Question

Two trains leaves a railway station at the same time. the first train travels due west and second train due north. if the speed of first train is 5 km/h more than the second train, then after 2 hours they are 50 km apart. find average speed of each train.

Answer 1

Two trains leave a railway station at the same time. The first train travels due west and the second train travels due north. The first train travels 5 km/hr faster than the second train. If after two hours, they are 50 km apart, find the speed of first train.

OPTION

1) 10 km/hr
2) 15 km/hr
3) 20 km/hr
4) 25 km/hr
5) 30 km/hr
6) 35 km/hr
7) 40 km/hr
8) 45 km/hr
9) 50 km/hr
10) None of these

Solution

Let the speed of the second train = x km/hr
The speed of the first train = x+5 km/hr
Distance covered after two hours by the first train is 2(x+5) km.
Distance covered by the second train after two hours is 2x km.
(2x)^2 + 4(x+5)^2=(50)^2 (Using pythagoras theorem)
Solve 
We get x=-20 and x=15
The speed of the first train = 15+5=20km/hr
option 3

Correct Option: 3 Best Solution (0)

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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