Solve the followings Questions.

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

Answer:

(i)2x² – 7x + 3 = 0

First we divide equation by 2,

(ii) 2x²+ x – 4 = 0

Dividing equation by 2,

(iii) 4x²+4√3x+3 = 0

Dividing equation by 4,

(iv) 2x²+ x + 4 = 0

Dividing equation by 2,

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

Answer:

(i) 2x² – 7x + 3 = 0

Comparing quadratic equation 2x² – 7x + 3 = 0 with general form ax² + bx + c = 0, we get a = 2, b = -7 and c = 3

Putting these values in quadratic formula

(ii) 2x²+ x – 4 = 0

Comparing quadratic equation 2x²+ x – 4 = 0 with the general form ax² + bx + c = 0 , we get a = 2, b = 1 and c = −4

Putting these values in quadratic formula ,

(iii) 4x²+4√3x+3 = 0

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

Putting these values in quadratic formula ,

(iv) 2x²+ x + 4 = 0

Comparing quadratic equation 2x²+ x + 4 = 0 with the general form ax² + bx + c = 0, we get a = 2, b = 1 and c = 4

Putting these values in quadratic formula

3. Find the roots of the following equations:

(i) 

(ii)

Answer:

(i)

Comparing equation with general form ax² + bx + c = 0,

We get a = 1, b = −3 and c = −1

Using quadratic formula

(ii)

⇒ x(x -2) -1 (x – 2) = 0

⇒ (x -1)(x – 2) = 0

⇒ x = 2, 1

4. The sum of reciprocals of Rehman’s ages (in years) 3 years ago and 5 years from now is 13. Find his present age.

Answer:

Let present age of Rehman = x years

Age of Rehman 3 years ago = (x − 3) years.

Age of Rehman after 5 years = (x + 5) years

According to the given condition:

It is given that the sum of the reciprocals of Rehman’s ages 3 years ago and 5 years from now is 1/3.

⇒ 3 (2x + 2) = (x − 3) (x + 5)

⇒ 6x + 6 = x² + 2x – 15

⇒x² – 4x -21 = 0

⇒ x(x – 7) + 3(x – 7) = 0

⇒ (x – 7)(x + 3) = 0

⇒ x = 7, -3

However, age cannot be negative.

Therefore, Rehman’s present age 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.

Answer:

Let Shefali’s marks in Mathematics = x

Let Shefali’s marks in English = 30 − x

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

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

According to given condition:

(x + 2) (27 − x) = 210

⇒ – x² + 25x + 156 = 0

⇒ x(x – 12) -13(x – 12) = 0

⇒ (x – 12)(x – 13) = 0

⇒ x = 12, 13

If the marks in Maths are 12, then marks in English will be 30 – 12 = 18

If the marks in Maths are 13, then marks in English will be 30 – 13 = 17

Therefore, her 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.

Answer:

Let the shorter side of the rectangle be x m.

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

Diagonal of rectangle =

It is given that the diagonal of the rectangle = (x+30)m

⇒ x² + (x + 30)² = (x + 60)²

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

⇒ x² – 60x – 2700 = 0

⇒ x(x – 90) + 30(x -90)

⇒ (x – 90)(x + 30) = 0

⇒ x = 90, -30

However, side cannot be negative. Therefore, the length of the shorter side will be 90 m.

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

Therefore, length of sides are 90 and 120 in 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.

Answer:

Let the larger and smaller number be x and y respectively.

According to the 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

However, the larger number cannot be negative as 8 times of the larger number will be negative and hence, the square of the smaller number will be negative which is not possible.

Therefore, the larger number will be 18 only.

x = 18

∴ y² = 8x = 8 × 18 = 144

⇒ y = 

∴ Smaller number = ±12

Therefore, the numbers are 18 and 12 or 18 and – 12.

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/hr more, it would have taken 1 hour less for the same journey. Find the speed of the train.

Answer:

Let the original speed of train be x km/hr. Then,

Increased speed of the train = (x + 5)km/hr

Time taken by the train under usual speed to cover 360 km =

Time taken by the train under increased speed to cover 360 km = 

⇒ 1800 = x² + 5x

⇒ x² + 5x – 1800 = 0

⇒ x² – 40x + 45x -1800 = 0

⇒ x(x – 40) + 45(x – 40) = 0

⇒(x – 40)(x + 45) = 0

So, either x – 40 = 0

x = 40

Or

x + 45 = 0

x = -45

But, the speed of the train can never be negative.

Hence, the original speed of train is x = 40 km/hr

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.

Answer:

Let the time taken by the smaller pipe to fill the tank be 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 = 

It is given that the tank can be filled in   hours by both the pipes together. Therefore,

⇒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 in this case, the time taken by the larger pipe will be negative, which is logically not possible.

Therefore, time taken individually by the smaller pipe and the larger pipe will be 25 and 25 – 10 =15 hours respectively.

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 two trains.

Answer:

Let the speed of the passenger train be x km/hr. Then,

Speed of the express train = (x + 11)km/hr

Time taken by the passenger train to cover 132 km between Mysore to Bangalore = hours

Time taken by express train to cover 132 km = 

According to the given condition,

⇒ 1452 = x² + 11x

⇒ x² – 33x + 44x -1452 = 0

x(x – 33) + 44(x – 33) = 0

(x – 33)(x + 44) = 0

So, either

x – 33 = 0

x = 33

Or

x + 44 = 0

x = -44

But, the speed of the train can never be negative.

Thus, when x = 33 then speed of express train

= x + 11

= 33 + 11

= 44

Hence, the speed of the passenger train is x = 33 km/hr

and the speed of the express train is x = 44 km/hr respectively.

11. Sum of areas of two squares is 468 m2. If, the difference of their perimeters is 24 metres, find the sides of the two squares.

Answer:

Let the side of the first square be ‘a’ m and that of the second be ′A′ m.

Area of the first square = a² sq m.

Area of the second square = A² sq m.

Their perimeters would be 4a and 4A respectively.

Given 4A – 4a = 24

A – a = 6              ……(1)

A²  + a²  = 468 ……….(2)

From (1), A = a + 6

Substituting for A in (2), we get

(a + 6)² + a² = 468

a² + 12a + 36 + a² = 468

2a² + 12a + 36 = 468

a² + 6a + 18 = 234

a² + 6a – 216 = 0

a² + 18a – 12a – 216 = 0

a(a + 18) – 12(a + 18) = 0

(a – 12)(a + 18) = 0

a = 12, – 18

So, the side of the first square is 12 m. and the side of the second square is 18 m.

CHAPTER NAME	OLD NCERT	NEW NCERT
Real Numbers	EXERCISE 1.1
	EXERCISE 1.2	1.1	CLICK HERE
	EXERCISE 1.3	1.2	CLICK HERE
	EXERCISE 1.4
Polynomials	EXERCISE 2.1	2.1	CLICK HERE
	EXERCISE 2.2	2.2	CLICK HERE
	EXERCISE 2.3
	EXERCISE 2.4
Pair of Linear Equations in Two Variables	EXERCISE 3.1
	EXERCISE 3.2	3.1	CLICK HERE
	EXERCISE 3.3	3.2	CLICK HERE
	EXERCISE 3.4	3.3	CLICK HERE
	EXERCISE 3.5
	EXERCISE 3.6
	EXERCISE 3.7
Quadratic Equations	EXERCISE 4.1	4.1	CLICK HERE
	EXERCISE 4.2	4.2	CLICK HERE
	EXERCISE 4.3
	EXERCISE 4.4	4.3	CLICK HERE
Arithmetic Progressions	EXERCISE 5.1	5.1	CLICK HERE
	EXERCISE 5.2	5.2	CLICK HERE
	EXERCISE 5.3	5.3	CLICK HERE
	EXERCISE 5.4	5.4 (Optional)	CLICK HERE
Triangles	EXERCISE 6.1	6.1	CLICK HERE
	EXERCISE 6.2	6.2	CLICK HERE
	EXERCISE 6.3	6.3	CLICK HERE
	EXERCISE 6.4
	EXERCISE 6.5
	EXERCISE 6.6
Coordinate Geometry	EXERCISE 7.1	7.1	CLICK HERE
	EXERCISE 7.2	7.2	CLICK HERE
	EXERCISE 7.3
	EXERCISE 7.4
Introduction to Trigonometry	EXERCISE 8.1	8.1	CLICK HERE
	EXERCISE 8.2	8.2	CLICK HERE
	EXERCISE 8.3
	EXERCISE 8.4	8.3	CLICK HERE
Some Applications of Trigonometry	EXERCISE 9.1	9.1	CLICK HERE
Circles	EXERCISE 10.1	10.1	CLICK HERE
	EXERCISE 10.2	10.2	CLICK HERE
Construction
Areas Related to Circles	EXERCISE 12.1
	EXERCISE 12.2	11.1	CLICK HERE
	EXERCISE 12.3
Surface Areas and Volumes	EXERCISE 13.1	12.1	CLICK HERE
	EXERCISE 13.2	12.2	CLICK HERE
	EXERCISE 13.3
	EXERCISE 13.4
	EXERCISE 13.5
Statistics	EXERCISE 14.1	13.1	CLICK HERE
	EXERCISE 14.2	13.2	CLICK HERE
	EXERCISE 14.3	13.3	CLICK HERE
	EXERCISE 14.4
Probability	EXERCISE 15.1	14.1	CLICK HERE
	EXERCISE 15.2