Question

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

Answer 1

$\rule{200}{2}$

$\Huge\bigstar\:\tt\underline\red{GIVEN}$

▪ $\sf\orange{1st\:train\:travels\:towards\:west}$

▪ $\sf\orange{2nd\:train\:travels\:towards\:north}$

▪ $\small\sf\orange{1st\:train\:travels\:5km/hr\:faster\:than\:2nd\:train}$

▪ $\sf\orange{After \:2\:hours\:they\:are\:50km}$

$\rule{200}{2}$

$\Huge\bigstar\:\tt\underline\red{TO\:FIND}$

▪ $\sf\orange{Average\:speed\:of\:each\:train}$

$\rule{200}{2}$

$\Huge\bigstar\:\tt\underline\red{SOLUTION}$

$\huge\dagger\:\:\sf\underline\pink{Let}$

• $\sf\blue{Speed\:of\:slower\:train\:=\:x\:kmph}$

• $\sf\blue{Then\:speed\:of\:faster\:train\:=\:x+5 \:kmph}$

$\sf\underline\gray{Distance\:=\:Speed\:×\:Time}$

$\sf\underline\orange{Distance\: travelled\:by\:the\:1st\:train}:$

$\Large\leadsto\:\:\: \: \sf \purple {2(x\:+\:5)}$

$\Large\leadsto\:\:\: \: \sf \green {2x\:+\:10}$

$\sf\underline\orange{Distance\: travelled\:by\:the\:2nd\:train}:$

$\Large\leadsto\:\:\: \: \sf \purple {2.x \: = \: 2x}$

$\bf\underline\red{By\: Pythagoras \: theorem}$

$\Large\leadsto\:\:\: \: \sf \purple {(2x)^2 + (2x+10)^2 = 50^2}$

$\Large\leadsto\:\:\: \: \sf \green {4x^2 + (4x^2 + 40x + 100) = 2500}$

$\Large\leadsto\:\:\: \: \sf \purple {4x^2 +4x^2 + 40x + 100 = 2500}$

$\Large\leadsto\:\:\: \: \sf \green {8x^2 + 40x - 2400 = 0}$

$\Large\leadsto\:\:\: \: \sf \purple {x^2 + 5x - 300 = 0}$

$\Large\leadsto\:\:\: \: \sf \green {x^2 + 20x - 15 - 300 = 0}$

$\Large\leadsto\:\:\: \: \sf \purple {x(x+20)-15(x+20)= 0}$

$\Large\leadsto\:\:\: \: \sf \green {(x+20)\:\:(x-15)=0}$

$\Large\leadsto\:\:\: \: \sf \purple {x-15=0\:\:x+20=0}$

$\Large\leadsto\:\:\: \: \sf \green {x=15 \:(or)\: -20}$

$\sf\gray{But\:x\: can't\:be\: negative}$

$\Large\dagger\:\sf\underline\red{Speed\:of\:the\:slower\:train}\:$

$\Large\leadsto\:\:\: \: \sf \underline\purple {x\:=\:15\:kmph}$

$\Large\dagger\:\sf\underline\red{Speed\:of\:the\:faster\:train}:$

$\Large\leadsto\:\:\: \: \sf \green {x+5}$

$\Large\leadsto\:\:\: \: \sf \purple {15+5}$

$\Large\leadsto\:\:\: \: \sf \underline\purple {20kmph}$

$\rule{200}{2}$

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