- RD Chapter 8- Quadratic Equations Ex-8.1
- 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-8.8
- 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-8.3 |
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-8.8 |
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 first number = x

Then second number = x + 1

Their product = 306

x (x + 1) = 306

=> x^{2} + x – 306 = 0

Required quadratic equation will be x^{2} +x – 306 = 0

**Answer
2** :

No. of marbles John and Jivanti have = 45

Let number of marbles John has = x

Then number of marbles Jivanti has = 45 -x

Every one lost 5 marbles, then John’s marbles = x – 5

and Jivanti’s marbles = 45 – x – 5 = 40 – x

According to the condition,

(x – 5) (40 – x) = 128

=> 40x – x^{2} – 200 + 5x – 128 = 0

=> -x^{2} + 45x – 328 = 0

=> x^{2} – 45x + 328 = 0

**Answer
3** :

Let the number of toys in a day = x

Cost of each toy = x – 55

on a particular cost of production = Rs. 750

x (x – 55) = 750

=> x^{2} – 55x – 750 = 0

Hence required quadratic equation will be x^{2} –55x – 750 = 0

**Answer
4** :

Let the base of a right triangle = x

Its height = x – 7

and hypotenuse = 13 cm

=> By Pythagoras Theorem

(Hypotenuse)^{2} = (Base)^{2} + (Height)^{2}

(13 )^{2} = x^{2} +(x – 7)^{2}

=> 169 = x^{2} + x^{2} – 14x + 49

=> 2x^{2} – 14x + 49 – 169 = 0

=> 2x^{2} – 14x – 120 = 0

=> x^{2} – 7x – 60 = 0 (Dividing by2)

Hence required quadratic equation will be x^{2} –7x – 60 = 0

**Answer
5** :

Distance between Mysore and Bangalore = 132 km

Let average speed of passenger train=x km/ hr

Then average speed of express train = (x + 11) km/hr

**Answer
6** :

Total distance = 360 km

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

Time taken = 360/x

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