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A person on tour has 4200 for his expenses. If he extends his tour for 3 days, hehas to cut down his daily expenses by 70. Find the original duration of the tour. |
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Answer» Let the original duration of stay be n.Let the daily expense be x.Then, xn = 4200=> x = (4200 / n)Also, ( x - 70 ) ( n + 3) = 4200=> nx - 70n + 3x -210 = 4200=> 4200 - 70n + 3x - 210 = 4200=> 70n - 3x + 210 = 0=> 70n - 3*(4200/n) + 210 = 0=> 70n^2 - 12600 + 210 n = 0=> n^2 - 180 + 3n = 0.=> n^2 + 3n - 180 = 0=> n^2 + 12n - 9n - 180 = 0=> n(n+12) - 9(n+12) = 0=> (n+12)(n-9) = 0=> n = 9.ORIGINAL DURATION OF STAY = 9 DAYS. PLEASE HIT THE LIKE BUTTON |
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