A person on tour has $R s 4200$ for his expenses his tour for $3$ days, he has $18$ cut down his daily expenses by $R s . 70$ . Find the original duration of the tour.

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Solution

$L e t t h e o r i g i n a l d u r a t i o n o f t o u r b e x d a y s$
$A m o u n t w i t h t h e p e r s o n i s R s .4200$
$D a i l y e x p e n s e s = R s . \frac{4200}{x}$
$G i v e n t h a t, t o u r e x t e n d e d f o r 3 m o r e d a y s$

$H e n c e t o t a l n u m b e r o f d a y s = (x + 3) d a y s$

$D a i l y e x p e n s e s = R s . \frac{4200}{(x + 3)}$

$B y t h e g i v e n p r o b l e m, w e h a v e$

$\frac{4200}{x} - \frac{4200}{x + 3} = 70$

$4200 (\frac{1}{x} - \frac{1}{x + 3}) = 70$

$60 [\frac{x + 3 - x}{x (x + 3)}] = 1$

$180 = x^{2} + 3 x$

$x^{2} + 3 x - 180 = 0$

$x^{2} + 15 x - 12 x - 180 = 0$

$x (x + 15) - 12 (x - 15) = 0$

$(x + 15) (x - 12) = 0$

$H e n c e x = 12 a s x c a n n o t b e n e g a t i v e$
$T h u s t h e o r i g i n a l d u r a t i o n o f t o u r i s 12 d a y s .$

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