19, then find the values of 5y -2x and
Solution:

Quesntion8. Find the values of x and y in the following rectangle

Quesntion9. Solve the following pairs of equations

Solution:

Hence, the required values of x and y are 1.2 and 2.1, respectively

Hence, the required values of x and y are 6 and 8, respectively

Hence, the required values of x and y are 3 and 2, respectively.

(v) Given pair of linear equations is
          43x + 67y = -24     …(i)
And   67x + 43y = 24      …(ii)
Now taking 43 × (i) + 67 ×(ii), we get

Hence, the required values of x and y are 1 and -1, respectively.
(vi) Given pair of linear equations is

Hence, the required values of x and y are a² and b², respectively.



Now, multiplying both sides of equation (i) by LCM (10, 5) = 10, we get
x + 2y – 10 = 0
Þ x + 2y = 10 … (iii)
Again, multiplying both sides of (iv) by LCM (8, 6) = 24, we get
3x + 4y = 360 … (iv)
Now, (iv) – 2 × (iii) gives:

Hence, the solution of the pair of equations is x = 340, y = – 165 and the required value of lis
Quesntion11. By the graphical method, find whether the following pair of equations are consistent or not. If consistent, solve them.
(i) 3x + y + 4 = 0, 6x – 2y + 4 = 0
(ii) x –2y = 6, 3x –6y = 0
(iii) x + y = 3, 3x + 3y = 9
Solution:
(i) Given pair of equations is:
 3x+ y + 4= 0 and 6x – 2y + 4 = 0
On comparing with a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 we have
and c₁ = 4  and a₂ = 6, b₂ = –2 and c₂ = 4

So, the given pair of linear equations has a unique solution and thereby it is consistent.    
We have,  3x + y + 4 = 0
Þ y = – 4 – 3x                           ….(i)
If x = 0, y = – 4
x = –1,  y = – 1
x = – 2, y = 2                    
x
0
-1
-2
y
-4
-1
2
Points
B
C
A
And 6- 2+ 4 = 0                     …..(ii)
Þ 2y = 6x + 4
Þ y = 3x + 2                                                                      
If x = 0, y = 2
x = –1, y = – 1
x = 1, y = 5
x
-1
0
1
y
-1
2
5
Points
C
Q
P
Plotting (i) and (ii) as per the respective values of x and y, we get two lines AB and PQ respectively that intersect at C (-1.-1).

Thus, the given pair of linear equations has no solution, i.e., inconsistent.
(iii) Given pair of equations is:
x + y = 3 and 3x + 3y = 9
On comparing with a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 we have
a₁ = 1, b₁ = 1 and c₁, = – 3 and a₂ = 3, b₂ = 3 andc₂ = – 9                         

So, the given pair of lines is coincident. Therefore, these lines have. Hence, the given pair of linear equations has infinitely many solutions, i.e., consistent.
Now, x + y = 3             ….(i)
Or      y = 3 – x
If x = 0, y = 3
x = 3, y = 0
x
0
3
y
3
0
Points
A
B
Also 3x + 3y = 9           ….(ii)
Or    3y = 9 – 3x 

Plotting (i) and (ii) for respective set of values for x and y, we get two lines AB and CD respectively, that are coincident.
Exercise 3.4
Long Answer Type Questions (Q. No. 1 to 6)
Quesntion1. Graphically, solve the following pair of equations
2x + y = 6 and 2x – y + 2 = 0
Find the ratio of the areas of the two triangles formed by the lines representing these equations with the X-axis and the lines with the Y-axis.
Solution:
First equation is: 2x + y = 6
⟹ y = 6 – 2x
If x = 0, y = 6
And if x = 3, y = 0
x

0

3
y

6

0
Points

B

A
Second equation is: 2x – y + 2= 0,
⟹ y = 6 – 2x
If x = 0, y = 2
And if x = −1, y = 0
x
0
-1
y
2
0
Points
D
C
Plotting 2x + y = 6 and 2x – y + 2= 0, as shown below, we obtain two lines AB and CD respectively intersecting at point, E (1, 4).

Hence, the ratio of the areas of the two triangles DACE and DBDE  is 4:1
Quesntion2. Determine graphically, the vertices of the triangle formed by the lines
y = x, 3y = x and x + y = 8.
Solution:
First equation is: y = x
If x = 1, y = 1
If x = 0, y = 0
If x = 2, y = 2
x
0
1
2
y
0
1
2
Points
0
A
8
Second equation is: x = 3y,
If x = 0, y = 0
If x = 3, y = 1
If x = 6, y = 2 
x
0
3
6
y
0
1
2
Points
0
C
D
Third equation is: x+ y = 8
⟹ y = 8 –x
Ifx= 0,y= 8
If x = 8, y = 0
If x = 4, y = 4
x
0
4
8
y
8
4
0
Points
P
Q
R
Plotting y = x, 3y = x and x + y = 8, we get three lines OQ, OD and PR respectively, as shown below.  

We see that lines OQ and OD intersect the line PR on Q and D, respectively.
So, DOQD is formed by these lines. Hence, the vertices of the DOQD formed by the given lines are O(0, 0), Q(4, 4) and D(6,2).
Quesntion3. Draw the graphs of the equations x = 3, x = 5 and 2x – y-4 = 0. Also find the area of the quadrilateral formed by the lines and the X-axis.
Solution:
First equation is: 2x – y– 4 = 0
⟹ y = 2x – 4
Ifx= 0,y= -4
If x = 2, y = 0
If x = 4, y = 4
x
0
2
y
-4
0
Points
P
Q
Plotting x = 3 and x = 5 and 2x – y– 4 = 0, we obtain three lines I, II and III respectively forming a quadrilateral ABCD with X-axis as shown below:

Quadrilateral ABCD is a trapezium with AB = OB – OA = 5 – 3 = 2, AD = 2 and BC = 6
Now, Area of trapezium ABCD = 1/2 × distance between parallel lines (AB)× (AD + BC) = 1/2 × 2 × (6 + 2) = 8sq units
Quesntion4. The cost of 4 pens and 4 pencils boxes is Rs.100. Three times the cost of a pen is Rs.15 more than the cost of a pencil box. Form the pair of Linear equations for the above situation. Find the cost of a pen and a pencil box.
Solution:
Let cost of a pen be Rs. x
And cost of a pencil box be Rs. y.
According to question,
4x + 4y = 100
⟹ x + y = 25…(i)
Also, 3x = y + 15
⟹ 3x – y = 15                          …(ii)
On adding equations (i) and (ii), we get
4x = 40
⟹ x = 10
By substituting x = 10, in equation (i) we get
y = 25 -10 = 15
Hence, the cost of a pen is Rs. 10 and that of a pencil box is Rs.15.
Quesntion5. Determine, algebraically, the vertices of the triangle formed by the lines 3 x – y = 3; 2x – 3y = 2 and x+ 2y = 8.
Solution:
Given equations are:
3x – 2y = 3                  …(i)
2x – 3y = 2                  …(ii)
x + 2y = 8                    …(iii)
Let lines (i), (ii) and (iii) represent the sides of a DABC i.e., AB, BC and CA, respectively.
On solving lines (i) and (ii), we will get the intersecting point B, solving (ii) and (iii) we will get C and solving (iii) and (i), will give point A.


⟹ y = 6 – 3
⟹ y = 3
So, the coordinate of point or vertex A is (2, 3).
Hence, the vertices of the ΔABC formed by the given lines are A (2, 3), B (1, 0) and C (4, 2)
Quesntion6. Ankita travels 14km to her home partly by rickshaw and partly by bus She takes half an hour, if she travels 2km by rickshaw and the remaining distance by bus. On the other hand, if she travels 4km by rickshaw and the remaining distance by bus, she takes 9min longer. Find the speed of the rickshaw and of the bus.
Solution:
Let the speed of the rickshaw and the bus are x and y km/h, respectively.



Hence, the speed of rickshaw is 10 km/h and that of bus is 40km/h.
Exercise 3.4
Long Answer Type QuestionsQuesntion (Q. No. 7 to 13)
7. A person, rowing at the rate of 5km/h in still water, takes thrice as much time in going 40km upstream as in going 40km downstream. Find the speed of the stream.
Solution:
Let the speed of the stream be x km/h.
Given, speed of person, rowing in still water = 5 km/h
The speed of a person rowing in downstream = (5 + x) km/h
and the speed of a person has rowing in upstream = (5 − x)km/h
Now, time taken to cover 40km downstream,

Hence, the speed of the stream is 2.5km/h.
Quesntion8. A motorboat can travel 30km upstream and 28km downstream in 7h. It can travel 21km upstream and return in 5h. Find the speed of the boat in still water and the speed of the stream.
Solution:
Let speed of the motorboat in still water = x km/h
And speed of the stream = y km/h
Then, speed of the motorboat in downstream = (x + y)km/h
And speed of the motorboat in upstream = (x – y) km/h.
Time taken by motorboat to travel 30 km upstream,




Hence, the speed of the motorboat in still water is 10km/h and the speed of the stream is 4km/h.
Quesntion9. A two-digit number is obtained by either multiplying the sum of the digits by 8 and then subtracting 5 or by multiplying the difference of the digits by 16 and then adding 3. Find the number.
Solution:
Let the two-digit number be = 10 x + y

Now, put the value of y in equation (i), we get
2x –7 × 3 = -5
⟹ 2x = 21 – 5 = 16
⟹ x = 8
Hence, the required two-digit number = 10x + y = 10 × 8 + 3 = 80 + 3 = 83
Quesntion10. A railway half ticket cost half the full fare but the reservation charges are the same on a half ticket as on a full ticket. One reserved first class ticket from the stations A to B costs Rs.2530. Also, one reserved first class ticket and one reserved first class half ticket from stations A to B costs Rs.3810. Find the full first class fare from stations A to B and also the reservation charges for a ticket.
Solution:
Let cost of full class fare be Rs x
Thus cost of half first class fare = Rs. x/2 , respectively
Again let reservation charges per ticket be Rs. y.

⟹ y = 2530-2500
⟹ y = 30
Hence, full first class fare from stations A to B is Rs. 2500 and the reservation for a ticket is Rs. 30.
Quesntion11. A shopkeeper sells a saree at 8% profit and a sweater at 10% discount thereby, getting a sum Rs.1008. If she had sold the saree at 10% profit and the sweater at 8% discount, she would have got Rs.1028 then find the cost of the saree and the list price (price before discount) of the sweater.
Solution:
Let the cost price of the saree be Rs. x
And list price of the sweater be Rs. y.


Therefore, the cost price of the saree and the list price (price before discount) of the sweater are Rs. 600 and Rs.400.
Quesntion12. Susan invested certain amount of money in two schemes A and B, which offer interest at the rate of 8% per annum and 9% per annum, respectively. She received Rs. 1860 as annual interest. However, had she interchanged the amount of investments in the two schemes, she would have received Rs. 20 more as annual interest. How much money did she invest in each scheme?
Solution:
Let the money invested in scheme A = Rs. x
Aand the money invested in scheme B = Rs. y
Case I:
Interest at the rate of 8% per annum on scheme A+ Interest at the rate of 9% per annum on scheme B = Total amount received

Case II:
Interest at the rate of 9% per annum on scheme A + Interest at the rate of 8% per annum on scheme B = Rs. 20 more than annual interest

Hence, the amount invested in schemes A and B are Rs.12000 and Rs. 10,000 respectively.
Quesntion13. Vijay had some bananas and he divided them into two lots A and B. He sold the first lot at the rate of Rs. 2 for 3 bananas and the second lot at the rate of Rs. 1 per banana and got a total of Rs. 400. If he had sold the first lot at the rate of Rs. 1 per banana and the second lot at the rate of Rs. 4 for 5 bananas, his total collection would have been Rs. 460. Find the total number of bananas he had.
Solution:
Let the number of bananas in lot A = x
and the number of bananas in lot B = y
Case I:
Cost of the first lot at the rate of Rs. 2 for 3 bananas + Cost of the second lot at the rate of Rs. 1 per banana = Amount received

Case II:
Cost of the first lot at the rate of Rs. 1 per banana + Cost of the second lot at the rate of Rs. 4 for 5 bananas = Amount received

Hence, Total number of bananas = x + y = 300 + 200 = 500 bananas.