Question

Two trains leave a railway station at the same time .The first train travel towards west and the second train towards north The first train travels 7 km per hour faster then the second train. If after 10 hrs they are 130 km apart find the avg speed of both the trains​

Answer 1

Answer:

Let the speed of the second train be x km hr. Then, the speed of the first train is (x + 5) km/hr. Let O be the position of the railway station from which the two trains leave. Hence, the speed of the second train is 15 km/hr and the speed of the first train is 20 km/hr.

Let the average spped of the second train be x km/h.

Thus, the speed of the first train will be (x+5) km/h.

Distance travelled by the first train in two hours =2(x+5) km/h

Distance travelled by the second train in two hours =2x km/h

Using pythagoras theorem, we have

4x

2

+4(x+5)

2

=50

2

4x

2

+4x

2 +40x+100=2500

8x

2

+40x−2400=0

x

2

+5x−300=0

(x−15)(x+20)=0

x=15,−20

Ignore the negative value of the speed. We have,

Speed of the second train be 15 km/h.

Spped of the first train be (15+5)= 20 km/h

hope this is help you☺ friend

Answer 2

$\large\underline{\sf{Solution-}}$

↝Let the speed of second train be 'x' km per hour.

So,

↝Speed of first train is (x + 7) km per hour.

↝Let O be the initial position from where two trains start.

↝ Let after 10 hours, first train is at A due west and second train is at B due North.

We know,

↝ Distance = Speed × Time.

So,

↝ Distance travelled by first train in 10 hours at the speed of (x + 7) km per hour = OA = 10 × (x + 7) = 10(x + 7) km

and

↝ Distance travelled by second train in 10 hours at the speed of (x) km per hour = OB = 10 × x= 10x km

Now, it is given that after 10 hours, they are 130 km apart.

↝ It implies, AB = 130 km

↝ In right angle triangle OAB,

↝ Using Pythagoras Theorem,

$\rm :\longmapsto\: {AB}^{2} = {OA}^{2} + {OB}^{2}$

$\rm :\longmapsto\: {130}^{2} = {(10(x + 7))}^{2} + {(10x)}^{2}$

$\rm :\longmapsto\:16900 = 100 {(x + 7)}^{2} + {100x}^{2}$

$\rm :\longmapsto\:169 = {(x + 7)}^{2} + {x}^{2}$

$\rm :\longmapsto\:169 = {x}^{2} + 49 + 14x + {x}^{2}$

$\rm :\longmapsto\:169 = 2{x}^{2} + 49 + 14x$

$\rm :\longmapsto \: 2{x}^{2} + 49 + 14x - 169 = 0$

$\rm :\longmapsto \: 2{x}^{2} + 14x - 120 = 0$

$\rm :\longmapsto \: {x}^{2} + 7x - 60 = 0$

$\rm :\longmapsto \: {x}^{2} + 12x - 5x- 60 = 0$

$\rm :\longmapsto\:x(x + 12) - 5(x + 12) = 0$

$\rm :\longmapsto\:(x + 12)(x - 5) = 0$

$\rm :\longmapsto\:x = 5 \: \: \: or \: \: \: x = - 12 \: \red{ \{rejected \}}$

Hence,

↝ Speed of first train = x + 7 = 12 km per hour

and

↝ Speed of second train = x = 5 km per hour.

Basic Concept Used :-

Writing Systems of Equation from Word Problem.

1. Understand the problem.

• Understand all the words used in stating the problem.

• Understand what you are asked to find.

2. Translate the problem to an equation.

• Assign a variable (or variables) to represent the unknown.

• Clearly state what the variable represents.

3. Carry out the plan and solve the problem.