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Question Number 204158 by siyathokoza last updated on 07/Feb/24

Commented by siyathokoza last updated on 07/Feb/24

$HI PROF . . .$ $Please explain this word problem$
Commented by siyathokoza last updated on 07/Feb/24


	Answered by deleteduser1 last updated on 07/Feb/24

	$s_{m} = \frac{d_{m}}{t_{m}} = \frac{65}{t_{m}}; s_{s} = s_{m} - 2 = \frac{d_{s}}{t_{s}} = \frac{65}{t_{m} + \frac{2}{3}} = \frac{65 \times 3}{3 t_{m} + 2}$ $\Rightarrow \frac{65}{t_{m}} = \frac{65 \times 3 + 6 t_{m} + 4}{3 t_{m} + 2} \Rightarrow 3 t_{m}^{2} + 2 t_{m} - 65 = 0 \Rightarrow t_{m} = \frac{13}{3}$ $t_{m} = \frac{13}{3} h r s \Rightarrow s_{m} = 15 k m / h r \Rightarrow s_{s} = 13 k m / h r$
	Answered by Rasheed.Sindhi last updated on 07/Feb/24

	$S p e e d o f S o u l : x k m / h o u r (s a y)$ $\Rightarrow S p e e d o f M a r k : (x + 2) k m / h o u r$ $T i m e t a k e n i n 65 k m s :$ $S a u l : t = \frac{s}{v} = \frac{65}{x} h o u r s$ $M a r k : t = \frac{65}{x + 2} h o u r s$ $\frac{65}{x} - \frac{65}{x + 2} = \frac{2}{3} [40 m i n u t e s = \frac{2}{3} h o u r s]$ $65 (3) (x + 2) - 65 (3) (x) = 2 (x) (x + 2)$ $195 x + 390 - 195 x = 2 x^{2} + 4 x$ $2 x^{2} + 4 x - 390 = 0$ $x^{2} + 2 x - 195 = 0$ $(x - 13) (x + 15) = 0$ $x = 13$ ${S o u l}^{'} s s p e e d : 13 k m / h$ ${M a r k}^{'} s s p e e d : 13 + 2 = 15 k m / h$
	Commented by Rasheed.Sindhi last updated on 07/Feb/24

	$T y p o s i r, c o r r e c t e d n o w, T h a n k s!$
	Commented by deleteduser1 last updated on 07/Feb/24

	$S u b s t i t u t e t h e s p e e d b a c k, y o u {w o n}^{'} t g e t$ $\frac{65}{x ({S a u l}^{'} s s p e e d)} - \frac{65}{(x + 2) ({M a r k}^{'} s s p e e d)} = \frac{2}{3}$
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