Model 5 Train Vs Both platform and a man/a pole Section-Wise Topic Notes With Detailed Explanation And Example Questions

Answer: (a)

Let the speed of trains be x

and y metre/sec respectively,

${100 + 95}/{x - y} = 27$

$x - y = 195/27 = 65/9$ ...(i)

Again,

$195/{x + y} = 9$

$x + y = 195/9$ ...(ii)

By equation (i) + (ii)

$2x = 65/9 + 195/9 = 260/9$

$x = 260/{2 × 9} = 130/9$ m/sec.

= $(130/9 × 18/5)$ kmph = 52 kmph

From equation (ii),

$y = 195/9 - 130/9 = 65/9$ m/sec.

= $65/9 × 18/5$ = 26 kmph

Answer: (c)

Rule 10 and Rule 1,

Let the length of the train be x metres.

Speed of train in crossing boy = $x/30$

Speed of train in crossing platform = ${x +110}/40$

According to the question,

${x +110}/40 =x/30$

${x +110}/4 = x/3$

4x = 3x + 330

x = 330 metres

Answer: (a)

Rule 10 and Rule 1,

Let length of train be x m

Speed of train = ${x + 264}/20$

Also, speed of train = $x/8$

Obviously, $x/8 = {x + 264}/20$

$x/2 = {x + 264}/5$

5x = 2x + 528

5x - 2x = 528

x = 528 ÷ 3 = 176 m