Two friends set off by train at dawn to visit each other. The two friends caught sight of each other through the window as the trains passed in opposite direction on adjacent tracks− it was 12^(oo) hours. The friends helplessly reached their destinations. If the first of them reached their destination at 16^(oo) and the second at 21^(oo) , what time did dawn break on that day ? Help me, please


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Question Number 215523 by Ismoiljon_008 last updated on 09/Jan/25

$T w o f r i e n d s s e t o f f b y t r a i n a t d a w n t o v i s i t$ $e a c h o t h e r . T h e t w o f r i e n d s c a u g h t s i g h t o f$ $e a c h o t h e r t h r o u g h t h e w i n d o w a s t h e t r a i n s$ $p a s s e d i n o p p o s i t e d i r e c t i o n o n a d j a c e n t t r a c k s -$ $i t w a s 12^{o o} h o u r s . T h e f r i e n d s h e l p l e s s l y r e a c h e d$ $t h e i r d e s t i n a t i o n s . I f t h e f i r s t o f t h e m r e a c h e d$ $t h e i r d e s t i n a t i o n a t 16^{o o} a n d t h e s e c o n d a t 21^{o o},$ $w h a t t i m e d i d d a w n b r e a k o n t h a t d a y ?$ $H e l p m e, p l e a s e$
Commented by nikif99 last updated on 09/Jan/25

Commented by mr W last updated on 10/Jan/25

$g r e a t t r a n s l a t i o n f o r t h e u n c l e a r$ $q u e s t i o n!$
	Answered by nikif99 last updated on 09/Jan/25

	$L e t t t h e d a w n .$ $[12 : 00] \Rightarrow v_{1} = \frac{a}{12 - t}, v_{2} = \frac{s - a}{12 - t}$ $[16 : 00] \Rightarrow v_{1} = \frac{s}{16 - t}, [16 : 00 - 12 : 00] \Rightarrow v_{1} = \frac{s - a}{4}$ $[21 : 00] \Rightarrow v_{2} = \frac{s}{21 - t}, [21 : 00 - 12 : 00] \Rightarrow v_{2} = \frac{a}{9}$ $\frac{v_{1}}{v_{2}} = \frac{9 (s - a)}{4 a} = \frac{a}{s - a} = \frac{21 - t}{16 - t} (1)$ $(1) \Rightarrow 9 {(s - a)}^{2} = 4 a^{2} \Rightarrow 5 a^{2} - 18 s a + 9 s^{2} = 0 \Rightarrow$ $a_{1, 2} = \frac{18 s \pm \sqrt{18^{2} s^{2} - 4 \times 5 \times 9 s^{2}}}{2 \times 5} = 3 s o r \frac{3 s}{5}$ $(1) \Rightarrow a (16 - t) = (s - a) (21 - t) \Rightarrow$ $37 a - 2 a t - 21 s + s t = 0 \Rightarrow t = \frac{37 a - 21 s}{2 a - s}$ $f o r a_{1} = 3 s \Rightarrow t = 18 : 00 h r e j e c t e d$ $f o r a_{2} = \frac{3 s}{5} \Rightarrow t = 6 : 00 a c c e p t e d$
	Commented by Ismoiljon_008 last updated on 10/Jan/25

	$t h a n k y o u v e r y m u c h$
	Answered by mr W last updated on 10/Jan/25

	Commented by mr W last updated on 10/Jan/25

	$t h i s i s t h e t i m e d i s t a n c e d i a g r a m$ $f o r b o t h t r a i n s .$ $\frac{x}{4} = \frac{4 + 5}{x}$ $\Rightarrow x^{2} = 4 \times 9 = 36 \Rightarrow x = 6 h$ $? = 12 - 6 = 6$ $i . e . t h e d a w n b r e a k s a t 6 o^{'} c l o c k .$
	Commented by mr W last updated on 10/Jan/25

	$a d d i t i o n a l q u e s t i o n :$ $s a y b o t h t r a i n s t a k e a p a u s e o f$ $2 h o u r s a n d t h e n g o b a c k . a t w h a t$ $t i m e c a n t h e f r i e n d s ‘ ‘ {s e e}^{″} e a c h$ $o t h e r a g a i n t h r o u g h t h e w i n d o w s ?$
	Commented by mr W last updated on 10/Jan/25

	Commented by mr W last updated on 10/Jan/25

	$s a y t h e y ‘ ‘ {s e e}^{″} e a c h o t h e r y h o u r s$ $a f t e r 23^{00} .$ $\frac{y}{15 - y} = \frac{5 - y}{5 + y}$ $\Rightarrow y = 3$ $? = 23^{00} + 3 = 2^{00}$ $i . e . t h e f r i e n d s s e e e a c h o t h e r a g a i n$ $a t 2 o^{'} c l o c k o n n e x t d a y .$
	Commented by mr W last updated on 10/Jan/25

	$f o l l o w i n g i s a n e x a m p l e h o w a t i m e$ $d i s t a n c e d i a g r a m f o r r e a l t r a i n$ $t r a f f i c m a y l o o k l i k e .$
	Commented by mr W last updated on 10/Jan/25

	Commented by Ismoiljon_008 last updated on 10/Jan/25

	$t h a n k y o u v e r y m u c h$
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