Question

From a station, to trains start at the same time. One train moves in West direction and other in North direction. First train moves 5 km/h faster than the second train. If after two hours, distance between the two trains is 50 km, find the average speed of each train.

Answer 1

Answer:

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Step-by-step explanation:

Solution:

Suppose, ‘A’ train moves in a West direction and ‘B’ in North direction.

Suppose, ‘A’ train moves in a West direction moves 5 km/hr faster than ‘B’ in North direction.

Let speed of train B= x km/hr

Speed of train A= x km/hr + 5 km/hr= x+5 km/hr

Now, we find the distance travelled by trains in 2 hours.

Distance covered by train A= Speed × Time

= (x+5) km/hr × 2hrs

= (2x+10) m

Distance travelled by train A is (2x+10) m.

Distance covered by train B= Speed × Time

= (x) km/hr × 2 hrs

= (2x) m

Distance travelled by train A is (2x) m.

Now, after two hours distance between these two trains is 50 km.

2x + (2x+10) = 50

4x+10 = 50

4x = 40

x = 10 km/hr

Speed of train B is 10 km/hr.

Speed of train A= x+5 = 10+5 = 15 km/hr

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Answer 2

$\huge \sf \color{purple}{\underline{\underline{Answer}}} :$

Speed of first train is 20 Km/h and speed of 2nd train is 15 Km/h

$\huge \sf \color{purple}{\underline{\underline{Solution}}}:$

➣ Let the 2nd train travel at X km/h

➣Then, the speed of a train is (5 +x) Km/hour.

➣ let the two trains live from station M.

➣ Distance travelled by first train in 2 hours

$\sf \boxed{\colorbox{skyblue}{Distance=speed×time}}$

= MA = 2(x+5) Km.

➣ Distance travelled by second train in 2 hours

= MB = 2x Km

$\sf \color{brown}{By \: Phythagoras \: theorem }$ AB²= MB²+MA²

⟹ 50²=(2(x+5)²+(2x)²

⟹ 2500 = (2x+10)² + 4x²

⟹8x² + 40x - 2400 = 0

⟹x² + 5x - 300 = 0

⟹x² + 20x -15x - 300 = 0

⟹x(x + 20) - 15(x + 20) = 0

⟹ (x + 20)(x -15) = 0

$\sf \boxed{\colorbox{lightgreen}{x=15 \: or \: -20}}$

Taking x = 15 , the speed of second train is 15 Km/h and speed of first train is 20 Km/h