In this page quadratic equation solution20 we are going to see solution of the word problems of the topic quadratic equation.

**Question 27**

A passenger train takes 3 hours less than a slow train for journey of 600 km. If the speed of the slow train is 10 km/hr less than that of the passenger train, find the speed of two trains.

**Solution:**

Here speed of slow train is compared with speed of passenger train

Let “x” be the speed of the of the passenger train

speed of the slow train is 10 km/hr less than that of the passenger train

So “x - 10” be the speed of the slow train

Distance has to be covered = 600 km

Time = Distance/speed

Let T1 be the time taken by passenger train

Let T2 be the time taken by the slow train

The differences of time taken by both trains are 3 hours

T1 = 600/x

T2 = 600/(x - 10)

T2 – T1 = 3 hours

[600/(x – 10)]– [600/x] = 3

600[ (1/(x - 10) - (1/x)] = 3

600/ (3) [(x - x + 10)/x(x – 10)] = 1

200 [10/x² – 10 x] = 1

2000/ (x² – 10 x) = 1

2000 = x² – 10 x

x² – 10 x - 2000 = 0

x² – 50 x + 40 x - 2000 = 0

x (x – 50) + 40 (x – 50) = 0

(x + 40) (x – 50) = 0

x + 40 = 0 x – 50 = 0

x = - 40 x = 50

Here x represents the speed of the train. So we should not take the negative value - 40 for x.

= x- 10

= 50- 10

= 40

Therefore the speed of passenger train = 50 km/hr

Speed of slow train = 40 m/hr

**Verification:**

Time = Distance/speed

Time taken by the passenger train to cover 600 km = 600/50

= 12 hours

Time taken by the slow train to cover 600 km = 600/40

= 15 hours

difference of time taken = 15 - 12 = 3 hours

quadratic equation solution20

- Back to worksheet
- Factoring a Quadratic Equation
- Factoring Worksheets
- Framing Quadratic Equation From Roots
- Framing Quadratic Equation Worksheet
- Remainder Theorem
- Relationship Between Coefficients and roots
- Roots of Cubic equation
- Roots of Polynomial of Degree4
- Roots of Polynomial of Degree5