- RD Chapter 8- Quadratic Equations Ex-8.1
- RD Chapter 8- Quadratic Equations Ex-8.2
- RD Chapter 8- Quadratic Equations Ex-8.3
- RD Chapter 8- Quadratic Equations Ex-8.4
- RD Chapter 8- Quadratic Equations Ex-8.5
- RD Chapter 8- Quadratic Equations Ex-8.6
- RD Chapter 8- Quadratic Equations Ex-8.7
- RD Chapter 8- Quadratic Equations Ex-8.9
- RD Chapter 8- Quadratic Equations Ex-8.10
- RD Chapter 8- Quadratic Equations Ex-8.11
- RD Chapter 8- Quadratic Equations Ex-8.12
- RD Chapter 8- Quadratic Equations Ex-8.13

RD Chapter 8- Quadratic Equations Ex-8.1 |
RD Chapter 8- Quadratic Equations Ex-8.2 |
RD Chapter 8- Quadratic Equations Ex-8.3 |
RD Chapter 8- Quadratic Equations Ex-8.4 |
RD Chapter 8- Quadratic Equations Ex-8.5 |
RD Chapter 8- Quadratic Equations Ex-8.6 |
RD Chapter 8- Quadratic Equations Ex-8.7 |
RD Chapter 8- Quadratic Equations Ex-8.9 |
RD Chapter 8- Quadratic Equations Ex-8.10 |
RD Chapter 8- Quadratic Equations Ex-8.11 |
RD Chapter 8- Quadratic Equations Ex-8.12 |
RD Chapter 8- Quadratic Equations Ex-8.13 |

**Answer
1** :

Let the speed of stream = x km/h

and speed of boat in still water = 8 km/h

Distance covered up stream = 15 km

and downstream = 22 km

Total time taken = 5 hours

According to the conditions,

**Answer
2** :

Let the original speed of the train = x km/h

Then, the increased speed of the train = (x + 5) km/h [by given condition]

and distance = 360 km

According to the question,

**Answer
3** :

Total journey = 200 km

Let the speed of fast train = x km/hr

Then speed of slow train = (x – 10) km/hr

According to the condition,

=> x (x – 50) + 40 (x – 50) = 0

=> (x – 50) (x + 40) = 0

Either x – 50 = 0, then x = 50

or x + 40 = 0, then x = -40 but it is not possible being negative

Speed of the fast train = 50 km/hr

and speed of the slow train = 50 – 40 = 10 km/hr

**Answer
4** :

Total journey = 150 km

Let the usual speed of the train = x km/hr

According to the condition,

=> x (x + 30) – 25 (x + 30) = 0

=> (x + 30) (x – 25) = 0

Either x + 30 = 0, then x = -30 but it is not possible being negative

or x – 25 = 0, then x = 25

Usual speed of the train = 25 km/hr

**Answer
5** :

Distance = 150 km

Let the speed of the person while going = x km/hr

Then the speed while returning = (x + 10) km/hr

According to the condition,

=> x (x + 30) – 20 (x + 30) = 0

=> (x + 30) (x – 20) = 0

Either x + 30 = 0, then x = -30 which is not possible being negative

or (x – 20) = 0 then x = 20

Usual speed of the man while going = 20 km/hr

**Answer
6** :

Distance = 1600 km

Let usual speed of the plane = x km/hr

Increased speed = (x + 400) km/hr

According to the condition,

is not possible being negative or x – 800 = 0, then x = 800

Usual speed of the plane = 800 km/hr

**Answer
7** :

Distance = 1200 km

Let usual speed of the aeroplane = x km/hr

Increased speed = (x + 100) km/hr

According to the condition,

=> x (x + 400) – 300 (x + 400) = 0

=> (x + 400) (x – 300) = 0

Either x – 300 = 0, then x = 300

or x + 400 = 0, then x = -400 which is not possible being negative

Usual speed of the plane = 300 km/hr

**Answer
8** :

Let its original average speed be x km/h. Therefore

63x + 72x+6 = 3

**Answer
9** :

Distance to be covered = 90 km

Let uniform-original speed = x km/h

Increased speed = (x + 15) km/hr

According to the condition,

=> x2 + 60x – 45x – 2700 = 0

=> x (x + 60) – 45 (x + 60) = 0

=> (x + 60)(x – 45) = 0

Either x + 60 = 0, then x = -60 which is not possible being negative

or x – 45 = 0, then x = 45

Original speed of the train = 45 km/hr

**Answer
10** :

Total distance = 360 km

Let uniform speed of the train = x km/hr

Increased speed = (x + 5) km/hr

According to the condition,

=> x (x + 45) – 40 (x + 45) = 0

=> (x + 45) (x – 40) = 0

Either x + 45 = 0, then x = -45 but it is not possible being negative

or x – 40 = 0, then x = 40

Speed of the train = 40 km/hr

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