- 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-8.8
- 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-8.13

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-8.8 |
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-8.13 |

**Answer
1** :

Let B can do the work in = x days

A will do the same work in = (x – 10) days

A and B both can finish the work in = 12 days

According to the condition,

=> x (x – 4) – 30 (x – 4) = 0

=> (x – 4) (x – 30) = 0

Either x – 4 = 0, then x = 4

or x – 30 = 0, then x = 30

But x = 4 is not possible

B can finish the work in 30 days

**Answer
2** :

Two pipes can fill the .reservoir in = 12 hours

Let first pipe can fill the reservoir in = x hrs

Then second pipe will fill it in = (x – 10) hours

Now according to the condition,

=> x² – 10x = 24x – 120

=> x² – 10x – 24x + 120 = 0

=> x² – 34x + 120 = 0

=> x² – 30x – 4x + 120 = 0

=> x (x – 30) – 4 (x – 30) = 0

=> (x – 30) (x – 4) = 0

Either x – 30 = 0, then x = 30

or x – 4 = 0 but it is not possible as it is < 10

The second pipe will fill the reservoir in = x – 10 = 30 – 10 = 20 hours

**Answer
3** :

Two taps can fill the tank in = 9(3/8) = 75/8 hr

Let smaller tap fill the tank in = x hours

Then larger tap will fill it in = (x – 10) hours

According to the condition,

Smaller tap can fill the tank in = 25 hours

and larger tap can fill the tank in = 25 – 10 = 15 hours

**Answer
4** :

=> 9x (x – 20) + 25 (x – 20) = 0

=> (x – 20) (9x + 25) = 0

Either x = – 20 = 0, then x = 20 or 9x + 25 = 0 then 9x = -25

=> x = −25/9 but it is not possible being negative

x = 20

Time taken by the two pipes = 20 minutes and 20 + 5 = 25 minutes

**Answer
5** :

Let pipe of larger diameter can fill the tank = x hrs

and pipe of smaller diameter can fill in = y hrs

=> 26x + 80 = x² + 10x

=> x² + 10x – 26x – 80 = 0

=> x² – 16x – 80 = 0

=> x² – 20x + 4x – 80 = 0

=> x (x – 20) + 4 (x – 20) = 0

=> (x – 20) (x + 4) = 0

Either x – 20 = 0, then x = 20

or x + 4 = 0, then x = – 4 which is not possible

x = 20 and y = 10 + x = 10 + 20 = 30

Larger pipe can fill the tank in 20 hours and smaller pipe can fill in 30 hours.

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