Question

A person on tour has 4200 rupees for his expenses . If he extend his tour for 3 day he has to cut down daily expenses by 70 rupees find the original duration of tour

Answer 1

Hi !

Original duration of tour = x days

Expense for one day [ originally] = 4200/x Rs

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When  the tour is extended for 3 days :-
New duration of tour is "x +3" days
Expenditure for 1 day [ new] = 4200/x+3 Rs

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A.T.Q .

4200/x - 4200/x +3 = 70

4200 [ 1/x - 1/x+3] =70

4200 [ x + 3 - x / x² + 3x ] = 70

4200 [ 3/x² + 3x ] = 70

12600/x² + 3x = 70

70(x² + 3x ) = 12600

x² + 3x = 180

x² + 3x - 180 = 0

solving by quadratic  formula :-

x = -b±√b² - 4ac   / 2a

x = -3 ± √ 3²  - 4× -180 /2
   = -3 ± √ 9 + 720  /2 
   = -3± √729  /2
   = -3± 27 /2
   = -3 + 27 /2
   = 24/2 = 12

Original duration of tour = 12 days

Answer 2

Let the original duration of the tour be x days.

A/C,

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

$\longrightarrow$$\sf\red{4200((x + 3) - x) = 70 \times x \times (x + 3)}$

$\longrightarrow$$\sf\red{180 = x(x + 3)}$

$\longrightarrow$$\sf\red{{x}^{2} + 3x - 180 = 0}$

$\longrightarrow$$\sf\red{(x + 15)(x - 12) = 0}$

$\longrightarrow$$\sf\red{x = - 15 \: or \: x = 12}$

As the number of days cannot be negative, x ≠ -15 ,so x = 12

Hence, the original duration of the tour = 12 days