NCERT Solutions Maths Class 10 Chapter 4

NCERT Solutions Maths Class 10 Chapter 4

 

NCERT Solutions Maths Class 10 Exercise 4.3

 

1. Find the roots of the following quadratic equations, if they exist, by the method of completing square:

(i) 2x2 − 7x + 3 = 0

(ii) 2x2 +  x − 4 = 0

(iii) 4x2 + 4√3x + 3 = 0

(iv) 2x2 + x + 4 = 0

 

Solution:  (i) 2x2 − 7x + 3 = 0

Following the procedure of completing square, we first divide the given equation by 2 to make the coefficient of x2 equal to 1.

⇒ 

We divide middle term of the equation by 2x, we get 

We add and subtract square of 7/4 from the equation 


 

[Since (a b)2 = a2 2ab + b2]

 

 

⇒ 

Taking square root on both sides, we get

 

 

Therefore, x = 1/2, 3

 

(ii) 2x2 +  x − 4 = 0

Following the procedure of completing square, we first divide the given equation by 2 to make the coefficient of x2 equal to 1.

⇒ 

We divide middle term of the equation by 2x, we get x/2 ÷ 2x = 1/4

We add and subtract square of 1/4 from the equation

⇒ 

 

[Since (a + b)2 = a2 + 2ab + b2]

 

 

Taking square root on both sides, we get

 

 

Therefore, 

 

(iii) 4x2 + 4√3x + 3 = 0

Following the procedure of completing square, we first divide the given equation by 4 to make the coefficient of x2 equal to 1.

⇒ 

We divide middle term of the equation by 2x, we get √3x ÷ 2x = √3/2

We add and subtract square of √3/2 from the equation 

 

 

 [Since (a + b)2 = a2 + 2ab + b2]

 

 

 

 

(iv) 2x2 + x + 4 = 0

Following the procedure of completing square, we first divide the given equation by 2 to make the coefficient of x2 equal to 1.

⇒ 

We divide middle term of the equation by 2x, we get x/2 ÷ 2x = 1/4

We add and subtract square of 1/4 from the equation 

 

 

[Since (a + b)2 = a2 + 2ab + b2]

 

 

⇒ 

Taking square root on both sides, right hand side does not exist because square root of negative number does not exist.

Therefore, there is no solution for the quadratic equation 2x2 + x + 4 = 0.

 

2. Find the roots of the following quadratic equations by applying 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) 2x2 − 7x + 3 = 0

Comparing the equation 2x2 − 7x + 3 = 0 with the general form of a quadratic equation ax2 + bx + c = 0, we get

a = 2, b = −7 and c = 3

Using the quadratic formula, we have

⇒ 

Putting the values of a, b and c, we get

⇒ 

 

 

 

 = 3, 1/2

 

(ii) 2x2 + x − 4 = 0

Comparing the equation 2x2 + x − 4 = 0 with the general form of a quadratic equation ax2 + bx + c = 0, we get

= 2, = 1 and = −4

Using the quadratic formula, we have

⇒  

Putting the values of a, b and c, we get

⇒ 

 

 


(iii) 4x2 + 4√3x + 3 = 0

Comparing the equation 4x2 + 4√3x + 3 = 0 with the general form of a quadratic equation ax2 + bx + c = 0, we get

= 4, 4√3 and = 3

Using the quadratic formula, we have

 ⇒ 

Putting the values of a, b and c, we get

⇒ 

 

 

A quadratic equation has two roots. Here, both the roots are equal.

Therefore, 

 

(iv) 2x2 + x + 4 = 0

Comparing the equation 2x2 + x + 4 = 0 with the general form of a quadratic equation ax2 + bx + c = 0, we get

= 2, = 1 and = 4

Using the quadratic formula, we have

 ⇒ 

Putting the values of a, b and c, we get

⇒ 
 

Since the square root of negative number is not defined.

Therefore, the quadratic equation 2x2 + x + 4 = 0 has no solution.

 

 

3. Find the roots of the following equations:

(i) 

(ii) 

Solution:  (i) 

 

x2 − 1 = 3x

x2 − 3x − 1 = 0

Comparing the equation x− 3x − 1 = 0 with the general form of a quadratic equation ax2 + bx + c = 0, we get

= 1, = −3 and = −1

Using the quadratic formula, we have

⇒ 

Putting the values of a, b and c, we get

⇒ 

 

 


(ii) 

 

 

−30 = x2 − 7x + 4x − 28

 x2 − 3x + 2 = 0

Comparing the equation x− 3x + 2 = 0 with the general form of a quadratic equation ax2 + bx + c = 0, we get

= 1, = −3 and = 2

Using the quadratic formula, we have

 ⇒ 

Putting the values of a, b and c, we get

⇒ 

 

 

 = 2, 1

 

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

 

Solution:  Let the present age of Rehman be x years.

3 years ago, the age of Rehman = (− 3) years.

After 5 years, the age of Rehman = (+ 5) years.

According to the question, we have

⇒ 

 

3(2+ 2) = (− 3) (+ 5)

6+ 6 = x2 − 3x + 5x – 15

6+ 6 = x2 + 2x − 15

 x2 + 2x − 6x – 15 − 6 = 0

 x2 − 4x – 21 = 0

Comparing the equation x2 − 4x – 21 = 0 with the general form of a quadratic equation ax2 + bx + c = 0, we get

= 1, = −4 and = −21

Using the quadratic formula, we have

Putting the values of a, b and c, we get

 

 

 

 = 7, −3

Here, = 7    [We discard x = −3, because the age cannot be negative.]

Therefore, the present age of Rehman is 7 years.

 

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 Shefali’s marks in Mathematics be x.

Then, Shefali’s marks in English is 30 – x.

If she had got 2 marks more in Mathematics, her marks in Mathematics would be = + 2

If she had got 3 marks less in English, her marks in English would be = 30 – − 3 = 27 − x

According to the question, we have

(+ 2) (27 − x) = 210

 27xx2 + 54 − 2x = 210

 x2 − 25x + 156 = 0

Comparing the equation x2 − 25x + 156 = 0 with the general form of a quadratic equation ax2 + bx + c = 0, we get

= 1, = −25 and = 156

Using the quadratic formula, we have

Putting the values of a, b and c, we get

⇒ 

 

 

 

 = 13, 12

Therefore, Shefali’s marks in Mathematics = 13 or 12

Thus, Shefali’s marks in English = 30 – = 30 – 13 = 17

Or Shefali’s marks in English = 30 – = 30 – 12 = 18

Hence, Shefali’s marks in Mathematics and English are (13, 17) or (12, 18).

 

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 the shorter side of the rectangular field be x metres.

Then, the diagonal of the rectangular field = (+ 60) metres

And the longer side of the rectangular field = (+ 30) metres

According to Pythagoras theorem, we have

(x + 60)2 = (x + 30)2 + x2

 x2 + 3600 + 120x = x2 + 900 + 60x + x2         [Since (a + b)2 = a2 + b2 + 2ab]

 x2 – 60x – 2700 = 0

Comparing the equation x2 – 60x – 2700 = 0 with the general form of a quadratic equation ax2 + bx + c = 0, we get

= 1, = −60 and = −2700

Using the quadratic formula, we have

Putting the values of a, b and c, we get

⇒ 

 

 

 

 = 90, –30

 = 90                  [We discard –30, because the length cannot be in negative.]

Therefore, the length of the shorter side = 90 metres

And the length of the longer side = + 30 = 90 + 30 = 120 metres

Hence, the sides of the rectangular field are 90 metres and 120 metres.

 

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 the smaller number be and the larger number be y.

According to the given condition, we have

y2 x2 = 180 … (1)

It is also given that the square of smaller number is 8 times the larger number.

x2 = 8y … (2)

Putting the value of x2 from equation (2) in equation (1), we get

y2 – 8y = 180

 y2 – 8y – 180 = 0

Comparing the equation y2 – 8y – 180 = 0 with the general form of a quadratic equation ay2 + by + c = 0, we get

= 1, = −8 and = −180

Using the quadratic formula, we have

⇒ 

Putting the values of a, b and c, we get

⇒ 

 

 

 

 = 18, −10

Putting the values of y in equation (2), we get

x2 = 8y

x2 = 8y = 8 × 18 = 144

 = ±12

And, x2 = 8y = 8 × −10 = −80            

[No real solution for x, because square root of negative number cannot be found.]

Therefore, two numbers are (12, 18) or (−12, 18).

 

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:  Let the speed of the train be x km/h.

Time taken to cover 360 km = 360/x hour

If the speed had been 5 km/h more, the train would have taken 1 hour less for the same journey.

According to the given condition, we have

⇒ 

 
 

360 × 5 = x2 + 5x

 x2 + 5x − 1800 = 0

Comparing the equation x2 + 5x − 1800 = 0 with the general form of a quadratic equation ax2 + bx + c = 0, we get

= 1, = 5 and = −1800

Using the quadratic formula, we have

Putting the values of a, b and c, we get

 

 

 

 = 40, −45

[We discard = −45, because the speed of the train cannot be negative.]

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


9. Two water taps together can fill a tank in 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 the time taken by tap of smaller diameter to fill the tank be x hours.

Then, the time taken by tap of larger diameter to fill the tank = (– 10) hours.

It means that the tap of smaller diameter fills (1/x)th part of tank in 1 hour … (1)

And, the tap of larger diameter fills part of tank in 1 hour … (2)

When both the taps are opened together, they fill the tank in 75/8 hours.

In 1 hour, they fill (8/75)th part of the tank … (3)

From equations (1), (2) and (3), we have

⇒ 

 

75(2− 10) = 8(x2− 10x)

150– 750 = 8x2 − 80x

8x2 − 80x − 150x + 750 = 0

8x2 − 230x + 750 = 0

Dividing the above equation by 2, we get

4x2 − 115x + 375 = 0

Comparing the equation 4x2 − 115x + 375 = 0 with the general form of a quadratic equation ax2 + bx + c = 0, we get

= 4, = −115 and = 375

Using the quadratic formula, we have

Putting the values of a, b and c, we get

 

 

 

 

 = 25, 3.75

The time taken by tap of larger diameter to fill the tank = – 10 = 3.75 – 10 = −6.25 hours

Time cannot be in negative. Therefore, we ignore this value.

The time taken by tap of larger diameter to fill the tank = – 10 = 25 – 10 = 15 hours

Hence, the time taken by tap of larger diameter to fill the tank is 15 hours and by tap of smaller diameter is 25 hours.

 

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 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 the passenger train be x km/h.

Then, the average speed of the express train = (+ 11) km/h

Time taken by the passenger train to travel 132 km = 132/x hours

Time taken by the express train to travel 132 km =  132/(x + 11) hours

According to the question, we have

⇒ 

 

 

132(11) = x(+ 11)

 1452 = x2 + 11x

x2 + 11x – 1452 = 0

Comparing the equation x2 + 11x – 1452 = 0 with the general form of a quadratic equation ax2 + bx + c = 0, we get

= 1, = 11 and = −1452

Using the quadratic formula, we have

 

Putting the values of a, b and c, we get

 

 

 

 

 

 = 33, −44

Therefore, the speed of the passenger train = 33 km/h

[We discard −44, because the speed cannot be in negative.]

And, the speed of the express train = + 11 = 33 + 11 = 44 km/h

 

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 perimeter of the first square be x metres.

Then, the perimeter of the second square = (+ 24) metres

The length of side of the first square = x/4 metres  

[Since the perimeter of a square = 4 × length of a side]

The length of side of the second square = (+ 24)/4 metres

Area of the first square = side × side = x/4 × x/4 = x2/16 sq. m

Area of the second square = 

According to the question, we have

 
 

 

2x2 + 576 + 48x = 468 × 16

2x2 + 48x + 576 = 7488

2x2 + 48x – 6912 = 0

x2 + 24x – 3456 = 0

Comparing the equation x2 + 24x – 3456 = 0 with the general form of a quadratic equation ax2 + bx + c = 0, we get

= 1, =24 and = −3456

Using the quadratic formula, we have

 

Putting the values of a, b and c, we get

 

 

 

 

 = 48, −72

Therefore, the perimeter of the first square = 48 metres

[We discard x = −72, because the perimeter of a square cannot be negative.]

And, the perimeter of the second square = + 24 = 48 + 24 = 72 metres

Therefore, the side of the first square = 

And, the side of the second square = 

Hence, the sides of the two squares are 12 m and 18 m.

Please do not enter any spam link in the comment box.

Post a Comment (0)
Previous Post Next Post