Math, asked by mnshkmr027, 9 months ago

pls give me solution anyone​

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Answered by RvChaudharY50
40

उतर :-

→ माना ट्रेन की चाल = y m/s

→ माना ट्रेन की लंबाई = z मीटर

तब ,

→ प्लेटफॉर्म को पार करने में कुल दूरी तय की = (z + 200)

→ पार करने में समय लगा = 30 सेकंड

→ चाल = y = (दूरी / समय) = (z+200/30)

→ 30y = z + 200

→ 30y - z = 200 ----------------- Eqn(1).

इसी प्रकार :-

→ आदमी को पार करने में दूरी तय = ट्रेन की लंबाई = z मीटर

→ समय = 20 सेकंड

→ उसी दिशा में आदमी की चाल = 6 * (5/18) = (5/3) m/s.

→ तब कुल चाल = (y - 5/3)

→ चाल = (दूरी / समय)

→ (y - 5/3) = z/20

→ (3y - 5)/3 = z/20

→ 20(3y - 5) = 3z

→ 60y - 100 = 3z

→ 60y - 3z = 100 ----------------- Eqn(2).

पहली Eqn. को 3 से गुणा करके , उसमे से दूसरी Eqn. घटाने पर :-

→ 3(30y - z) - (60y - 3z) = 3*200 - 100

→ 90y - 3z - 60y + 3z = 600 - 100

→ 30y = 500

→ y = (500/30)

y = (50/3) m/s.

अत :-

→ x = (50/3) * (18/5)

→ x = 10 * 6

→ x = 60 किमी / घंटा (d) .

(अच्छा सवाल l)

Answered by Anonymous
23

{\huge{\bf{\red{\underline{Question:}}}}}

A train traveling at a speed of 6:36 pm x km / h crossed a 200 meter long platform in 30 seconds and a person walking in the same direction at a speed of 6 km / h in 20 seconds Left it behind. What is the value of x?

{\bf{\blue{\underline{Now:}}}}

  • Let the the speed of train be = u
  • and length be = w

At platform,

  • Distance = w + 200
  • Time = 30 sec
  • Speed = D/t = (w+200)/30

 : \implies{\sf{ 30u = w + 200}} \\ \\

 : \implies{\sf{ 30u -w = 200......(1)}} \\ \\

__________________________________

When a person walking in the same direction,

  • Distance = w
  • Time = 20 sec
  • Speed = w/20

 : \implies{\sf{ u  -  \frac{5}{3} =  \frac{w}{20} }} \\ \\

 : \implies{\sf{   \frac{3u - 5}{3} =  \frac{w}{20} }} \\ \\

 : \implies{\sf{   20({3u - 5}) =  3w}} \\ \\

 : \implies{\sf{   60u - 100 =  3w}} \\ \\

 : \implies{\sf{   60u +3w= 100......(2)}} \\ \\

___________________________________

By elimination method,

30u - w = 200× 3

60u +3w = 100×-1

___________________________________

90u -3w = 600

-60u +3w = -100

_____________

30u. =500

_____________

 : \implies{\sf{u =  \frac{500}{30}  }} \\ \\

 :  \implies{\sf{u =  \frac{50}{3} }} \\ \\

Now,

 : \implies{\sf{x =  \frac{50}{3}  \times  \frac{18}{5} }} \\ \\

 : \implies{\sf{x =  10 \times 6}} \\ \\

 : \implies \boxed{\sf{x =  60km.hr}} \\ \\

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