43Pair of Linear Equations in Two Variables
MORE POINTS TO REMEMBER
System of linear equations: A pair of linear equations in two variables is said to be a
system of two linear equations in two variables.
For example, 2x + 3y = 5 and x + 2y = 3 are system of simultaneous linear equations in two
variables.
Solution of system of linear equations: A pair of values of x and y satisfying each one of
the equations in a given system of pair of linear equations in x and y is called solution of
the given system.
For example, x = –1 and y = 2 is solution of system of linear equations x + 2y = 3, 4x + 3y = 2,
as these value of x and y satisfy each of them
LHS = –1 + 2 × 2 = –1 + 4 = 3 = RHS
and LHS = 4 × (–1) + 3 × 2 = –4 + 6 = 2 = RHS
Consistent system of linear equations: A system of linear equations is said to be consistent,
if it has at least one solution.
Inconsistent system of linear equation: A system of linear equations is said to be
inconsistent, if it has no solution.
For example, 2x + 3y == 69 3 is inconsistent because there is no value of x and y which satisfy
4x + 6y
both equations of above system.
Multiple Choice Questions [1 mark]
Choose and write the correct option in the following questions.
1. If a pair of linear equations has infinitely many solutions, then the lines representing them
will be
(a) parallel (b) intersecting or coincident
(c) always intersecting (d) always coincident
2. If the lines given by 3 x + 2 ky = 2 and 2x + 5y + 1 = 0 are parallel, then value of k is
–45 (b) 25 15 3 [NCERT Exemplar]
(a) 4 2
(c) (d)
3. If am ≠ bl, then the system of equations, ax + by = c, lx + my = n
(a) has a unique solution (b) has no solution
(c) has infinitely many solutions (d) may or may not have a solution
4. If x = a, y = b is the solution of the equations x – y = 2 and x + y = 4, then the values of a and
b are, respectively [NCERT Exemplar]
(a) 3 and 5 (b) 5 and 3 (c) 3 and 1 (d) –1 and –3
5. 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. The present ages (in years) of the son and the father are, respectively.
[NCERT Exemplar]
(a) 4 and 24 (b) 5 and 30 (c) 6 and 36 (d) 3 and 24
44 Xam idea Mathematics–X
6. The pair of equations 5 x – 15y = 8 and 3x – 9 y = 24 has [NCERT Exemplar]
5
(a) one solution (b) two solutions (c) infinite solutions (d) no solution
7. The sum of the digits of a two-digit number is 9. If 27 is added to it, the digit of number get
reversed. The number is [NCERT Exemplar]
(a) 25 (b) 72 (c) 63 (d) 36
8. The value of k for which the system of equations 2x + ky = 12, x + 3y – 4 = 0 are inconsistent is
(a) 21 (b) 1 (c) 6 (d) 4
4 6 21
9. The pair of equations x = a and y = b graphically represents lines which are
(b) intersecting at (b, a) (c) coincident [NCERT Exemplar]
(a) parallel (d) intersecting at (a, b)
10. The pair of equations y = 0 and y = –7 has [NCERT Exemplar]
(a) one solution
(c) infinitely many solutions (b) two solutions
(d) no solution
11. If a pair of linear equations is consistent, then the lines will be [NCERT Exemplar]
(a) parallel (b) always coincident
(c) intersecting or coincident (d) always intersecting
12. Graphically, the pair of equations [NCERT Exemplar]
6x – 3y + 10 = 0; 2x – y + 9 = 0 represents two lines which are
(a) intersecting at exactly one point (b) intersecting at exactly two points
(c) coincident (d) parallel
13. If 2x – 3y = 7 and (a + b)x – (a + b – 3)y = 4a + b represent coincident lines, then a and b
satisfy the equation
(a) a + 5b = 0 (b) 5a + b = 0
(c) a – 5b = 0 (d) 5a – b = 0
14. Five years ago, A was thrice as old as B and ten years later, A shall be twice as old as B. What
is the present age of A?
(a) 20 (b) 50 x (c) 60 (d) 40
a y
15. The area of the triangle formed by the line + b = 1 with the coordinate axes is
(a) ab (b) 2ab (c) 1 ab (d) 1 ab
2 4
16. One equation of a pair of dependent linear equations is – 5x +7y = 2. The second equation can be
(a) 10x + 14y + 4 = 0 (b) –10x – 14y + 4= 0
(c) –10x + 14y + 4 = 0 (d) 10x – 14y = – 4
17. A pair of linear equations which has a unique solution x = 3, y = –2 is
(a) x + y=1 (b) 2x + 5y + 4 = 1 (c) 2x – y = 1 (d) x – 4y = 14
2x – 3y =12 4x + 10y + 8 = 0 3x + 2y = 0 5x – y = 13
18. The pair of equations x +2y + 5 = 0 and – 3x – 6y + 1 = 0 have [NCERT Exemplar]
(a) a unique solution (b) exactly two solutions
(c) infinitely many solutions (d) no solution
45Pair of Linear Equations in Two Variables
19. Gunjan has only ™ 1 and ™ 2 coins with her. If the total number of coins that she has is 50 and
the amount of money with her is ™ 75, then the number of ™ 1, and ™ 2 coins are respectively
(a) 25 and 25 (b) 15 and 35 (c) 35 and 15 (d) 35 and 20
20. For what value of k, do the equations 3x – y + 8 = 0 and 6x – ky = – 16 represent coincident
lines? [NCERT Exemplar]
(a) 1 (b) – 1 (c) 2 (d) –2
2 2
21. The value of c for which the pair of equation cx – y = 2 and 6x – 2y = 3 will have infinitely many
solutions is [NCERT Exemplar]
(a) 3 (b) –3 (c) –12 (d) no value
Answers 2. (c) 3. (a) 4. (c) 5. (c) 6. (c)
8. (c) 9. (d) 10. (d) 11. (c) 12. (d)
1. (d) 14. (b) 15. (c) 16. (d) 17. (a) 18. (d)
7. (d) 20. (c) 21. (d)
13. (c)
19. (a)
Fill in the Blanks [1 mark]
Complete the following statements with appropriate word(s) in the blank space(s).
1. An equation whose degree is one is known as a _________________ equation.
2. A pair of linear equations is _________________ if it has no solution.
3. If a pair of linear equations has a solution, either a unique or infinitely many, then it is said to be
_________________.
4. Every solution of a linear equation in two variables is a point on the _________________
representing it.
5. If a pair of linear equations has infinitely many solutions, then its graph is represented by a pair
of _________________ lines.
6. A pair of ________________ lines represent the pair of linear equations having no solution.
7. A pair of linear equations has _________________ solution(s) if it is represented by intersecting
lines graphically.
8. If a pair of linear equations is consistent, then the lines will be ________________.
9. Graphically, the pair of equations 6x – 3y + 10 = 0; 2x – y + 9 = 0 represents two lines which are
________________.
10. If two lines represent a pair of inconsistent linear equations then both lines must be
________________.
11. The graph of the equation x – y = 0 passes through the intersection of x-axis and ________________.
x y
12. The coordinates of the point where x-axis and the line 2 + 3 = 1 intersect are ________________.
13. If we draw x = a and y = b graphically then we get lines intersecting at ________________.
14. One of the common solution of ax + by = c and y axis is ________________.
15. For two linear equations a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0, a1 = b1 = c1 is the condition
for ________________. a2 b2 c2
46 Xam idea Mathematics–X
Answers
1. linear 2. inconsistent 3. consistent 4. line 5. coincident 6. parallel
7. unique 11. y-axis
12. (2, 0) 183.. i(na,tebr) secting1o4r. c(o0i,ncbcid) ent 9. parallel 10. parallel
15. infinite solutions [1 mark]
Very Short Answer Questions
1. If the lines given by 4x + 5ky = 10 and 3x + y + 1 = 0 are parallel, then find value of k.
Sol. Since the given lines are parallel.
\ 4 = 5k ! – 10 i.e., k = 4 .
3 1 1 15
2. Given the linear equation 2x + 3y – 8 = 0. Write another linear equation in two variables such
that the geometrical representation of the pair so formed are parallel lines.
Sol. For parallel lines, we have
a1 = b1 ! c1
a2 b2 c2
⇒ 2 = 3 ! –8 [Here, a1 = 2, b1 = 3, c1 = – 8]
a2 b2 c2
⇒ a2 should be equal to 2 and c2 can take any value.
b2 3
∴ For the lines to be parallel another possible linear equation can be 4x + 6y + 5 = 0.
However, more such equations are possible.
3. Do the equations 4x + 3y – 1 = 5 and 12x + 9y = 15 represent a pair of coincident lines?
Sol. \ HerGei,v en 14e2q=ua39tio !ns165do ni o.et.,r epre saae12n=t abbp12 a!ir c1 [NCERT Exemplar]
c2
of coincident lines.
4. Find the co-ordinate where the line x – y = 8 will intersect y-axis.
Sol. The given line will intersect y-axis when x = 0.
\ 0 – y = 8 ⇒ y = –8
Required coordinate is (0, –8).
5. Write the number of solutions of the following pair of linear equations:
x + 2y – 8 = 0, 2x + 4y = 16
Sol. Here, a1 = 1 , b1 = 2 = 1 , c1 = –8 = 1
a2 2 b2 4 2 c2 –16 2
Since a1 = b1 = c1
a2 b2 c2
\ The given pair of linear equations has infinitely many solutions.
Short Answer Questions-I [2 marks]
1. Is the following pair of linear equations consistent? Justify your answer.
2ax + by = a, 4ax + 2by – 2a = 0; a, b ≠ 0
Sol. Yes,
Here, aa12 = 2a = 1 , b1 = b = 1 , cc12 = –a = 1
4a 2 b2 2b 2 – 2a 2
Pair of Linear Equations in Two Variables 47
a1 = b1 = c1
a2 b2 c2
\ The given system of equations is consistent.
2. Find c if the system of equations [CBSE 2019 (30/1/1)]
cx + 3y + (3 – c) = 0; 12x + cy – c = 0 has infinitely many solutions.
Sol. Given system of equations,
cx + 3y + (3 – c) = 0 and 12x + cy – c = 0
For infinite solutions, we know that 3 3–c
a1 b1 c1 c c –c
a2 = b2 = c2 ⇒ 12 = =
Now, c = 3 and 3 = 3–c
12 c c –c
⇒ c2 = 36 – 3c = (3 – c) c
⇒ –3 = 3 – c
⇒ c=3+3=6
c=6
⇒ c = ! 6 ⇒
\ c = 6, for infinitely many solutions.
3. Find the value(s) of k so that the pair of equations x + 2y = 5 and 3x + ky + 15 = 0 has a unique
solution. [CBSE 2019 (30/2/1)]
Sol. Given equation:
x + 2y = 5 and 3x + ky + 15 = 0
⇒ x + 2y – 5 = 0 and 3x + ky + 15 =0
For unique solution
a1 b1
a2 ! b2 ⇒ 1 ! 2
3 k
Hence, k ! 6; k is any real value except 6.
4. Solve the following pair of linear equations. [CBSE 2019 (30/3/3)]
3x + 4y = 10 and 2x – 2y = 2
Sol. Given equations:
3x + 4y = 10 … (i)
2x – 2y = 2 … (ii)
Multiply (i) by 2 and (ii) by 3, we get
6x + 8y = 20
6x – 6y = 6
– +–
14y = 14
⇒ y = 1 ∴ 2x – 2(1) = 2 ⇒ 2x = 4 ⇒ x = 2
On comparing the ratios a1 , b1 and c1 , find out whether the following pair of linear equations
a2 b2 c2
is consistent or inconsistent. (Q. 5 and 6)
5. 32 x + 5 y = 7; 6. 4 x + 2y = 8
3 3
9x – 10y = 14 2x + 8y = 12 [NCERT Exemplar]
Sol. 5. We have, 3 x+ 5 y=7 …(i)
2 3
48 Xam idea Mathematics–X
9x – 10y = 14 …(ii)
Here, a1 = 3 , b1 = 5 , c1 =7
2 3
a2 = 9, b2 = –10, c2 = 14
Thus, =aa12 2=×3 9 1 =, bb12 3× (5–=10) – 1
6 6
Hence, a1 ≠ b1 . So, it has unique solution and is consistent.
a2 b2
Sol. 6. We have, 4 x + 2y=8 …(i)
3
2x + 3y = 12 …(ii)
4
Here, a1 = 3 , b1 = 2, c1 = 8
and a2 = 2, b2 = 3, c2 = 12
Thus, a1 = 3 4 2 = 2 ; b1 = 2 ; c1 = 8 = 2
a2 # 3 b2 3 c2 12 3
Since a1 = b1 = c1 , so equations (i) and (ii) represent coincident lines.
a2 b2 c2
Hence, the pair of linear equations is consistent with infinitely many solutions.
On comparing the ratios a1 , b1 and c1 , find out whether the lines representing the following
a2 b2 c2
pair of linear equations intersect at a point, are parallel or coincident: (Q. 7 to 9).
7. 5x – 4y + 8 = 0 8. 9x + 3y + 12 = 0
7x + 6y – 9 = 0 18x + 6y + 24 = 0
9. 6x – 3y + 10 = 0
2x – y + 9 = 0 [NCERT]
Sol. 7. We have, 5x – 4y + 8 = 0 …(i)
7x + 6y – 9 = 0 …(ii)
Here, a1 = 5, b1 = –4, c1 = 8
and, aaa212==775, b2 = 6, c2bb12==––9
equations and
Here, 4 = – 2
. 6 3
b1
Since a1 ≠ b2 So, (i) and (ii) represent intersecting lines.
a2
Sol. 8. We have, 9x + 3y + 12 = 0 …(i)
18x + 6y + 24 = 0 …(ii)
Here, a1 = 9, b1 = 3, c1 = 12
and a2 = 18, b2 = 6, c2 = 24
a1 = 9 = 1 ; b1 = 3 = 1 ; c1 = 12 = 1
a2 18 2 b2 6 2 c2 24 2
Here, a1 = b1 = c1 , so equations (i) and (ii) represent coincident lines.
a2 b2 c2
49Pair of Linear Equations in Two Variables
Sol. 9. We have, 6x – 3y + 10 = 0 …(i)
2x – y + 9 = 0 …(ii)
Here, a1 = 6, b1 = –3, c1 = 10
and a2 = 2, b2 = –1, c2 = 9
Since, a1 = 6 = 3, b1 = – 3 = 3, c1 = 10
a2 2 b2 – 1 c2 9
a1 = b1 ≠ c1
a2 b2 c2
So, equations (i) and (ii) represent parallel lines.
10. A result of global warming is that the ice of some glaciers is melting. Twelve years after the ice
disappears and tiny plants called lichen, start to grow on the rocks.
Each lichen grows approximately in the shape of a circle.
The relationship between the diameter of this circle and the age of the lichen can be
approximated with the formula:
d = 7.0× (t – 12), for t ≥ 12
Where d represents the diameter of the lichen in millimetre, and t represents the number of
years after the ice has disappeared.
(a) Using the formula, calculate the diameter of the lichen, 16 years after the ice disappeared.
Show your calculation.
(b) Reshma measured the diameter of some lichen and found it was 35 millimetre. How many
years ago did the ice disappear at this spot? Show your calculation.
Sol. (a) By given formula,
d = 7.0× (t – 12), for t ≥ 12
⇒ d = 7.0× (16 – 12)
⇒ d = 7.0× 4
⇒ d = 7.0 × 2
⇒ d = 14 mm
(b) By given formula,
d = 7.0× (t – 12), for t ≥ 12
⇒ 35 = 7.0× (t – 12)
⇒ 5= t – 12
⇒ 25= t – 12
⇒ t = 25 + 12
⇒ t = 37years
⇒ The age of lichen is 37 years.
⇒ The ice disappeared at this spot 37 years ago.
Short Answer Questions-II [3 marks]
1. Solve: ax + by = a – b and bx – ay = a + b
Sol. The given system of equations may be written as
ax + by – (a – b) = 0
bx – ay – (a + b) = 0
50 Xam idea Mathematics–X
By cross-multiplication, we have
b x – b) = a –y – b) = a 1
– (a – (a b
– a – (a + b) b – (a + b) b –a
⇒ x = –y = 1
⇒ b # {– (a + b)} – (–a) # {– (a – b)} a # {– (a+ b)} – b # {– (a – b)} –a2 – b2
⇒
⇒ – b (a+ x a (a – b) = – a (a+ –y b (a – b) = 1 b2)
b) – b) + – (a2 +
– x a2 = – – y b2 = 1 b2)
b2 – a2 – – (a2 +
y
x b2 ) = (a2 b2 ) = 1 b2 )
– (a2 + + – (a2 +
⇒ x=– (a2+ b2 ) =1 and y= (a2+ b2 ) ) = –1
– (a2 + b2 ) – (a2 + b2
Hence, the solution of the given system of equations is x = 1, y = –1
2. Solve the following linear equations:
152x – 378y = –74 and –378x + 152y = –604 [NCERT]
Sol. We have, 152x – 378y = –74 ...(i)
...(ii)
–378x + 152y = –604
Adding equation (i) and (ii), we get
152x – 378y = –74
–_3__7_8_x_+__1__5_2_y_=__–_6_0_4__
–226x – 226y = –678
⇒ –226(x +y) = –678
⇒ x+y = – 678
⇒ x+y – 226
= 3 …(iii)
Subtracting equation (ii) from (i), we get
152x – 378y = –74
–378x + 152y = –604
+– +
530x – 530y = 530
⇒ x – y = 1 …(iv)
Adding equations (iii) and (iv), we get
x + y = 3
x– y=1 ⇒ x = 2
2x = 4
Putting the value of x in (iii), we get
2 + y = 3 ⇒ y = 1
Hence, the solution of given system of equations is x = 2, y = 1.
51Pair of Linear Equations in Two Variables
3. Solve for x and y
b x üüabü y a2 b2 x y ab
a
Sol. We have, b x + a y= a2 + b2 …(i)
a b
x + y = 2ab …(ii)
Multiplying (ii) by b/a, we get
b x + b y= 2b2 …(iii)
a a
Subtracting (iii) from (i), we get
c a – b m y = a2 + b2 – 2b2 ⇒ d a2a–bb2 ny=(a2 – b2)
b a
⇒ y=(a2 – b2) # ab ⇒ y = ab
(a2 – b2)
Putting the value of y in (ii), we get
x + ab = 2ab ⇒ x = 2ab – ab ⇒ x = ab
\ x = ab, y = ab
4. (i) For which values of a and b does the following pair of linear equations have an infinite
number of solutions?
2x + 3y = 7
(a – b)x + (a + b)y = 3a + b – 2
(ii) For which value of k will the following pair of linear equations have no solution?
3x + y = 1
(2k – 1)x + (k – 1)y = 2k + 1 [NCERT]
Sol. (i) We have, 2x + 3y = 7 …(i)
…(ii)
(a – b) x + (a + b) y = 3a + b – 2
Here, a1 = 2, b1 = 3, c1 = 7
and a2 = a – b, b2 = a + b, c2 = 3a + b – 2
For infinite number of solutions, we have
a1 b1 c1
a2 = b2 = c2 ⇒ 2 b = 3 = 7 – 2
a– a+b 3a + b
Now, 2 b = 3
a– a+b
⇒ 2a + 2b = 3a – 3b ⇒ 2a – 3a = –3b – 2b
⇒ – a = – 5b …(iii)
…(iv)
\ a = 5b
Again, we have
3 = 7 – 2 ⇒ 9a + 3b – 6 = 7a + 7b
a+b 3a + b
⇒ 9a – 7a + 3b – 7b – 6 = 0 ⇒ 2a – 4b – 6 = 0 ⇒ 2a – 4b = 6
⇒ a – 2b = 3
Putting a = 5b in equation (iv), we get
5b – 2b = 3 or 3b = 3 i.e., b= 3 =1
3
Putting the value of b in equation (iii), we get a = 5(1) = 5
52 Xam idea Mathematics–X
Hence, the given system of equations will have an infinite number of solutions for a = 5
and b = 1.
(ii) We have, 3x + y = 1 ⇒ 3x + y – 1 = 0 …(i)
(2k – 1) x + (k – 1) y = 2k + 1
⇒ (2k – 1) x + (k – 1) y – (2k + 1) = 0 …(ii)
Here, a1 = 3, b1 = 1, c1 = –1
a2 = 2k – 1, b2 = k – 1, c2 = – (2k + 1)
For no solution, we must have
a1 = b1 ≠ c1 ⇒ a1 = b1 ≠ 1
Now, a2 b2 c2 a2 b2 2k +1
3 = k 1 1 ⇒ 3k – 3 = 2k – 1
2k – 1 –
⇒ 3k – 2k = 3 – 1 ⇒ k=2
Hence, the given system of equations will have no solutions for k = 2.
5. Find whether the following pair of linear equations has a unique solution. If yes, find the
solution.
7x – 4y = 49 and 5x – 6y = 57
Sol. We have, 7x – 4y = 49 ...(i)
and 5x – 6y = 57 ...(ii)
Here,
a1 = 7, b1 = –4, c1 = 49
a2 = 5, b2 = –6, c2 = 57
So,
a1 = 7 , b1 = −4 = 2
a2 5 b2 −6 3
Since, a1 ≠ b1
a2 b2
So, system has a unique solution.
Multiply equation (i) by 5 and equation (ii) by 7 and subtract
35x – 20y = 245
–35x +– 42y =– 399
22y = –154 ⇒ y = –7
Put y = –7 in equation (ii)
5x –6(–7) = 57 ⇒ 5x = 57 – 42 ⇒ x = 3
Hence, x = 3 and y = –7.
6. Solve the following pair of equations for x and y.
ax2 − b2 = 0; a2b + b2a = a+ b, x ≠ 0, y ≠ 0
y x y
Sol. a2 – b2 = 0 ...(i)
x y
a2 b + b2 a = a + b ...(ii)
x y
Multiply equation (i) by a and adding to equation (ii)
a2 a – b2 a + a2 b + b2 a = 0 + ^a + bh
x y x y
53Pair of Linear Equations in Two Variables
⇒ a3 + a2 b = a+b ⇒ a2 ^ a + bh = a+b & x = a2 (a + b) = a2
x x x a+b
Putting the value of x in equation (i), we get
a2 – b2 =0 ⇒ 1– b2 = 0 ⇒ b2 =1 ⇒ y = b2
a2 y y y
Hence, x = a2, y = b2.
7. The larger of two supplementary angles exceeds the smaller by 18°. Find the angles.
[CBSE 2019 (30/2/1)]
Sol. Let the larger supplementary angle be x.
∴ Other angle = 180° – x
According to question,
x = 180° – x + 18°
2x = 198°
x = 99°
∴ Supplementary angles are 99°, 81°.
8. A father's age is three times the sum of the ages of his two children. After 5 years his age will
be two times the sum of their ages. Find the present age of father. [CBSE 2019 (30/1/1)]
Sol. Let father's age be x years and sum of ages of his children's ages be y years.
∴ x = 3y ...(i)
After 5 years
Father’s age = (x + 5) years
Sum of his children’s age = y + 5 + 5 = (y + 10) years
∴ According to question,
x + 5 = 2(y + 10)
⇒ x + 5 = 2y + 20
⇒ 3y + 5 = 2y + 20 [From (i)] ⇒ 3y – 2y = 20 – 5
⇒ y = 15
Putting y = 15 in equation (i), we get
x = 3y = 3 × 15 = 45
∴ Father’s age = 45 years
9. A fraction becomes 1 when 2 is subtracted from the numerator and it becomes 1 when 1 is
3 2
subtracted from its denominator. Find the fraction. [CBSE 2019 (30/1/1)]
Sol. Let numerator of fraction be x and denominator be y.
x
∴ Fraction = y
According to question,
x–2 = 1 ⇒ 3x – 6 = y
y 3 ⇒ 3x – y = 6
…(i)
Again,
x = 1 ⇒ 2x = y – 1
y –1 2 ⇒ 2x – y = –1
…(ii)
From equation (i) and (ii), we have
54 Xam idea Mathematics–X
3x – y = 6
2x – y = –1
–+ +
x=7
⇒ Now, putting x = 7 in equation (i), we have
3 × 7 – y = 6 ⇒ 21 – 6 = y
⇒ 15 = y ⇒ y = 15
∴ Required fraction is 7 .
15
10. Solve the following pairs of equations by reducing them to a pair of linear equations:
(i) 7x – 2y = 5 (ii) 1 y + 1 y = 1
xy 3x+ 3x - 4
8x+xy7y =15 1 y) – 1 y) = –1 [NCERT]
2(3x + 2(3x – 8
Sol. (i) We have
7x – 2y =5 ⇒ 7x – 2y =5 ⇒ 7 – 2 =5
xy xy xy y x
And, 8x +7y =15 ⇒ 8x + 7y =15 ⇒ 8 + 7 =15
Let xy and xy xy y x
1 = u 1 = v
y x
7u – 2v = 5 …(i)
8u + 7v = 15 …(ii)
Multiplying (i) by 7 and (ii) by 2 and adding, we have
49u – 14v = 35
16u + 14v = 30
65u = 65
∴ u= 65 =1
65
Putting the value of u in equation (i), we have
7 × 1 – 2v = 5 ⇒ –2v = 5 – 7 = – 2
\ ⇒
–2v = –2 v= – 2 =1
Here ⇒ – 2
u=1 ⇒ 1 =1 y=1
y
and v=1 ⇒ 1 =1 ⇒ x=1
x
Hence, the solution of given system of equations is x = 1, y = 1.
(ii) We have, 1 + 1 y = 3
3x + y 3x – 4
1 y) – 1 y) = – 1
Let 2 (3x + 2 (3x – 8
1 y = u and 1 =v
3x + 3x – y
We have, u + v = 3 …(i)
4
55Pair of Linear Equations in Two Variables
u – v = – 1
⇒ 2 2 8
u– v =– 1 ⇒ u–v= – 2 = – 1
2 8 8 4
\ u–v=– 1 …(ii)
4
Adding (i) and (ii), we have
u+v= 3
4
u –v= – 1
4
2u = 3 – 1 = 3–1 = 2
⇒ 4 4 4 4
u= 2 = 1 \ u = 1
4#2 4 4
Now putting the value of u in equation (i), we have
1 +v= 3 ⇒ v = 3 – 1 = 3 – 1 = 2 = 1 ⇒ v = 1
Here, 4 4 4 4 4 4 2 2
and
u= 1 ⇒ 1 y = 1 ⇒ 3x + y = 4 …(iii)
4 3x + 4
v = 1 ⇒ 1 y = 1 ⇒ 3x – y = 2 ...(iv)
2 3x – 2
Now, adding (iii) and (iv), we have
3x + y = 4
3x – y = 2
6x = 6
\ x = 6 =1
6
Putting the value of x in equation (iii), we have
3×1+y=4
⇒ y=4–3=1
Hence, the solution of given system of equations is x = 1, y = 1.
11. Mohan and Sohan are two farmers in a village, both have their own plot for agriculture,
adjacent to each other shown below:
Mohan’s Sohan’s
plot plot
100 m 150 m
Mohan is the owner of rectangular plot whose perimeter is 100 metres and Sohan is also the
owner of a rectangular plot whose perimeter is 150 m and this plot has length twice that of
Mohan and breadth is 5 metres more than that of Mohan. Now answer the following questions.
56 Xam idea Mathematics–X
(i) Write the linear equations for both the plot.
(ii) Write the dimension of Mohan’s plot.
(iii) Write the dimension of Sohan’s plot.
Sol. Let x m and y m be the length and breadth of the rectangular plot of Mohan.
Therefore, 2x m be the length and (y + 5) m be the breadth of the rectangular plot of Sohan.
(i) Perimeter of the plot of Mohan = 100 m
⇒ 2(x + y) = 100 ⇒ x + y = 50
and, perimeter of the plot of Sohan = 150 m
2(2x + y + 5) = 150 ⇒ 2x + y + 5 = 75 ⇒ 2x + y = 70
(ii) We have, pair of linear equations
x + y = 50 ...(i)
2–x +– y = –70 ...(ii)
On subtracting – x = –20
⇒ x = 20 m
Putting the value of x in equation (i), we get
20 + y = 50 ⇒ y = 50 – 20 = 30
∴ y = 30 m
∴ Dimension of plot Mohan have 20 m × 30 m.
(iii) Now, length for Sohan’s plot = 2x = 2 × 20 = 40 m
and breadth for Sohan’s plot = y + 5 = 30 + 5 = 35 m
∴ Dimension of Sohan’s plot = 40 m by 35 m.
12. Vijay studying in class X participated in a periodic test having multiple choice questions.
Vijay scored 40 marks in the test by getting 3 marks for each right answer and losing 1 mark
for each wrong answer. Had 4 marks been awarded for each correct answer and 2 marks been
deducted for each incorrect answer, then Vijay would have scored 50 marks?
How many questions were there in the test?
Sol. Let Vijay answered x correct and y incorrect.
Therefore, total number of questions = x + y
Therefore According to question, 3x – y = 40 ...(i)
Also, 4x – 2y = 50
⇒ 2 (2x – y) = 50
⇒ 2x – y = 25 ...(ii)
From equations (i) and (ii), we have
3x – y = 40
– 2x+ y = 25
On subtracting x = 15
Putting x = 15 in equation (ii), we get
2 × 15 – y = 25 ⇒ –y = 25 – 30 = –5
⇒ y = 5
∴ Total number of questions = 15 + 5 = 20
57Pair of Linear Equations in Two Variables
Long Answer Questions [5 marks]
1. Form the pair of linear equations in this problem, and find its solution graphically :
10 students of Class X took part in a Mathematics quiz. If the number of girls is 4 more than
the number of boys, find the number of boys and girls who took part in the quiz. [NCERT]
Sol. Let x be the number of girls and y be the number of boys.
According to question, we have
x = y + 4
⇒ x – y = 4 …(i)
Again, total number of students = 10
Therefore, x + y = 10 …(ii)
From equation (i), we have the following table:
x0 4 7
y –4 0 3
From equation (ii), we have the following table: 10 7
x0
y 10 0 3
Plotting this, we have
Fig. 3.1
Here, the two lines intersect at point (7, 3) i.e., x = 7, y = 3.
So, the number of girls = 7
and number of boys = 3.
58 Xam idea Mathematics–X
2. Show graphically the given system of equations
2x + 4y = 10 and 3x + 6y = 12 has no solution.
Sol. We have, 2x + 4y = 10
⇒ 4y = 10 – 2x ⇒ y= 5 – x
Thus, we have the following table: 2
x1 3 5
0
y2 1
Plot the points A(1, 2), B(3, 1) and C(5, 0) on the graph paper. Join A, B and C and extend it on
both sides to obtain the graph of the equation 2x + 4y = 10.
We have, 3x + 6y = 12
⇒ 6y = 12 – 3x ⇒ y= 4– x
2
Thus, we have the following table:
x2 0 4
y1 2 0
Plot the points D (2, 1), E (0, 2) and F (4, 0) on the same graph paper. Join D, E and F and extend
it on both sides to obtain the graph of the equation 3x + 6y = 12.
Fig. 3.2
We find that the lines represented by equations 2x + 4y = 10 and 3x + 6y = 12 are parallel. So,
the two lines have no common point. Hence, the given system of equations has no solution.
3. Solve the following pairs of linear equations by the elimination method and the substitution
method:
(i) 3x – 5y – 4 = 0 and 9x = 2y + 7 (ii) x + 2y = −1 and x − y = 3 [NCERT]
2 3 3
Sol. (i) We have, 3x – 5y – 4 = 0 ⇒ 3x – 5y = 4 …(i)
Again, 9x = 2y + 7 ⇒ 9x – 2y = 7 …(ii)
59Pair of Linear Equations in Two Variables
By elimination method:
Multiplying equation (i) by 3, we get
9x – 15y = 12 …(iii)
Subtracting (ii) from (iii), we get
9x – 15y = 12
– 9x+– 2y =– 7
–13y = 5
⇒ y= – 5
13
Putting the value of y in equation (ii), we have
9x – 2c – 5 m= 7 ⇒ 9x + 10 =7 ⇒ 9x =7 – 10
13 13 13
⇒ 9x = 91 – 10 ⇒ 9x = 81 ⇒ x = 9
13 13 13
Hence, the required solution is x = 9 , y= – 5
13 13
By substitution method:
Expressing x in terms of y from equation (i), we have
4 + 5y
x = 3
Substituting the value of x in equation (ii), we have
9 # d 4 + 5y n – 2y = 7
3
⇒ 3 × (4 + 5y) – 2y = 7
⇒ 12 + 15y – 2y = 7 ⇒ 13y = 7 – 12
\ y= – 5
13
Putting the value of y in equation (i), we have
3x − 5× − 5 = 4 ⇒ 3x + 25 =4
13 13
⇒ 3x = 4 – 25 ⇒ 3x = 27
13 13
\ x = 9
13
Hence, the required solution is x = 9 , y=– 5 .
13 13
(ii) We have, x + 2y = –1 ⇒ 3x + 4y = – 1
2 3 6
\ 3x + 4y = –6 ...(i)
...(ii)
and x– y =3 ⇒ 3x – y =3
3 3
\ 3x – y = 9
By elimination method:
Subtracting (ii) from (i), we have 15
5
5y = –15 or y= – = – 3
60 Xam idea Mathematics–X
Putting the value of y in equation (i), we have
3x + 4 × (–3) = – 6 ⇒ 3x – 12 = – 6
\ 3x = – 6 + 12 ⇒ 3x = 6
\ x = 6 =2
3
Hence, solution is x = 2, y = – 3.
By substitution method:
Expressing x in terms of y from equation (i), we have
x= – 6– 4y
3
Substituting the value of x in equation (ii), we have
3#d – 6– 4y n – y=9 ⇒ – 6 – 4y – y = 9 ⇒ – 6 – 5y = 9
3
\ –5y = 9 + 6 = 15
y= 15 = –3
–5
Putting the value of y in equation (i), we have
3x + 4 × (–3) = – 6 ⇒ 3x – 12 = – 6
\ 3x = 12– 6 = 6 \ x = 6 =2
3
Hence, the required solution is x = 2, y = –3.
Form the pair of linear equations in the following problems and find their solutions (if they
exist) by any algebraic method (Q. 4 to 7):
4. A part of monthly hostel charges is fixed and the remaining depends on the number of days
one has taken food in the mess. When a student A takes food for 20 days, she has to pay `1000
as hostel charges whereas a student B, who takes food for 26 days, pays `1180 as hostel
charges. Find the fixed charges and the cost of food per day. [NCERT]
Sol. Let the fixed charge be `x and the cost of food per day be `y.
Therefore, according to question,
x + 20y = 1000 …(i)
x + 26y = 1180 …(ii)
Now, subtracting equation (ii) from (i), we have
x + 20y = 1000
x + 26y = 1800
–– –
– 6y = – 180
y= –180 = 30
–6
Putting the value of y in equation (i), we have
x + 20 × 30 = 1000 ⇒ x + 600 = 1000 ⇒ x = 1000 – 600 = 400
Hence, fixed charge is `400 and cost of food per day is `30.
5. Yash scored 40 marks in a test, getting 3 marks for each right answer and losing 1 mark
for each wrong answer. Had 4 marks been awarded for each correct answer and 2 marks
been deduced for each incorrect answer, then Yash would have scored 50 marks. How many
questions were there in the test? [NCERT]
Sol. Let x be the number of questions of right answer and y be the number of questions of wrong answer.
\ According to question,
61Pair of Linear Equations in Two Variables
3x – y = 40 … (i)
…(ii)
and 4x – 2y = 50
or 2x – y = 25
Subtracting (ii) from (i), we have
3x – y = 40
2x – y = 25
–+ –
x =15
Putting the value of x in equation (i), we have
3 × 15 – y = 40 ⇒ 45 – y = 40
\ y = 45 – 40 = 5
Hence, total number of questions is x + y i.e., 5 + 15 = 20.
6. Places A and B are 100 km apart on a highway. One car starts from A and another from B at the
same time. If the cars travel in the same direction at different speeds, they meet in 5 hours. If they
travel towards each other, they meet in 1 hour. What are the speeds of the two cars? [NCERT]
Sol. Let the speed of two cars be x km/h and y km/h respectively.
Case I: When two cars move in the same direction, they will meet each other at P after 5 hours.
Fig. 3.3
The distance covered by car from A = 5x (Distance = Speed × Time)
and distance covered by the car from B = 5y
\ 5x – 5y = AB = 100 ⇒ x – y = 100
5
\ x – y = 20 …(i)
Case II: When two cars move in opposite direction, they will meet each other at Q after one hour.
Fig. 3.4
The distance covered by the car from A = x
The distance covered by the car from B = y
\ x + y = AB = 100 ⇒ x + y = 100 …(ii)
Now, adding equations (i) and (ii), we have
2x = 120 ⇒ x= 120 = 60
2
Putting the value of x in equation (i), we get
60 – y = 20 ⇒ – y = – 40 \ y = 40
Hence, the speeds of two cars are 60 km/h and 40 km/h respectively.
7. The area of a rectangle gets reduced by 9 square units, if its length is reduced by 5 units and
breadth is increased by 3 units. If we increase the length by 3 units and the breadth by 2 units,
the area increases by 67 square units. Find the dimensions of the rectangle. [NCERT]
Sol. Let the length and breadth of a rectangle be x and y respectively.
Then area of the rectangle = xy
According to question, we have
62 Xam idea Mathematics–X
(x – 5) (y + 3) = xy – 9 ⇒ xy + 3x – 5y – 15 = xy – 9
⇒ 3x – 5y = 15 – 9 = 6 ⇒ 3x – 5y = 6 …(i)
Again, we have
(x + 3) (y + 2) = xy + 67 ⇒ xy + 2x + 3y + 6 = xy + 67
⇒ 2x + 3y = 67 – 6 = 61 ⇒ 2x + 3y = 61 …(ii)
Now, from equation (i), we express the value of x in terms of y as
x = 6 + 5y
3
Substituting the value of x in equation (ii), we have
2 # d 6 + 5y n+ 3y = 61 ⇒ 12 +10y + 9y = 61
3 3
171
⇒ 19y = 183 – 12 = 171 ⇒ y = 19 = 9
Putting the value of y in equation (i), we have
3x – 5 × 9 = 6 ⇒ 3x = 6 + 45 = 51
\ x= 51 =17
3
Hence, the length of rectangle = 17 units and breadth of rectangle = 9 units.
8. Roohi travels 300 km to her home partly by train and partly by bus. She takes 4 hours if
she travels 60 km by bus and the remaining by train. If she travels 100 km by bus and the
remaining by train, she takes 10 minutes longer. Find the speed of the train and the bus
separately. [NCERT]
Sol. Let the speed of the bus be x km/h and speed of the train be y km/h.
According to question, we have
60 + 240 = 4
x y
And 100 + 200 = 4 + 10 = 4 + 1 = 25 ⇒ 100 + 200 = 25
x y 60 6 6 x y 6
Now, let 1 =u and 1 = v
x y
\ 60u + 240v = 4 …(i)
100u + 200v = 25 …(ii)
6
Multiplying equation (i) by 5 and (ii) by 6 and subtracting, we have
300u + 1200v = 20
6 00u +1200v = 25
−−
−
−300u =−5
u = −5 = 1
−300 60
Putting the value of u in equation (i), we have
60 # 1 + 240v= 4 ⇒ 240v = 4 – 1 = 3
60
\ v = 3 = 1
240 80
Now, u= 1 ⇒ 1 = 1 \ x = 60
60 x 60
63Pair of Linear Equations in Two Variables
and v= 1 ⇒ 1 = 1 \ y = 80
80 y 80
Hence, speed of the bus is 60 km/h and speed of the train is 80 km/h.
9. The sum of a two digit number and the number formed by interchanging its digits is 110. If
10 is subtracted from the first number, the new number is 4 more than 5 times the sum of the
digits in the first number. Find the first number.
Sol. Let the digits at unit and tens places be x and y respectively.
Then, number = 10y + x ...(i)
Number formed by interchanging the digits = 10x + y
According to the given condition, we have
(10y + x) + (10x + y) = 110
⇒ 11x + 11y = 110 ⇒ x + y – 10 = 0 ...(ii)
Again, according to question, we have
(10y + x) – 10 = 5 (x + y) + 4
⇒ 10y + x – 10 = 5x + 5y + 4
⇒ 10y + x – 5x – 5y = 4 + 10
5y – 4x = 14 or 4x – 5y + 14 = 0 ...(iii)
By using cross-multiplication in equations (ii) and (iii), we have
–y
1 # 14 x # (–10) = 1 # 14 – 4# (–10) = 1# (– 1 1# 4
– (– 5) 5) –
–y –y
⇒ 14 x 50 = 14 + 40 = – 1 4 & – x = 54 = 1
– 5– 36 –9
⇒ x= – 36 and y= – 54 ⇒ x = 4 and y = 6
–9 –9
Putting the values of x and y in equation (i), we get
First number = 10 × 6 + 4 = 64.
10. Jamila sold a table and a chair for ™1050, thereby making a profit of 10% on the table and 25%
on the chair. If she had taken a profit of 25% on the table and 10% on the chair she would have
got ™1065. Find cost price of each. [NCERT Exemplar]
Sol. Let cost price of table be ™x and the cost price of the chair be ™y.
The selling price of the table, when it is sold at profit of 10% = ™ c x + 10x m =™ 110x
100 100
The selling price of the chair when it is sold at a profit of 25% = ™ e y + 25y o = ™ 125y
100 100
125y
So, 110x + 100 =1050 ...(i)
100
When the table is sold at a profit of 25%
Its selling price = ™ c x + 25 x m = ™ 125 x
100 100
When the chair is sold at a profit of 10%
Selling price = ™ d y+ 10y n = ™ 110y
100 100
So, 125 x + 110y =1065 ...(ii)
100 100
From equation (i) and (ii), we get
64 Xam idea Mathematics–X
110x + 125y = 105000
and 125x + 110y = 106500
On adding and subtracting these equations, we get
235x + 235y = 211500
and 15x – 15y = 1500
i.e., x + y = 900 ...(iii)
...(iv)
x – y = 100
Solving equation (iii) and (iv), we get
x = 500, y = 400
So, the cost price of the table is ™500 and the cost price of the chair is ™400.
11. In a residential school of Delhi, Vijay and Subhash are studying in class X. The rule of fee
structure of hostel charges of this school is as follow:
A part of monthly hostel charges is fixed and the remaining depends on the number of days
one has taken food in the mess. When Vijay takes food for 20 days then he has to pay ™1000 as
hostel charges whereas Subhash takes food for 26 days and has to pay ™1180 as hostel charges.
(i) Find the fixed charge.
(ii) Find the cost of food per day.
Sol. Let fixed charge be ™x and charge per day for the food in the mess be ™y.
∴ Hostel charge for Vijay = x + 20y
1000 = x + 20y
∴ x + 20y = 1000 ...(i)
Also, hostel charge for Subhash = x + 26y
⇒ 1180 = x + 26y
∴ x + 26y = 1180 ...(ii)
From equation (i) and (ii), we have
x + 20y = 1000
– x + 26y = 1180 ∴ y = 30
On subtracting – 6y = –180
Putting the value of y = 30 in equation (i), we get
x + 20 × 30 = 1000 ⇒ x = 1000 – 600 = 400
∴ x = 400
Thus, (i) Fixed charge = ™400
(ii) Cost of food per day = ™30.
HOTS [Higher Order Thinking Skills]
1. 8 men and 12 boys can finish a piece of work in 10 days while 6 men and 8 boys can finish it
in 14 days. Find the time taken by one man alone and that by one boy alone to finish the work.
Sol. Let one man alone can finish the work in x days and one boy alone can finish the work in y days.
Then, One day work of one man = 1 , One day work of one boy = 1
x y
\ One day work of 8 men = 8 , One day work of 12 boys = 12
x y
Since 8 men and 12 boys can finish the work in 10 days
10c 8 + 12 m =1 ⇒ 80x + 120 =1 ...(i)
x y x y
65Pair of Linear Equations in Two Variables
Again, 6 men and 8 boys can finish the work in 14 days
\ 14c 6 + 8 m =1 ⇒ 84 + 112 =1 ...(ii)
x y x y
Put 1 = u and 1 = v in equations (i) and (ii), we get
x y
80u + 120v – 1 = 0 and 84u + 112v – 1 = 0
By using cross-multiplication, we have
120 # (–1) u = –v = 1 84 #120
– 112 # (–1) 80 # (– 1) –84 # (–1) 80 #112 –
⇒ u = – –v = 8960 1 ⇒ u = –v = 1
–120 +112 80 +84 – 10080 –8 4 –1120
⇒ u= –8 = 1 and v= –4 = 1
– 1120 140 –1120 280
We have, u = 1 ⇒ 1 = 1 ⇒ x = 140
140 x 140
and v= 1 ⇒ 1 = 1 ⇒ y = 280.
280 y 280
Hence, one man alone can finish the work in 140 days and one boy alone can finish the work in
280 days..
2. A boat goes 30 km upstream and 44 km downstream in 10 hours. In 13 hours, it can go 40 km
upstream and 55 km downstream. Determine the speed of the stream and that of the boat in
still water. [CBSE 2019(30/1/1)]
Sol. Let the speed of the boat be x km/h and speed of the stream be y km/h.
\ Speed of boat in upstream = (x – y) km/h
and speed of boat in downstream = (x + y) km/h
\ According to question,
30 + 44 = 10 ...(i)
x– y x+y
and 40 + 55 = 13 ...(ii)
x– y x+y
(i) × 4 – (ii) × 3 gives
x 1 y (176 – 165) = 1
+
⇒ 11 = x + y ⇒ x + y = 11 ...(iii)
Putting the value of x + y = 11 in equation (i), we have
30 + 44 = 10 & 30 + 4 = 10
x– y 11 x– y
⇒ 30 = 10 – 4 = 6 ⇒ x – y = 5 ...(iv)
x– y x=8
From equation (iii) and (iv), we have
x + y = 11
x– y=5
2x = 16 ⇒
66 Xam idea Mathematics–X
Putting x = 8 in equation (iii), we have
8 + y = 11 ⇒ y = 11 – 8 = 3 ⇒ y = 3
\ Speed of boat is 8 km/h and speed of stream is 3 km/h.
3. Students of a class are made to stand in rows. If one student is extra in each row, there would
be 2 rows less. If one student is less in each row, there would be 3 rows more. Find the number
of students in the class.
Sol. Let total number of rows be y and total number of students in each row be x.
\ Total number of students = xy
Case I: If one student is extra in each row, there would be two rows less.
Now, number of rows = (y – 2)
Number of students in each row = (x + 1)
Total number of students = number of rows × number of students in each row
xy = (y – 2)(x + 1) ⇒ xy = xy + y – 2x – 2
⇒ xy – xy – y + 2x = –2 ⇒ 2x – y = –2 …(i)
Case II: If one student is less in each row, there would be 3 rows more.
Now, number of rows = (y + 3)
and number of students in each row = (x – 1)
Total number of students = number of rows × number of students in each row
\ xy = (y + 3)(x – 1) ⇒ xy = xy – y + 3x – 3
– 3x + y = – 3
xy – xy + y – 3x = –3 ⇒ …(ii)
On adding equations (i) and (ii), we have
2x – y = – 2
– 3x + y = – 3
– x=–5
or x=5
Putting the value of x in equation (i), we get
2(5) – y = –2 ⇒ 10 – y = –2
– y = –2 – 10 ⇒ – y = – 12 or y = 12
\ Total number of students in the class = 5 × 12 = 60.
4. Draw the graph of 2x + y = 6 and 2x – y + 2 = 0. Shade the region bounded by these lines and
x-axis. Find the area of the shaded region.
Sol. We have, 2x + y = 6, ⇒ y = 6 – 2x
When x = 0, we have y = 6 – 2 × 0 = 6
When x = 3, we have y = 6 – 2 × 3 = 0
When x = 2, we have y = 6 – 2 × 2 = 2
When x = 1, we have y = 6 – 2 × 1 = 4
Thus, we get the following table:
x0 3 2 1
y6 0 2 4
Now, we plot the points A(0, 6), F(1, 4) C(2, 2) and B(3, 0) on the graph paper. We join A, B and
C and extend it on both sides to obtain the graph of the equation 2x + y = 6.
67Pair of Linear Equations in Two Variables
We have, 2x – y + 2 = 0 ⇒ y = 2x + 2
When x = 0, we have y = 2 × 0 + 2 = 2
When x = –1, we have y = 2 × (–1) + 2 = 0
When x = 1, we have y = 2 × 1 + 2 = 4
Thus, we have the following table:
x 0 –1 1
4
y2 0
Now, we plot the points D(0, 2), E(–1, 0) and F(1, 4) on the same graph paper. We join D, E and
F and extend it on both sides to obtain the graph of the equation 2x – y + 2 = 0.
Fig. 3.5
It is evident from the graph that the two lines intersect at point F(1, 4). The area enclosed by the
given lines and x-axis is shown in Fig. 3.5.
Thus, x = 1, y = 4 is the solution of the given system of equations. Draw FM perpendicular from
F on x-axis.
Clearly, we have
FM = y-coordinate of point F(1, 4) = 4 and BE = 4
\ Area of the shaded region = Area of DFBE
⇒ Area of the shaded region = 1 (base × height) = 1 (BE × FM)
2 2
= c 1 # 4 # 4m = 8 sq. units.
2
68 Xam idea Mathematics–X
5. 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 distance covered
by the train. [NCERT]
Sol. Let actual speed of the train be x km/h and actual time taken be y hours.
Then, distance covered = speed × time = xy km ... (i)
Case I: When speed is (x + 10) km/h, then time taken is (y – 2) hours
\ Distance covered = (x + 10)(y – 2)
⇒ xy = (x + 10)(y – 2) [From (i)]
⇒ xy = xy – 2x + 10y – 20 ⇒ 2x – 10y = –20
⇒ x – 5y = –10 ...(ii)
Case II: When speed is (x – 10) km/h, then time taken is (y + 3) hours.
\ Distance covered = (x – 10)(y + 3)
⇒ xy = (x – 10)(y + 3) [From (i)]
⇒ xy = xy + 3x – 10y – 30
⇒ 3x – 10y = 30 ...(iii)
Multiplying equation (ii) by 2 and subtracting it from (iii), we get
3x – 10y = 30
– 2x +– 10y = +–20
x = 50
Putting x = 50 in equation (ii), we get
50 – 5y = –10
⇒ 50 + 10 = 5y ⇒ y = 12
\ Distance covered by the train = xy km = 50 × 12 km = 600 km
PROFICIENCY EXERCISE
QQ Objective Type Questions: [1 mark each]
1. Choose and write the correct option in each of the following questions.
(i) For which value (s) of p will the lines represented by the following pair of linear equations
be parallel 3x – y – 5= 0 and 6x – 2y – p = 0
(a) all real values except 10 (b) 10
(c) 5 (d) 1
2 2
24
(ii) The pair of equations 5 x – 15 y = 8 and 3x – 9y = 5 has
(a) one solution (b) two solutions (c) infinite solutions (d) no solution
(iii) 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
69Pair of Linear Equations in Two Variables
(iv) Graphically, the pair of equations 6x – 3y + 10 = 0; 2x – y + 9 = 0; represent two lines which
are
(a) intersecting at exactly one point (b) intersecting at exactly two points
(c) coincident (d) parallel
(v) A pair of linear equation is consistent if their graph lines will be
(a) intersecting or parallel (b) intersecting or coincident
(c) coincident or parallel (d) can't say
2. Fill in the blanks.
(i) If the system of equations; kx – 5y = 2, 6x + 2y = 7 has no solution then k = ___________ .
(ii) Two lines representing two linear equations having no solution are always ___________ .
(iii) Every linear equation in two variables has ___________ solution (s).
(iv) a1 _=__bb_12__=__cc_12_;_folirnetws.o equations a1x +b1y + c1 = 0 and a2x +b2y + c2 = 0 is the condition
a2
for
(v) Pair of lines represented by the equations 2x + y – 3 = 0 and 4x + ky + 6 = 0 will be parallel
if k = ______________.
QQ Very Short Answer Questions: [1 mark each]
3. If x = a, y = b is the solution of the pair of equations x – y = 2 and x + y = 4, find the value of a
and b. [CBSE 2018 (C) (30/1)]
4. For the pair of equations lx + 3y = –7, 2x – 6y = 14 to have infinitely many solutions, find the
value of l.
5. Write the number of solutions of the following pair of linear equations:
3x – 7y = 1 and 6x – 14y – 3 = 0 1 3
2 2
6. How many solutions does the pair of equations x + 2y = 3 and x+ y – =0 have?
7. Find the value of k for which the system of equations 2x + y – 3 = 0 and 5x + ky + 7 = 0 has no
solution.
8. Find the value of k for which the system of equations 2x + 3y = 7 and 8x + (k + 4)y – 28 = 0 has
infinitely many solutions.
9. Find the value of k for which the system of equations kx – y = 2, 6x – 2y = 3, has a unique
solution.
10. Do the following equations represent a pair of coincident lines?
–2x – 3y = 1, 6y + 4x = – 2 [NCERT Exemplar]
QQ Short Answer Questions–I: [2 marks each]
11. ABCD is a rectangle. Find the value of x or y.
C
D x+y
14 cm x–y
A 30 cm B
12. Find the value of k for which the following pair of linear equations have infinitely many solutions.
2x + 3y = 7, (k + 1) x + (2k – 1) y = 4k +1 [CBSE 2019 (30/1/2)]
13. Find the value (s) of k so that the pair of equations x + 2y = 5 and 3x + ky + 15 = 0 has a unique
solutions. [CBSE 2019 (30/2/1)]
70 Xam idea Mathematics–X
14. For what value of k, will the following pair of equations have infinitely many solutions?
2x + 3y = 7 and (k + 2) x – 3 (1 – k) y = 5k + 1 [CBSE 2019 (30/2/2)]
15. Sumit is 3 times as old as his son. Five years later, he shall be two and a half time as old as his son.
How old is sumit at present? [CBSE 2019 (30/2/3)]
16. Solve the following pair of linear equations as: [CBSE 2019 (30/3/1)]
3x – 5y = 4
2y + 7 = 9x
17. Find the solution of the pair of equations:
3x + 8 = – 1, 1 – 2 = 2, x, y ! 0 [CBSE 2019 (30/2/3)]
y x y
18. Find the value(s) of k for which the pair of equations
)3kxx ++26yy =3 has unique solution. [CBSE 2019 (30/4/2)]
=10
19. For what value of k, does the system of linear equations 2x + 3y = 7 and (k – 1) x + (k + 2) y = 3k
have an infinite number of solution? [CBSE 2019 (30/5/1)]
20. Find the relation between p and q if x = 3 and y = 1 is the solution of the pair of equations
x – 4y + p = 0 and 2x + y – q – 2 = 0. [CBSE 2019 (C) (30/1/1)]
21. Is the pair of equations x – y = 5 and 2y – x = 10 inconsistent? Justify your answer.
22. Find the value of k for which the lines (k + 1)x + 3ky + 15 = 0 and 5x + ky + 5 = 0 are coincident.
23. If the system of equations 2x + 3y = 7 and (a + b)x + (2a – b)y = 21 has infinitely many solutions,
then find a and b.
24. On comparing the ratios a1 , b1 and c1 , find out whether the linear equations 4x – 5y = 8 and
a2 b2 c2
3x – 15 y = 6 are consistent or inconsistent.
4
25. For which value(s) of k will the pair of equations kx + 3y = k – 3, 12x + ky = k have no solution?
26. If x = a and y = b is the solution of the pair of equation x – y = 5 and x + y = 3, find a and b.
27. Find the value of k for which the system of equations x + 3y – 4 = 0 and 2x + ky = 7 is inconsistent.
QQ Short Answer Questions–II: [3 marks each]
28. A part of monthly hostel charges in a college hostel are fixed and the remaining depends on the
number of days one has taken food in the mess. When a student A takes food for 25 days, he has
to pay ™ 4,500, whereas a student B who takes food for 30 days, has to pay ™ 5,200. Find the
fixed charges per month and the cost of food per day.
29. Find the solution of the pair of equations x + y – 1= 0 and x + y =15. Hence, find l, if
y = lx + 5. 10 5 8 6
30. Find the values of a and b for which the following pair of equations have infinitely many solutions:
(i) 2x + 3y = 7 and 2ax + ay = 28 – by
(ii) 2x + 3y = 7, (a – b)x + (a + b)y = 3a + b – 2
(iii) 2x – (2a + 5)y = 5, (2b + 1)x – 9y = 15
31. The angles of a triangle are x, y and 40°. The difference between the two angles x and y is 30°.
Find x and y.
71Pair of Linear Equations in Two Variables
32. The angles of a cyclic quadrilateral ABCD are ∠A = (2x + 4)°, ∠B = (y + 3)°, ∠C = (2y + 10)°,
∠D = (4x – 5)°. Find x and y and hence the values of the four angles.
33. Solve each of the following systems of equations by the method of cross-multiplication:
(i) ax + by = a – b (ii) 2(ax – by) + a + 4b = 0
bx – ay = a + b 2(bx + ay) + b – 4a = 0
(iii) 57 + x 6 y = 5 (iv) mx – ny = m2 + n2
x+ y –
x3+8 y + 21 = 9 x + y = 2m
x– y
34. The age of the father is twice the sum of the ages of his two children. After 20 years, his age will
be equal to the sum of the ages of his children. Find the age of the father.
35. There are some students in the two examination halls A and B. To make the number of students
equal in each hall, 10 students are sent from A to B. But if 20 students are sent from B to A,
the number of students in A becomes double the number of students in B. Find the number of
students in the two halls.
36. Solve the following pairs of equations:
(i) x + y = 4 (ii) x + y = a+ b
3 4 a b
56x – 8y = 4 ax2 + by2 = 2, a, b ! 0
x+y
(iii) 2 + 3 = 2 (iv) xy = 2
x y
4x –9 = – 1 x x–y y = 6
y
(v) 3x –f y+ 7 p– 8 = 0 (vi) 2 + 3 = 9
11 x y xy
2y + x +711 = 10 4x + 9y = 2x1y , x ! 0, y ! 0
37. The larger of two supplementary angles exceeds thrice the smaller by 20 degrees. Find them.
38. Solve graphically each of the following systems of linear equations. Also, find the coordinates of
the points where the lines meet the axis of y:
(i) x + 2y – 7 = 0 (ii) 3x + 2y = 12
2x – y – 4 = 0 5x – 2y = 4
y
39. If 3x + 7y = –1 and 4y – 5x + 14 = 0, then find the values of 3x – 8y and x – 2 .
40. Two numbers are in the ratio of 1 : 3. If 5 is added to both the numbers, the ratio becomes 1 : 2.
Find the numbers.
41. A’s age is six times B’s age. Four years hence, the age of A will be four times B’s age. Find the
present ages, in years, of A and B.
42. Draw the graph of the equations x – y + 1 = 0 and 3x + 2y – 12 = 0. Using this graph, find the
values of x and y which satisfy both the equations. [CBSE 2019 (30/3/1)]
QQ Long Answer Questions: [5 marks each]
43. Draw the graphs of the equations y = –1, y = 3 and 4x – y = 5. Also, find the area of the
quadrilateral formed by the lines and the y-axis.
72 Xam idea Mathematics–X
44. Solve the following system of linear equations graphically and shade the region between the two
lines and x-axis.
(i) 3x + 2y – 4 =0 (ii) 3x + 2y – 11 = 0
2x – 3y – 7 = 0 2x – 3y + 10 = 0
45. Ankita travels 14 km to her home partly by rickshaw and partly by bus. She takes half an hour if
she travels 2 km by rickshaw, and the remaining distance by bus. On the other hand, if she travels
4 km by rickshaw and the remaining distance by bus, she takes 9 minutes longer. Find the speed
of the rickshaw and of the bus. [NCERT Exemplar]
46. The sum of a two digit number and the number obtained by reversing the order of its digits is
165. If the digits differ by 3, find the number.
47. Two years ago, a father was five times as old as his son. Two years later, his age will be 8 more than
three times the age of the son. Find the present ages of father and son.
48. The sum of the numerator and denominator of a fraction is 3 less than twice the denominator.
If the numerator and denominator are decreased by 1, the numerator becomes half the
denominator. Determine the fraction.
49. A two digit number is 4 times the sum of its digits and twice the product of the digits. Find the
number.
50. A part of monthly hostel charges in a college are fixed and the remaining depend on the number
of days one has taken food in the mess. When a student A takes food for 15 days, he has to pay
`1200 as hostel charges whereas a student B, who takes food for 24 days, pays `1560 as hostel
charges. Find the fixed charge and the cost of food per day.
51. Points A and B are 70 km apart on a highway. A car starts from A and another car starts from
B simultaneously. If they travel in the same direction, they meet in 7 hours, but if they travel
towards each other, they meet in one hour. Find the speed of the two cars.
52. A man travels 600 km partly by train and partly by car. If he covers 400 km by train and the rest
by car, it takes him 6 hours and 30 minutes. But, if he travels 200 km by train and the rest by car,
he takes half an hour longer. Find the speed of the train and that of the car.
53. 2 women and 5 men can together finish a piece of embroidery in 4 days, while 3 women and 6
men can finish it in 3 days. Find the time taken by 1 woman alone, and that taken by 1 man alone
to finish the embroidery.
54. Yash scored 35 marks in a test, getting 2 marks for each right answer and losing 1 mark for each
wrong answer. Had 3 marks been awarded for each correct answer and 2 marks been deducted
for each incorrect answer, then Yash would have scored 50 marks. How many questions were
there in the test?
55. 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.
56. The car hire charges in a city comprise of a fixed charges together with the charge for the distance
covered. For a journey of 12 km, the charge paid is `89 and for a journey of 20 km, the charge
paid is `145. What will a person have to pay for travelling a distance of 30 km?
57. A takes 6 days less than B to do a work. If both A and B working together can do it in 4 days, how
many days will B take to finish it?
73Pair of Linear Equations in Two Variables
Answers
1. (i) (a) (ii) (c) (iii) (d) (iv) (d) (v) (b)
2. (i) – 15 (ii) parallel (iii) infinite (iv) coincident (v) 2 5
2
3. a = 3, b = 1 4. – 1 5. No solution 6. Infinite 7. k =
8. k = 8 9. k ≠ 3 10. Yes 11. 2x2=, 8193 , y = –5
12. k = 5 13. k ≠ 6 14. k = 4 15. 45 years 16. 13
17. x = 1, y = – 2
18. k ≠1 (the pair of equations have unique solution for all real value of k except 1
b1
19. k = 7 20. q = 5p 21. No, since a1 ≠ b2 so it has a unique solution
a2
22. 14 23. a = 5, b=1 24. consistent 25. k = –6 26. a = 4, b = –1
27. 6 28. fixed charge = ™ 1000; cost of food per day = ™140
29. x = 340, y = –165, l = –1 30. (i) a = 4, b = 8 (ii) a = 5, b = 1 (iii) a = –1, b = 5
2 2
31. x = 85°, y = 55° 32. x = 33, y = 50, ∠A = 70°, ∠B = 53°, ∠C = 110°, ∠D = 127°
33. (i) x = 1, y = –1 (ii) x = –1 , y = 2 (iii) x = 11, y = 8 (iv) x = m + n, y = m – n
2
34. 40 years 35. 100 students in Hall A, 80 students in Hall B
36. (i) x = 6, y = 8 (ii) x = a2, y = b2 (iii) x = 4, y = 9 (iv) x = –1 , y = 1 (v) x = 3, y = 4
2 4
(vi) x = 1, y = 3 37. 40°, 140°
5
38. (i) x = 3, y = 2, (0, 3.5), (0, –4) (ii) x = 2, y = 3, (0, 6) and (0, –2) 39. 14, – 2
40. 5 and 15 41. 36 years and 6 years 42. x = 2, y = 3 43. 6 square units
44. (i) x = 2, y = –1 (ii) x = 1, y = 4 45. 10 km/h, 40 km/h 4 46. 69 or 96
7
47. Father’s age = 42 years, Son’s age =10 years 48. 49. 36
50. `600, `40 51. Speed of car from point A = 40 km/ h and from point B = 30 km/h
52. 100 km/h, 80 km/h 53. One man in 36 days, One woman in 18 days
54. 25 55. 36 student 56. `215 57. 12 days
SELF-ASSESSMENT TEST
Time allowed: 1 hour Max. marks: 40
Section A
1. Choose and write the correct option in the following questions. (4 × 1 = 4)
(i) The value of k for which the lines (k + 1) x + 3ky + 15 = 0 and 5x +ky +5 = 0 are coincident is
(a) 14 (b) 2 (c) –14 (d) –2
(ii) If x = a and y= b is the solution of the equations x – y = 2 and 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
(iii) Two numbers are in the ratio 1 : 3. If 5 is added to both the numbers, the ratio becomes
1 : 2. The numbers are
(a) 4 and 12 (b) 5 and 15 (c) 6 and 18 (d) 7 and 21
74 Xam idea Mathematics–X
(iv) The pair of linear equations 4x + 6y = 9 and 2x + 3y = 6 has
(a) no solution (b) many solution
(c) two solutions (d) one solution
2. Fill in the blanks. (3 × 1 = 3)
(i) If the system of equations 2x + 3y = 5, 4x + ky = 10 has infinitely many solutions, then k is
_____________.
(ii) The equations x + 2y = 3 and 2x + 4y + 7 = 0 represent the pair of _______________ lines.
(iii) If the pair of linear equations in two variables is consistent, then the lines represented by
two equations is either _______________ or _______________.
3. Solve the following questions. (3 × 1 = 3)
(i) ™4900 was divided among a group of 150 children. If each girl gets ™50 and each boy gets
™25 then find the number of boys in the group.
(ii) If a x – b y = b – a and b x – a y = 0 then what is the value of x – y?
(iii) If 6 + 12 = 7 and 2 + 3 = 2 then find the solution of the pair of equations.
x y x y
Section B
QQ Solve the following questions. (3 × 2 = 6)
4. If the system of equations 4x + y = 3 and (2k – 1)x + (k – 1)y = 2k + 1 is inconsistent, then find k.
5. If x + 1 is a factor of 2x3 + ax2 + 2bx + 1, then find the values of a and b given that 2a – 3b = 4.
6. Draw the graph of the pair of equations x – 2y = 4 and 3x + 5y = 1. Write the vertices of the
triangle formed by these lines and the y-axis. Also find the area of this triangle.
QQ Solve the following questions. (3 × 3 = 9)
7. Half the perimeter of a rectangular garden, whose length is 4 m more than its width is 36 m.
Find the dimensions of the garden.
8. Find the value of a and b for which the following pair of linear equation has infinite number of
solutions.
2x + 3y = 7
2ax + (a + b)y = 28
OR [CBSE 2019 (30/3/3)]
Solve the following pair of linear equations:
3x + 4y = 10
2x – 2y = 2
9. A person, rowing at the rate of 5 km/h in still water, takes thrice as much time in going 40 km
upstream as in going 40 km downstream. Find the speed of the stream. [NCERT Exemplar]
QQ Solve the following questions. (3 × 5 = 15)
10. The cost of 4 pens and 4 pencil boxes is `100. Three times the cost of a pen is `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. [NCERT Exemplar]
11. Determine graphically, the vertices of the triangle formed by the lines y = x, 3y = x, x + y = 8.
[NCERT Exemplar]
75Pair of Linear Equations in Two Variables
12. 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 `1860 as annual interest.
However, had she interchanged the amount of investment in the two schemes, she would have
received `20 more as annual interest. How much money did she invest in each scheme?
[NCERT Exemplar]
Answers
1. (i) (a) (ii) (c) (iii) (b) (iv) (a)
2. (i) 6 (ii) parallel (iii) intersecting; coincident
3. (i) 104 (ii) b – a (iii) 2, 3
4. 3 5. a = 5, b = 2 6. (0, –2), (0, 1 ), (2, –1); 11 sq. units
2 5 5
7. Length = 20 m, Breadth = 16 m 8. α = 4, β = 8 OR x = 2, y = 1 9. 2.5 km/h
10. `10, `15 11. (0, 0), (4, 4), (6, 2)
12. Scheme A → `12,000, Scheme B → `10,000
zzz
76 Xam idea Mathematics–X
Quadratic 4
Equations
BASIC CONCEPTS – A FLOW CHART
Quadratic Equations 77
MORE POINTS TO REMEMBER
If ax2 + bx + c is factorizable into a product of two linear factors, then the roots of the
quadratic equation ax2 + bx + c = 0 can be found by equating each factors to zero.
For example, x2 – 5x + 6 = 0 ⇒ x2 – 3x – 2x + 6 = 0
⇒ x(x – 3) – 2 (x – 3) = 0 ⇒ (x – 3) (x – 2) = 0
⇒ x – 3 = 0 or x – 2 = 0 ⇒ x = 3 or x = 2
⇒ x = 3, 2 are roots of x2 – 5x + 6 = 0.
Sridharacharya’s Formula: The following formula used to find the roots a, b of quadratic
equation ax2 + bx + c = 0, a ≠ 0, is given by a = – b+ b2 – 4ac and b = – b – b2 – 4ac are
2a 2a
called Sridharachaya’s formula as it was first given by an ancient Indian Mathematician
Sridharacharya in 1025 AD.
Multiple Choice Questions [1 mark]
Choose and write the correct option in the following questions.
1. Which of the following is a quadratic equation? 2 [NCERT Exemplar]
(a) x2 + 2x + 1 = (4 – x)2 + 3 (b) – 5n
2x2 = (5 – x) d2x –
(c) (k + 1)x2+ 3 x = 7 , where k = – 1 (d) x3 – x2 = (x – 1)3
2
2. Which of the following is not a quadratic equation? [NCERT Exemplar]
(a) 2(x – 1)2 = 4x2 – 2x + 1 (b) 2x – x2 = x2 + 5
(c) ^ 2 x + 3h2 + x2 = 3x2 – 5x (d) (x2 + 2x)2 = x4 + 3 + 4x3
3. Which of the following equations has 2 as a root? [NCERT Exemplar]
(a) x2 – 4x + 5 = 0 (b) x2 + 3x – 12 = 0
(c) 2x2 – 7x + 6 = 0 (d) 3x2 – 6x – 2 = 0
4. Which of the following equations has the sum of its roots as 3? [NCERT Exemplar]
(a) 2x2 – 3x + 6 = 0 (b) – x2 + 3x – 3 = 0
(c) 2 x2 – 3 x + 1 = 0 (d) 3x2 – 3x + 3 = 0
2
5. Value(s) of k for which the quadratic equation 2x2 – kx + k = 0 has equal roots is
[NCERT Exemplar]
(a) 0 (b) 4 (c) 8 (d) 0 and 8
6. The quadratic equation 2x2 – 5 x + 1 = 0 has [NCERT Exemplar]
(a) two distinct real roots
(c) no real root (b) two equal real roots
(d) more than two real roots
7. The two consecutive odd positive integers, sum of whose squares is 290 are
(a) 13, 15 (b) 11, 13 (c) 7, 9 (d) 5, 7
8. (x2 + 1)2 – x2 = 0 has (b) two real roots
(d) one real root
(a) four real roots
(c) no real roots
78 Xam idea Mathematics–X
9. The quadratic equation with real co-efficients whose one root is 5+ 5 , is
2
(a) x2 – 5x + 5 = 0 (b) x2 + 5x – 5 = 0
(c) x2 – 5x – 5 = 0 (d) x2 – 5x +1 = 0
10. Root of the equation x2 – 0.09 = 0 is
(a) 0.3 (b) 0.03 (c) no root (d) none of these
11. If ax2 + bx + c = 0 has equal roots, then the value of b is
(a) ! ac (b) ! 2 ac (c) ac (d) none of these
12. If 1 is a root of the equation x2 + kx – 5 = 0, then the value of k is
2 4
1 1
(a) 2 (b) – 2 (c) 4 (d) 2
13. If the list price of a toy is reduced by ™ 2, a person can buy 2 toys more for ™ 360. The original
price of the toy is
(a) ™18 (b) ™ 20 (c) ™ 19 (d) ™ 21
14. If the equation x2 – kx +9 = 0 does not possess real roots, then
(a) – 6 < k < 6 (b) k > 6 (c) k < – 6 (d) k = ± 6
15. Which of the following is a quadratic equation?
(a) (4 – x) (3x + 1) = 2 – 3x2 (b) (x + 3)2 – 5= x2 + 9
(c) (k – 4) x2 – 3x = 9k – 4 , (k = 4) (d) (x + 1)3 = x3 – 5
16. Which of the following equations has – 1 as a root?
(a) x2 + 3x – 10 = 0 (b) x2 – x – 12 = 0
(c) 3x2 – 2x – 5 = 0 (d) 9x2 + 24x + 16 = 0
17. If the difference of roots of the quadratic equation x2 + kx + 12 = 0 is 1, the positive value of k is
(a) – 7 (b) 7 (c) 4 (d) 8
18. The quadratic equation whose one rational root is 3 + 2 is
(a) x2 – 7x + 5 (b) x2 + 7x + 6 = 0
(c) x2 – 7x + 6 (d) x2 – 6x + 7 = 0
19. Which of the following equations has two distinct real roots? [NCERT Exemplar]
(a) 2x2 – 3 2x + 9 = 0 (b) x2 + x – 5 = 0
4
(c) x2 + 3x + 2 2 = 0 (d) 5x2 – 3x + 1 = 0
20. Which of the following equations has no real roots? [NCERT Exemplar]
(a) x2 – 4x + 3 2 = 0 (b) x2 + 4x – 3 2 = 0
(c) x2 – 4x – 3 2 = 0 (d) 3x2 + 4 3 x + 4 = 0
Answers
1. (d) 2. (c) 3. (c) 4. (b) 5. (d) 6. (c)
7. (b) 8. (c) 9. (a) 10. (a) 11. (b) 12. (a)
13. (b) 14. (a) 15. (d) 16. (c) 17. (b) 18. (d)
19. (b) 20. (a)
Quadratic Equations 79
Fill in the Blanks [1 mark]
Complete the following statements with appropriate word(s) in the blank space(s).
1. A real number α is said to be a _______________ of the quadratic equation ax2 + bx + c = 0, if
aα2 + bα + c = 0.
2. For any quadratic equation ax2 + bx + c = 0, b2 – 4ac, is called the _______________ of the
equation.
3. A polynomial of degree 2 is called the _______________ polynomial.
4. If the discriminant of a quadratic equation is zero, then its roots are _______________ and
_______________ .
5. If the discriminant of a quadratic equation is greater than zero, then its roots are
_______________and _______________ .
6. A quadratic equation does not have any real roots if the value of its discriminant is
_______________ zero.
7. The roots of a quadratic equation is same as the _______________ of the corresponding quadratic
polynomial.
8. If the equation x2 + x – 5 = 0 then product of its two roots is _______________ .
9. The equation of the form ax2 + bx = 0 will always have _______________ roots.
10. If the product ac in the quadratic equation ax2 + bx + c is negative, then the equation cannot
have _______________ roots.
11. _______________ is a root of quadratic equation x2 –32 0,.0va4lu=e 0o.f k is _______________.
12. If one root of quadratic equation 6x2 – x–k=0 is
13. If one root of a quadratic equation is – b+ b2 – 4ac , other root is _______________.
2a
14. If the coefficient of x2 and constant term of a quadratic equation have _______________ signs
then the quadratic equation has real roots.
15. If 2 is a zero of the quadratic polynomial p(x) then 2 is a root of the quadratic equation
_______________.
Answers
1. root 2. discriminant 3. quadratic 4. real, equal 5. real, distinct 6. less than
11. 0.2 12. 2
7. zeros 8. – 5 9. real 10. non-real
[1 mark]
13. –b – b2 – 4ac 14. opposite 15. p(x) = 0
2a
Very Short Answer Questions
1. What will be the nature of roots of quadratic equation 2x2 – 4x + 3 = 0? [CBSE 2019 (30/2/1)]
Sol. D = b2 – 4ac = 42 – 4 × 2 × 3 = 16 – 24 = –8 < 0
Since D < 0
Hence, roots are not real.
2. If x = 3 is one root of the quadratic equation x2 – 2kx – 6= 0, then find the value of k.
[CBSE 2018]
Sol. x = 3 is a root of the equation x2 – 2k x – 6 = 0.
32 – 2k (3) – 6 = 0 ⇒ 9 – 6k – 6 = 0
1
⇒ 3 – 6k = 0 ⇒ 3 = 6k ⇒ k = 2 .
80 Xam idea Mathematics–X
3. Find the value of k for which the quadratic equation kx (x – 2) + 6 = 0 has two equal roots.
[CBSE 2019 (30/4/1)]
Sol. Given equation kx2 –2kx + 6 = 0
For two equal roots
D = 0
⇒ b2 – 4ac = 0
⇒ 4k2 – 4k × 6 = 0
⇒ 4k (k – 6) = 0 ⇒ k = 6 [k ≠ 0, as if k = 0 then the given equation is not a valid equation.]
4. If a and b are the roots of the equation x2 + ax – b = 0, then find a and b.
Sol. Sum of the roots = a+b=– B = – a
A
C
Product of the roots = ab = A =– b
⇒ a + b = – a and ab = – b
⇒ 2a = – b and a = –1 ⇒ b = 2 and a = – 1
5. Show that x = – 2 is a solution of 3x2 + 13x + 14 = 0.
Sol. Put the value of x in the quadratic equation,
LHS = 3x2 + 13x + 14 = 3(–2)2 + 13(–2) + 14
= 12 – 26 + 14 = 0 = RHS
Hence, x = – 2 is a solution.
6. Write the discriminant of the quadratic equation: (x + 5)2 = 2(5x – 3).
[CBSE 2019 (30/3/1), (30/3/3)]
Sol. Given equation:
(x + 5)2 = 2(5x – 3)
x2 + 25 + 10x = 10x – 6
⇒ x2 + 31 = 0
⇒ Here a = 1, b = 0, c = 31
D = b2 – 4ac
= – 4 × 1 × 31 = –124
∴ Discriminant = –124
Short Answer Questions-I [2 marks]
1. State whether the equation (x + 1)(x – 2) + x = 0 has two distinct real roots or not. Justify your
answer.
Sol. (x + 1)(x – 2) + x = 0 ⇒ x2 – x – 2 + x = 0 ⇒ x2 – 2 = 0
D = b2 – 4ac = 0 – 4(1) (–2) = 8 > 0
\ Given equation has two distinct real roots.
2. Is 0.3 a root of the equation x2 – 0.9 = 0? Justify.
Sol. If 0.3 is a root of the equation x2 – 0.9 = 0, then
x2 – 0.9 = (0.3)2 – 0.9 = 0.09 – 0.9 ≠ 0
Hence, 0.3 is not a root of given equation.
Quadratic Equations 81
3. Find the value of k, for which x = 2 is a solution of the equation kx2 + 2x – 3 = 0.
[CBSE 2019 (30/5/1)]
Sol. If x = 2 is a solution of kx2 + 2x – 3 = 0, then
k(22) + 2(2) – 3 = 0
⇒ 4k + 4 – 3 = 0 ⇒k= –1
⇒ 4k = –1 4
4. Find the value of k for which the equation x2 + k(2x + k − 1)+ 2 = 0 has real and equal
roots. [CBSE Delhi 2017]
Sol. Given quadratic equation: x2 + k(2x + k – 1) + 2 = 0
⇒ x2 + 2kx + (k2 – k + 2) = 0
For equal roots, b2 – 4ac = 0
⇒ 4k2 – 4k2 + 4k – 8 = 0
⇒ 4k = 8 ⇒ k = 2
5. If –5 is a root of the quadratic equation 2x2 + px – 15 = 0 and the quadratic equation
p(x2 + x) + k = 0 has equal roots, then find the value of k.
[CBSE (F) 2014, (AI) 2016]
Sol. Since – 5 is a root of the equation 2x2 + px – 15 = 0
\ 2(–5)2 + p(–5) – 15 = 0
⇒ 50 – 5p – 15 = 0 or 5p = 35 or p = 7
Again p(x2 + x) + k = 0 or 7x2 + 7x + k = 0 has equal roots.
\ D = 0 or 49 – 4 × 7k = 0 ⇒ k= 49 = 7
i.e., b2 – 4ac = 0 28 4
6. Does there exist a quadratic equation whose co-efficients are rational but both of its roots are
irrational? Justify your answer.
Sol. Yes, x2 – 4x + 1 = 0 is a quadratic equation with rational co-efficients.
Its roots are 4! (– 4)2 – 4 #1#1 = 4 ! 12 =2! 3 , which are irrational.
2 2
7. Solve the quadratic equation 2x2 + ax – a2 = 0 for x. [CBSE Delhi 2014]
Sol. 2x2 + ax – a2 = 0
Here, a = 2, b = a and c = – a2
Using the formula,
x= – b! b2 – 4ac , we get
2a
x= – a! a2 – 4 # 2 # (– a2) = – a ! 9a2 = – a ! 3a
2#2 4 4
x = – a + 3a = a , x= – a– 3a = – a ⇒ x= a , – a
4 2 4 2
8. Find the roots of the quadratic equation 2 x2 + 7x + 5 2 = 0 . [CBSE Delhi 2017]
Sol. The given quadratic equation is
2 x2 + 7x + 5 2 = 0
By applying mid term splitting, we get
2 x2 + 2x + 5x + 5 2 = 0
82 Xam idea Mathematics–X
⇒ 2 x (x + 2) + 5 (x + 2) = 0 ⇒ ( 2 x + 5) (x + 2) = 0
⇒
x= –5 , – 2 or –5 2 , – 2
2 2
9. If x =b. 32 an d x = –3 ar e root s of the quadratic equation ax2 + 7x + b = 0, find the values of a
and [CBSE Delhi 2016]
Sol. Let us assume the quadratic equation be Ax2 + Bx + C = 0.
Sum of the roots = – B
A
⇒ –7 = 2 –3 & a=3
a 3
C
Product of the roots = A
⇒ b = 2 × (−3) ⇒ b = −2 ⇒ b = 3× (−2) ⇒ b = −6
a 3 a
Short Answer Questions-II [3 marks]
1. A two digit number is four times the sum of the digits. It is also equal to 3 times the product
of digits. Find the number. [CBSE (F) 2016]
Sol. Let the ten’s digit be x and unit’s digit = y
Number = 10x + y
\ 10x + y = 4(x + y) ⇒ 6x = 3y ⇒ 2x = y ...(i)
Again 10x + y = 3xy
10x + 2x = 3x (2x) ⇒ 12x = 6x2 [From equation (i)]
⇒ 6x2 – 12x = 0 ⇒ 6x(x – 2) = 0 ⇒ x = 2 (rejecting x = 0)
From (i), 2x = y ⇒ y = 4
\ The required number is 24.
2. Find the value of p, for which one root of the quadratic equation px2 – 14x + 8 = 0 is 6 times
the other. [CBSE (AI) 2017]
Sol. Let the roots of the given equation be a and 6a.
Thus the quadratic equation is (x – a) (x – 6a) = 0
⇒ x2 – 7ax + 6a2 = 0 ...(i)
Given equation can be written as
14 8
x2 – p x+ p = 0 ...(ii)
Comparing the co-efficients in (i) & (ii) 7α = 14 and 6α2 = 8
p p
7α = 14 ⇒ α= 2
p p
2
and 6α2 = 8 ⇒ 6×e 2 = 8 ⇒ p = 6×4 ⇒p=3
p p o p 8
3. If ad ≠ bc, then prove that the equation
(a2 + b2) x2 + 2(ac + bd)x + (c2 + d2) = 0 has no real roots. [CBSE (AI) 2017]
Sol. The given quadratic equation is (a2 + b2)x2 + 2(ac + bd)x + (c2 + d2) = 0.
D = b2 – 4ac
Quadratic Equations 83
= 4(ac + bd)2 – 4(a2 + b2) (c2 + d2)
= – 4(a2d2 + b2c2 – 2abcd) = – 4(ad – bc)2
Since ad ≠ bc and (ad – bc)2 > 0
Therefore D < 0
Hence, the equation has no real roots.
4. Find the roots of the following quadratic equation by factorisation:
1
2x2 – x + 8 = 0 [NCERT]
Sol. We have, 2x2 − x + 1 = 0
8
⇒ 16x2 – 8x +1 =0
⇒ 8
16x2 – 8x + 1 = 0
⇒ 16x2 – 4x – 4x + 1 = 0
⇒ 4x (4x – 1) – 1(4x – 1) = 0
⇒ (4x – 1) (4x – 1) = 0
So, either 4x – 1 = 0 or 4x – 1 = 0
x = 1 or x = 1
4 4
1 1
Hence, the roots of the given equation are 4 and 4 .
5. Find the roots of the following quadratic equation by applying the quadratic formula.
4x2 + 4 3 x + 3 = 0 [NCERT]
Sol. We have, 4x2 + 4 3 x + 3 = 0
Here, a = 4, b = 4 3 and c = 3
Therefore, D = b2 – 4ac = (4 3)2 – 4 × 4 × 3 = 48 – 48 = 0
D = 0, roots exist and are equal.
Thus, x= – b! D = –4 3 ! 0 = –3
2a 2 ×4 2
Hence, the roots of given equation are –3 and –3 .
2 2
6. Using quadratic formula solve the following quadratic equation:
p2x2 + (p2 – q2) x – q2 = 0
Sol. We have, p2x2 + (p2 – q2) x – q2 = 0
Comparing this equation with ax2 + bx + c = 0, we have
a = p2, b = p2 – q2 and c = – q2
\ D = b2 – 4ac = (p2 – q2)2 – 4 × p2 × (– q2)
= (p2 – q2)2 + 4p2q2 = (p2 + q2)2 > 0
So, the given equation has real roots given by
84 Xam idea Mathematics–X
a= – b+ D = – (p2 – q2) +(p2 + q2) = 2q2 = q2
2a 2p2 2p2 p2
and b= – b– D = – (p2 – q2) – (p2 + q2) = – 2p2 = –1
2a 2p2 2p2
q2
Hence, roots are p2 and –1.
7. Find the nature of the roots of the following quadratic equation. If the real roots exist, find them:
3x2 – 4 3 x + 4 = 0 [NCERT]
Sol. We have, 3x2 – 4 3 x + 4 = 0
Here, a = 3, b = – 4 3 and c = 4
Therefore, D = b2 – 4ac = (– 4 3)2 – 4 × 3 × 4 = 48 – 48 = 0
Hence, the given quadratic equation has real and equal roots.
Thus, x= – b! D = – (– 4 3)! 0 = 23
2a 2#3 3
Hence, equal roots of given equation are 2 3 , 23 .
3 3
8. Find the value of k for the following quadratic equation, so that they have two equal roots.
kx (x – 2) + 6 = 0 [NCERT]
Sol. We have, kx(x – 2) + 6 = 0 ⇒ kx2 – 2kx + 6 = 0
Here, a = k, b = – 2k, c = 6
For equal roots, we have
D = 0 ⇒ (–2k)2 – 4 × k × 6 = 0
i.e., b2 – 4ac = 0 ⇒ 4k (k – 6) = 0
⇒ 4k2 – 24k = 0
Either 4k = 0 or k – 6 = 0 ⇒ k = 0 or k = 6
But k ≠ 0 (because if k = 0 then given equation will not be a quadratic equation).
So, k = 6.
9. If the roots of the quadratic equation (a – b) x2 + (b – c) x + (c – a) = 0 are equal, prove that
2a = b + c. [CBSE (F) 2017]
Sol. Since the equation (a – b) x2 + (b – c) x + (c – a) = 0 has equal roots, therefore discriminant
D = (b – c)2 – 4(a – b) (c – a) = 0
= b2 + c2 – 2bc – 4(ac – a2 – bc + ab) = 0
= b2 + c2 – 2bc – 4ac + 4a2 + 4bc – 4ab = 0
= 4a2 + b2 + c2 – 4ab + 2bc – 4ac = 0
= (2a)2 + (– b)2 + (– c)2 + 2(2a) (– b) + 2(– b) (– c) + 2 (– c) 2a = 0
= (2a – b – c)2 = 0 ⇒ 2a – b – c = 0 ⇒ 2a = b + c.
Hence proved
10. If the equation (1 + m2) x2 + 2mcx + c2 – a2 = 0 has equal roots, show that c2 = a2 (1 + m2).
[CBSE Delhi 2017]
Sol. The given equation is (1 + m2) x2 + (2mc) x + (c2 – a2) = 0
Quadratic Equations 85
Here, A = 1 + m2, B = 2mc and C = c2 – a2
Since the given equation has equal roots, therefore D = 0 ⇒ B2 – 4AC = 0.
⇒ (2mc)2 – 4(l + m2) (c2 – a2) = 0
⇒ 4m2c2 – 4(c2 – a2 + m2c2 – m2a2) = 0
⇒ m2c2 – c2 + a2 – m2c2 + m2a2 = 0 [Dividing throughout by 4]
⇒ – c2 + a2 (1 + m2) = 0 ⇒ c2 = a2 (1 + m2)
Hence proved
11. If sin q and cos q are roots of the equation ax2 + bx + c = 0, prove that a2 – b2 + 2ac = 0.
Sol. Sum of the roots = −B ⇒ sin q + cos q = –b ...(i)
A a
Product of the roots = C ⇒ sin q . cos q = c ...(ii)
A a
Now, we have, sin2 q + cos2 q = 1 –b 2 c
a a
⇒ (sin q + cos q)2 – 2sin q cos q = 1 ⇒ c m – 2 . =1
⇒ b2 – 2c =1 or b2 – 2ac = a2
a2 a
⇒ a2 – b2 + 2ac = 0
12. Determine the condition for one root of the quadratic equation ax2 + bx + c = 0 to be thrice
the other.
Sol. Let the roots of the given equation be a and 3a.
Then sum of the roots = a + 3a = 4a = –b ...(i)
a
c
Product of the roots = (a)(3a) = 3a2 = a ...(ii)
From (i), a = – b
4 a
2 3b2
(ii) ⇒ 3c –b c ⇒ 16a2 = c
4a m= a a
⇒ 3b2 = 16ac, which is the required condition.
13. Solve for x: 2c 2x – 1 m – 3c x+ 3 m = 5; x ! – 3, 1 [CBSE (F) 2014]
x+ 3 2x – 1 2
Sol. 2c 2x – 1 m – 3c x+3 m = 5 ⇒ c 4x – 2 m – c 3x + 9 m = 5
x+ 3 2x – 1 x+ 3 2x – 1
(4x – 2)(2x – 1) – (3x + 9)(x + 3) = 5(x + 3)(2x – 1)
(8x2 – 4x – 4x + 2) – (3x2 + 9x + 9x + 27) = 5(2x2 – x + 6x – 3)
8x2 – 8x + 2 – 3x2 – 18x – 27 = 10x2 + 25x – 15
5x2 – 26x – 25 = 10x2 + 25x – 15
5x2 + 51x + 10 = 0
5x2 + 50x + x + 10 = 0
5x (x + 10) + 1 (x + 10) = 0
(5x + 1) (x + 10) = 0
5x + 1 = 0 or x + 10 = 0
x = –1 or x = – 10
86 Xam idea Mathematics–X 5
14. Solve for x: x2 + 5x – (a2 + a – 6) = 0 [CBSE (F) 2015]
Sol. x2 + 5x – (a2 + a – 6) = 0
x2 + 5x – (a2 + 3a – 2a – 6) = 0
x2 + 5x – [a(a + 3) –2 (a + 3)] = 0
\ x2 + 5x – (a – 2) (a + 3) = 0
x2 + (a + 3)x – (a – 2) x – (a – 2) (a + 3) = 0
x[x + (a + 3)] – (a – 2) [x + (a + 3)] = 0
[{x + (a + 3)} {x – (a – 2)}] = 0
\ x = – (a + 3) or x = (a – 2) ⇒ x = – (a + 3), (a – 2)
Alternative method
x2 + 5x – (a2 + a – 6) = 0
\ x= –5! 52 – 4 # 1 # [– (a2 + a – 6)]
2#1
= –5! 25 + 4a2 + 4a – 24 = –5! 4a2 + 4a + 1
2 2
= –5! ^2ah2 + 2.^2ah.1 + 12 = –5! ^2a + 1h2
2 2
= – 5 ! (2a + 1) = – 5 + 2a + 1 , – 5 – 2a –1 = 2a – 4, –2a – 6 = (a – 2), – (a + 3)
2 2 2 2 2
15. Solve for x : (x 1 – 2) + (x – 1 – 3) = 2 , x ! 1, 2, 3. [CBSE (AI) 2016]
– 1) (x 2) (x 3
Sol. 1 – 2) + 1 = 2 ⇒ (x – 3) + (x – 1) = 2
(x – 1) (x (x – 2) (x – 3) 3 (x – 1) (x – 2) (x – 3) 3
⇒ 3(x – 3 + x – 1) = 2(x – 1)(x – 2)(x – 3) ⇒ 3(2x – 4) = 2(x – 1)(x – 2)(x – 3)
⇒ 3 × 2(x – 2) = 2(x – 1)(x – 2)(x – 3) ⇒ 3 = (x – 1)(x – 3) i.e., x2 – 4x = 0
⇒ x(x – 4) = 0 \ x = 0, x = 4
16. If the roots of the equation (c2 – ab)x2 – 2(a2 – bc)x + b2 – ac = 0 in x are equal, then show that
either a = 0 or a3 + b3 + c3 = 3abc.
[CBSE (AI) 2017]
Sol. For equal roots D = 0
Therefore 4(a2 – bc)2 – 4(c2 – ab) (b2 – ac) = 0
⇒ 4[a4 + b2c2 – 2a2bc – b2c2 + ac3 + ab3 – a2bc] = 0
⇒ a(a3 + b3 + c3 – 3abc) = 0
⇒ Either a = 0 or a3 + b3 + c3 = 3abc
17. If the roots of the quadratic equation (x – a) (x – b) + (x – b) (x – c) + (x – c) (x – a) = 0 are equal,
then show that a = b = c. [CBSE (F) 2017]
Sol. Given (x – a) (x – b) + (x – b) (x – c) + (x – c) (x – a) = 0
⇒ x2 – ax – bx + ab + x2 – bx – cx + bc + x2 – cx – ax + ac = 0
⇒ 3x2 – 2(a + b + c)x + ab + bc + ca = 0
Now, for equal roots D = 0
⇒ B2 – 4AC = 0
⇒ 4(a + b + c)2 – 12(ab + bc + ca) = 0
Quadratic Equations 87
⇒ 4a2 + 4b2 + 4c2 + 8ab + 8bc + 8ca – 12ab – 12bc – 12ca = 0
⇒ 2[2a2 + 2b2 + 2c2 – 2ab – 2bc – 2ca] = 0
⇒ 2[(a2 + b2 – 2ab) + (b2 + c2 – 2bc) + (c2 + a2 – 2ca)] = 0
⇒ [(a – b)2 + (b – c)2 + (c – a)2] = 0
⇒ a – b = 0, b – c = 0, c – a = 0
⇒ a = b, b = c, c = a ⇒ a = b = c
18. A farmer plants apple trees in a square pattern. In order to protect the apple trees against the
wind he plants conifer trees all around the orchard. Here you see a diagram of this situation
where you can see the pattern of apple trees and conifer trees for any number (n) rows of apple
trees:
= conifer
= apple tree
(a) Complete the table:
n Number of apple trees Number of conifer trees
118
24
3
4
5
(b) There are two formulae you can use to calculate the number of apple trees and the number
of conifer trees for the pattern described above:
Number of apple trees = n2
Number of conifer trees = 8n
Where n is the number of rows of apple trees.
There is a value of n for which the number of apple trees equals the number of conifer
trees. Find the value of n and show your method of calculating this.
(c) Suppose the farmer wants to make a much larger orchard with many rows of trees. As the
farmer makes the orchard bigger, which will increase more quickly: the number of apple
trees or the number of conifer trees? Explain how you found your answer.
Sol.: (a) Completed table as below:
n Number of apple trees Number of conifer trees
118
2 4 16
3 9 24
4 16 32
5 25 40
88 Xam idea Mathematics–X
(b) According to question,
number of apple trees = number of conifer trees
⇒ n2 = 8n
⇒ n2 – 8n = 0 ⇒ n(n – 8) = 0
⇒ n = 0 or (n – 8) = 0
⇒ n = 8, here n ≠ 0
(c) The number of apple trees increase more quickly than the number of conifer trees. We can
decide it because apple trees are increasing linearly.
Long Answer Questions [5 marks]
1. Using quadratic formula, solve the following equation for x: abx2 + (b2 – ac) x – bc = 0
Sol. We have, abx2 + (b2 – ac) x – bc = 0
Here, A = ab, B = b2 – ac, C = – bc
\ x= – B! B2 – 4AC
2A
⇒ x = – (b2 – ac) ! (b2 – ac)2 – 4 (ab) (– bc) ⇒ x = – (b2 – ac) ! (b2 – ac)2 + 4ab2 c
2ab 2ab
⇒ x = – (b2 – ac) ! (b4 – 2ab2 c + a2 c2 + 4ab2 c
2ab
⇒ x= – (b2 – ac) ! (b2 + ac)2 ⇒ x = – (b2 – ac) ! (b2 + ac)
2ab 2ab
⇒ x = – (b2 – ac) + (b2 + ac) or x = – (b2 – ac) – (b2 + ac)
2ab 2ab
x = 2ac or x = – 2b2 ⇒ x = c or x = –b
2ab 2ab b a
2. Find the value of p for which the quadratic equation
(2p + 1)x2 – (7p + 2)x + (7p – 3) = 0 has equal roots. Also find these roots. [CBSE Delhi 2014]
Sol. Since the quadratic equation has equal roots, D = 0 i.e., b2 – 4ac = 0
In (2p + 1)x2 – (7p + 2)x + (7p – 3) = 0
Here, a = (2p + 1), b = – (7p + 2), c = (7p – 3)
\ (7p + 2)2 – 4(2p + 1)(7p – 3) = 0
⇒ 49p2 + 4 + 28p – (8p + 4) (7p – 3) = 0
⇒ 49p2 + 4 + 28p – 56p2 + 24p – 28p + 12 = 0
⇒ –7p2 + 24p + 16 = 0 ⇒ 7p2 – 24p – 16 = 0
⇒ 7p2 – 28p + 4p – 16 = 0 ⇒ 7p(p – 4) + 4(p – 4) = 0
⇒ (7p + 4)(p – 4) = 0 ⇒ p = – 4 or p = 4
7
4
For p = – 7
d2 # d –4 n + 1 nx2 – d7 # d –4 n + 2nx + d7 # d –4 n – 3 n = 0
⇒ 7 7 7
⇒
–1 x2 + 2x – 7 = 0 ⇒ x2 – 14x + 49 = 0
7
x2 – 7x – 7x + 49 = 0 ⇒ x(x – 7) – 7(x – 7) = 0
Quadratic Equations 89
⇒ (x – 7)2 = 0 ⇒ x = 7, 7
For p = 4, (2 × 4 + 1)x2 – (7 × 4 + 2)x + (7 × 4 – 3) = 0
⇒ 9x2 – 30x + 25 = 0 ⇒ 9x2 – 15x – 15x + 25 = 0
⇒ 3x(3x – 5) –5(3x – 5) = 0 ⇒ (3x – 5) (3x – 5) = 0
⇒ x = 5 , 5
3 3
1
3. The sum of the reciprocals of Rehman’s age (in years) 3 years ago and 5 years from now is 3 .
Find his present age. [NCERT]
Sol. Let the present age of Rehman be x years.
So, 3 years ago, Rehman’s age = (x – 3) years
And 5 years from now, Rehman’s age = (x + 5) years
Now, according to question, we have
1 + 1 = 1
x–3 x+5 3
⇒ x+5+x–3 = 1 ⇒ 2x + 2 = 1
(x – 3) (x + 5) 3 (x – 3) (x + 5) 3
⇒ 6x + 6 = (x – 3) (x + 5) ⇒ 6x + 6 = x2 + 5x – 3x – 15
⇒ x2 + 2x – 15 – 6x – 6 = 0 ⇒ x2 – 4x – 21 = 0
⇒ x2 – 7x + 3x – 21 = 0
⇒ x(x – 7) + 3(x – 7) = 0
⇒ (x – 7) (x + 3) = 0 ⇒ x = 7 or x = –3
But x ≠ –3 (age cannot be negative)
Therefore, present age of Rehman = 7 years. 1
10
4. The difference of two natural numbers is 5 and the difference of their reciprocals is . Find
the numbers. [CBSE Delhi 2014]
Sol. Let the two natural numbers be x and y such that x > y.
According to the question,
Difference of numbers, x – y = 5 ⇒ x = 5 + y ...(i)
Difference of the reciprocals,
1 – 1 = 1 ...(ii)
y x 10
Putting the value of (i) in (ii) 5+y– y
y (5 + y)
1 – 1 = 1 ⇒ = 1
y 5+y 10 10
⇒ 50 = 5y + y2 ⇒ y2 + 5y – 50 = 0
y(y + 10) – 5(y + 10) = 0
⇒ y2 + 10y – 5y – 50 = 0 ⇒
y=5
⇒ (y – 5) (y + 10) = 0
\ y = 5 or y = –10
y is a natural number. \
Putting the value of y in (i), we have
x = 5 + 5 ⇒ x = 10
The required numbers are 10 and 5.
90 Xam idea Mathematics–X
5. The sum of the squares of two consecutive odd numbers is 394. Find the numbers.
[CBSE (F) 2014, Delhi 2017 (C)]
Sol. Let the two consecutive odd numbers be x and x + 2.
⇒ x2 + (x + 2)2 = 394 ⇒ x2 + x2 + 4 + 4x = 394
⇒ 2x2 + 4x + 4 = 394 ⇒ 2x2 + 4x – 390 = 0
⇒ x2 + 2x – 195 = 0 ⇒ x2 + 15x – 13x – 195 = 0
⇒ x(x + 15) – 13(x + 15) = 0 ⇒ (x – 13) (x + 15) = 0
⇒ x – 13 = 0 or x + 15 = 0 ⇒ x = 13 or x = –15
Hence, the numbers are 13 and 15 or –15 and –13.
6. In a class test, the sum of Shefali’s marks in Mathematics and English is 30. Had she got 2
marks more in Mathematics and 3 marks less in English, the product of her marks would have
been 210. Find her marks in the two subjects. [NCERT]
Sol. Let Shefali’s marks in Mathematics be x.
Therefore, Shefali’s marks in English is (30 – x).
Now, according to question,
(x + 2) (30 – x – 3) = 210 ⇒ (x + 2) (27 – x) = 210
⇒ 27x – x2 + 54 – 2x = 210 ⇒ 25x – x2 + 54 – 210 = 0
⇒ 25x – x2 – 156 = 0 ⇒ – (x2 – 25x + 156) = 0
⇒ x2 – 25x + 156 = 0 ⇒ x2 – 13x – 12x + 156 = 0
⇒ x(x – 13) – 12(x – 13) = 0 ⇒ (x – 13) (x – 12) = 0
Either x – 13 = 0 or x – 12 = 0
⇒ x = 13 or x = 12
∴ Shefali’s marks in Mathematics = 13, marks in English = 30 – 13 = 17
or Shefali’s marks in Mathematics = 12, marks in English = 30 – 12 = 18.
7. A train travels 360 km at a uniform speed. If the speed has been 5 km/h more, it would have
taken 1 hour less for the same journey. Find the speed of the train. [CBSE 2019 (30/5/1)]
Sol. Let the uniform speed of the train be x km/h.
Then, time taken to cover 360 km = 360 h
x
Now, new increased speed = (x + 5) km/h
So, time taken to cover 360 km = 360 h
x+5
According to question, 360 – 360 = 1
x x+5
⇒ 360c 1 – 1 m = 1 ⇒ 360 (x + 5 – x) = 1
x + x (x + 5)
⇒ x 5
\ 360 # 5
x (x + 5) = 1 ⇒ 1800 = x2 + 5x
x2 + 5x – 1800 = 0 ⇒ x2 + 45x – 40x – 1800 = 0
⇒ x(x + 45) – 40 (x + 45) = 0 ⇒ (x + 45) (x – 40) = 0
Either x + 45 = 0 or x – 40 = 0
\ x = – 45 or x = 40
But x cannot be negative, so x ≠ – 45
Therefore, x = 40
Hence, the uniform speed of train is 40 km/h.
Quadratic Equations 91
8. The sum of the areas of two squares is 468 m2. If the difference of their perimeters is 24 m,
find the sides of the two squares.
Sol. Let x be the length of the side of first square and y be the length of side of the second square.
Then, x2 + y2 = 468 ...(i)
Let x be the length of the side of the bigger square.
4x – 4y = 24
⇒ x – y = 6 or x = y + 6 ...(ii)
Putting the value of x in terms of y from equation (ii) in equation (i), we get
(y + 6)2 + y2 = 468
⇒ y2 + 12y + 36 + y2 = 468 or 2y2 + 12y – 432 = 0
⇒ y2 + 6y – 216 = 0 ⇒ y 2 + 18y – 12y – 216 = 0
⇒ y(y + 18) –12(y + 18) = 0 ⇒ (y + 18)(y – 12) = 0
Either y + 18 = 0 or y – 12 = 0
⇒ y = –18 or y = 12
But, sides cannot be negative, so y = 12
Therefore, x = 12 + 6 = 18
Hence, sides of two squares are 18 m and 12 m.
9. Seven years ago Varun’s age was five times the square of Swati’s age. Three years hence,
Swati’s age will be two-fifth of Varun’s age. Find their present ages.
Sol. Seven years ago, let Swati’s age be x years. Then, seven years ago Varun’s age was 5x2 years.
\ Swati’s present age = (x + 7) years
Varun’s present age = (5x2 + 7) years
Three years hence,
Swati’s age = (x + 7 + 3) years = (x + 10) years
Varun’s age (5x2 + 7 + 3) years = (5x2 + 10) years
According to the question,
x + 10 = 2 (5x2 + 10) ⇒ x + 10 = 2 × 5 (x2 + 2)
5 5
⇒ x + 10 = 2x2 + 4 ⇒ 2x2 – x – 6 = 0
⇒ 2x2 – 4x + 3x – 6 = 0 ⇒ 2x (x – 2) + 3(x – 2) = 0
⇒ (2x + 3) (x – 2) = 0
⇒ x – 2 = 0 or 2x + 3 = 0
⇒ x = 2 [ 2x + 3 ≠ 0 as x > 0]
Hence, Swati’s present age = (2 + 7) years = 9 years
and Varun’s present age = (5 × 22 + 7) years = 27 years
10. A train covers a distance of 300 km at a uniform speed. If the speed of the train is increased
by 5 km/hour, it takes 2 hours less in the journey. Find the original speed of the train.
[CBSE (AI) 2017]
Sol. Let the original speed of the train = x km/h.
Therefore, time taken to cover 300 km = 300 hours ...(i)
x
When its speed is increased by 5 km/h, then time taken by the train to cover the distance of
300 km = 300 hours ...(ii)
x+5
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