NCERT Solutions Maths Class 10 Exercise 3.5
1. Which of the following pairs of linear
equations has unique solution, no solution, or infinitely many solutions? In
case there is a unique solution, find it by using cross multiplication method.
(i) x – 3y – 3 = 0
3x –
9y – 2 = 0
(ii) 2x + y = 5
3x +
2y = 8
(iii) 3x − 5y = 20
6x −
10y = 40
(iv) x − 3y – 7 =
0
3x −
3y – 15 = 0
Solution: (i) x − 3y – 3 = 0 ….. (1)
3x −
9y – 2 = 0 ….. (2)
Comparing
equation x − 3y – 3 = 0
with a1x + b1y + c1 = 0 and 3x − 9y – 2 = 0
with a2x + b2y + c2 = 0, we get
a1
= 1, b1 = –3, c1 =
–3, a2 = 3, b2 = –9, c2 = –2
Now, a1/a2 = 1/3, b1/b2
= 1/3 and c1/c2 = 3/2
Here, a1/a2
= b1/b2 ≠ c1/c2
It
means that the two lines are parallel.
Therefore,
there is no solution for the given equations.
(ii) 2x + y = 5 ….. (1)
3x + 2y = 8 ….. (2)
Comparing equation 2x + y = 5 with a1x + b1y + c1
= 0 and 3x + 2y = 8
with a2x + b2y + c2 = 0, we get
a1 = 2, b1 = 1, c1
= –5, a2 = 3, b2
= 2, c2 = –8
Now, a1/a2 = 2/3 and b1/b2
= 1/2
Here, a1/a2 ≠ b1/b2
This means that there is a unique solution for the
given equations.
Using cross-multiplication method,
⇒ x = 2 and y = 1
(iii) 3x − 5y = 20 ….. (1)
6x − 10y = 40 ….. (2)
Comparing
equation 3x − 5y = 20
with a1x + b1y + c1 = 0 and 6x − 10y = 40 with a2x
+ b2y + c2 = 0, we get
a1
= 3, b1 = –5, c1 = –20, a2 = 6, b2 =
– 10, c2 = –40
Now, a1/a2 = 1/2, b1/b2
= 1/2 and c1/c2 = 1/2
Here, a1/a2
= b1/b2 = c1/c2
It
means the lines coincide with each other.
Hence,
there are infinitely many solutions to the given equations.
(iv) x −
3y – 7 = 0 ….. (1)
3x − 3y – 15 = 0 ….. (2)
Comparing
equation x − 3y – 7 = 0 with
a1x + b1y + c1 = 0 and 3x − 3y – 15 = 0
with a2x + b2y + c2 = 0, we get
a1
= 1, b1 = –3, c1 = –7, a2 = 3, b2
= –3, c2 = –15
Now, a1/a2 = 1/3 and b1/b2
= 1
Here, a1/a2 ≠ b1/b2
It
means that we have a unique solution for the given equations.
⇒ x = 4 and y = –1
2. (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
Solution: (i) 2x + 3y =
7 ….. (1)
(a − b)x +
(a + b)y = 3a + b –
2 ….. (2)
Comparing
equation 2x + 3y – 7 = 0
with a1x + b1y + c1 = 0 and (a − b)x + (a + b)y − 3a – b + 2 = 0
with a2x + b2y + c2 = 0, we get
a1
= 2, b1 = 3, c1 = −7, a2 = (a – b),
b2 = (a + b) and
c2 = 2 – b – 3a
Linear equations have infinitely many solutions
if a1/a2
= b1/b2 = c1/c2
⇒ 2a + 2b = 3a − 3b and 6 − 3b − 9a = −7a − 7b
⇒ a = 5b ….. (1) and 2a = 4b + 6 ….. (2)
Putting
a = 5b from equation (1)
in equation (2), we get
2(5b) = 4b + 6
⇒ 10b − 4b = 6
⇒ 6b = 6
⇒ b = 1
Putting
the value of b in equation (1), we get
a = 5b = 5(1) = 5
Therefore, a = 5 and b = 1.
(ii) 3x + y =
1 …… (1)
(2k − 1)x + (k −
1)y = 2k + 1 ….. (2)
Comparing
equation 3x + y – 1 = 0
with a1x + b1y + c1 = 0 and (2k − 1)x + (k − 1)y −2k – 1 = 0, with a2x + b2y
+ c2 = 0, we get
a1
= 3, b1 = 1 and c1
= –1, a2 = (2k –1), b2 = (k –1) and c2 = −2k − 1
Linear
equations have no solution if a1/a2 = b1/b2
≠ c1/c2
⇒ 3(k − 1) = 2k – 1
⇒ 3k – 3 = 2k − 1
⇒ k = 2
3. Solve the following pair of linear
equations by the substitution and cross-multiplication methods:
8x + 5y = 9
3x + 2y = 4
Solution: Substitution Method:
8x + 5y = 9 … (1)
3x + 2y = 4 … (2)
From equation (1),
we have
5y = 9 − 8x
⇒ y = (9 − 8x)/5
Putting this value of y in equation (2), we get
3x + 2(9 − 8x)/5 = 4
⇒ 3x + (18 − 16x)/5 = 4
⇒ 3x – 16x/5
= 4 – 18/5
⇒ 15x − 16x = 20 – 18
⇒ x = −2
Putting the value of x in equation (1),
we get
8(−2) + 5y = 9
⇒ 5y = 9 + 16 = 25
⇒ y = 5
Therefore, x = −2
and y = 5.
Cross-multiplication Method:
8x + 5y = 9 … (1)
3x + 2y = 4 … (2)
⇒ x = −2 and y = 5
4. Form the pair of linear equations in the
following problems and find their solutions (if they exist) by any algebraic
method:
(i)
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 Rs 1000 as hostel charges whereas a student B, who takes
food for 26 days, pays Rs 1180 as hostel charges. Find the fixed charges and
the cost of food per day.
(ii)
A fraction becomes 1/3 when 1 is subtracted from the
numerator and it becomes 1/4 when 8 is added to its denominator. Find the
fraction.
(iii)
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 deducted for each incorrect answer, then Yash would
have scored 50 marks. How many questions were there in the test?
(iv)
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?
(v)
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.
Solution:
(i) Let the fixed
monthly charge is Rs x and the charge
of food for one day be Rs y.
According
to the given conditions, we have
x + 20y = 1000 … (1)
and x + 26y = 1180 … (2)
Subtracting
equation (1) from equation (2), we get
6y = 180
⇒ y = 30
Putting
the value of y in equation (1), we
get
x + 20(30) = 1000
⇒ x = 1000 – 600
⇒ x = 400
Therefore, the
fixed monthly charge is Rs 400 and the charge of food for one day is Rs 30.
(ii) Let the numerator of the required
fraction be x and the denominator be y.
According
to the given conditions, we have
⇒ 3x – 3 = y … (1) 4x = y + 8 … (2)
⇒ 3x – y = 3 … (1) 4x – y = 8 … (2)
Subtracting
equation (1) from equation (2), we get
4x – y − (3x − y) = 8 – 3
⇒ x = 5
Putting
the value of x in equation (1), we
get
3(5) – y = 3
⇒ 15 – y = 3
⇒ y = 12
Therefore, the numerator
is 5 and the denominator is 12.
Thus,
fraction = 5/12
(iii) Let the number of correct
answers be x and the number of wrong
answers be y.
According
to the given conditions, we have
3x – y = 40 … (1)
And, 4x − 2y = 50 … (2)
From equation (1), y = 3x − 40
Putting
this value of y in equation (2), we get
4x – 2(3x − 40) = 50
⇒ 4x − 6x + 80 = 50
⇒ −2x = −30
⇒ x = 15
Putting
the value of x in equation (1), we
get
3(15) – y = 40
⇒ 45 – y = 40
⇒ y = 45 – 40
⇒ y = 5
Therefore,
the number of correct answers is 15 and the number of wrong answers is 5.
Total number of questions = x + y = 15 + 5 = 20
(iv)
Let the speed of the
car which starts from point A be x km/h and the speed of the car which starts
from point B be y km/h.
When
the cars travel in the same direction, relative speed = (x − y) km/h
When
the cars travel in the opposite directions, relative speed = (x + y)
km/h
According
to the first condition, we have
T
= D/S
5
= 100/(x − y) (Assuming x > y)
⇒ 5x − 5y = 100
⇒ x – y = 20 … (1)
According
to the second condition, we have
T
= D/S
1 = 100/(x
+ y)
⇒ x + y = 100 … (2)
Adding
equations (1) and (2), we get
2x = 120
⇒ x = 60 km/h
Putting
the value of x in equation (1), we get
60 – y = 20
⇒ y = 60 – 20
⇒ y = 40 km/h
Therefore,
the speed of the car starting from point A is 60 km/h.
And,
the speed of the car starting from point B is 40 km/h.
(v) Let the length of the rectangle be
x units and the breadth of the
rectangle be y units.
Area = xy square units.
According to the first condition, we have
xy – 9 = (x − 5) (y + 3)
⇒ xy – 9 = xy +
3x − 5y – 15
⇒ 3x − 5y = 6 … (1)
According to the second condition, we have
xy + 67 = (x + 3) (y + 2)
⇒ xy + 67 = xy +
2x + 3y + 6
⇒ 2x + 3y = 61 … (2)
From
equation (1), we get
3x = 6 + 5y
⇒ x = (6 + 5y)/3
Putting
this value of x in equation (2), we
get
2(6 + 5y)/3 + 3y = 61
⇒ 12 + 10y + 9y = 183
⇒ 19y = 171
⇒ y = 9 units
Putting
the value of y in equation (2), we
get
2x + 3(9) = 61
⇒ 2x = 61 – 27 =
34
⇒ x = 17 units
Therefore,
the length is 17 units and the breadth is 9 units.