**Selina Concise Class 10 Math Chapter 6 ****Solving (simple) problems (Based on Quadratic Equations) Exercise 6C Solutions**

** Solving (simple) problems **

**(Based on Quadratic Equations) **

**Exercise 6C**

**(Q1) The speed of an ordinary train is x km per hr and that of an express train is (x + 25) km per hr. **

**(i) Find the time taken by each train to cover 300 km **

**(ii) If the ordinary train takes 2hrs more than the express train; **

**Calculate speed of the express train **

**Solution: **

Given that the speed of an ordinary train is x km per hr and the speed of an express train is (x + 25) km per hr.

(i) The time taken by each train to cover 300 km

Means distance = 300 km

We know that,

Speed = distance/time

or time = distance/speed

∴ Ordinary train – express train

= 2

The time taken by ordinary train to cover 300 km = 300/x hrs

and the time taken by express train to cover 300km = 300/x+25 hrs

If the ordinary train takes 2 hrs more than the express train to cover the distance of 300 kms

∴ ordinary train + express train = 2

300/x – 300/x + 25 = 2

300 (x + 25) – 300x/x (x + 25) = 2

300x + 7500 – 300x/x^{2} + 25x = 2

7500/x^{2} + 25x = 2

7500 = 2 (x^{2} + 25x)

= 2x^{2} + 50x

7500 = 2x^{2} + 50x

2x^{2} + 50x – 7500 = 0

Divide by ‘2’

2x^{2}/2 + 50x/2 – 7500/2 = 0

x^{2} + 25x – 3750 = 0

x^{2} + 75x – 50x – 3750 = 0

x (x + 75) – 50 (x + 75) = 0

(x + 75) (x- 50) = 0

x + 75 = 0 or x – 50 = 0

x = -75 or x = 50

∴ We know that the speed cannot be negative.

∴ x = -75 is not solution.

∴ x = 50 is the only solution

∴ The speed of the express train = (x + 25) km/hr

= 50 + 25

= 75 km/hr

∴ The speed of the ordinary train x = 50 km/hr

** **

**(Q2) If the speed of a car is increased by 10km/hr, it takes 18 minutes less to cover a distance of 36km, Find the speed of the car. **

**Solution: **

Let us consider the speed of the car is x km/hr.

Given that the distance is 36km

We know that,

Time = distance/speed

∴ The time taken to cover a distance of 36km = 36/x hrs

If the speed of a car is increased by 10 km per hr.

∴ (x + 10) km/hr

∴ The time taken to cover a distance of 36km = 36/(x+10) hrs

From question we have write,

36/x – 36/(x + 10) = 18/60

36 (x + 10) – 36x/x (x + 10) = 18/60

36x + 360 – 36x/x^{2} + 10x = 18/60

360/x^{2} + 10x = 18/60

60 (360) = 18 (x^{2} + 10x)

21, 600 = 18x^{2} + 180x

Dividing both sides by 18,

21,600/18 = 18x^{2}/18 + 180x/18

1200 = x^{2} + 10x

x^{2} + 10x – 1200 = 0

x^{2} + 40x – 30x – 1200 = 0

x (x + 40) – 30 (x + 40) = 0

(x + 40) (x – 30) = 0

x + 40 = 0 or x – 30 = 0

x = -40 or x = 30

But the speed of the car cannot be negative

∴ x = 30 is the only solution

∴ The speed of the car is 30 km/hr.

** **

**(Q3) If the speed of an aeroplane is reduced by 40 km/hr, it takes 20 minutes more t cover 1200 km. Find the speed of the aeroplane. **

**Solution: **

Let us assume that the initial speed is x km/hr

∴ If the speed of an aroplane is reduced by 40 km/hr.

∴ (x – 40) km/hr

It takes 20 minutes more to cover 1200 km.

1200/x – 40 – 1200/x = 20/60 (Cover minute in hour)

1200 [1/x – 40 – 1/x] = 1/3

1200 [x – (x – 40)/x (x – 40)] = 1/3

1200 [x – x + 40/x^{2} – 40x] = 1/3

1200 [40/x^{2} – 40x] = 1/3

1200×40×3 = x^{2} – 40x

1200×120 = x^{2} – 40x

144000 = x^{2} – 40x

x^{2} – 40x – 144000 = 0

x^{2} – 400x + 360x – 14400 = 0 {144000}

x (x – 400) + 360 (x – 400) = 0

(x – 400) (x + 360) = 0

x – 400 = 0 or x + 360 = 0

x = 400 or x = -360

But the speed cannot be negative

∴ x = 400 is the solution

∴ The speed of the aeroplane is 400 km/hr.

** **

**(Q4) A car covers a distance of 400 km at a certain speed, had the speed been 12km/h more, the time taken for the journey would have been 1 hour 40 minutes less. Find the original speed of the car. **

**Solution: **

Let us assume that the original speed of the car is x km/hr

A car covers a distance of 400 km at a certain speed.

(x + 12) km/h

∴ Total distance = 400 km.

∴ The time taken for the journey would have been 1 hour 40 minutes less.

∴ Tim = 1 hour 40 min

= 1 + 40/60 (Convert into hour)

= 1 + 2/3

= 3+2/3

= 5/3 hr.

∴ The original speed of the car is x km/hr.

and the distance = 400 km

∴ 400/x

∴ A car covers a distance of 400 km at certain speed.

400/x + 2

∴ 400/x – 400/x + 12 = 5/3

400 [1/x – 1/x+12] = 5/3

400 [x + 12 – x/x (x + 12)] = 5/3

400 [12/x^{2} + 12x] = 5/3

400×12×3 = 5 (x^{2} + 12x)

400×36 = 5x^{2} + 60x

14,400 = 5x^{2} + 60x

Divide by 5, on both sides, 2,880 = x^{2} + 12x

x^{2} + 12x – 2880 = 0

x^{2} + 60x – 48x – 2880 = 0

x (x + 60) – 48 (x + 60) = 0

(x + 60) (x – 48) = 0

x + 60 = 0 or x – 48 = 0

x = -60 or x = 48

∴ But the speed is not negative

∴ x = 48 is the only solution.

∴ The original speed of the car is 48 km/hr.

** **

**(Q5) A girl goes to her friend’s house, which is at a distance of 12km. She covers half of the distance at a speed of x km/hr and the remaining distance at a speed of (x + 2) km/hr. If she takes 2 hrs 30 minutes to cover the whole distance, find ‘x’ **

**Solution: **

Total distance 12km

∴ She covers half of the distance at a speed of x km/hr.

6/x

The remaining distance at a speed of (x + 2) km/hr

6/x + 2

∴ 6/x + 6/x + 2 = 2 ½

= 4+1/2

6 (x + 2) + 6x/x (x+ 2) = 5/2

6x + 12+6x/x^{2} + 2x = 5/2

12x + 12/x^{2} + 2x = 5/2

12x + 12/x^{2} + 2x = 5/2

2 (12x + 12) = 5 (x^{2} + 2x)

24x + 24 = 5x^{2} + 10x

5x^{2} + 10x – 24x – 24 = 0

5x^{2} – 14x – 24 = 0

5x^{2} – 20x + 6x – 24 = 0

5x (x – 4) + 6 (x – 4) = 0

(x – 4) (5x + 6) = 0

x – 4 = 0 or 5x + 6 = 0

5x = -6

x = 4

x = -6/5

∴ The speed cannot be negative

∴ x = 4 is the only solution.

**Here is your solution of Selina Concise Class 10 Math Chapter 6 Solving (simple) problems (Based on Quadratic Equations) **

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