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These NCERT Solutions for Class 10 Maths Chapter Ex 4.4 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Exercise 4.4

Question 1.
Find the nature of the roots of the following quadratic equations. If the real roots exist, find them:
(i) 2x² – 3x + 5 = 0
(ii) 3x² – 4√3x + 4 = 0
(iii) 2x²-6x + 3 = 0
Solution:
(i) We have,
2x² – 3x + 5 = 0
Here, a = 2, b = – 3 and c = 5
∴ Discriminant (D) = b² – 4ac = (- 3)² – 4 x 2 x 5 = 9 – 40 = – 31
∴ D= – 31
Here, discriminant (D) < 0 Therefore, the given quadratic equation has no real roots.

(ii) We have, 3x² – 4\(\sqrt{3x}\) + 4 = 0
Here, a = 3, b= – 4\(\sqrt{3x}\) and c = 4
∴ Discriminant (D) = b² – 4ac = \((-4 \sqrt{3 x})^{2}\) x 4 = 48 – 48
∴ Discriminant (D) = 0
Therefore, the given quadratic equation has two equal real roots.

(iii) We have, 2x² – 6x + 3 = 0
Here, a = 2,b = – 6, c = 3
∴ Discriminant (D) = b² – 4ac = (- 6)2 – 4 x 2 x 3 = 36 – 24 = 12
∴ D= 12 Discriminant (D) > 0
Therefore, the given quadratic equation has two distinct real roots. Distinct roots \(\sqrt{3x}\)

Question 2.
Find the value of k for each of the following quadratic equations, so that they have two real equal roots.
(i) 2x² + kx + 3 = 0
(ii) kx(x – 2) + 6 = 0
Solution:
(i) We have,
2x² + kx + 3 = 0
Here, a = 2, b = k and c = 3
∴ Discriminant (D) = b² – 4ac = k² – 4 x 2 x 3
D = k² – 24
But we have given that equation has two real and equal roots.
∴ D = 0
k² – 24 = 0
k² = 24
∴ k = \(\sqrt{24}\)
k = ±2\(\sqrt{6}\)

(ii) We have,
kx (x – 2) + 6 = 0
or, kx² – 2kx + 6 = 0
Here, a = k, b = – 2k and c = 6
∴ Discriminant (D) = b²- 4ac
= (- 2k)² – 4 x k x 6
D= 4k² – 24k
But we have given that equation has two real equal roots.
∴ D = 0
4k² – 24k = 0
or, 4k(k – 6) = 0
So, 4k = 0 or k – 6 = 0
Therefore, k = 0 or k = 6

Question 3.
Is it possible to design a rectangular mango grove whose length is twice its breadth and the area is 800m2 ? If so, find its length and breadth.
Solution:
Let breadth of rectangular mango grove = x
∴ length of rectangular mago grove = 2x
∴ Area of rectangular mango grove = l x b
or 800 = x × 2x [∴ Area of rectangle = Length x Breadth]
or 800 = 2x²
or x = \(\frac { 800 }{ 2 }\)
∴ x = \(\sqrt{400}\) = ± 20
Since, dimensions cannot be in negative, so we reject the value of x = – 20.
∴ Breadth of rectangular mango grove = x = 20 m
and length of rectangular mango grove = 2x – 2 x 20 = 40m

Question 4.
Is the following situation possible? If so, determine their present ages. The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48.
Solution:
Let the present age of one friend be x
∴ the present age of other friend be 20 – x
Four years ago age of first friend = x – 4
and four years ago age of second friend = (20 – x) – 4 = 16 – x
∴ According to question,
(x – 4) (16 – x) = 48
or 16x – x² – 64 + 4x = 48
or – x² + 20x – 64 = 48
or x² – 20x + 64 + 48 = 0
or x² – 20x + 112 = 0
Here, a 1,b = -20 and c = 112
∴ Discriminant (D) b² – 4ac
= (-20)2 – 4 x 1 x 112
= 440 – 448 = -48
∴ D<0
So, the equation has no any real roots.
Therefore, the given situation is not possible.

Question 5.
It is possible to design a rectangulat park of perimeter 80 m and area 400 m²? If so find its length and breadth.
Solution:
Let the length of the rectangular park = x
∴ the breadth of the rectangular park = (40 – x)
∴ Area of rectangular park = Length x Breadth
or 400 = x(40 – x)
or 400 = 40x – x²
or x² – 40x + 400 = 0
Here, a = 1, b = -40 and c = 400
∴ Discriminant (D) = b² – 4ac
= (- 40)² – 4 x 1 x 400
= 1600 – 1600
D = 0
Therefore, equation has two real and equal roots.
\(\begin{aligned}
x &=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a} \\
&=\frac{-(-40) \pm \sqrt{0}}{2 \times 1}=\frac{40}{2}=20
\end{aligned}\)
So, the length of rectangular park = x = 20 m
and the breadth of rectangular park = (40 -x)
= 40 – 20 = 20 m.

These NCERT Solutions for Class 10 Maths Chapter Ex 4.3 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Exercise 4.3

Question 1.
Find the roots of the following quadratic equations, if they exist, by the method of completing the square:
(i) 2x² – 7x + 3 = 0
(ii) 2x² + x – 4 = 0
(iii) 4x² + 4√3x + 3 = 0
(iv) 2x² + x + 4 = 0
Solution:
(i) We have,

Adding and subtracting (\(\frac { 7 }{ 4 }\))² in equation (i), we get,

(ii) We have,

Adding and subtracting (\(\frac { 1 }{ 4 }\))² in equation (i), we get,

(iii) We have,

Adding and subtracting \(\left(\frac{\sqrt{3}}{2}\right)^{2}\) in equation (i), we get,

(iv) We have,

Adding and subtracting (\(\frac { 1 }{ 4 }\))² in equation (i), we get,

So, roots do not exist.

Question 2.
Find the roots of the quadratic equations by applying the quadratic formula.
(i) 2x² – 7x + 3 = 0
(ii) 2x² + x – 4 = 0
(iii) 4x² + 4√3x + 3 = 0
(iv) 2x² + x + 4 = 0
Solution:
(i) We have,
2x² – 7x + 3 = 0
Here a = 2, b = – 7, and c = 3
∴ By using quadratic formula, we get,

(ii) We have,
2x² + x – 4 = 0
Here a = 2, b = 1, and c = – 4
∴ By using quadratic formula, we get,

(iii) We have,
4x² + 4√3x + 3 = 0
Here a = 4, b = 4√3 and c = 3
∴ By using quadratic formula, we get,

(iv) We have,
2x² + x + 4 = 0
Here a = 2, b = 1, and c = 4
∴ By using quadratic formula, we get,

Here, b² – 4ac < 0
So, the given equation has no any real roots.

Question 3.
Find the roots of the following equations:
(i) x – \(\frac { 1 }{ x }\) = 3 ; x ≠ 0
(ii) \(\frac { 1 }{ x+4 }\) – \(\frac { 1 }{ x-7 }\); x ≠ – 4, 7
Solution:

Question 4.
The sum of the reciprocals of Rehman’s ages, (in years) 3 years ago and 5 years from now is \(\frac { 1 }{ 3 }\) Find his present age.
Solution:
Let the present age of Rehman be x years
3 years ago Rehman’s age was = (x – 3) years
5 years from now Rehman’s age will be = (x + 5) years
∴ According to question

Using the quadratic formula

Since, age cannot be in negative,
so we ignore the value of x = – 3
Therefore, x = 7
So, present age of Rehman = x = 7 years.

Question 5.
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 their marks would have been 210. Find her marks in the two subjects.
Solution:
Let the marks secured by Shefali in Mathematics = x Then,
According to question,
(x + 2) x (30 – x – 3) = 210
or (x + 2) x (27 – x) = 210
or 27x – x² + 54 – 2x – 210
or 25x – x² + 54 = 210
or x² – 25x – 54 + 210 = 0
or x² – 25x + 156 = 0
Here, a = 1, b = – 25 and c = 156
Using the quadratic formula

Therefore, if Shefali’s marks in Mathematics = x = 13
Then, Shefail’s marks in English = 30 – x = 30 = 17
and if Shefail’s marks in Mathematics = x = 12
Then, Shefail’s marks in Mathematics = 30 – x = 30 – 12 = 18

Question 6.
The diagonal of a rectangular field is 60 metres more than the shorter side. If the longer side is 30 metres more than the shorter side, find the sides of the field.

Solution:
Let ABCD be a rectangular field.
Let AB, its shorter side, be x metres And BC, the longer side be (x + 30) metres
Diagonal AC of the rectangular field be x + 60 metres.
By Pythagoras theorem, we have,
(AC)² = (AB)² + (BC)²
∴ (x – 60)² = x² + (x + 30)²
or x² + 120x + 3600 = x² + x² + 60x + 900
or x² + 120x + 3600 = 2x² + 60x + 900
2x² + 60x + 900 – x² – 120x – 3600 = 0
or x² – 60x – 2700 = 0
Here, a = 1, b = – 60 and c = – 2700
Using the quadratic formula, we get

Since, dimension cannot be in negative,
so we ignore the value of x = – 30
Therefore, x = 90
∴ Length of shorter side of rectangular field = x = 90 metres
and length of longer side of rectangular = x + 30 = 90 + 30 = 120 metres.

Question 7.
The difference of squares of two numbers is 180. The square of the smaller number is 8 times the larger number. Find the two numbers.
Solution:
Let x be the larger number.
So, the square of the smaller number = 8x According to the second condition, the square of the larger number – the square of the smaller number = 180
x² – 8x = 180
⇒ x² – 8x – 180 = 0
⇒ x² – 18x + 10x – 180 = 0
⇒ x (x – 18) + 10 (x – 18) = 0
⇒ (x – 18) (x + 10) = 0
⇒ x = 18 or x = – 10
x = – 10 does not exist, so it is neglected.
So, x = 18 (the larger number)
Smaller number = \(\sqrt{8x}\)
⇒ \(\sqrt{8×18}\) = \(\sqrt{144}\) = ± 12
∴ The smaller and larger numbers are ±12and 18 respectively.

Question 8.
A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less for the same journey. Find the speed of the train.
Solution:
Total distance travelled = 360 km
Let uniform speed be x km/h
Then, increased speed = (x + 5) km/h
According to question,

Since, speed cannot be in negative, so we reject the value of x = – 45
∴ x = 40
So, the usual speed of train = x km/h = 40 km/h.

Question 9.
Two water taps together can fill a tank in 9\(\frac { 3 }{ 8 }\) hours. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.
Solution:
Let larger pipe fills the tank in x hours and the smaller pipe fills the tank in y hours.
The tank filled by the larger pipe in 1 hour = \(\frac { 1 }{ x }\)
and part of tank filled by larger tap in 1 hour = \(\frac { 1 }{ x-10 }\)
Total = 9\(\frac { 3 }{ 8 }\) hours = \(\frac { 75 }{ 8 }\)
∴ According to questions,

Using the quadratic formula

So, if smaller tap take 3.75 hour to fill the tank then time taken by larger tap = 3.75 = 10 = – 6.25 hour
But we know that time cannot be in negative The value of x = 3.75 is neglected.
Therefore, time taken by the smaller tap to fill the tank = x = 25 hours.
and time taken by the larger tap to fill the bank = x – 10 = 25 – 10 = 15 hours.

Question 10.
An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bengaluru (without taking into consideration the time they stop at intermediate stations). If the average speed of the express train is 11 km/h more than that of the passenger train, find the average speed of the two trains.
Solution:
Let the average speed of passenger train be x km/h
Then, average speed of express train = x + 11 km/h
Total distance from Mysore to Bangalore is 132 km.
Therefore, time taken by the passenger train to cover the distance of 132 km = \(\frac { 132 }{ x }\)h
According to question,

Using the quadratic formula

But we know that speed of train cannot be in negative.
∴ The value of x = – 44 is neglected.
∴ x = 33
Therefore, the speed of passenger train = x = 33 km/h
and the speed of express train = x + 11 = 33+ 11 = 44 km/h.

Question 11.
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.
Solution:
Let the sides of smaller square = x
The sides of larger square = \(\sqrt{468-x^{2}}\)
[∴ Area of square = (side)²]
Now, perimeter of smaller square = 4x
and perimeter of larger square = 4 x \(\sqrt{468-x^{2}}\)
According to question,
4 x \(\sqrt{468-x^{2}}\) = 24
or 4 x \(\sqrt{468-x^{2}}\) = 24 + 4x
or \(\sqrt{468-x^{2}}\) = \(\frac{4(6+x)}{4}\)
or \(\sqrt{468-x^{2}}\) = 6 + x
Squaring both sides, we get,
468 – x² = (6 + x²)
or 468 – x² = 36 + x² + 12x
[∴ (a + b)² = a² + b² – 2ab]
or 36 + x² + 12x + x² – 468 = 0
or 2x² + 12x – 432 = 0
or x² + 6x – 216 = 0
Here, a = 1, b = 6 and c = – 216
Using the quadratic fomula,

But we know that length of the side of square cannot be in negative.
So, the value of x = – 18 is neglected. Therefore, x = 12
So, length of the sides of smaller square = x = 12 m
and length of the sides of larger square = \(\sqrt{468-x^{2}}\)
= \(\sqrt{468-(12)^{2}}\)
= \(\sqrt{468-144}\)
= \(\sqrt{324}\)
= 18 m

These NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.6 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.6

Question 1.
Solve the following pairs of equations by reducing them to a pair of linear equations:

Solution:

By cross multiplication method

By cross multiplication method

By cross multiplication method

By cross multiplication method


solving for u and v by cross multiplication method:

By cross multiplication method:

By cross multiplication method:

(viii) We have

(i) and (iii), we get
P + q = – \(\frac { 3 }{ 4 }\) … (iii)
\(\frac { 1 }{ 2 }\)P – \(\frac { 1 }{ 2q }\) = \(\frac { -1 }{ 8 }\) … (iv)
By eliminating method Adding equations (iii) and (iv) we get
q = \(\frac { 1 }{ 2 }\)
But P = \(\frac { 1 }{ 3x+y }\)
and q = \(\frac { 1 }{ 3x – y }\)
\(\frac { 1 }{ 3x+y }\) = \(\frac { 1 }{ 4 }\)
and
3x + y = 4 … (v)
3x – y = 2 … (vi)
By eliminating method
Adding equations (v) and (vi) we get
6x = 6
x = 1

Question 2.
Formulate the following problems as a pair of equations, and hence find their solutions:
(i) Ritu can row’ downstream 20 km in 2 hours, and upstream 4 km in 2 hours. Find her speed of rowing in still water and the speed of the current.

(ii) 2 women and 5 men can together finish an embroidery work in 4 days, while 3 women and 6 men can finish it in 3 days. Find the time taken by 1 woman alone to finish the work, and also that taken by 1 man alone.

(iii) Roohi travels 300 km to her home partly by train and partly by bus. She takes 4 hours if she travels 60 km by train and the remaining by bus. If she travels 100 km by train and the remaining by bus, she takes 10 minutes longer. Find the speed of the train and the bus separately.
Solution:
(i) Let Ritu’s speed in still water = x km/h
Speed of current = y km/h
During downstream, speed = (x + y) km/h
During upstream, speed = (x – y) km/h
In the first case
The time taken in hour be f, then

By eliminating method
Adding equations (i) and (ii), we get
2x = 12
x = 6
Putting this value in equation (ii), we get
We get
– y = 2 – 6
y = 4
Then x = 6 and y = 4 where x and y are respectively speed (in km/h) of rowing and current.

(ii) Let the number of days taken by one woman by n and the number of days taken by one man by m.
According to question
\(\frac { 2 }{ n }\) + \(\frac { 5 }{ m }\) = \(\frac { 1 }{ 4 }\) … (i)
\(\frac { 3 }{ n }\) + \(\frac { 6 }{ m }\) = \(\frac { 1 }{ 3 }\) … (i)
Putting \(\frac { 1 }{ n }\) = p and \(\frac { 1 }{ m }\) = q in equations (i) and (ii) we get
2p + 5q = \(\frac { 1 }{ 4 }\) … (iii)
3p + 6q = \(\frac { 1 }{ 3 }\) … (iv)
or 8p + 20q – 1 = 0 … (iii)
9p + 18q -1 = 0 … (iv)
By cross multiplication method

(iii) Let the speed of the train be u km/h
and the speed of the bus be v km/h
In the first case
\(\frac { 60 }{ u }\) + \(\frac { 240 }{ v }\) = 4 … (i)
In the second case
\(\frac { 100 }{ u }\) + \(\frac { 200 }{ v }\) = \(\frac { 25 }{ 6 }\) … (ii)
Putting \(\frac { 1 }{ u }\) = p and \(\frac { 1 }{ v }\) = q in equations (i) and (ii)
we get
60p + 240q = 4 … (iii)
100p + 200p = \(\frac { 25 }{ 6 }\) … (iv)
or 60p + 240p – 4 = 0 … (v)
600p + 120q – 25 = 0 … (vi)
By using elemination method multiply equation (ii) by 10 we get
600p + 2400 – 40 = 0 … (vii)
Substract equation (iv) from equations (v)
1200q = 15
1200q = 15
q = \(\frac { 15 }{ 12000 }\)
q = \(\frac { 1 }{ 80 }\)
Putting this value in equations (iv)
P = \(\frac { 1 }{ 60 }\)
But p = \(\frac { 1 }{ u }\) and q = \(\frac { 1 }{ v }\)
\(\frac { 1 }{ u }\) = \(\frac { 1 }{ 60 }\) and \(\frac { 1 }{ v }\) = \(\frac { 1 }{ 80 }\)
u = 60 and v = 80

These NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.7

Question 1.
The age of two friends Ani and Biju differ by 3 years. Ani’s father Dharam is twice as old as Ani and Biju is twice as old as his sister Cathy. The ages of Cathy and Dharam differ by 30 years. Find the ages of Ani and Biju.
Solution:
Let the age of Ani by x then the age of Biju is x – 3 and if the age of Ani’s father Dharam be y.
The age of Cathy = \(\frac { 1 }{ 2 }\) the age of Biju then the age of Cathy is \(\frac { x-3 }{ 2 }\).
According to question

The age of Ani is 19 years and the age of Biju = x – 3
the age of Biju = 19 – 3 = 16
Hence the age of Ani is 19 years and the age of Biju is 16 years.

Question 2.
One says, “Give me a hundred, friend! I shall then become twice as rich as you”. The other replies, “If you give me ten, I shall be six times as rich as you”. Tell me what is the amount of their (respective) capital?
Solution:
Let the two friends have ₹ x and ₹ y.
According to the first condition:
One friend has an amount = ₹(x + 100)
Other has an amount = ₹ (y – 100
∴  (x + 100) =2 (y – 100)
⇒  x + 100 = 2y – 200
⇒ x – 2y = – 300       …(i)
According to the second condition:
One friend has an amount = ₹(x – 10)
Other friend has an amount =₹ (y + 10)
∴  6(x – 10) = y + 10
⇒ 6x – 60 = y + 10
⇒    6x-y = 70                                        …(ii)
Multiplying (ii) equation by 2 and subtracting the result from equation (i), we get:
x – 12x = – 300 – 140
⇒ -11x = -440
⇒  x = 40
Substituting x = 40 in equation (ii), we get
6 x 40 – y = 70
⇒ -y   = 70- 24
⇒  y   = 170
Thus, the two friends have ₹ 40 and ₹ 170.

Question 3.
A train covered a certain distance at a uniform speed. If the train would have been 10 km/h faster, it would have taken 2 hours less than the scheduled time. And, if the train were slower by 10 km/h, it would have taken 3 hours more than the scheduled time. Find the distance covered by the train.
Solution:
Let the original speed of the train be x km/h
and the time taken to complete the journey be y hours.            ‘
Then the distance covered = xy km

Case I : When speed = (x + 10) km/h and time taken = (y – 2) h
Distance = (x + 10) (y – 2) km
⇒   xy = (x + 10) (y – 2)
⇒ 10y – 2x = 20
⇒  5y – x = 10
⇒ -x + 5y = 10   …(i)

Case II : When speed = (x – 10) km/h and time taken = (y + 3) h
Distance = (x – 10) (y + 3) km
⇒  xy = (x – 10) (y + 3)
⇒ 3x- 10y = 30    …(ii)
Multiplying equation (i) by 3 and adding the result to equation (ii), we get
15y – 10y = 30
⇒ 5y = 60
⇒   y   = 12
Putting y = 12 in equation (ii), we get
3x- 10 x 12= 30
⇒ 3x  = 150
⇒ x = 50
∴  x = 50 and y = 12
Thus, original speed of train is 50 km/h and time taken by it is 12 h.
Distance covered by train = Speed x Time
=  50 x 12 = 600 km.

Question 4.
The students of a class are made to stand in rows. If 3 students are extra in a row, there would be 1 row less. If 3 students are less in a row, there would be 2 rows more. Find the number of students in the class.
Solution:
Let the number of rows be x and the number of students in each row be y.
Then the total number of students = xy
Case I : When there are 3 more students in each row
Then the number of students in a row = (y + 3)
and the number of rows = (x – 1)
Total number of students = (x – 1) (y + 3)
∴ (x – 1) (y + 3) = xy
⇒  3x  – y = 3 … (i)

Case II : When 3 students are removed from each row
Then the number of students in each row = (y-3)
and the number of rows = (x + 2)
Total number of students = (x + 2) (y – 3)
∴  (x + 2) (y – 3) = xy
⇒ – 3x + 2y = 6 … (ii)
Adding the equations (i) and (ii), we get
– y + 2y = 3 + 6
⇒ y = 9
Putting y = 9 in the equation (ii), we get
– 3x + 18 = 6
– 3x + 18 = 6
⇒ x = 4
∴ x = 4 and y = 9
Hence, the total number of students in the class is 9 x 4 = 36.

Question 5.
In a ∆ABC, ∠C = 3 ∠B = 2(∠A + ∠B). Find the three angles.
Solution:
Let ∠A = x° and ∠B = y°.
Then ∠C = 3∠B = (3y)°.
Now ∠A + ∠B + ∠C = 180°
⇒ x + y + 3y = 180°
⇒ x + 4y = 180° …(i)
Also, ∠C = 2(∠A + ∠B)
⇒ 3y – 2(x + y)
⇒ 2x – y = 0° …(ii)
Multiplying (ii) by 4 and adding the result to equation (i), we get:
9x = 180°
⇒ x = 20°
Putting x = 20 in equation (i), we get:
20 + 4y = 180°
⇒ 4y = 160°
⇒  y =  \(\frac { 160 }{ 40 }\)  = 40°
∴ ∠A = 20°, ∠B = 40° and ∠C = 3 x 40° = 120°.

Question 6.
Draw the graphs of the equations 5x – y = 5 and 3x – y = 3. Determine the coordinates of the vertices of the triangle formed by these lines and the y-axis.
Solution:
We have

Hence the vertices of the triangle are (1,0), (0, – 3), (0, – 5).

Question 7.
Solve the following pairs of linear equations:

Solution:
(i) The given equations are
px + qy = p – q  …(1)
qx – py = p + q …(2)
Multiplying equation (1) by p and equation (2) by q and then adding the results, we get:
x(p² + q²)
x = 1
Putting this value in equation (1),
we get p + qy = p – q
qy = – q
y = – 1

(ii) We have
ax + by =c …(i)
bx + ay = 1 + c …(ii)
Multiply equation (i) by b and equation (ii) by a, we get
abx + b²y = bc … (iii)
abc + a²y = a + bc … (iv)
Subtract equation (ii) by equation (iii) we get
(b² – a²)y = bc – a – ac
(b² – a²)y = c(b – a) – a

(iii) We have
\(\frac { x }{ a }\) – \(\frac { y }{ b }\) = 0 … (i)
ax + by = a² + b²
Multiply equation (i) by ab we get
bx – ay = 0 … (iii)
Multiply equation (iii) by b and equation (ii) by (a), we get
b²x – aby = 0 … (iv)
a²x + aby = a² + ab² … (v)
Adding equation (iv) and (v), we get
(a² + b²)x = a³ + ab²
(a² + b²)x = a(a² + b²)
x = a
Putting this value in equation (v), we get y = b

(iv) We have
(a – b) x + (a + b) y = a² – 2ab – b²
(a + b)(x + y) = a² + b²
The above system of equations can be written as
(a – b) x + (a + b) y – (a² – 2ab – b)² = 0
ax + ay + bx + by – (a² + b²) = 0
(a – b) x + (a + b) y – (a² – 2ab – b²) = 0
(a + b)x + (a + b)y – (a² + b²) = 0
Applying the cross multiplication method, we get

(v) The given equations may be written as:
76x – 189y = – 37 … (1)
– 189x + 76y = – 302  … (2)
Multiplying equation (1) by 76 and equation (2) by 189, we get:
5776x – 14364y = – 2812  … (3)
– 35721x + 14364y = -57078 … (4)
Adding equations (3) and (4), we get:
5776x – 35721x = – 2812 – 57078
⇒ – 29945x = – 59890
⇒  x = 2
Putting x = 2 in equation (1), we get:
76 x 2 – 189y = – 37
⇒ 152 – 189y = – 37
⇒ – 189y = – 189
⇒ y = 1
Thus, x = 2 and y = 1 is the required solution.

Question 8.
ABCD is a cyclic quadrilateral (see figure). Find the angles of the cyclic quadrilateral.

Solution:
In a cyclic quadrilateral sum of the opposite angle is 180°,
So, ∠A + ∠C = 180°
4y + 20 – 4x = 180°
4y – 4x = 160° … (i)
∠B + ∠D= 180°
3y – 5 + 5 – 7x = 180° … (ii)
3y – 7x = 180°
Multiply equation (i) by 3 and equation (ii) by 4 then subtract equation (ii) from equation (i) we get,
16x = – 240
x = – 15
Putting this value ie equation we get
y = 25
∠A = 4y + 20
∠A = 120
∠B = 3y – 5
∠B = 70°
∠C = – 4x = 60°
∠D = 110°

These NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.4

Question 1.
Solve the following pair of linear equations by the elimination method and the substitution
(i) x + y = 5 and 2x – 3y = 4
(ii) 3x + 4y = 10 and 2x – 2y = 2
(iii) 3x – 5y – 4 = 0 and 9x = 2y + 7
(iv) x/2 + 2y/3 = -1 and x – y/3 = 3
Solution:
(i) By Elimination Method:
Fquations are x + y = 5
and 2x – 3y = 4
Multiply equation (i) by 2 and subtract equation (ii) from it, we have

(ii) By Elimination method:
Equations are 3x + 4y = 10
and 2x – 2y = 2
Multiplying equation (ii) by 2 and adding to equation (i), we

(iii) By Elimination Method:

(iv) By Elimination Method:

Substracting equation (i) from (iii) we get
5y = – 15
⇒ y = – 3
Subtracting the value of y = – 3 in (ii), we get
3x – (- 3) = 9
3x + 3 = 9
3x = 6 x = 2
∴ x = 2 and y = 3
Substituting method
From equation (ii) we have
3x – y = 9
⇒ 3x = 9 + y
⇒ x = \(\frac { 9 + y }{ 3 }\)
Substituting the value of x = equation, (i) we get
3(\(\frac { 9 + y }{ 3 }\)) + 4y = – 6
⇒ 9 + y + 4y = – 6
⇒ 5y = – 15
⇒ y = – 3
Again, substituting the value of y = -3 in equation (i) we get
3x – (-3) = 9
⇒ 3x + 3 = 9
⇒ 3x = 6
⇒ x = 2
∴ x = 2 and y = – 3

Question 2.
Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method:
(i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes – if we only add 1 to the denominator. What is the fraction₹

(ii) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?

(iii) The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.

(iv) Meena went to a bank to withdraw ₹ 2000. She asked the cashier to give her ₹ 50 and ₹ 100 notes only. Meena got 25 notes in all. Find how many notes of ₹ 50 and ₹ 100 she received.

(v) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid ₹ 27 for a book kept for seven days, while Susy paid ₹ 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.
Solution:
(i) Let numerator be x and denominator be y.
\(\frac { x + 1 }{ y – 1 }\) = 1
x – y = – 2 … (i)
According to question’s second condition
\(\frac { x }{ y + 1 }\) = \(\frac { 1 }{ 2 }\)
2x – y = 1 … (i)
Substract equation (i) from equation we eliminate y
x = 3
Substitute this value in equation (i) we get
3 – y = – 2
y = 5
Hence the fraction is \(\frac { 3 }{ 5 }\)

(ii) Let the present age of Nuri be x
and let the present age of Sonu be y.
According to questions first condition
(x – 5) = 3 (y – 5)
x – 3y = – 10 … (i)
According to eq. question’s second condition
(x + 10) = 2 (y+ 10)
x – 2y = 10 … (ii)
Substracting equation (i) from equation (ii)
We eliminate for x.
y = 20
Putting this value in equation (i)
x – 60 = – 10
x = 50
Present age of Nuri is 50 years old and Sonu is 20 years old.

(iii) Let the ten’s digit of the number be x
and the unit’s digit of the number by y.
According to question’s first condition
x + y = 9 … (i)
According to question’s second condition
9(10 x + y) = 2(10y + x)
90x + 9y – 20y + 2x
88x = 11y
8x – y = 0 (dividing by 11)
8x – y = 0 … (ii)
Adding equation (i) and (ii) we eleminate for y, we get
9x = 9
x = 1
Putting the value in equation (i)
1 + y = 9
y = 8
Hence the units digit is 8 and the ten’s digit is 1. Then the number is 18.

(iv) Let the notes of ₹ 50 be x
and let the notes of ₹ 100 be y.
According to question’s first condition
50x + 100y = 2000
x + 2 y= 40 … (i)
According to question second condition
x + y = 25
Substract equation (i) from equation (ii) we eleminate for x.
– y = – 15
y = 15
Putting this value in equation (i)
x + 30 = 40
x = 40
Hence ₹ 50 notes is 10 and the notes of ₹ 100 is 15.

(v) Let the fixed charge be ₹ x and let the additional charge per day be ₹ y.
According to questions first condition
x + 4y = 27 … (i)
According to questions second condition
x + 2y = 21 … (ii)
Eliminating x for equation (i) and (ii)
We subtract equation (ii) from equation (1)
2y = 6
y = 3
Putting this value in equation (ii)
x + 6 = 21
x = 21 – 6
x = 15
Hence the fixed charge is ₹ 15 and the additional charge per day is ₹ 3.

These NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.5 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.5

Question 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 – Sy = 20
6x – 10y = 40

(iv) x – 3y – 7 = 0
3x – 3y – 15 = 0
Solution:

(iii) Equations are 3x – 5y = 20 and 6x – 10y = 40
Here,

So the linear equation has infinitely many solution.

(iv) We have

Question 2.
(i) for which values of a and b does the following pair of linear equation 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 equation have no solution.
3x + y = 1
(2k – 1)x + (k – 1)y = 2k + 1
Solution:
(i) We have
2x + 3y – 7 = 0 … (i)
(a – b) x + (a + b) y – (3a + b – 2) = 0 … (ii)
Here, a₁ = 2, b₁ = 3, c₁ = 7
a₂ = (a – b), b₂ = a + b, c₃ = – (3a + b – 2)

Substracting equation (iv) from equation (iii) we eleminute a
– 4b = – 4
b = 1
Putting this value in equation (iv)
a – 5 = 0
d = 5
Hence a = 5 and h = 1 are the values when equation gives infinite many solution.

(ii) We have
3x + y – 1 = 0 … (i)
(2k – 1)s + (k – 1)y – 2k + 1 = 0 … (ii)
a₁ = 3, b₂ = 1
a₂ = (2k – 1) b₂ = (k – 1)
For no solution

k = 2 is the value when equation has no solution.

Question 3.
Solve the following pair of linear equations by the substitution and cross-multiplication methods:
8x + 5y = 9
3x + 2y = 4
Solution:
Equations are

By cross multiplication Method :
Equations are

Question 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 ₹ 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.

(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 x be the fixed charge of the food and y be the charge of food per day.
According to question’s first condition
x + 20y = 1000 … (i)
According to question’s second condition
x + 26y = 1180 … (ii)
By elimination method
Subtract equation (i) from equation (ii)
We get
6y = 180
y = 30
Putting this value is equation (i), we get
x + 20 x 30 = 1000
x = 400

(ii) Let the numerator of the fraction be x and the denominator of the fraction by y.
According to equation first condition
\(\frac { x – 1 }{ y }\) = \(\frac { 1 }{ 3 }\)
3x – y = 3 … (i)
According to question’s second condition
\(\frac { x }{ y + 8}\) = \(\frac { 1 }{ 4 }\)
4x – y = 8 … (ii)
By elemination method
Substracted equation (i) from equation (ii)
x = 5
Putting this value in equation (i)
y = 12
Hence the fraction is \(\frac { 5 }{ 12 }\)

(iii) Let the number of right answers be x and the number of wrong answers by y.
According to question
3x – y = 40 … (i)
4x – 2y = 50
2x – y = 25 … (ii)
By elimination method
Substracted equation (ii) from equation (i)
x = 15
Putting this value in equation (i)
45 – y = 40
y = 5
So the right questions are 15 are wrong questions are 5. The total questions are 20.

(iv) Let the speed of one car be u km/h.
and the speed of another car be v km/h.
According to question
u – v = \(\frac { 100 }{ 5 }\)
u – v = 20
and u + v = 100
By elimination method
Adding both equations
2u = 120
u = 60 km/h
Putting this value in equation in (i)
v = 40 km/h
Hence the speed of one car is 60 km/h and another is 40 km/h.

(v) Let the length of the rectangle be x unit.
and the breadth of the rectangle be y unit.
According to question
In first case
Area of rectangle = x × y
(x – 5) (y + 3) = xy – 9
or 3x – 5y – 6 = 0
In second case
(x + 3) (y – 2) = xy + 67
or 2x + 3y – 61 = 0
By cross multiplication method.