NCERT Solutions class 10 maths chapter 4 – Pair of Linear Equations in Two Variables

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
Solutions: (i) 2x² – 7x + 3 = 0
⇒ 2x² – 7x = – 3
Dividing by 2 on both sides, we get
⇒ x² - - =0
⇒ x² - 2× x =
On adding ()² to both sides , we get
⇒ (x)² - 2× x + ()² = ()² -
⇒ (x - )² = () – ()
⇒(x- )² =
⇒(x- )² = ±
⇒ x = ±
⇒ x = + or x = –
⇒ x = or x =
⇒ x = 3 or x =
(ii) 2x² + x – 4 = 0
⇒ 2x² + x = 4
Dividing both sides of the equation by 2
⇒ x² +x/2 = 2
Now on adding (1/4)² to both sides
⇒ (x)² + 2 × x × 1/4 + (1/4)² = 2 + (1/4)²
⇒ (x + 1/4)² = 33/16
⇒ x + 1/4 = ± √33/4
⇒ x = ± √33/4 – 1/4
⇒ x = ± √33-1/4
Therefore, either x = √33-1/4 or x = -√33-1/4
(iii) 4x² + 4√3x + 3 = 0
⇒ (2x)² + 2 × 2x × √3 + (√3)² = 0
⇒ (2x + √3)² = 0
⇒ (2x + √3) = 0 and (2x + √3) = 0
Therefore, either x = -√3/2 or x = -√3/2.
(iv) 2x² + x + 4 = 0
⇒ 2x² + x = -4
Dividing both sides of the equation by 2,
⇒ x² + 1/2x = 2
⇒ x² + 2× x 1/4 = -2
By adding (1/4)² to both sides of the equation,
⇒ (x)² + 2x × 1/4 + (1/4)² = (1/4)² – 2
⇒ (x + 1/4)² = 1/16 – 2
⇒ (x + 1/4)² = -31/16
the square of numbers cannot be negative.
Therefore, there is no real root for the given equation, 2x² + x + 4 = 0.

Question 2

Find the roots of the quadratic equations given in Q.1 above by applying the quadratic formula.

Answer 2

(i) 2x² – 7x + 3 = 0
On comparing the given equation with ax² + bx + c = 0
a = 2, b = -7 and c = 3
⇒ x = (7±√(49 – 24))/4
⇒ x = (7±√25)/4
⇒ x = (7±5)/4
⇒ x = (7+5)/4 or x = (7-5)/4
⇒ x = 12/4 or 2/4
∴ x = 3 or 1/2
(ii) 2x² + x – 4 = 0
On comparing the given equation with ax² + bx + c = 0
a = 2, b = 1 and c = -4
⇒x = -1±√1+32/4
⇒x = -1±√33/4
∴ x = -1+√33/4 or x = -1-√33/4
(iii) 4x² + 4√3x + 3 = 0
On comparing the given equation with ax² + bx + c = 0,
a = 4, b = 4√3 and c = 3
⇒ x = -4√3±√48-48/8
⇒ x = -4√3±0/8
∴ x = -√3/2 or x = -√3/2
(iv) 2x² + x + 4 = 0
On comparing the given equation with ax² + bx + c = 0,
a = 2, b = 1 and c = 4
⇒ x = -1±√1-32/4
⇒ x = -1±√-31/4
the square of a number can never be negative. Therefore, there is no real solution for the given equation.

Question 3

Find the roots of the following equations:
(i) x-1/x = 3, x ≠ 0
(ii) 1/x+4 – 1/x-7 = 11/30, x = -4, 7
Solution: (i) x-1/x = 3
⇒ x² – 3x -1 = 0
On comparing the given equation with ax² + bx + c = 0,
a = 1, b = -3 and c = -1
⇒ x = 3±√9+4/2
⇒ x = 3±√13/2
∴ x = 3+√13/2 or x = 3-√13/2
(ii) 1/x+4 – 1/x-7 = 11/30
⇒ x-7-x-4/(x+4)(x-7) = 11/30
⇒ -11/(x+4)(x-7) = 11/30
⇒ (x+4)(x-7) = -30
⇒ x² – 3x – 28 = 30
⇒ x² – 3x + 2 = 0
by factorization method
⇒ x² – 2x – x + 2 = 0
⇒ x(x – 2) – 1(x – 2) = 0
⇒ (x – 2)(x – 1) = 0
⇒ x = 1 or 2

Question 4

The sum of the reciprocals of Rehman’s ages, (in years) 3 years ago and 5 years from now is 1/3. Find his present age.

Answer

Let present age of Rehman is x years.

Three years ago, Rehman’s age was (x – 3) years.

Five years after, his age will be (x + 5) years.

the sum of the reciprocals of Rehman’s ages 3 years ago and after 5 years is equal to 1/3.

∴ 1/x-3 + 1/x-5 = 1/3

(x+5+x-3)/(x-3)(x+5) = 1/3

(2x+2)/(x-3)(x+5) = 1/3

⇒ 3(2x + 2) = (x-3)(x+5)

⇒ 6x + 6 = x² + 2x – 15

⇒ x² – 4x – 21 = 0

⇒ x² – 7x + 3x – 21 = 0

⇒ x(x – 7) + 3(x – 7) = 0

⇒ (x – 7)(x + 3) = 0

⇒ x = 7, -3

age cannot be negative.

Therefore, Rehman’s present age is 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.

Answer

Let the marks of Shefali in Maths x.

Then, the marks in English will be 30 – x.

According to question

(x + 2)(30 – x – 3) = 210

(x + 2)(27 – x) = 210

⇒ -x² + 25x + 54 = 210

⇒ x² – 25x + 156 = 0

⇒ x² – 12x – 13x + 156 = 0

⇒ x(x – 12) -13(x – 12) = 0

⇒ (x – 12)(x – 13) = 0

⇒ x = 12, 13

when x=12,marks in Maths then marks in
English 30 – 12 = 18 and
the marks in Maths are 13, then marks in
English 30 – 13 = 17.

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.

Answer

Let the shorter side of the rectangle be x m.

Then, larger side of the rectangle = (x + 30) m

the length of the diagonal is = x + 30 m

⇒ x² + (x + 30)² = (x + 60)²

⇒ x² + x² + 900 + 60x = x² + 3600 + 120x

⇒ x² – 60x – 2700 = 0

⇒ x² – 90x + 30x – 2700 = 0

⇒ x(x – 90) + 30(x -90) = 0

⇒ (x – 90)(x + 30) = 0

⇒ x = 90, -30

side of the field cannot be negative. So the length of the shorter side will be 90 m.

and the length of the larger side will be (90 + 30) m = 120 m.

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.

Answer

Let the larger and smaller number be x and y,

According to question

x² – y² = 180 and y² = 8x

⇒ 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, -10

Now x = 18

∴ y² = 8x = 8 × 18 = 144

⇒ y = ±√144 = ±12

∴ Smaller number = ±12

Again x= -10 and y²=[8 x (-10)]= 80 which is not possible

So the numbers are 18 and 12 or 18 and -12.

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.

Answer

Let the uniform speed of the train be x km/hr.

Time taken to cover 360 km = 360/x hr.

According question,

⇒ (x + 5)(360-1/x) = 360

⇒ 360 – x + 1800-5/x = 360

⇒ x² + 5x + 10x – 1800 = 0

⇒ x(x + 45) -40(x + 45) = 0

⇒ (x + 45)(x – 40) = 0

⇒ x = 40, -45

As speed cannot be negative.

So, the speed of train is 40 km/h.

Question 9

Two water taps together can fill a tank in 9 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.

Answer

Let the time taken by the smaller pipe to fill the tank = x hr.

Time taken by the larger pipe = (x – 10) hr

Part of tank filled by smaller pipe in 1 hour = 1/x

Part of tank filled by larger pipe in 1 hour = 1/(x – 10)

the tank can be filled in 9 = 75/8 hours by both the pipes together.

so,

1/x + 1/x-10 = 8/75

x-10+x/x(x-10) = 8/75

⇒ 2x-10/x(x-10) = 8/75

⇒ 75(2x – 10) = 8x² – 80x

⇒ 150x – 750 = 8x² – 80x

⇒ 8x² – 230x +750 = 0

⇒ 8x² – 200x – 30x + 750 = 0

⇒ 8x(x – 25) -30(x – 25) = 0

⇒ (x – 25)(8x -30) = 0

⇒ x = 25, 30/8

Time taken by the smaller pipe cannot be 30/8 = 3.75 hours, as the time taken by the larger pipe will become negative, which is not possible.

So, time taken by each tap to fill the tank separetely are 25 – 10 =15 hours

Question 10

An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore (without taking into consideration the time they stop at intermediate stations). If the average speeds of the express train is 11 km/h more than that of the passenger train, find the average speed of the two trains.

Answer

Let the average speed of passenger train = x km/h.

Average speed of express train = (x + 11) km/h

time taken by passenger train to cover 132 km is =

h

According to question

(132/x) – (132/(x+11)) = 1

132(x+11-x)/(x(x+11)) = 1

132 × 11 /(x(x+11)) = 1

⇒ 132 × 11 = x(x + 11)

⇒ x² + 11x – 1452 = 0

⇒ x² + 44x -33x -1452 = 0

⇒ x(x + 44) -33(x + 44) = 0

⇒ (x + 44)(x – 33) = 0

⇒ x = – 44, 33

Speed cannot be negative.

Hence , the speed of the passenger train will be 33 km/h and the average speed of the express train will be 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.

Answer

suppose sides of the two squares be x m and y m.

And area of the squares will be x² and y² respectively.

and their perimeter will be 4x and 4y respectively

According to question ,

x² + y² =468

and difference of their perimeter =24

4x – 4y = 24

x – y = 6

x = y + 6

x² + y² = 468

⇒ (6 + y²) + y² = 468

⇒ 36 + y² + 12y + y² = 468

⇒ 2y² + 12y + 432 = 0

⇒ y² + 6y – 216 = 0

⇒ y² + 18y – 12y – 216 = 0

⇒ y(y +18) -12(y + 18) = 0

⇒ (y + 18)(y – 12) = 0

⇒ y = -18, 12

As the side of a square cannot be negative.

Hence, the sides of the squares are 12 m and (12 + 6) m = 18 m.